Topics from Geometry
1
SEVERAL TOPICS FROM GEOMETRY FRANZ ROTHE
2010 Mathematics subject classification: 51-01; 51A05; 51A30; 51Exx; 51Gxx; 51Fxx; 51M09; . Keywords and phrases: Instructional exposition, General theory and projective geometries, Desarguesian and Pappian geometries, Finite geometry and special incidence structures, Ordered geometries, Metric geometry, Elementary problems in hyperbolic and elliptic geometries, Gaussian curvature.
Contents
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1
A Synopsis of Euclid
12
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Book I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book II. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book III. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book IV. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book VI. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book XI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book XII. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Book XIII. Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
On the History of Translations and Editions of Euclid’s Elements
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I
Neutral Geometry
2
Hilbert’s Axioms of Geometry 2.1 Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 David Hilbert’s axiomatization of Euclidean geometry . . . . . . 2.2.1 Introduction from Hilbert’s Foundations of Geometry . . . . 2.2.2 Hilbert’s axioms . . . . . . . . . . . . . . . . . . . . . . . 2.3 Importance and Impact of Hilbert’s Foundations of Geometry . . 2.3.1 Hilbert’s Foundations in Comparison with Euclid’s Elements 2.3.2 The Impact of Hilbert’s Foundations of Geometry . . . . . . 2.3.3 Drawbacks and Lacuna of Hilbert’s Foundations . . . . . . 2.4 Frege’s Critique and Hilbert’s answer . . . . . . . . . . . . . . . 2.5 About the consistency proof for geometry . . . . . . . . . . . . . 2.6 General remark about models in mathematics . . . . . . . . . . . 2.7 What is completeness? . . . . . . . . . . . . . . . . . . . . . . . 2.8 More metamathematical considerations . . . . . . . . . . . . . .
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23 23 25 26 26 30 30 32 32 33 34 35 35 36
Incidence Geometry 3.1 Elementary propositions about incidence planes . . . . . . . . . 3.2 Finite incidence geometries . . . . . . . . . . . . . . . . . . . 3.3 Affine incidence planes . . . . . . . . . . . . . . . . . . . . . . 3.4 Introduction of coordinates . . . . . . . . . . . . . . . . . . . . 3.5 Finite coordinate planes . . . . . . . . . . . . . . . . . . . . . 3.6 Projective incidence planes . . . . . . . . . . . . . . . . . . . . 3.7 The Fano Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Projective plane with coordinates . . . . . . . . . . . . . . . . 3.9 Finite affine and projective incidence planes . . . . . . . . . . . 3.10 Elementary propositions for three-dimensional incidence spaces 3.11 Three-dimensional Euclidean incidence geometry . . . . . . . .
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38 38 40 47 49 50 52 55 58 64 65 68
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The Theorems of Desargues and Pappus 4.1 Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . . . . 4.2 Theorem of Desargues and related theorems in projective setting 4.3 Tiling in perspective view . . . . . . . . . . . . . . . . . . . . 4.4 The Prime Power Conjecture about Non-Desarguesian Planes . 4.5 Hilbert’s investigation about the Theorem of Desargues . . . . . 4.6 The Mouton plane . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Theorem of Pappus and related theorems . . . . . . . . . . . . 4.8 Theorem of Hessenberg . . . . . . . . . . . . . . . . . . . . . 4.9 Relations to the "Little Theorems" . . . . . . . . . . . . . . . .
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72 72 73 81 86 87 88 91 91 93
The Axioms of Order and Their Consequences 5.1 Order of points on a line . . . . . . . . . 5.2 Bernays’ Lemma . . . . . . . . . . . . . 5.3 Plane separation . . . . . . . . . . . . . 5.4 Four-point and n-point Theorems . . . . 5.5 Angles . . . . . . . . . . . . . . . . . . 5.6 Space separation . . . . . . . . . . . . . 5.7 Interior and exterior of a triangle . . . . .
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96 96 99 100 104 110 115 117
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Topics from Geometry
5.8 5.9 5.10 5.11 6
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Convexity . . . . . . . . . . . . . . . . . Topology of the ordered incidence plane . Left and right, orientation . . . . . . . . The restricted Jordan Curve Theorem . .
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120 122 126 131
Finite Affine and Projective Incidence Planes and Latin Squares 6.1 Latin squares . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Latin squares from finite fields . . . . . . . . . . . . . . . . 6.3 Finite Non-Desarguesian planes . . . . . . . . . . . . . . . 6.4 A note on projective spaces . . . . . . . . . . . . . . . . .
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134 134 136 140 144
Congruence of Segments, Angles and Triangles 7.1 Congruence of segments . . . . . . . . . . 7.2 Some elementary triangle congruences . . 7.3 Congruence of angles . . . . . . . . . . . 7.4 SSS congruence . . . . . . . . . . . . . . 7.5 Right, acute and obtuse angles . . . . . . . 7.6 Constructions with Hilbert tools . . . . . . 7.7 Remarks about angles . . . . . . . . . . . 7.8 Orientated angles . . . . . . . . . . . . . . 7.9 The exterior angle theorem . . . . . . . . . 7.10 Congruence of z-angles . . . . . . . . . . 7.11 Consequences of the exterior angle theorem 7.12 SSA congruence . . . . . . . . . . . . . . 7.13 Reflection . . . . . . . . . . . . . . . . . . 7.14 Applied problems . . . . . . . . . . . . . . 7.15 Independence of the SAS-axiom . . . . . . 7.16 The Moulton plane . . . . . . . . . . . . . 7.17 Restriction of SAS congruence . . . . . . .
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147 147 152 157 163 167 173 181 185 187 191 193 209 213 216 220 222 223
Measurement and Continuity 8.1 The Archimedean axiom . . . . . . . . . . 8.2 Axioms related to completeness . . . . . . 8.2.1 Cantor’s axiom . . . . . . . . . 8.2.2 Dedekind’s axiom . . . . . . . . 8.2.3 Hilbert’s axiom of completeness
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227 227 234 234 235 238
Legendre’s Theorems 9.1 The First Legendre Theorem . . . 9.2 The Second Legendre Theorem . 9.3 The alternative of two geometries 9.4 What is the natural geometry? . .
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240 240 242 247 250
10 Neutral Geometry of Circles and Continuity 10.1 Immediate consequences of neutral geometry . . . . . . . . . . . . 10.2 The tangent is the limiting position of a secant . . . . . . . . . . . 10.3 Mutual placement of two circles . . . . . . . . . . . . . . . . . . . 10.4 Continuity principles for circles . . . . . . . . . . . . . . . . . . . 10.5 Continuity principles for circles are independent of Hilbert’s axioms 10.6 Derivation of continuity principles from Dedekind’s axiom . . . . .
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255 255 261 264 266 272 273
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11 Neutral Triangle Geometry 11.1 Introduction from neutral geometry . . . . . . . . . . . 11.2 The circum-circle . . . . . . . . . . . . . . . . . . . . . 11.3 Interlude about points of congruent distance from a line 11.4 The midpoint triangle and its altitudes . . . . . . . . . . 11.4.1 Immediate consequences for the altitudes . . . . . 11.5 The Hjelmslev Line . . . . . . . . . . . . . . . . . . . 11.6 The in-circle and the ex-circles . . . . . . . . . . . . . . 11.7 The Hjelmslev quadrilateral in neutral geometry . . . . 11.8 Limitations of neutral triangle geometry . . . . . . . . .
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278 278 279 285 285 288 288 289 294 298
12 Towards a Natural Axiomatization of Geometry 12.1 The Uniformity Theorem . . . . . . . . . . 12.2 Some strange polygons . . . . . . . . . . . 12.3 Defect and AAA congruence . . . . . . . . 12.4 A hierarchy of planes . . . . . . . . . . . . 12.5 Wallis’ axiom . . . . . . . . . . . . . . . . 12.6 Proclus’ Theorem . . . . . . . . . . . . . . 12.7 More about Aristole’s axiom . . . . . . . .
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302 302 307 310 312 312 314 316
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13 Area in neutral geometry 13.1 Equidecomposable and equicomplementable figures 13.2 The winding number . . . . . . . . . . . . . . . . . 13.3 Area of rectilinear figures . . . . . . . . . . . . . . 13.4 A standard equicomplementable form for a figure . . 13.5 The role of the Archimedean axiom . . . . . . . . .
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319 319 322 323 327 329
14 A Simplified Axiomatic system of Geometry—my own Suggestion 14.1 A simplified axiomatization of geometry . . . . . . . . . . . 14.2 Fundamental constructions with Euclidean tools . . . . . . . 14.3 Euclidean tools are at least as strong as Hilbert tools . . . . . 14.4 Birkhoff’s set of postulates . . . . . . . . . . . . . . . . . . .
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331 331 334 336 342
II
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Euclidean Geometry
15 Some Euclidean Geometry of Circles 15.1 Thales’ Theorem . . . . . . . . . . . . . . . 15.2 Rectangles and the converse Thales Theorem 15.3 Construction of tangents to a circle . . . . . 15.4 A bid of philosophy . . . . . . . . . . . . . 15.5 Common tangents of two circles . . . . . . . 15.6 Angles in a circle . . . . . . . . . . . . . . . 15.7 On the nature of the tangent . . . . . . . . . 15.8 The two circle lemma . . . . . . . . . . . .
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345 345 353 358 358 360 363 372 373
Topics from Geometry
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16 Simple Euclidean Geometry 16.1 Five constructions of the parallel . . . . . . . . . . . . 16.2 Dividing a segment into any number of congruent parts 16.3 Some triangle constructions . . . . . . . . . . . . . . 16.4 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . .
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376 376 379 383 388
17 Pappus’ and Pascal’s Theorems 392 17.1 Pappus’ Theorem in Euclidean geometry . . . . . . . . . . . . . . . . . . . . . 392 17.2 Pascal’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 18 Arithmetic of Segments—Hilbert’s Road from Geometry to Algebra 18.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Construction of the field of segment arithmetic . . . . . . . . . 18.2.1 Commutativity . . . . . . . . . . . . . . . . . . . . . . . 18.2.2 The distributive law . . . . . . . . . . . . . . . . . . . . . 18.2.3 The associative law . . . . . . . . . . . . . . . . . . . . . 18.2.4 A direct proof of commutativity of segment arithmetic . . 18.2.5 Alternative proof of associativity . . . . . . . . . . . . . .
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401 401 402 404 405 406 407 409
19 Similar Triangles 19.1 Basic properties from Euclid 19.2 Some exercises . . . . . . . 19.3 Secants in a circle . . . . . 19.4 The lens equation . . . . . .
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411 411 415 422 424
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426 426 428 430 433 434
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435 435 437 440 441 442 444
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20 The Pythagorean and related Theorems 20.1 The leg and the altitude theorems . . . . . . . . . . . 20.2 The Pythagorean theorem . . . . . . . . . . . . . . . 20.3 Some number theory inspired by Pythagoras’ theorem 20.4 Some (sandbox) applications . . . . . . . . . . . . . . 20.5 The parallelogram equation . . . . . . . . . . . . . . 21 Trigonometry 21.1 The law of sines . . . . . . . . . 21.2 The law of cosines . . . . . . . . 21.3 The addition theorem for tangent 21.4 Unwound angles . . . . . . . . . 21.5 Extension to unwound angles . . 21.6 Mollweid’s formulas . . . . . . .
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22 Measurement of the Circle 445 22.1 Approximate measurement of a circular arc . . . . . . . . . . . . . . . . . . . . 445 22.2 Arc length and area of a circular sector . . . . . . . . . . . . . . . . . . . . . . 449
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23 Area in Euclidean Geometry 23.1 Equidecomposable and equicomplementable figures 23.2 The Theorem of Pythagoras and related results . . . 23.3 Dudeney’s dissection problem . . . . . . . . . . . . 23.4 Euclidean area . . . . . . . . . . . . . . . . . . . . 23.5 The role of the Archimedean axiom . . . . . . . . . 23.6 Some uniqueness results for justification . . . . . . 23.7 About the volume of polyhedra . . . . . . . . . . . 23.8 Heron’s formula for the area of a triangle . . . . . . 23.9 Algebraic relations for the pieces of a triangle . . .
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453 453 454 459 465 468 469 472 473 474
24 Coordinate Planes—Descartes’ Road from Algebra to Geometry 24.1 A Hierarchy of Cartesian planes . . . . . . . . . . . . . . . 24.2 About Archimedean fields . . . . . . . . . . . . . . . . . . 24.3 Congruence in a Pythagorean plane . . . . . . . . . . . . . 24.4 Transfer of an angle . . . . . . . . . . . . . . . . . . . . . 24.5 Enough rigid motions exist in a Pythagorean plane . . . . . 24.6 Euclidean fields and intersection properties of circles . . . .
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476 476 479 479 481 483 486
25 Standard Euclidean Triangle Geometry 25.1 The circum-center . . . . . . . . . . . . . . . . 25.2 Double and half size triangles . . . . . . . . . . 25.3 The centroid . . . . . . . . . . . . . . . . . . . 25.4 The orthocenter . . . . . . . . . . . . . . . . . . 25.5 The in-circle and the three ex-circles . . . . . . . 25.6 The road to the orthocenter via the orthic triangle 25.7 The Euler line . . . . . . . . . . . . . . . . . .
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492 492 494 496 496 498 498 500
26 Harmonic Points 26.1 The Theorems and Menelaus and Cevá 26.2 The circle of Apollonius . . . . . . . . 26.3 An application to electrostatics . . . . . 26.4 The perspective view . . . . . . . . . .
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502 502 506 513 515
27 Advanced Euclidean Geometry 27.1 A Euclidean egg . . . . . . . . . . . . . . . . . . 27.2 The egg built from inside . . . . . . . . . . . . . . 27.3 A triangle construction using the sum of two sides 27.4 Archimedes’ Theorem of the broken chord . . . . 27.5 The Theorem of Collignon . . . . . . . . . . . . . 27.6 Vectors and special quadrilaterals . . . . . . . . . 27.7 The Theorem of Ptolemy . . . . . . . . . . . . . . 27.8 The quadrilateral of Hjelmslev . . . . . . . . . . .
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520 520 522 524 526 532 533 535 539
28 The Regular Pentagon 28.1 The Euclidean construction with the Golden Ratio 28.2 Relation between the sides of pentagon and 10-gon 28.3 The construction with Hilbert tools . . . . . . . . 28.4 Variants of the Euclidean construction . . . . . . . 28.5 A false pentagon . . . . . . . . . . . . . . . . . .
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543 543 545 549 553 554
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29 Circles, Tangents, Power and Inversion 29.1 The equipower line of two circles . . . . . . . . 29.2 Common tangents of two circles . . . . . . . . . 29.3 Definition and construction of the inverted point 29.4 The gear of Peaucollier . . . . . . . . . . . . . . 29.5 Invariance properties of inversion . . . . . . . .
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557 557 562 563 564 566
30 A Glimpse at Elliptic Geometry 30.1 Elliptic geometry is derived from spherical geometry 30.2 The conformal model . . . . . . . . . . . . . . . . . 30.3 Falsehood of the exterior angle theorem . . . . . . . 30.4 Area of a spherical triangle . . . . . . . . . . . . . . 30.5 Does Pythagoras’ imply the parallel postulate? . . . 30.5.1 Main results obtained by calculus . . . . . 30.5.2 Examples for the "lunes of Pythagoras" . 30.6 The stereographic projection . . . . . . . . . . . . . 30.6.1 The stereographic projection is an inversion . . 30.6.2 Many congruent angles . . . . . . . . . . . . .
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572 572 573 575 577 580 580 580 585 585 587
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31 Euclidean Constructions with Restricted Means 592 31.1 Constructions by straightedge and unit measure . . . . . . . . . . . . . . . . . . 592 31.2 Tools equivalent to straightedge and compass . . . . . . . . . . . . . . . . . . . 597 31.3 Hilbert tools and Euclidean tools differ in strength . . . . . . . . . . . . . . . . 601 32 Trisection of an Angle and the Delian Problem 32.1 Trisection by Archimedes . . . . . . . . . . . . . 32.2 Trisection by Nicomedes . . . . . . . . . . . . . . 32.3 Trisection with Nicolson’s angle, and by origami . 32.4 Construction of the cubic root by two-marked ruler 32.5 Definition and equations of the conchoid . . . . . 32.6 The conchoid and the construction of the cube root 32.7 Duplication of the cube by two-marked ruler . . . 32.8 Duplication of the cube and the curve of Agnesi . 32.9 A close look at Nicomedes’ trisection . . . . . . . 32.10 Trisection and the conchoid . . . . . . . . . . . . 32.11 Archimedes trisection yields more solutions . . . . 32.12 Archimedes’ trisection and the conchoid . . . . .
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604 604 606 609 613 615 617 617 622 623 624 627 629
33 The Heptagon 33.1 The construction . . . . . . . . . . 33.2 Trigonometric calculations . . . . . 33.3 Using complex numbers . . . . . . 33.4 Geometric proof . . . . . . . . . . 33.5 A false but almost regular heptagon 33.6 Plemelj’s construction . . . . . . .
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632 632 633 635 636 637 638
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34 Geometric Constructions and Field Extensions 34.1 A reminder about polynomials . . . . . . 34.2 Algebraic numbers . . . . . . . . . . . . 34.3 The dimension of a field extension . . . . 34.4 The field of algebraic numbers . . . . . . 34.5 Minimal fields . . . . . . . . . . . . . . 34.6 Towers in the constructible field . . . . . 34.7 Totally real extensions . . . . . . . . . . 34.8 Duplication of a cube . . . . . . . . . . . 34.9 Angle trisection . . . . . . . . . . . . . . 34.10 Regular polygons . . . . . . . . . . . . . 34.11 The 17-gon construction . . . . . . . . . 34.12 A short paragraph about Fermat numbers 34.13 Eisenstein’s irreducibility criterium . . . 34.14 The constructible polygons . . . . . . . .
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641 641 642 643 645 647 648 649 652 653 655 655 659 660 663
35 The Lunes of Hippocrates 35.1 The historic lunes . . . . . . . . . . . . 35.2 Historic remark . . . . . . . . . . . . . 35.3 Some historic and less historic exercises 35.4 The lune equation . . . . . . . . . . . 35.5 Vieta’s and Euler’s lunes . . . . . . . .
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666 666 673 674 678 685
36 Transcendental Numbers and the Constructibility of the Lunes 36.1 About transcendental numbers . . . . . . . . . . . . . . . 36.2 Which lunes are algebraic, which constructible? . . . . . 36.3 Irreducibility of the lune equation . . . . . . . . . . . . . 36.4 Tschebychev polynomials . . . . . . . . . . . . . . . . .
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688 688 691 692 693
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37 Alhazen’s problem 698 37.1 Elementary cases of Alhazen’s problem . . . . . . . . . . . . . . . . . . . . . . 698 37.2 Trigonometric solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703 38 The Quadrix of Hippias 705 38.1 The quadratrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 38.2 Constructible points on the quadratrix . . . . . . . . . . . . . . . . . . . . . . . 705 39 Advanced Euclidean Triangle Geometry 39.1 Morley’s Theorem . . . . . . . . 39.2 The Euler line . . . . . . . . . . 39.3 Proof of Euler’s Theorem . . . . 39.4 Proof of the Nine-Point Theorem 39.5 Proof of Feuerbach’s Theorem . . 39.6 Additional questions . . . . . . . 39.7 The Simson line . . . . . . . . .
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711 711 713 714 716 718 724 726
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9
Hyperbolic Geometry
731
40 Hyperbolic Geometry in the Poincaré Model 40.1 Inversion . . . . . . . . . . . . . . . . . . . . 40.2 Points and lines in the Poincaré model . . . . . 40.3 Introduction of metric properties . . . . . . . . 40.4 The angle of parallelism . . . . . . . . . . . . 40.5 Hyperbolic reflection . . . . . . . . . . . . . . 40.6 Proof of the SAS axiom via the Poincaré model 40.7 Their are enough rigid motions . . . . . . . . . 40.8 Horocycle . . . . . . . . . . . . . . . . . . . . 40.9 Circles and hypercircles . . . . . . . . . . . . 40.10 We have obtained all circle-like curves . . . . 40.11 Towards the Klein model . . . . . . . . . . . .
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732 732 732 736 738 741 743 745 747 749 751 752
41 Geometric Constructions in the Poincaré Disk 41.1 Basic constructions from neutral geometry 41.2 Typically hyperbolic constructions . . . . . 41.3 Circle constructions . . . . . . . . . . . . 41.4 Triangle constructions . . . . . . . . . . . 41.5 The altitudes and the orthocenter . . . . . .
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754 754 759 760 764 767
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771 771 776 782 790 794
42 Hyperbolic Geometry in Klein’s Model 42.1 Setup of Klein’s model . . . . . . 42.2 Angle of parallelism . . . . . . . 42.3 Projective nature of Klein’s model 42.4 Engel’s Theorem . . . . . . . . . 42.5 The Hjelmslev quadrilateral . . .
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43 Hyperbolic Triangle Geometry 43.1 Hyperbolic trigonometry . . . . . . . . . . . . 43.2 The orthocenter . . . . . . . . . . . . . . . . . 43.3 About the circum-circle . . . . . . . . . . . . 43.4 Thales’ Theorem in hyperbolic geometry . . . 43.5 The centroid . . . . . . . . . . . . . . . . . . 43.6 The in-circle, ex-circles, and the orthic triangle
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797 797 800 802 805 809 810
44 The Poincaré Half-Plane Model 44.1 Poincaré half-plane and Poincaré disk . . 44.2 The Euler-Lagrange equation . . . . . . 44.3 The curve of minimal hyperbolic length . 44.4 The minimum of hyperbolic length . . . 44.5 Some useful reflections in the half-plane .
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813 813 815 817 818 819
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45 Hilbert’s Axiomatization of Hyperbolic Geometry 45.1 Some basic facts about semi-hyperbolic planes . . . . . . . . 45.1.1 The equivalence relation of limiting parallelism . . . . . 45.1.2 Limiting triangles . . . . . . . . . . . . . . . . . . . . . 45.2 The hyperbolic parallel postulate . . . . . . . . . . . . . . . . 45.3 Construction of the common perpendicular . . . . . . . . . . 45.4 The enclosing line . . . . . . . . . . . . . . . . . . . . . . . 45.5 The three reflections Theorem . . . . . . . . . . . . . . . . . 45.6 Construction of Hilbert’s field of ends . . . . . . . . . . . . . 45.6.1 Addition of ends . . . . . . . . . . . . . . . . . . . . 45.6.2 Multiplication of ends . . . . . . . . . . . . . . . . . 45.6.3 The distributive law . . . . . . . . . . . . . . . . . . 45.7 Reconstruction of the complex field and the half-plane model 45.8 The angle unboundedness postulate . . . . . . . . . . . . . .
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821 821 822 827 829 830 832 834 839 839 840 841 842 844
46 Gauss’ Differential Geometry and the Pseudo-Sphere 46.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 46.2 About Gauss’ differential geometry . . . . . . . . . . . 46.3 Riemann metric of the Poincaré disk . . . . . . . . . . . 46.4 Riemann metric of Klein’s model . . . . . . . . . . . . 46.5 A second proof of Gauss’ remarkable theorem . . . . . 46.6 Principal and Gaussian curvature of rotation surfaces . . 46.7 The pseudo-sphere . . . . . . . . . . . . . . . . . . . . 46.8 Poincaré half-plane and Poincaré disk . . . . . . . . . . 46.9 Embedding the pseudo-sphere into Poincaré’s half-plane 46.10 Embedding the pseudo-sphere into Poincaré’s disk . . . 46.11 About circle-like curves . . . . . . . . . . . . . . . . . 46.12 Mapping the boundaries . . . . . . . . . . . . . . . . .
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846 846 846 847 851 854 857 859 862 863 864 865 867
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References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
Solomon Feferman Anita B. Feferman, Alfred Tarski: Life and Logic, Cambridge University Press, 2004. Michael Artin, Algebra, second ed., Prentice Hall, 2011. Alan Baker, Transcendental Number Theory, Cambridge University Press, 1990. R. Baldus and F. Löbell, Nichteuklidische Geometrie, Hyperbolische Geometrie der Ebene, Sammlung Göschen 970, Berlin, 1964. Arthur Baragar, Constructions using a compass and twice-notched straightedge, The American Mathematical Monthly 109 (2002), 151–164. Heinrich Behnke, Rückblick auf die Geschichte der Mathematischen Annalen, Math. Ann. (1973), I–VII. Paul Bernays, Betrachtungen über das Vollständigkeitsaxiom und verwandter Axiome, Math. Zeitschrift 63 (1955), 219–292. Edward B. Burger and Robert Tubbs, Making Transcendence Transparent, Springer, 2004. Barry A. Cipra, Computer search solves an old math problem, Science 242 (1988), 1507–1508. Th. Clausen, Vier neue mondförmige Flächen, deren Inhalt quadrirbar ist, Crelle 21 (1840), 375–376. H.S.M. Coxeter and S.L. Greitzer, Geometry Revisited, ninth printing, New Mathematical Library 19, The Mathematical Asociation of America, 1967. H.E. Dudeney, The Canterbury Puzzles, and other curios problems, Nelson, London, 1929. E.H.Moore, Trans. Math. Soc. (1902). G. Feigl, Über die elementaren Anordnungssätze der Geometrie, Jahresbericht der Deutschen Math.-Vereinigung 33 (1924). F.R.Moulton, A simple non-desarguesian plane geometry, Trans. Math. Soc. (1902). Chaim Goodman-Strauss, Compass and straightedge in the Poincaré disk, The American Mathematical Monthly 108 (2001), 38–49.
Topics from Geometry [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40]
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Marvin J. Greenberg, Euclidean and Non-Euclidean Geometry, fourth ed., W.H. Freeman and Company, New York, 2008. , Old and new results in the foundations of elementary plane Euclidean and Non-Euclidean geometries, The American Mathematical Monthly 117 (2010), 198–219. Robin Hartshorne, Geometry: Euclid and Beyond, second printing, Springer, 2002. , Non-Euclidean iii.36, The American Mathematical Monthly 110 (2003), 495–502. David Hilbert, Foundations of Geometry, 2nd English ed., Open Court, La Salle, 1971. , Grundlagen der Geometrie, 14. Auflage, B.G.Teubner, Stuttgart, Leibzig, 1999. Ross Honsberger, Episodes in nineteenth and twentieth century Euclidean geometry, New Mathematical Library, vol. 37, The Mathematical Association of America, 2005. Roger A. Johnson, Advanced Euclidean Geometry, second printing, Dover Publications, Inc., Mineola, New York, 2007. Wilbur Richard Knorr, The Ancient Tradition of Geometric Problems, Dover Publications, 1945/1993. J. Kürschàk, Das Streckenabtragen, Math. Ann. 55 (1902). Stefan Mykytiuk and Abe Shenitzer, Four significant axiomatic systems and some of the issues associated with them, The American Mathematical Monthly 102 (1995), 62–67. David Park, The Grand Contraption— the world in myth, number, and chance, second printing , Princeton University Press, 2005. M. M. Postnikov, The problem of squarable lunes, The American Mathematical Monthly 107 (2000), 645–651. Kenneth H. Rosen (editors), Handbook of Discrete and Computational Mathematics, CRC Press, 2000. A. Rosenthal, Math. Ann. 71. Alexander Ryba and Joseph Stern, Equimodular polynomials and the tritangency theorems of Euler, Feuerbach and Guinand, The American Mathematical Monthly 118 (2011), 217–228. James R. Smart, Modern Geometries, fifth ed., Brooks/Cole Publishing Company, 1997. John Stillwell, The Four Pillars of Geometry, Springer, 2005. Rüdiger Thiele, Hilbert’s twenty-fourth problem, The American Mathematical Monthly 110 (2003), 1–24. N. Tschebotarow, Über quadrierbare Kreisbogenzweiecke, Math. Zeitschrift 39 (1934), 161–175. H. Tverberg, A proof of the Jordan Curve Theorem, Bull. Lond. Math. Soc. 12 (1980), 34–38. B.L. van der Waerden, Science Awakening I, Kluwer Academic Publishers, 1954. O. Veblen and J.H. Maclagan-Wedderburn, Non-desarguesian and non-pascalian geometries, Transactions of the American Mathematical Society 8 (1907), 379–388. Robert S. Wolf, A Tour through Mathematical Logic, the carus mathematical monographs 30, Mathematical Association of America, 2005.
Franz Rothe, Department of Mathematics and Statistics University of North Carolina at Charlotte Charlotte, NC 28223 e-mail:
[email protected] January 2, 2017
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F. Rothe
0. A Synopsis of Euclid Only the very important items are explained below. 0.1.
Book I
Definitions 1. A point is that which has no part. 2. A line is length without breadth. 3. — — 4. A straight line lies evenly with its points. 5. — — 6. — — 7. — — 8. A plane angle is the inclination of two lines. 9. — — 10. When two adjacent angles are equal, they are right angles. 11. — — 12. — — 13. — — 14. — — 15. A circle is a line all of whose points are equidistant from a point. 16. — — 17. — — 18. — — 19. — — 20. A triangle with two equal sides is isosceles. 21. — — 22. — — 23. Parallel straight lines are lines in the same plane that do not meet, no matter how far extended in either direction.
Topics from Geometry
13
Postulates 1. To draw a line through two points. 2. To extend a given line. 3. To draw a circle with given center through a given point. 4. All right angles are equal. 5. If a line crossing two other lines makes the sum of the interior angles on the same side less than two right angles, then these two lines will meet on that side when extended far enough. Common Notions 1. Things equal to the same thing are equal to each other. 2. Equals added to equals are equal. 3. Equals subtracted from equals are equal. 4. Things which coincide are equal. 5. The whole is greater than the part. Propositions 1. To construct an equilateral triangle on a given segment. 2. To draw a segment equal to a given segment at a given point. 3. To cut off a smaller segment from a larger segment. 4. Side-angle-side (SAS) congruence for triangles. 5. The base angles of an isosceles triangle are equal. 6. If the base angles are equal, the triangle is isosceles. 7. It is not possible to put two triangles with equal sides on the same side of a segment. 8. Side-side-side (SSS) congruence for triangles. 9. To bisect an angle. 10. To bisect a segment. 11. To construct a perpendicular to a line at a given point on the line. 12. To drop a perpendicular from a point to a line not containing the point. 13. A line standing on another line makes angles with sum equal to two right angles. 14. — —
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15. Vertical angles are equal. 16. The exterior angle of a triangle is greater than either opposite interior angle. 17. Any two angles of a triangle have a sum less than two right angles. 18. If one side of a triangle is greater than another, then the angle opposite to the greater side is greater than the angle opposite to the smaller side. 19. If one angle of a triangle is greater than another, then the side opposite to the greater angle is greater than the side opposite to the smaller angle. 20. The sum of any two sides of a triangle is greater than the third side. 21. — — 22. To construct a triangle from all three sides given— provided the sum of any two sides is greater than the remaining side. 23. To produce a given angle at a given point and side. 24. For two triangles with two pairs of equal sides, the triangle with the greater angle between them has the greater opposite side. 25. For two triangles with two pairs of equal sides, the triangle with the greater third side has the greater angle between the two matched sides. 26. Angle-side-angle (ASA) and angle-angle-side (AAS) congruence for triangles. 27. If the two alternate interior angles between a transversal and two lines are equal, then these two lines are parallel. 28. If the exterior angle between a transversal and the first line and the opposite interior angle between the transversal and the second line have sum two right angles, then the two lines are parallel. 28a. If the two interior angles between a transversal and two lines have sum two right angles, then the two lines are parallel. 29. A transversal of two parallel lines makes equal alternate interior angles with the two parallels. 30. Lines parallel to the same line are parallel. 31. To draw a line parallel to a given line through a given point. 32. The sum of angles of a triangle is two right angles. 32a. An exterior angle equals the sum of the two opposite interior angles of a triangle. 33. The lines joining the endpoints of two parallel segments are parallel and the segments cut out by the first two parallels are congruent. 34. The opposite sides and angles of a parallelogram are equal.
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35. Parallelograms on the same base and in the same parallels have equal areas. 36. Parallelograms on equal bases and in the same parallels have equal areas. 37. Triangles on the same base and in the same parallels have equal areas. 38. Triangles on equal bases and in the same parallels have equal areas. 39. Triangles with equal areas on the same base lie in the same parallels. 40. Triangles with equal areas on equal bases lie in the same parallels. 41. The area of a parallelogram is twice the area of a triangle on the same base and in the same parallels. 42. To construct a parallelogram with a given angle and area equal to the area of a given triangle. 43. The triangles on opposite sides of a diagonal of a parallelogram are equal. 44. To construct a parallelogram with given side and angle and area equal to the area of a given triangle. 45. To construct a parallelogram with a given angle and area equal to the area of a given figure. 46. To construct a square on a given segment. 47. (Theorem of Pythagoras) The square on the hypothenuse is equal in area to the sum of the squares on the legs of the triangle. 48. If the sum of the areas of squares on two sides of a triangle is equal to the area of the square on the third side, then the triangle is right. 0.2.
Book II. Propositions
1. The rectangle contained by two lines is the sum of the rectangles contained by one and the segments of the other. 4. The area of the square on the whole segment is equal to the sum of the areas on the two parts of the segment plus twice the area of the rectangle on these two parts. 5. The area of the square on the first longer part of a segment is equal to the area of the rectangle with the difference of the parts and the entire segment as sides, plus the area of the square on the second part. 1 6. The area of a rectangle with one side a longer segment and an added piece, the second side the added piece plus the area of the square on half of the longer segment add up to the area of the square on the half plus the added piece. 2 11. To cut a longer segment so that the rectangle on the whole and the part is equal to the square on the other part. 3 14. To construct a square with area equal to that of a given figure. 1 2 3
p2 = [p − (c − p)]c + (c − p)2 2 2 (c + p)p + 2c = 2c + p Solve cx = (c − x)2 for the part x.
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0.3.
F. Rothe
Book III. Propositions
1. To find the center of a circle. 2. The segment joining two points of a circle lies inside the circle. 5. If two circles intersect, they do not have the same center. 6. If two circles are tangent, they do not have the same center. 10. Two circles can intersect in at most two points. 11. 12. If two circles are tangent, their centers lie in a line with the point of tangency. 16. The line perpendicular to a diameter at its endpoint is tangent to a circle, and the "angle between the tangent line and the circle" 1 is less than any "rectilineal angle". 17. To draw a tangent to a circle from a point outside the circle. 18. A tangent line to a circle is perpendicular to the radius at the point of tangency. 19. The perpendicular to a tangent line at the point of tangency will pass through the center of a circle. 20. The angle at the center is twice the angle at a point of the circumference subtending a given arc of a circle. 21. Two angles from points of a circle subtending the same arc are equal. 22. The opposite angles of a quadrilateral in a circle have sum equal to two right angles. 31. The angle in a semicircle is a right angle. 32. The angle between a tangent line and a chord of a circle is equal to the angle on the arc cut off by the chord. 35. If two chords cut each other inside the circle, the rectangle on the segments on one chord is equal in area to the rectangle on the segments on the other chord. 36. From a point outside a circle, let a tangent and a secant be drawn. The square on the segment on the tangent is equal in area to the rectangle formed by the two segments from the point outside of the circle to the endpoints of the secant. 37. From a point outside a circle are drawn two lines cutting or touching the circle. If the square on the segment on one line is equal in area to the rectangle formed by the two segments from the point outside to the endpoints of the secant on the other line, then the first line is tangent to the circle. 1
The notion in quotes is not well defined by today’s standard.
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0.4.
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Book IV. Propositions
1. To inscribe a given segment in a circle. 2. To inscribe a triangle, equiangular to a given triangle, in a circle. 3. To circumscribe a triangle, equiangular to a given triangle, around a circle. 4. To inscribe a circle in a triangle. 5. To circumscribe a circle around a triangle. 10. To construct an isosceles triangle whose base angles are twice the vertex angle. 11. To inscribe a regular pentagon in a circle. 12. To circumscribe a regular pentagon around a circle. 15. To inscribe a regular hexagon in a circle. 16. To inscribe a regular 15-sided polygon in a circle. 0.5.
Book V
Definitions 4. Magnitude are said to have a ratio if either one, being multiplied, an exceed the other. 5. Four magnitudes a, b, c, d are in the same ratio a : b equal to c : d, if and only if for any whole numbers m and n either one of these three cases occurs:
0.6.
both
ma > nb and mc > nd,
both
ma = nb and mc = nd,
both
ma < nb and mc < nd.
Book VI. Propositions
1. The areas of triangles of the same height are in the same ratio as their bases. 2. A line is parallel to the base of a triangle if and only if it cuts the two other sides in the same ratio—proportionately. 3. A line from the vertex of a triangle to the opposite side bisects the angle if and only if it cuts the opposite side in proportion to the remaining sides of the triangle. 4. The sides of equiangular triangles are proportional. 5. If the sides of two triangles are proportional, their corresponding angles are pairwise equal. 6. If two triangles have one pair of equal angles, and the sides containing these pair are proportional, then the triangles are similar. 8. The altitude from the right angle of a right triangle divides it into two triangles similar to each other and to the whole triangle.
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12. To find a fourth proportional for three given segments. 13. To find a mean proportional between two given segments. 16. Four segments are proportional if and only if the rectangle on the extremes is equal in area to the rectangle on the means. 30. To cut a segment in extreme and mean ratio.
1
31. Any figure on the hypothenuse of a right triangle has area equal to the sum of the areas of similar figures on the legs of the triangle. 0.7.
Book X
1. Given two unequal quantities, if one subtracts from the greater one the smaller one, and repeats this process enough times, there will remain a quantity lesser than the smaller of the two original quantities. 117. 0.8.
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The diagonal of a square is incommensurable with its side. Book XI
Definitions 25. A cube is a polyhedron made of six equal squares. 26. An octahedron is a polyhedron made of eight equal equilateral triangles. 27. An icosahedron is a polyhedron made of twenty equal equilateral triangles. 28. An dodecahedron is a polyhedron made of twelve equal regular pentagons. Propositions 21. The plane angles in a solid angle make a sum less than four right angles. 28. A parallelepiped is bisected by its diagonal plane. 29. 30. Parallelepipeds on the same base and of the same height are equal in volumes. 31. Parallelepipeds on equal bases and of the same height are equal in volumes. 0.9.
Book XII. Propositions
2. The areas of circles are in the same ratio as the squares on their diameters. 3. A pyramid is divided into two pyramids and two prisms. 5. Pyramids of the same height on triangular bases have volumes in the ratio of the areas of the bases. 6. A prism with triangular bases is divided into three triangular pyramids with equal volumes. 1 2
Solve c : x = x : p for x. not in Heath, but in Commandino
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0.10.
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Book XIII. Propositions
7. If at least three angles of an equilateral pentagon are equal, the pentagon is regular. 10. In a circle, the square on the side of the inscribed pentagon is equal in area to the square on the side of the inscribed hexagon plus the square on the inscribed decagon. 3 13. To inscribe a tetrahedron in a sphere. 14. To inscribe a octahedron in a sphere. 15. To inscribe a cube in a sphere. 16. To inscribe an icosahedron in a sphere. 17. To inscribe a dodecahedron in a sphere. 18. (Postscript) Besides these five figures there is no other contained by equal regular polygons.
3
a25 = R2 + a210
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F. Rothe
1. On the History of Translations and Editions of Euclid’s Elements "Jüngling, merke dir in Zeiten, Wo sich Geist und Sinn erhöht: Dass die Muse zu g e l e i t e n , Doch zu l e i t e n nicht versteht." Johann Wolfgang Goethe The highly acclaimed Elements is simply a huge collection—divided into 13 books—of 465 propositions from plane and solid geometry. Today, it is generally agreed that relatively few of these theorems were Euclid’s own invention. Rather, by compiling, editing, and perfecting the known body of Greek mathematics, he created a superbly organized treatise. The Elements were so successful and revered that they thoroughly obliterated all preceding works of its kind. They have become a lasting contribution to mankind. It is a misfortune that no copy of Euclid’s Elements has been found that actually dates from the author’s own time. Hence his writings have had to be reconstructed from numerous recensions, commentaries, and remarks by other writers. Proclus makes it clear that the Elements were highly valued in Greece and refers to numerous commentaries on it. Among the most important must have been those by Heron (100 B.C. - A.D.100), Porphyry and Pappus (third century A.D.). Modern editions of the work are based upon a revision that was prepared by the Greek commentator Theon of Alexandria (end of the fourth century A.D). He lived almost 700 years after the time of Euclid. Until 1808, Theon’s revision was the oldest edition of the Elements known. In 1808, Napoleon ordered valuable manuscripts to be taken from Italian libraries and send to Paris. At that occasion, Francois Peyrard found, in the Vatican library, a tenth-century copy of an edition of Euclid’s Elements that predidates Theon’s recension. The historians J.L.Heiberg and Thomas L. Heath have used principally this manuscript. A study of this older edition and a careful sifting of citations and remarks made by early commentators indicate that the introductory material of Euclid’s original treatise undoubtedly underwent some editing in the subsequent revisions. But the propositions and their proofs have remained as Euclid wrote them, except for minor additions and deletions. There are also Arabic translations of Greek works and Arabic commentaries, presumably based on Greek manuscripts no longer available. The first really satisfying Arabic translation of the Elements was done by Tâbit ibn Qorra (826-901). These, too, have been used to decide what was in Euclid’s original. But the Arabic translations and revisions are on the whole inferior to the Greek manuscripts. The first complete Latin translation of the Elements were not made from the Greek but from Arabic. In the eighth century, a number of Byzantine manuscripts of Greek works were translated by the Arabians. In 1120, the English scholar Adelard of Barth (ca. 1075-1160) made a Latin translation of the Elements from one of the older Arabian translations. Other Latin translations were made from the Arabic by Gherardo of Cremona (1114-1187) and, 150 years after Adelard, by Johannes Campanus. Euclid’s Elements were widely read, translated and changed by people all over the world. From 1601 to 1607, Mattes Ricci (1552-1610) from Italy and Hsü Kuang-ching (1562-1634) translated the first six books into Chinese. This played a significant role in the subsequent development of mathematics in China. The first printed edition of the Elements was made at Venice in 1482 and contained Campanus’ translation. This very rare book was beautifully executed. It was the first mathematical book of any
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consequence to be printed. In 1574, Clavius (1537-1612) published an edition that is valuble for its extensive scholia. An important Latin translation from the Greek was made by Commandino in 1572. This translation served as a basis for many subsequent translations, including the very influential work by Robert Simson. From the latter, many English editions were derived. The first complete English translation of the Elements was the monumental Billingsley translation issued in 1570. Nicolaus Mercator (ca. 1620-1687) published an edition, along with his works in trigonometry, astronomy and other topics. Adrien-Marie Legendre (1752-1833) is known for his very popular Éléments de géométrie, in which he attempted a pedagogical improvement of Euclid’s Elements by considerably rearranging and simplifying many of the propositions. He used some algebra not in the original. The high school versions most widely used during our century are patterned on Legendre’s modification of Euclid’s work. Euclid’s Elements is to geometry what the Bible is to Christianity. No work, except the Bible, has been more widely used, edited, or studied than Euclid’s Elements. Over one thousand editions of Euclid’s Elements have appeared since the first one printed in 1482. For more than two millennia, this work has dominated all teaching of geometry. Abraham Lincoln is credited with learning logic by studying Euclid’s Elements, as his biographer Carl Sandburg tells. In his autobiography, Bertrand Russell (1872-1970) penned this remarkable recollection: "At the age of eleven, I began Euclid, with my brother as tutor. This was one of the great events of my life, as dazzling as first love." Probably no work has exercised a greater influence on scientific thinking.
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Part I
Neutral Geometry January 2, 2017
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2. Hilbert’s Axioms of Geometry 2.1. Logic This subsection is too short to do justice to its topic. A comprehensive introduction into mathematical logic may be found in the book "A Tour through Mathematical Logic" by Robert S. Wolf, see [40]. By a statement is meant any sentence for which it is meaningful to consider its truth or falsehood. 1 Thus any statement can be either true or false,— but a statement is not automatically assumed to be true. Often, I shall put quotation marks around statements in order to remind the reader to this important remark. Let A and B denote any statements. The negation of a statement is true if and only if the statement itself is false. The negation of formula A is denoted by ¬A or not(A) or simply not A. By A ≡ B, we denote a true equivalence. Thus A ≡ B means that it is true that either formulas A and B are both true, or formulas A and B are both false. 2 A formal implication A ⇒ B is defined by a truth table—without assuming or presupposing any causal relation between the statements A and B. According to such a definition, the implication A ⇒ B becomes false if and only if statement A is true and statement B is false. Thus ¬(A ⇒ B) ≡ A and ¬B A ⇒ B ≡ ¬A or B The statements A ⇒ B and A ⇒ ¬B can both be true simultaneously. They are not negations of each other. Indeed (A ⇒ B) and (A ⇒ ¬B) ≡ A is false Further considerations of such a handling of true and false statements is the topic of propositional logic. Problem 2.1. Give the converse of the following sentences: "If Peter drives fast, he does not read the traffic signs." Answer. "If Peter does not read the traffic signs, then he drives fast." "If Tom reads the traffic signs, then he drives fast." Answer. "If Tom drives fast, then he reads the traffic signs." Problem 2.2. Give the negation of the following sentences in a clear and simple form. You are not supposed to simply say that the statement is false! "Bill drives slowly and he reads the traffic signs." Answer. "Either Bill drives fast or he does not read the traffic signs." A second possible answer. "If Bill drives slowly, he does not read the traffic signs." "If Peter drives fast, he does not read the traffic signs." Answer. "Peter drives fast and he reads the traffic signs." "If Tom reads the traffic signs, then he drives fast." 1 2
This is a rather ad hoc definition, but I cannot find any better one. It would be good to have a different notation for an arbitrary equivalence, which may be true or not.
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Answer. "Tom reads the traffic signs, but he does not drive fast." "If Paul drives fast, then he does not read the traffic signs." Answer. "Paul drives fast and he reads the traffic signs." For the formulation of—at least part of—substantial mathematics, one needs more: the predicate logic which deals with variables and propositional functions. Indeed the predicate logic, also called quantified logic, is the Swiss pocket knife of mathematical logic. In principle, it is not necessary to specify the meaning or kind of a variable. The important assumption is an infinite supply of values for any variable. The propositional functions are properties which may be true or false depending on the value of the variable x. They are denoted by P(x). Such a sentence P(x) is also called a predicate, because in the English language the property is grammatically a predicate. The universal quantification of a predicate P(x) is the statement "For all values of x, the predicate P(x) is true." This universal quantification of P(x) is denoted by ∀xP(x). The symbol ∀ is called the universal quantifier. The existential quantification of a predicate P(x) is the statement "There exist values of x for which the predicate P(x) is true." This existential quantification of P(x) is denoted by ∃xP(x). The symbol ∃ is called the existential quantifier. Problem 2.3. Find the pair of equivalent statements. Find all pairs of statement and its negation and mark them with matching color. How many pairs (colors) are there? Find the statements, of which the negation is not listed and encircle them with a closed line. 1. "If a triangle is isosceles, the base angles are congruent." 2. "All men like to drive." 3. "If a triangle is isosceles, one base angle is larger than the other one." 4. "Some men do not like to drive." 5. "It rains and the streets are dry." 6. "No woman likes to drive." 7. "Some women like to drive." 8. "There exists a triangle with sum of its angles equal to two right angles." 9. "Every triangle has sum of angles either less or more than two right angles." 10. "No triangle is equilateral." 11. "All triangles are equilateral." 12. "No point has three or more lines passing through it." 13. "There exists a point through which at most two lines pass." 14. "There exists a point through which three or more lines pass." 15. "Every point has at most two lines passing through it." 16. "If it rains, the streets are wet." Answer. Statements (12) and (15) are equivalent. Five pairs of statement and negation are [(2)(4)] [(5)(16)] [(6)(7)] [(8)(9)] [(12)(14)], or equivalently [(14)(15)]. Statements (1)(3)(10)(11)(13) are not negated.
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Remark. It helps to write some statements in the symbols from predicate logic. Let Pn denote the predicate "Through point P pass exactly n lines." Here are four statements in symbolic logic: 12.
"No point has three or more lines passing through it." ¬∃P∃n (n ≥ 3 ∧ Pn)
13.
"There exists a point through which at most two lines pass." ∃P∃n (n ≤ 2 ∧ Pn)
14.
"There exists a point through which three or more lines pass." ∃P∃n (n ≥ 3 ∧ Pn)
15.
"Every point has at most two lines passing through it." ∀P∃n (n ≤ 2 ∧ Pn)
The term "axiomatic method" was coined by Hilbert to describe part of his formalist program. The essentials were already developed in ancient times in Euclid’s Elements, and refined in Hilbert’s foundations of geometry of 1899. Euclid’s Elements is the oldest surviving work in which mathematical subjects were developed from scratch in a thorough, rigorous and axiomatic way. He puts his principles at the beginning of the Elements and names them common notions and postulates. In place of the common notion, today are put the logical axioms. In place of the postulates, one has the proper axioms, which are specific to the first-order language and the subject under consideration. The formalist program goes beyond the classical axiomatic approach by explicitly defining not only the language and axioms to be used, but also the rules of inference. In the usual approach to first-order logic, two features of the rules of inference are worth noting: • first, they are based entirely on logic; • second, they are the only way of generating new steps and and proving any theorem. Without rules of inference no mathematics can be done—any axiomatic system would be useless. Classically, postulates were supposed to be evident truths. But truth does not enter into the formalist viewpoint. Nevertheless, even today, to build any meaningful mathematical theory, the formulas taken to be axioms should be very few, and be based on as simple as possible principles. They should be justified on intrinsic and extrinsic reasons, in other words well motivated and needed for the consequential development. 2.2. David Hilbert’s axiomatization of Euclidean geometry Already in 1899, on the occasion of the inauguration of the Gauss-Weber monument in Göttingen, David Hilbert had published his first edition of Foundations of Geometry. This work is considered to be the first version of Euclidean geometry that is truly axiomatic, in the sense that there were no hidden appeals to spatial intuition. Too, Hilbert’s work has much contributed to a deeper understanding of the relation between geometric and algebraic structures. This work has been leading for the clear axiomatic way of doing mathematics in the twentieth century.
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2.2.1. Introduction from Hilbert’s Foundations of Geometry We shall be lead to several, apparently very simple, but nevertheless very deep and difficult problems. We shall be challenged by very new and—as I believe fruitful—problems, and see remarkable connections between the elements of arithmetic and geometry, gaining another insight into the unity of mathematics. From the preface of Hilbert’s lecture notes "Foundations of Euclidean Geometry" (1898/99) Thus all human knowledge begins with intuition, proceeds to notions, and ends with ideas. Hilbert’s preamble, citing Kant’s "Critique of Pure Reason"(1781) Geometry needs—similar to arithmetic—only a few simple basic principles for its consequential development. These basic principles are called the axioms of geometry. Starting with Euclid, the setup of the axioms of geometry and the investigation of their mutual connections has been the subject of many excellent treatises in the mathematical literature. The problem in question is basically a logical analysis of our spatial imagination. The present—meaning Hilbert’s—investigation is a new attempt to set up a complete and as simple as possible system of axioms. And, furthermore, to deduct from them the most important geometric theorems, such that the meaning and importance of the different axioms and their consequences become clear. Hilbert’s introduction to "Foundations of Geometry" (1899) 2.2.2. Hilbert’s axioms As suggested in this paragraph, Hilbert introduces the modern habit to develop the consequences of the different groups of axioms immediately after introducing them. But—because I guess it is more convenient for the reader—we shall now state the complete axiomatic system at once, at the beginning. I have included all axioms, even those only needed for three dimensional geometry. 0. Undefined elements and relations Elements: • A class of undefined objects called points, denoted by A, B, C, . . . . • A class of undefined objects called lines, denoted by a, b, c, . . . . • A class of undefined objects called planes, denoted by α, β, γ, . . . . Relations: • Incidence (being incident, lying on, containing) • Order (lying between) (for points on a line) • Congruence (for segments and angles) Remark. Planes are only needed to include three dimensional geometry.
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I. Axioms of incidence I.1 For two points A and B there exists a line that contains each of the points A, B. I.2 For two [different] points A and B there exists no more than one line that contains each of the points A, B. I.3a On any line, there exist at least two points. I.3b There exist at least three points that do not lie on a line. Remark. We did separate axiom I.3 into I.3a and I.3b, in order to stress there is no direct logical connection between the two sentences intended. Remark. Additional axioms I.4 through I.8 are only needed for three dimensional geometry. I.4 For any three points A, B, C [that do not lie on the same line], there exists a plane α that contains each of the points A, B, C. I.4a For every plane there exists a point which it contains. I.5 For any three points A, B, C that do not lie on the same line, there exists no more than one plane that contains each of the points A, B, C. I.6 If two different points A , B of a line a lie on a plane α, then every point of line a lies in the same plane α. Remark. In this case, we say that the line a lies in the plane α. I.7 If two planes α, β have a point A in common, then they have at least one more point B in common. I.8 There exist at least four points which do not lie in a plane. II. Axioms of order II.1 If a point B lies between a point A and a point C, then the points A, B, C are three distinct points of a line, and B lies between C and A. II.2 For two points A and C, there also exists at least one point D on the line AC such that C lies between A and D. II.3 Of any three points on a line there exists no more than one that lies between the other two. DEFINITIONS: Segment, point of segment, interior and exterior of segment, ray, half plane, triangle. 1 Remark. A segment AB is assumed to have two different endpoints A and B. II.4 (Pasch’ Axiom) Let A, B, C be three points that do not lie on a line and let a be a line in the plane ABC which does not meet any of the points A, B, C. If the line a passes through a point of the segment AB, it also passes through a point of the segment AC, or through a point of the segment BC. 1
lateron: quadrilateral, polygon, side of polygon, vertex of polygon, closed and open polygon, simple polygon.
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III. Axioms of congruence III.1 If A, B are two points on a line a, and A0 is a point on the same or another line a0 , then it is always possible to find a point B0 on a given side of the line a0 through A0 such that the segment AB is congruent to the segment A0 B0 . In symbols AB A0 B0 . III.1b For any two different points A and B, the segments AB and BA are congruent.
1
III.2 If a segment A0 B0 and a segment A00 B00 are congruent to the same segment AB, then segment A0 B0 is also congruent to segment A00 B00 . III.3 On a line, let AB and BC be two segments which except for B have no point in common. Furthermore, on the same or another line a0 , let A0 B0 and B0C 0 be two segments which except for B0 also have no point in common. In that case, if AB A0 B0 and BC B0C 0 , then AC A0C 0 DEFINITION: Angle. III.4a Let ∠(h, k) be an angle in a plane α and h0 a ray in a plane α0 that emanates from the point O0 . Then there exists in the plane α0 one (and only one) ray k0 such that the angle ∠(h, k) is congruent to the angle ∠(h0 , k0 ) and at the same time all interior points of the angle ∠(h0 , k0 ) lie on the given side of h0 . This means that ∠(h, k) ∠(h0 , k0 ) III.4b The ray k0 in (III.4a) is unique. III.4c Every angle is congruent to itself ∠(h, k) ∠(h, k) III.4d Every angle is congruent to the angle with the legs switched ∠(h, k) ∠(k, h) III.5 If for two triangles 4ABC and 4A0 B0C 0 the congruences AB A0 B0 , AC A0C 0 , ∠BAC ∠B0 A0C 0 hold, then the congruence ∠ABC ∠A0 B0C 0 is also satisfied. IV. Axiom of parallelism IV.1 Let a be any line and A a point not on a. Then there exists at most one line in the plane determined by line a and point A that passes through A and does not intersect a. 1
I have added the item (III.1b) to Hilbert’s axioms.
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V. Axioms of continuity V.1 (Axiom of Archimedes) If AB and CD are any segments, then there exists a number n such that n segments congruent to CD constructed contiguously from A, along a ray from A through B, will pass beyond B. V.2 (Axiom of completeness) An extension of a set of points on a line, with its order and congruence relations existing among the original elements as well as the fundamental properties of line order and congruence that follow from Axioms I-III and from V.1, is impossible. Definition 2.1 (Hilbert plane). A Hilbert plane is any model for two-dimensional geometry where Hilbert’s axioms of incidence (I.1)(I.2)(I.3a)(I.3b), order (II.1) through (II.4), and congruence (III.1) through (III.5) hold. Neither the axioms of continuity—Archimedean axiom and the axiom of completeness—nor the parallel axiom need to hold for an arbitrary Hilbert plane. Definition 2.2 (Pythagorean plane). A Pythagorean plane is a Hilbert plane for which the axiom of parallelism (IV.1) holds.
Figure 2.1. Logical relations of geometries
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2.3. Importance and Impact of Hilbert’s Foundations of Geometry In this essay, we address the following questions: • Which features in Hilbert’s Foundations of Geometry are new and different from Euclid’s Elements? • Which topics are still not included, and what are the results achieved elsewhere about these topics. Of course looking at first in this manuscript, but more important beyond at the worldwide research? • Which important questions are still open? • Which benefits for mathematics have been achieved by Hilbert’s Foundations of Geometry? 2.3.1. Hilbert’s Foundations in Comparison with Euclid’s Elements As already stressed in his introduction, Hilbert’s goal is the consequential development of geometry. This is already an important justification for the axioms themselves, especially since no claim of absolute truth of the content of the axioms is intended. Secondly the style of presentation is determined by this goal. The axioms are presented in five groups: axioms of incidence, axioms of order, axioms of congruence, the axiom of parallelism, and the axioms of continuity. For each group, the theorems spelling out their consequences are proved, and examples of different structure are given. Such a study is done on purpose separately for the different groups. This approach is now commonplace in algebra or topology, but it is different from the style of classical geometry texts. The role of the primary elements and relations is seen differently from Euclid. For Euclid, these were abstract entities given by nature. They are given without question, nevertheless they are still explained by phrases like "A point has no parts." Modern mathematicians point out that such a sentence poses more questions than it answers. What is the way out to avoid questioning every and even the most basic notions and how can one break the infinite chain of regress? The key point is the use of primary elements and relations. These entities cannot, may not, and need not to be defined. They get their meaning only via the way they are used in the axioms, proofs and theorems. To start this process, not only primary elements, but moreover primary relations, too, have to been postulated, in order to get the connections between the abstract objects. In the Foundations of Geometry the points, lines and planes are used as primary objects. They are connected, by the relations of incidence, order, congruence of segments and congruence of angles. Furthermore, equality is a relation from mathematical logic. Their exist quite a few different equivalence relation important for geometry: equality, congruence of segments, congruence of angles, similarity for triangles and other figures, having same area for figures, having same volume for three dimensional polyhedra. In sweeping simplicity, Euclid used the same word "equal" for all these relations,—and afterwards even seemed to have justified the properties of an equivalence relation simply by the use of the word "equal". On the contrary, Hilbert and his followers clearly distinguish these and still several further relations, use different words and symbols for them, and prove their properties. Among Hilbert’s five groups of axioms (incidence, order, congruence, parallelism, continuity), only the axioms of congruence and parallelism have a clear-cut counterpart in Euclid. Only Euclid’s first postulate "to draw a line between two points" refers to incidence. Hilbert clearly separates the questions of existence and uniqueness, by postulating them in the two different axioms (I.1) and (I.2). The axioms of incidence referring to three dimensional geometry have no correspondence in Euclid. There are hints to the Archimedean axiom, but the axioms of order are totally omitted. The axioms of order are a striking innovation based on the work of Pasch of 1880. They were totally omitted in Euclid’s Elements. Axiom (II.4) is still named Pasch’s axiom. Put into colloquial language, it tells that a line which intersects one side of triangle, intersect a second side, too. The axioms of order have been simplified in later editions of the Foundations of Geometry, taking advantage of work of E.H.Moore and Veblen.
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Hilbert introduces two axioms of continuity: (V.1) is the Archimedean axiom and (V.2) the axiom of completeness. The axioms of continuity do not appear in Euclid’s postulates. But the definition of same ratios a : b and c : d from Euclid’s book V (credited to Eudoxus) makes only sense, if one assumes the Archimedean axiom. The Archimedean axiom allows the measurement of segments and angles using real numbers. During the measurement process, a real number giving the length of a segment is produced, digit by digit in the form of a binary fraction. Since Hilbert, this axiom is also known as the axiom of measurement. The clear-cut understanding of continuity was only achieved by Dedekind and Cantor in the late nineteenth century. There are several axioms for completeness, with very similar implications, which nevertheless have slight but deep differences. It is hard to say which one of these alternatives is the most natural axiom. Even Hilbert has suggested different axioms of continuity in different editions of his foundations of geometry. My favorite is Cantor’s axiom, which occurs in the very first edition of Hilbert and in Nichteuklidische Geometrie, Hyperbolische Geometrie der Ebene by Baldus and Löbell [7], p.43. The axioms of congruence resemble more to Euclid’s Elements than the other groups, but even here we find important differences and innovations. Nowhere in Hilbert’s Foundations of Geometry, circle appear at all, indeed they are not even defined. Instead of Euclid’s straightedge and compass, the transfer of segments and angles becomes the basic tools for geometric constructions. These tools turn out to be a bid weaker than straightedge and compass, but suffice for a few fundamental constructions. More important, the SAS congruence is introduced as an axiom. Even more, Hilbert proves the independence of the SAS axiom. Euclid has tried to justify the SAS congruence by his principle of superposition. Because of the independence of the SAS axiom, the principle of superposition turns out to be at best a physical thought experiment, but cannot replace the SAS axiom. In a totally different approach, it is possible to use the motion of figures as a building block of geometry. But in this case, extra work is needed to clarify what kind of motions are allowed. An (not totally rigorous) attempt in this direction is Hadamard’s Leçons de Géometrie Elémentaire of 1901-1906. So far, we have seen that Hilbert has achieved to make the foundations of geometry rigorous, without any hidden appeal to intuition, but kept the spirit of Euclid’s Elements as much as possible. The investigations about the nature of axioms, are topics totally different from Euclid. In Hilbert’s Foundations of Geometry, the questions of consistency, categorial nature, and independence of his axioms are addressed. I think that only a person of Hilbert’s optimism could address such questions at that time. Now we know from the work of Gödel and Tarski, that consistency can only be proved for a too small part of mathematics. The most accessible topic is independence. Hilbert proves the independence of the SAS-axiom, the parallel axiom, and the Archimedean axiom. The independence of the parallel axiom is rather informally justified via the spherical geometry. In an appendix to the foundations, Hilbert gives a detailed axiomatic approach to hyperbolic geometry. Legendre’s theorems the angle sum of triangles in neutral geometry, as exposed in detail. Relative consistency is proved, once consistency of the real number system is taken for granted—which turned out to be the really deep unsolvable problem! Hilbert proves that his axiom system is categorial, once his axiom (V.2) of completeness is assumed, but states clearly that the system without this axiom is not categorial. Here are his own words: As one realizes, there are infinitely many geometries which satisfy the axiom groups I through IV and (V.1). On the other hand, there is only one—namely the Cartesian geometry—which satisfies the completeness axiom (V.2), too.
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2.3.2. The Impact of Hilbert’s Foundations of Geometry Hilbert’s work is considered to be the first version of Euclidean geometry that is truly axiomatic, in the sense that there were no hidden appeals to spatial intuition. But the Foundations of Geometry are much more than just a clarification of Euclid’s Elements. Clearly this is one goal of Hilbert. A second goal of equal importance is a deeper understanding of the relation between geometric and algebraic structures. Already in the introduction, Hilbert says: "We shall be challenged by very new and—as I believe fruitful—problems, and see remarkable connections between the elements of arithmetic and geometry, gaining another insight into the unity of mathematics." Such a claim is well justified. It was Hilbert who first established a clear correlation between geometric and algebraic structures. These investigation came out of projective geometry, which is a historic predecessor and Hilbert’s starting point for the Foundations. It turns out that in coordinate geometry • the Theorem of Pappus is equivalent to commutative multiplication of the coordinate field, • the Theorem of Desargues is equivalent to associative multiplication of the coordinate field Further results were obtained and are included in the latest edition of Hilbert’s foundations. Here are two examples: Hessenberg gave in 1904 a purely geometric proof that the Theorem of Pappus implies the Theorem of Desargues (see Theorem of Hessenberg 4.8). A simple example for a non-Desarguean projective plane was introduced by E. R. Moulton in the article [15] A simple non-desarguesian plane geometry, Trans. Math. Soc. (1902). The Moulton plane is useful to clarify the logical relations between different geometric structures. The separate investigations about parts of the axioms have become more and more detailed and refined. Further research has extended the correlations of algebra and geometry to more exotic structures. The article of Hubert Kiechle, Alexander Kreuzer and Heinrich Wefelscheid in the fourteenth edition [22] of Hilbert’s foundations from 1999 contains some relevant information. For some of these ideas, an accessible account with examples are given by John Stillwell [34] in his exposition The Four Pillars of Geometry, Springer, 2005. A totally new topic is finite incidence geometry. The connections to scheduling problems in computer science, large scale computation, and to sophisticated algebraic structures has lead to new research. Some results are indicated in the section on Finite Affine and Projective Incidence Planes and Latin Squares. Finally, we all know that the axiomatic method is now almost commonplace in modern mathematics. Were does the word "complete" for existence of limits of Cauchy sequences come from? Many mathematicians may not even realize that it comes from the axiom of completeness in Hilbert’s Foundations of Geometry. Here the axiomatic method is introduced in such a satisfactory way that it has been exemplary for the modern style of research and presentation in pure mathematics. Let me finish with this citation: We shall be lead to several, apparently very simple, but nevertheless very deep and difficult problems. (from Hilbert’s preamble) Thus all human knowledge begins with intuition, proceeds to notions, and ends with ideas. (from Kant’s "Critique of Pure Reason"(1781)) 2.3.3. Drawbacks and Lacuna of Hilbert’s Foundations I want to ask, which topics are still not included, and what are the results achieved elsewhere about these topics. Of course I look at first at the present manuscript, after that beyond at the worldwide research. The first subject I miss in Hilbert’s foundation, are the axioms and theorems dealing with circles. Researchers like Greenberg and Hartshorne have meanwhile introduced the line-circle
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intersection property 10.5 and the Circle-circle intersection property 10.6 as the relevant axioms about the intersection of circles with lines, and of circles with circles. Several interesting theorems can be investigated already in neutral geometry. There is the remarkable three-circle theorem about of the threefold intersection of their three common chords, which holds even in neutral geometry. By the way, the figure is depicted on the cover of Robin Hartshorne’s book [19] Geometry: Euclid and Beyond. Here, it would be nice to have a proof totally in neutral geometry. Presently, I know only a proof done at first in Euclidean geometry, and in a second step extended to hyperbolic geometry by means of the Poincaré disk model. The Uniformity Theorem 16 allows to classify all Hilbert planes into one of three types, depending solely on the angle sum of triangles. This is a sharpening of Legendre’s Second Theorem. Hilbert includes the two Legendre Theorems, but only mentions the more general Uniformity Theorem, giving credit to his student Max Dehn. Actually, the result may have been known earlier, A complete proof is contained in Robin Hartshorne’s book Geometry: Euclid and Beyond [19] and in the present manuscript. Greenberg has recently stressed the equivalence of the Euclidean parallel axiom with the angle sum of the triangle being two right angles together with Aristole’s Angle Unboundedness Axiom 12.2. Proclus’ Theorem 17 clarifies and sharpens very old ideas of Proclus about the parallel postulate. A corresponding result in hyperbolic geometry is much deeper and harder to prove. Here it would be a progress to have a proof totally in neutral geometry. The current proof mentioned by Greenberg [18] in his article Old and new results in the foundations of elementary plane Euclidean and Non-Euclidean geometries, is based on rather sophisticated elaborations of the Klein disk model of hyperbolic geometry. +++++++++++++++++ 2.4. Frege’s Critique and Hilbert’s answer The book Foundations of Geometry was discussed and critized by Frege in an exchange of letters between the two scholars. Between October 1895 and November 1903, we know of nine letters having been exchanged. In the letter of December 1899 Hilbert writes: I still have to talk about one objection. You tell me, my notions, for example "point", "between" have not been fixed uniquely—for example the notion "between" is grasped (gefasst) differently on p.20 and there a point is a pair of numbers—. Yes, it is indeed self evident that every theory is only a framework or scheme of notions together with the necessary relations between them, and the basic elements (Grundelemente) can be thought of in any arbitrary manner. When I imagine as my points [their relation of order, my lines] any arbitrary system of things, for example the system: love, law, chimney sweeper. . . , I imagine [this system] and then I think of all my axioms as connections between these things; henceforth my theorems, for example the theorem of Pythagoras, are valid about these things, too. In other words: every theory can always be applied to infinitely many systems of basic elements (Grundelemente). One has indeed only to apply a one-to-one invertible transformation and agree [to postulate] the same corresponding axioms for the transformed basic elements. In fact this state of affairs is used frequently, for example in projective geometry as principle of duality, and by me in proofs of independence.
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On another occasion, Hilbert made the point that his proofs should stay completely correct even if the words "point", "line", and "plane" were replaced throughout by "table", "chair", and "beer mug." In the two last letters of September 1900 and November 1903, Hilbert formulates his point of view concisely. In September 1900 Hilbert answers to Frege: It is really my opinion that a notion can only be logically clarified by its relations with other notions. These relations, formulated in distinct assertions, I call axioms. Hence I arrive at considering the axioms (together with the names for the notions) as the definitions of the notions. These convictions did not arise for me out of pure fancy, but I see myself pushed to them as a necessity for the rigorous and logical building of a theory. I have arrived at the conviction that subtle facts in mathematics and the natural sciences can be treated with certainty only in this way—otherwise one only turns oneself in cycles. 2.5. About the consistency proof for geometry Around 1900, Hilbert originally wanted to obtain a consistency proof for geometry based on the well known construction of the Cartesian geometry with real-number valued coordinates. Hence he called for a proof of consistency of the real number system, among his 23 Paris problems. Because of Gödel’s incompleteness theorem, it turns out to be provably impossible to prove consistency of the real number system. Today, it is clear to everybody that a consistent proof of geometry cannot be based on the coordinate geometry with real numbers as coordinates. The consistency of the axioms of geometry, without the axioms of continuity, has been proved by Alfred Tarski. A rather popular and readable account is contained in Feferman’s biography [1] of Tarski. Tarski’s consistency proof uses his earlier result that the theory of real-closed fields 1 is consistent. One needs an infinity number of axioms to characterize a real-closed field, but one can still use a language of first-order logic. Secondly, a model of geometry is constructed: this is simply the analytic geometry with a a real-closed field as coordinates. Both the axioms of a real-closed field, and the axioms of geometry (I),(II),(III),(IV) are formulated in a language of first-order logic. Question. Why is there no contradiction between Tarski’s proof of consistency of geometry and Gödel’s incompleteness theorem? Answer. The axiomatic of the Tarski geometry does not have enough power to define the Peano arithmetic. Indeed, even with any infinitely many axioms formulated in first-order arithmetic one can never define Peano arithmetic. Hence one cannot introduce a Gödel numbering for theorems and proofs nor obtain incompleteness via a diagonalization argument. Following Tarski’s approach, we stay within first-order logic, and cannot define the Peano arithmetic. No contradiction to Gödel’s incompleteness theorem arises. Question. Why is it impossible to prove consistency of geometry including the axioms of continuity? Answer. The situation becomes different, and indeed any hope for a consistency proof is doomed, once we introduce either one of the axioms of continuity (V.1) or (V.2). These axioms cannot be formulated in any language of first-order logic. Even more, they either need for their formulation the natural real numbers, or enough theory of sets as is needed for the construction of the real numbers. Hence Gödel’s incompleteness theorem applies. No consistency proof is possible neither for the axiomatic system of the real numbers, nor for the Cartesian geometry with real-valued coordinates, nor for Hilbert’s system of axioms (I),(I),III),(IV),(V). 1
A field is called real-closed if and only if every odd-order polynomial has a zero
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2.6. General remark about models in mathematics The meaning and usage of the notion "model" in the natural sciences on the one side, and mathematics on the other side, are different. For example, the astronomer says that Newton’s theory of gravitation is a model for the solar system. The mathematician says that the solar system is a model of Newton’s theory of gravitation. 2
In the natural sciences, one has to begin with a complicated reality. One wants to extract some salient features, and describe or explain them in hopefully more simple mathematical terms. The scientist calls such a mathematical description a model. On the other hand, the mathematician already begins with a theory, which often can be pinpointed by the appropriate axioms. In a second step, one looks for other examples, problems or contexts to which the theory fits. To achieve such a fit, the abstract objects and relations of the theory have to be interpreted as specific notions occurring in the example at hand. Being interpreted in the context of the example, one has to ask and check whether the objects and relations from the abstract theory actually satisfy the axioms originally postulated. That can turn out to be true or not, depending on the properties of the example chosen. Furthermore, it can turn out that the same theory fits for quite different contexts. In building a model, the mathematician has to think on two levels. The basic level is given by the specific example at hand and its accepted properties. In favorite circumstances, one needs only rudiments of set theory and the number system as accepted properties. One can think of this level as the "background ontology". The secondary level brings in the originally undefined objects and relations of an abstract, and often more modern, theory, which now get interpreted in terms of the basic level. If the axioms can be proved to hold for this specific example, we say that we have constructed a model for the abstract theory. In this way, one obtains a proof of relative consistency of the abstract theory. 2.7. What is completeness? As currently used in the foundations of mathematics, in mathematical research, and by the overall scientific community, the word completeness has several—I am sure at least five—different meanings. • In the foundations of mathematics "completeness" of a theory can mean "validity implies provability", which has has been confirmed, in the case of first-order logic by Gödel’s completeness theorem of first-order logic. But, on the other hand, the term "completeness" of a theory has the second, much more powerful, meaning that "either a statement or its contrary is always provable". This has been rejected for theories including the natural numbers by Gödel’s famous incompleteness theorem. The confusion is even made worse by the fact that the same famous logician proved these two important theorems very shortly one after the other. • Beyond its usage in model theory and mathematical logic, further meanings are given to the same word "completeness". Actually, we are no longer talking about lack or availability or proofs, but about lack or availability of elements occurring as limits. It is customary in mathematical research that the name "completeness" can refer to either axiom (V.2) from Hilbert’s foundations of geometry, and the corresponding axiom for the real number system,—or refer to the postulate stating convergence of every Cauchy sequence in a Hilbert space, Banach space or metric space. The latter is the notion of completeness occurring in real analysis and functional analysis. Even in that more narrow context, there are different axioms running under the name of completeness. For a final clarification, one needs to use names as Cauchy-completeness, or Dedekind-completeness. 2
Another consequence of these matters: A scientific model can only be falsified. A mathematical model can be verified
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Remark. It is worthwhile to realize that either Cauchy or Dedekind completeness property of the reals is a fundamentally more complex statement than the properties that define realclosed fields, and cannot be replaced by any list of first-order properties. No countable list of first-order properties can characterize the real numbers uniquely. • Furthermore, in the overall scientific community, the term "completeness of a theory" means that a theory includes all relevant notions and their logical connections—and thus does no longer refer to notions only within a single given theory. In this sense, the term is used by Hilbert in his introduction to the foundations of geometry, where he announces ". . . a new attempt to set up a complete and as simple as possible system of axioms." Too, in this sense, Einstein talks about his belief that quantum mechanics is incomplete. Abraham Pais [?] states in his Einstein biography on p.449: "From 1931 on, the issue for Einstein was no longer the consistency of quantum mechanics but rather its completeness." Here is my briefest attempt for a clarification: In the first case, completeness is related to lack versus availability of proofs In the second case, completeness is related to lack versus availability of elements occurring as limits. In the third case, completeness asks for availability of an entire theory. 2.8.
More metamathematical considerations
Remark (About the two meanings of completeness). A sentence from a theory is called valid if and only if the sentence holds in every structure which is a model of the theory. This definition has a precise meaning in mathematical logic, once one has gives precise definitions for the notions of—: language, sentence, theory, structure, model. Following this approach, Tarski has introduced into mathematical logic the notion of truth—also called validity. Mathematical logic provides an exact definition of the notion of a proof, and hence provability of a sentence, too. It is one of the basic problems of mathematical logic to investigate the relation of validity and provability. Is every provable sentence valid? Is every valid sentence provable? The first question turns out to be much easier to answer. Indeed, the theorem of soundness states that a sentence which can be proved is valid. The second question cannot be answered in general. A theory in which every valid sentence can be proved is called "complete". (I prefer the term model-complete). Gödel’s completeness theorem states that every first-order theory is complete. Thus he has given us a positive answer to the second question, at least in the context of first-order logic. This achievement was actually the topic of his Ph.D. thesis. By Gödel theorem any valid sentence from a first-order theory has a correct logical proof. A theory is called negation-complete if and only if for every sentence, either the sentence or its negation, but not both, can be proved in the theory. The most common meaning of completeness is indeed negation-completeness. This is a very strong requirement, so strong indeed that most of the time it turns out to be just wishful thinking. Here is a statement of the Turing-Church Theorem: The set of laws of logic in the language of Peano arithmetic is undecidable. The set of laws of logic in the language of any first-order language with at least one relation or function symbol of two or more variables is undecidable. The Turing-Church Theorem provides a negative answer to Hilbert’s "Entscheidungsproblem" (decision problem). The Gödel-Rosser Incompleteness Theorem is a consequence to the negative answer of the decision problem: There does not exist a negation-complete, axiomatizable extension of Peano arithmetic. Question. Why does there not arise a contradiction Gödel’s completeness theorem of first-order logic and the Gödel-Rosser Incompleteness Theorem?
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Answer. Here is my (perhaps awkward) attempt of a brief clarification or "enlightenment", following mainly Robert Wolf [40]. It is certainly not true that every sentence or its negation is provable in first-order logic. But Gödel’s completeness theorem 1 does not yield negation completeness of first-order logic. We have instead only the weaker result that every valid sentence is provable. Indeed validity is a rather strong model-theoretic assumption. Only under this strong assumption of validity of any statement does Gödel’s completeness theorem assure the existence of a proof for this statement. There does not occur a contradiction between Gödel’s completeness theorem— for which I would prefer the name "Realization Theorem"—and the Turing-Church Theorem. Remark (About second-order logic). The caveat in Gödel’s completeness theorem is its restriction to first-order logic. A lot of mathematics can actually be expressed in first-order logic, but firstorder logic is not powerful enough to express all mathematics and even less to express all scientific theory. By definition, first-order logic includes the propositional logic with connectives and, or, if. . . then, if and only if. First-order logic does include quantified logic and thus allows statements of the form "For all x a statement S (x) holds" and "There exists an x for which a statement S (x) holds. Here x is allowed to be any primary (undefined) element occurring in the respective theory, and S (x) is any statement in this theory in which may occur the element x. Too, first-order logic does include the notion of equality as a primary relation. Scientific theory and mathematics as a whole needs to include the natural numbers and even set theory. In terms of mathematical logic, this means that we have to include (at least) Peano arithmetic into the relevant theory and its language. Indeed Peano arithmetic is only the most barren way for an abstract theory about the natural numbers. It has been proved that Peano arithmetic cannot be stated purely within the restriction to first-order logic. Second-order logic allows quantifier to have as their range no longer single primary objects, but allows quantifier ranging over subsets of the primary objects, properties, or even axioms. In Peano arithmetics, the induction axiom—in each of its different forms—needs quantifiers of one of this types. All the different axioms of continuity need such quantifiers. Leibniz’ definition of equality as "Two objects are equal iff they behave equally in all circumstances" depends on second-order logic. Indeed, every really powerful axiomatic system contains somewhere second-order logic. I would not call these axioms "a flaw in the ointment" as Stefan Mykytiuk and Abe Shenitzer [27] seem to suggest in their nice article "Four significant axiomatic systems and some of the issues associated with them". The language to first-order logic has a real caveat—a serious restriction one cannot live with forever. Let me use a metapher used by Hilbert in another context: I think of this restriction being similarly embarrassing as taking away the telescope from an astronomer, or the microscope from a biologist—. So, in the end, practicable mathematics and actual mathematical research uses Peano arithmetic, continuity, and even more than that, and hence will need second-order logic.
1
I would prefer the name "Realization Theorem"
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3. Incidence Geometry 3.1.
Elementary propositions about incidence planes
Definition 3.1 (Incidence plane). Any model for which Hilbert’s axioms of incidence (I.1)(I.2)(I.3a)(I.3b) are valid is called an incidence plane. Lemma 3.1. Any two different lines of an incidence plane have either no or exactly one point in common. Problem 3.1. Give a short explanation to convince yourself that Lemma 3.1 does hold. Answer. Take any two different lines l , m. Assume towards a contradiction they intersect at two different points A , B. By axiom (I.2), there exists at most one line through these two points. Hence l = m, contradicting the assumption that l and m are two different lines. Hence two different lines intersect at most in one point. Definition 3.2 (Parallel lines). Two different lines which lie in one plane and do not intersect are called parallel. Question. Give at least two further useful formulations of Lemma 3.1. Answer. Here are three possible answers: • Any two different lines lying in a plane and not parallel, have a unique point of intersection. • If two lines have two or more points in common, they are equal. • Two different lines have at most one point in common. Lemma 3.2. In any incidence plane, there exist two different lines through every point. Proof. By axiom (I.3b), there exist at least three points that do not lie on a line. We call them A, B and C. Let any point P be given. The point P may be among the three points A, B, C or not. Thus the proof involves four points. • In the case that point P is one of the three points A, B, C, we draw the three lines AB, BC and CA. These are three different lines, and two of them go through the given point P. • In the case that point P is different from all three points A, B, C, we draw the three lines PA, PB and PC. At least two of them are different since A, B, C do not lie on a line. In both cases we have obtained two different lines through the arbitrary point P. For two dimensional geometry, all lines and points are assumed to lie in the same plane. Referring to the parallel postulates, we need not state that the parallel to a given line through a given point actually lies in this plane. This leads to the following a bid simpler formulation: Definition 3.3 (Euclidean Parallel Postulate). For every line l and for every point P lying not on l, there exists a unique parallel m to l through point P. But Hilbert’s parallel axiom (IV) is a weaker assertion, it postulates only the uniqueness of the parallel to a given line through a given point. Definition 3.4 (Hilbert’s Parallel Postulate). For every line l and for every point P lying not on l, there exists at most one parallel m to l through point P.
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Proposition 7.43 implies existence of a parallel once the axioms of order and congruence are assumed. Hence Euclid’s and Hilbert’s parallel postulate turn out to be equivalent in Hilbert planes. But in the case of incidence planes, such an equivalence does not hold. Definition 3.5. We say that a two-dimensional incidence geometry for which the Euclidean Parallel Postulate holds, has the Euclidean parallel property. Definition 3.6. We say that a two-dimensional incidence geometry has the elliptic parallel property iff any two lines do intersect—necessarily at a unique point. Definition 3.7. We say that a two-dimensional incidence geometry has the hyperbolic parallel property iff for every line l and point P not on line l, there exist at least two parallels to line l through point P. Lemma 3.3 (Proclus’ Lemma). In any incidence plane where Hilbert’s Parallel Postulate for plane geometry 3.4 holds, the following statements are valid: • a third line intersecting one of two parallel lines intersects the other one, too. • a third line parallel to one of two parallel lines, is parallel to the other one, too. Problem 3.2. Explain why Proclus’ Lemma is an easy consequence of the uniqueness of parallels. Convince yourself that, conversely, Proclus’ Lemma implies the uniqueness of parallels. Solution. Suppose towards a contradiction that the transversal t intersects one of the parallel lines l and m, but not the other one. We may assume that P is the intersection point of lines t and m. If lines t and l would not intersect, then t and m would be two different parallels of line l through point P. This contradicts the uniqueness of parallels. Conversely, we now assume Proclus’ Lemma to be true and check the uniqueness of the parallel to a given line l through a given point P not on line l. Let m and t be two parallels to line l through point P—these may be equal or different lines. The line t is a transversal intersecting one of the two parallel lines m k l at point P. Hence it intersects the second line l, too, contrary to the assumption. The only possibility left is that m = t. Hence the parallel to a given line through a given point is unique. Remark. Don’t fool yourself to think that all these words amount to a proof of the parallel postulate! All we have shown is: either both uniqueness of parallels and Proclus’ Lemma hold or neither uniqueness of parallels nor Proclus’ Lemma do hold. Problem 3.3. Given any incidence plane where Hilbert’s Parallel Postulate for plane geometry 3.4 holds. Convince yourself that the relation that two lines are either equal or parallel is an equivalence relation. Answer. For any two lines, let l ∼ m mean l = m or l k m. The relation ∼ has the three defining properties of an equivalence relation. Indeed this relation is reflexive: since l = l is a logical axiom. symmetric: since l = m implies m = l, and l k m implies m k l. transitive: assume that m ∼ l and l ∼ k. If two of the three lines are equal, we substitute equals to get m ∼ k. We now assume that all three lines are different. The two lines m and k are either parallel or they intersect at one point, by Hilbert’s first Proposition 3.6. If m and k would intersect at point P, the line l would have two different parallels through P, which is impossible in an affine plane. Therefore m and k are parallel, as to be shown.
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A relation that is reflexive, symmetric and transitive is called an equivalence relation. Corollary 1 (About uniqueness of parallels). For any incidence plane the following statements are equivalent: (a) uniqueness of parallels,—stated in Hilbert’s Parallel Postulate 3.4 for plane geometry,—holds; (b) any third line intersecting one of any two parallel lines intersects the other one, too; (c) any third line parallel to one of any two parallel lines, is parallel to the other one, too; (d) being equal or parallel defines an equivalence relation among the lines. 3.2.
Finite incidence geometries
Problem 3.4 (The four-point incidence geometries). Find all non isomorphic incidence geometries with four points. Which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold? Which one is the smallest affine plane?
Figure 3.1. There are two four-point incidence geometries.
Answer. There exist two non-isomorphic four-point geometries. (a) Six lines with each one two points. It has the Euclidean parallel property, and is the smallest affine plane. (b) There are four lines, one of which has three points. It has the elliptic parallel property. Problem 3.5 (The five-point incidence geometries). Find all non isomorphic incidence geometries with five points. Describe the properties of their points and lines. Which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold? Answer. There exist four different five-point geometries. (a) Ten lines have each two points. This model has the hyperbolic parallel property. (b) Exactly one line with three points. Altogether, there are eight lines. Neither parallel property holds. (c) Two intersecting lines with three points. Altogether, there are six lines. Neither parallel property holds.
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Figure 3.2. There are four five-point incidence geometries.
(d) There are five lines, one of which has four points. This model has the elliptic parallel property. Problem 3.6. (i) Which ones of the four- or five-point incidence geometries satisfy the statement: On every line there exist at least two points, and furthermore, there exist three points not on this line. (ii) Why is this statement different from Hilbert’s axioms? Answer. (i) The statement of the problem postulates existence of at least two points on any given line, additionally three extra points lying not this line. Thus the statement implies existence of five different points. Hence it does not hold for any four point incidence geometry. Moreover, it can only hold for a five point geometry, where every line has exactly two points. Indeed it holds only for the five-point handshake model. (ii) Hilbert’s axiom of incidence for plane geometry are (I.1)(I.2) and (I.3). Axiom (I.3) consists of two sentences: I.3a There exist at least two points on a line. I.3b There exist at least three points that do not lie on a line. Here part (I.3a) and (I.3b) do not refer to each other. Part (I.3a) is a universal statement about any line. But part (I.3b) is a purely existential statement about three points. On the other hand, the statement given in the problem is a different universal statement about a line. The first part repeats Hilbert’s (I.3a). The second part goes on requiring additionally existence three points which are not lying on this same line. Problem 3.7. Which ones of the incidence axioms (I.1), (I.2), (I.3), and which parallel property hold in the first model drawn on page 42. Answer. Every two points lie on a unique line, and every two lines intersect at a unique point. Hence axioms (I.1) and (I.2), and the elliptic parallel property hold. Axiom (I.3) does not hold. Before we proceed to six or more points, we take notice of two simple models of finite incidence geometry—the extreme ones in a sense—which exist for all numbers n ≥ 4 of points. Definition 3.8 (Hand-shake model). A hand shake model is an incidence geometry for which every line has exactly two points.
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Figure 3.3. Too simple, this is not even an incidence geometry.
Figure 3.4. A straight fan is an incidence geometry—but still no projective plane.
Definition 3.9 (Straight fan). A straight fan is an incidence geometry with all but one point lying on one line. Problem 3.8. How many lines does the hand-shake incidence geometry with n points have. How many lines does the straight fan with n points have. Why do there exist two or more non-isomorphic incidence geometries for any number n ≥ 4 of points. Answer. The hand-shake incidence geometry with n points has (n − 1) + (n − 2) + · · · + 1 =
(n − 1)n 2
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lines. We see this as follows. We can connect the first point to the other (n − 1) points. Disregarding this point, we connect the second point to (n − 2) different points, and so on. The last line to be drawn is between the (n − 1)-th and the n-th point. The straight fan with n points has n lines. There is one long line with n − 1 points, and only one point P does not lie on this line. There are n − 1 lines with two points each of which connects P to a different point on the long line. We see that for n ≥ 4, the hand-shake model with n points has more lines than the straight fan with n points. Hence there exist two or more non-isomorphic incidence geometries for any number n ≥ 4 of points. Problem 3.9. Which parallel property holds for the hand-shake model with 4 points. Which parallel property holds for the hand-shake model with n ≥ 5 points. Which parallel property holds for a straight fan. Answer. The hand-shake model with 4 points has the Euclidean parallel property. All the handshake models with n ≥ 5 points have the hyperbolic parallel property. For all straight fans, the elliptic parallel property holds. Problem 3.10 (Optional). Show that every finite incidence geometry has at least as many lines as points. http://en.wikipedia.org/wiki/De_Bruijn %E2%80%93Erd%C5%91s_theorem_%28incidence_geometry%29 http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem Definition 3.10. The dual of an incidence plane with elliptic parallel property is the plane where the points are the lines of the old one, and vice versa. Question. What is the dual of a straight fan with n points. Answer. The dual is isomorphic to the primal, with n points (old lines), and n lines (old points) having the corresponding incidence relations. Problem 3.11 (The six-point incidence geometries). Find all non-isomorphic incidence geometries with six points. Count how many non-isomorphic models do exist. For the models different from the handshake and straight fan, mark all three-point lines with different blue shades, and any four-point line in red. Describe the properties of their points and lines. Count the lines. Which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold? Answer. There are nine non-isomorphic six-point incidence planes: 1. The handshake model with 15 short lines. 2. The model with only one three-point line has 13 lines. 3. The model with only one four-point line has 1 + 1 + 2 · 4 = 10 lines. 4. The model with two parallel three-point lines has 2 + 3 · 3 = 11 lines. 5. The model with two intersecting three-point lines has 2 + 2 · 2 + 5 = 11 lines. 6. The model with a intersecting three-point line and four-point line has 2 + 2 · 3 = 8 lines. 7. The model with three intersecting three-point lines forming a "triangle" has 3 + 3 · 2 = 9 lines. 8. The model with three intersecting three-point lines forming a "triangle" and the in-circle a fourth three-point line has 4 + 3 = 7 lines.
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Figure 3.5. Enumerate the pencils for the two models (A) and (B).
9. The straight fan with 1 + 5 = 6 lines. Only the straight fan is elliptic. There is no Euclidean model. There are several mixed models, with are not hyperbolic neither. Problem 3.12. Given an incidence plane for which Hilbert’s Parallel Postulate is valid. As mentioned in the lecture, and stated in Corollary 1: being equal or parallel defines an equivalence relation among the lines. Each one of the equivalence classes of equal or parallel lines is still called a pencil. But, contrary to the situation for an affine plane, not all pencils need to consist of the same number of lines. Determine and enumerate the pencils for the two models (A) and (B) drawn in the figure on page 44. Put curely brackets around each pencil! Answer. Pencils of parallels for model (A): {a1, a3, a7} , {a2} , {a4} , {a5} , {a6}. Pencils of parallels for model (B): {b1, b3} , {b2, b4} , {b5} , {b6}. Definition 3.11 (Isomorphism of incidence planes). Two models for incidence planes are called isomorphic if and only if there exists a bijection φ between the points of the two planes, and a bijection ϕ between the lines of the two planes such that incidence is preserved. Remark. For incidence to be preserved, the two bijections are required to satisfy the assumption ϕ(PQ) = φ(P)φ(Q)
(3.1)
for all points P, Q from the first incidence plane. In other words, if the bijection of the points maps P 7→ φ(P) and Q 7→ φ(Q), then the bijection of the lines maps PQ 7→ φ(P)φ(Q). Problem 3.13. Given two incidence geometries, it is not obvious whether they are isomorphic. By corresponding labelling of the points in both geometries, show an isomorphism between the two six-point incidence geometries in the figure on page 45. Answer. The isomorphism is indicated in the figure on page 45.
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Figure 3.6. Two isomorphic six-point incidence geometries
Figure 3.7. The labelling of the points shows the isomorphism.
Figure 3.8. Name the points as to show the isomorphism.
Problem 3.14. Given two incidence geometries, it is not obvious whether they are isomorphic. By corresponding labelling of the points in both geometries, show an isomorphism between the two seven-point incidence geometries in the figure on page 45. Answer. The isomorphism is indicated in the figure on page 46.
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Figure 3.9. The names of the points show the two drawings of the Fano-plane are really isomorphic.
Problem 3.15 (A nine point incidence geometry). Find a model of incidence geometry with nine points, which satisfies (I3+) "Every line contains exactly three points." and for which the Euclidean parallel axiom does hold: "For every line l and every point P not on l, there exists exactly one parallel to the line l through P." It is enough to provide a drawing to explain the model. How many lines does the model have? Use colors for the lines. Choose clearly different colors for the lines in different directions, but give each set of three parallel lines different shades of nearby color. Answer. There is exactly one such model. It has 12 lines. Name any point A and let two lines through it consist of the points a = {A, B, C} and d = {A, D, G} (lines a and d are drawn horizontally and vertically). Let {D, E, F} be the parallel to line {A, B, C} through D. Two different parallels to a given line cannot intersect, because of the uniqueness of parallels. Hence the parallel to a through point G contains the remaining points and necessarily is {G, H, I}. The parallel to line {A, D, G} through B contains one of the points E, F and one of the points G, H. Possibly by exchanging names of those four points, we can assume that {B, E, H} is the parallel to line {A, D, G} through B. Finally, {C, F, I} is the parallel to line {A, D, G} through C. Up to now, we have mentioned and drawn six lines. Because of incidence axiom (I.1), there is a unique line through any two points. Hence there need to exist further lines in the model. As an example, we find the line g through points A and E. It cannot pass through any of the points B, C, D, F, H, I, because otherwise we get a line with four or more points on it. Hence g = {A, E, I} is a line. Similarly, one finds lines {H, E, C}, and finally {B, D, I},{A, F, H},{B, F, G},{C, D, I}. It does not matter that the last four lines cannot be drawn as straight Euclidean lines. Neither does it matter that they have more intersection points—those are not included as points of the model. The model contains 12 lines. It is the unique model satisfying all requirements.
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Figure 3.10. A nine-point incidence geometry
Remark. The model can be constructed using analytic geometry and arithmetic modulo 3. Take as "points" the ordered pairs (a, b) with a, b ∈ Z3 . As in analytic geometry, lines are given by linear equations ax + by + c = 0 with a, b, c ∈ Z3 and a, b are not both zero. One defines the slope of a line in the usual way. Lines are parallel if and only if they have the same slope. To find the parallel to a given line through a given point, one uses the point slope equation of a straight line. This procedure shows that the Euclidean parallel property holds. 3.3.
Affine incidence planes
Definition 3.12 (Affine plane). An affine plane is a set of points, and a set of lines, satisfying the axioms: A.1 Every two different points lie on exactly one line. A.2 If point P does not lie on a line l, there exists exactly one line m through the point P that does not intersect l. A.3 There exist three points that do not lie on a line. Problem 3.16. The reader should convince himself that in an affine plane 1. on every line lie at least two points;
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2.
in every point intersect at least two lines.
Proof. Let A, B, C be the three points not on a line which exist according to axiom (A.3). 1. Given is any line l. At most one of the three lines AB, BC and AC is parallel or equal to the given line l. We may assume that neither line AB nor line AC is parallel or equal to the given line l. Hence they intersect the given line in the points P = AB ∩ l and Q = AC ∩ l. • If these two points P , Q are different, we are ready. • We get a more involved argument in the special case that P = Q. In this case, we get A = P = Q. If the line l intersects the line BC, we get a second intersection point R = BC ∩ l on the line l and are ready again. • Still open is only the case that lines BC and l are parallel and the point A = P lies on the line l. Let m be the parallel to line c = AB through point C. Since m , BC k l, the line m is not parallel to l. Hence these two lines intersect at point R = m ∩ l , A, which is a second point on the line l. 2. Given is any point P. Assume that P , A, B, C. Since these three points do not lie on a line, we get from lines PA, PB and PC at least two different ones through point P. In the special case that P = A, we get two different lines AB and AC through point P. We argue similarly in the special cases where P = B or P = C.
Figure 3.11. In an affine plane, we find a second point on any given line.
Axiom (A.1) repeats Hilbert’s incidence axioms (I.1) and (I.2). Axiom (A.2) is the Euclidean parallel property. In problem 3.16, it is proved that on every line of an affine plane lie at least two points. Together with (A.3), we see that Hilbert’s incidence axiom (I.3) holds for an affine plane, too. Hence an affine plane is just an incidence plane for which the Euclidean parallel property holds.
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Corollary 2 (Characterizations of an affine plane). For any incidence plane the following statements are equivalent: (a) The incidence plane is an affine plane; (b) for every line l and for every point P lying not on l, there exists a unique parallel m to l through point P. Thus the Euclidean parallel postulate 3.3 holds. 3.4.
Introduction of coordinates
Definition 3.13 (The Cartesian plane). The "points" of the Cartesian plane are ordered pairs (x, y) of real numbers x, y, ∈ R. The "lines" of the Cartesian plane are given by equations ax + by + c = 0
(3.2)
with reals coefficients a, b, c, and a and b not both equal to zero. A "point lies on a line" if and only if the coordinate pair (x, y) satisfies the equation of the line. Two lines are equal if and only if they contain the same points. Remark. Because coefficients (a, b, c) and (λa, λb, λc) with λ , 0 give rise to the same line, the triples (a, b, c) are homogeneous coordinates. Theorem 3.1. The Cartesian plane is an affine plane. Because the construction of the affine plane from the real numbers uses only addition, subtraction, multiplication and division, replacing the real numbers by any finite or infinite field F and doing the same construction once more, leads to an affine plane, too. Thus one gets the following general definition. Definition 3.14 (The Cartesian plane over a field F). The "points" of the Cartesian plane are ordered pairs (x, y) of elements x, y ∈ F. The "lines" of the Cartesian plane of the field F are equations ax + by + c = 0 with coefficients a, b, c from the field, of which a and b are not both zero. A "point lies on a line" if and only if the coordinate pair (x, y) satisfies the equation of the line. Two lines are equal if and only if they contain the same points. Theorem 3.2 (The Cartesian plane over an arbitrary field). In a Cartesian plane over any field, there exist a unique line between any two points. There exists a unique parallel to a line through a given point. Hilbert’s axioms (I.1)(I.2)(I.3a)(I.3b) and the Euclidean parallel postulate (IV*) hold. Hence a Cartesian plane over a field is an affine plane. Remark. Any axioms may only be used without further justification, in the context that they refer to primary (undefined) elements and relations. The situation is different for the Cartesian plane introduced above, which is another important example for a model or structure. For a model or structure, the primary elements and relations are replaced by elements and relations which are explicitly defined. For a model one has to check whether the axioms are valid for these (new) elements and relations. Thus to check theorem 3.2, one has to give a justification, why the incidence axioms (I.1), (I,2), (I.3a), (I.3b) and the Euclidean parallel postulate hold for the "points", the "lines" and the "incidences" of the Cartesian plane specified above.
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Problem 3.17. Use your knowledge of College algebra, and give in a few words an explanation why Theorem 3.2 is valid. Answer. We begin by checking postulates (I.1) and (I.2). Given are two distinct points (x1 , y1 ) and (x2 , y2 ) through which we need a unique line. In the ususal way, one defines the slope m=
y2 − y1 x2 − x1
and puts m = ∞ in the special case that x1 = x2 . The line through point (x1 , y1 ) with slope m has the equation y = y1 + m(x − x1 ) (3.3) in the usual case of finite slope, and x = x1 = x2 for m = ∞. This equation of a line in this form is uniquely determined by two points (x1 , y1 ) , (x2 , y2 ), as the reader should check. Hence two lines through two distinct points contain the same points and are equal. To check axiom (I.3a), we use that each field contains the two different elements 0 , 1. On any given line with finite slope, we get two points on a line by putting x = 0, x = 1. We put y = 0, y = 1 for a vertical line. To check axiom (I.3b), we may for example use the three points (0, 0), (1, 0) and (1, 1). Hence we see that the Cartesian plane over any field is an incidence plane. To get an affine plane, we need still to check that the Euclidean parallel postulate (IV*) does hold. We see that two distinct lines y = mx + k and y = m0 x + k0 with different slopes m , m0 do intersect. But two distinct lines with the same slope have m = m0 and k , k0 ,— they do not intersect. Too, one may include vertical line in this statement. Given any line y = mx + k and point (x1 , y1 ), the parallel through the point is obtained by the point-slope formula (3.3). An easy reasoning shows that Euclidean parallel postulate (IV*) does hold in the case of a vertical line, too. 3.5.
Finite coordinate planes
Problem 3.18 (A nine point incidence geometry). Find an affine plane of order three. Answer. As a simple example, we see that the affine plane given by the figure on page 717 can be constructed as Z23 , using the field Z3 , where the arithmetic is calculation modulo 3. This becomes even more visible by repeating copies of the same figure, shifted by three units along the x and y axes in a double periodic way. Obviously, the model has order 3, since there are n2 = 9 points. Hence there are n2 + n = 12 lines. There is exactly one such model. The finite fields are well- known from modern algebra. For any order n = pr where p is a prime number, and r ≥ 1 any natural number, there exists exactly one finite field. It is the Galois field, denoted by Fn . Corollary 3 (Affine planes from finite fields). If the order n = pr is a prime power, there exists exactly an affine incidence plane of order pr . It has n2 = p2r points and n2 + n = p2r + pr lines. The points are ordered pairs (x, y) with arbitrary x, y ∈ Fn . In other words, the set of points is the Cartesian product Fn × Fn . Hence there are n2 = p2r points. The lines are the n vertical lines x = c with any c ∈ Fn , as well as the other lines with an equation of the form y = mx + k with arbitrary m, k ∈ Fn . Hence there are n2 + n = p2r + pr lines.
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Problem 3.19 (Scheduling problem I). Make a 6 day schedule for a school with 25 students. Each day the students are divided in a different way into 5 groups of 5 students. Never are two students in the same group more than one time during the week. Answer. We use the coordinate plane Z5 × Z5 . With addition and multiplication modulo 5, the set Z5 is a field, because 5 is a prime number. Hence the coordinate plane Z5 × Z5 is an affine plane with order 5. As explained by Proposition 3.4, the number of points on any line in a affine plane is given by the order, here n = 5. The lines can be partitioned into n + 1 = 6 pencils, each containing 5 parallel lines. Hence for each pencil, the set of parallel lines contained in the pencil, determines the groups of students for one weekday. For each of the 6 weekdays, we use parallel lines with a different slope 0, 1, 2, 3, 4, and finally on one day the vertical parallel lines. One can make a picture (see page 51) of the schedule by drawing the 5 × 5 pattern of dots separately for every day. The five parallel lines in every pattern are indicated by different symbols for their points. One needs curved lines to connect all five points of a line. 1 Clearly such a picture contains more insight than a bare-bone list.
Figure 3.12. A picture to explain the schedule for five groups of five students each, on six days.
Problem 3.20 (Scheduling problem II). Make a 5 day schedule for a school with 15 students. Each day the students are divided in a different way into 5 groups of 3 students. Never are two students in the same group more than one time during the week. Answer. We cut down the solution of the last problem and retain just the 15 students from three groups on the sixth weekday. For the remaining five days, we obtain five groups with three students, as required. Problem 3.21 (Scheduling problem III). Make a 7 day schedule for a school with 35 students. Each day the students are divided in a different way into 7 groups of 5 students. Never are two students in the same group more than one time during the week. 1
The figure leaves out these curve parts, the reader should put them in.
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3.6. Projective incidence planes Projective planes are incidence planes with elliptic parallel property. But an additional axiom is needed, in order to rule out uninteresting cases. A straight fan (see definition 3.9) is too simple for what one wants to be a projective plane. In a projective plane, one wants to build perspective and the corresponding mappings and measurements. As natural starting point of such an endeavor, one needs four points in a convenient configuration. I like to think of them, as the standpoint of the observer, a point some distance away from him, and two points on the horizon to the left and right. Definition 3.15 (Quadrangle and quadrilateral). Four points, no three of which lie on a line, are said to be a quadrangle. These four points are necessarily distinct. Four lines, no three of which intersect at a point, are said to be a quadrilateral. Again these four lines are necessarily distinct. Definition 3.16 (Projective plane). A projective plane is a class of points, and a class of lines satisfying the axioms: P.1 Every two distinct points lie on exactly one line. P.2 Every two distinct lines intersect at exactly one point. P.3 There exist four points of which no three lie on a line. Remark. The lines of a projective plane are not necessarily sets of points. Problem 3.22. Convince yourself that in every point of a projective plane intersect at least three lines. Secondly, convince yourself that on every line of a projective plane lie at least three points. Finally check that a quadrilateral exists. Answer. By axiom (P.3), there exist at least four points A, B, C, D of which no three lie on a line. Let any point P be given. In the case that point P is different from all four points A, B, C, D, we draw the lines PA, PB, PC and PD. At least three of them are different since no three points among A, B, C, D lie on a line. In the case that point P is one of the four points A, B, C, D, we get three different lines through the given point P among the six lines connecting A, B, C, D. In both cases we have obtained three different lines through the arbitrary point P. The proof confirming that three points lie on every line is done quite similarly, interchanging the roles of points and lines. Proposition 3.1 (The straight fan is the only exception). An incidence geometry with elliptic parallel property is either a straight fan (see definition 3.9), or a quadrangle exists. Hence, in the latter case, it is a projective plane. Proof. We assume the elliptic parallel property holds, but no quadrangle exists. We have to check that we get a straight fan. The three-point incidence geometry is a straight fan, as claimed. Hence we now assume at least four points to exist. Take any four points of the given incidence plane. Since no quadrangle exists, three among them lie on a line. We can assume that the three points A, B, C lie on the line l. By the axiom (I.3) for the incidence plane, there exists a further point P not on the line l. We claim there cannot exist two points Q , P neither of which lies on the line l. Assume towards a contradiction that the two points Q , P both do not lie on l. The lines PQ and l do intersect, either in one of the three points A, B, C, or still another point. We can assume
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the intersection point A0 is different from both B and C. Of the four points B, C, P, Q at least three lie on a line. This could be points P, Q, B or P, Q, C. Both cases are impossible, since the lines PQ and l = BC intersect only in point A0 , B, C. Hence there exists exactly one point P not on the line l. Thus the incidence geometry is a straight fan. Definition 3.17 (Pencil). An equivalence class of (equal or) parallel lines is called a pencil. Definition 3.18 (Improper elements and projective completion of an affine plane). The points and lines from the original affine plane are called proper points and proper lines. To each pencil corresponds an improper point. 1 The line through all improper points is called the improper line. 2 Both the proper and the improper points (lines) of an affine plane are the points (lines) of a projective plane. By adding the improper points and the improper line, one may construct from any affine plane a projective plane, called the projective completion. Problem 3.23. Convince yourself that the projective completion of an affine plane is indeed a projective plane. Starting from the axioms (A.1)(A.2)(A.3) for the affine plane, one needs to check the validity of axioms (P.1)(P.2)(P.3). Check of validity of axiom (P.1). Given are two distinct points A , B of the projective plane. We distinguish three cases, depending on whether these are proper or improper points. • In case that both A and B are proper points, existence and uniqueness of the line through A and B is postulated by axiom (A.1). • Assume now that only one of the two points is proper. We may assume that point A is proper, and B∞ is an improper point. By definition, the point B∞ is a pencil of parallel lines. Take any line b in that pencil. By axiom (A.2) there exists a unique parallel m to line b through point A. In the projective completion, the line m is the unique line through points A and B∞ . • Finally, we assume that both A∞ and B∞ are improper points. In the projective completion, the improper line l∞ is a line through points A∞ and B∞ . Any other line cannot go through both A∞ and B∞ , since the extension of each proper line contains only one improper point. We see that in all possible cases, there exists a unique line through points A and B. Hence axiom (P.1) is valid. Check of validity of axiom (P.2). Given are two distinct lines a , b of the projective plane. We distinguish three cases, depending on whether these are proper or improper lines. • Let both a and b be distinct proper lines, and assume they intersect at point P in the affine plane. They cannot have at second proper intersection point since by axiom (A.1) the line through two distinct points is unique. They cannot have any further improper intersection point, otherwise lines a and b would have to be parallel to two distinct pencils, which is impossible. • Let both a and b be distinct proper lines, and assume they do not intersect in the affine plane. Hence they belong to the same pencil P∞ of parallel lines. Extended to the projective plane, the two lines intersect at the improper point P∞ . They cannot have any further improper intersection point, otherwise lines a and b could not be parallel. • Assume now that only line a is proper, and b∞ is the improper line. The line a is contained in exactly one pencil A∞ of parallel lines. This pencil is the unique intersection of lines a and b∞ in the projective completion. 1
This name is given for visual reasons. We want to catch the vague idea of a "point very far away" in the direction pointed to by a pencil, and make it precise in terms of rudimental set theory. 2 I reserve the names points at infinity and line at infinity to hyperbolic geometry.
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We see that in all possible cases, there exists a unique intersection point of lines a and b. Hence axiom (P.2) is valid. Check of validity of axiom (P.3). Finally, to confirm (A.3), we obtain four points no three on which lie on a line. By axiom (A.3), there exist, in the affine plane, three points A, B, C not lying on a line. We draw the parallel m to line AB through point C. Let C∞ be the improper point for that direction. We draw the parallel n to line AC through point B. Let B∞ be the improper point for that direction. The lines m and n are not parallel since the respectively parallel lines AB k m and AC k n intersect at A. Hence the lines m and n intersect, in the affine plane, at a point D = m ∩ n. Of the four points A, D, B∞ , C∞ no three lie on a (projective) line, confirming axiom (P.3). Remark. Here is more simple of axiom (P.3): Finally, to confirm (P.3), we show existence of four points no three on which lie on a line. By axiom (A.3), there exist, in the affine plane, three points A, B, C not lying on a line. We draw the parallel m to line AB through point C, and the parallel n to line AC through point B. The lines m and n are not parallel since the respectively parallel lines AB and AC intersect at A. Hence the lines m and n intersect, in the affine plane, at a point D = m ∩ n. Of the four points A, B, C, D no three lie on a (projective) line, thus confirming axiom (P.3). Proposition 3.2 (Cutdown of a projective plane to an affine plane, and reconstruction of the projective plane). Given is a projective plane. Any line—and all points on it—are marked as improper. (a) The proper points and lines yield an affine plane. (b) Concatination of improper elements leads back to the original projective plane. Check of validity of axiom (A.1). We begin to check the validity of the axioms for an affine plane. Existence and uniqueness of a line through any two distinct points is postulated by axiom (P.1), confirming axiom (A.1). Check of validity of axiom (A.2). Next we check existence and uniqueness of a parallel to a given proper line l through the proper point P not lying on l. Let l∞ be the line marked as improper. Since l∞ , l, axiom (P.2) yields, in the projective plane, existence of a unique intersection point Q∞ = l∞ ∩ l. The line m = PQ∞ is retained in the affine plane. Since Q∞ = l ∩ m is an improper point, as seen in the affine plane, the lines l and m are parallel. Hence we have obtained the parallel m to the given line l through point P, as required. Any other parallel m0 k l through point P extended, goes through point Q∞ , too. By the uniqueness of the line through P and Q∞ , we conclude m = m0 . Hence uniqueness of parallels holds, too. Check of validity of axiom (I.3a). Let l∞ be the line marked as improper. The set of points P of the original projective plane is cut down to the set of points A = P \ l∞ of the affine plane. Each line of the projective plane, at least three points, as has been shown in the solution of Problem 3.22. For a line l , l∞ , only one of them, the point l ∩ l∞ , is improper. Hence there are at least two proper points on the line l. Check of validity of axiom (I.3b). Finally, to confirm (A.3), we obtain three points not lying on a line. By axiom (P.3), there exist, in the projective plane, four points A, B, C, D of which no three lie on a line. In the cases that none, or only one of them lies on the improper line l∞ , we are ready. Now assume that two of them, say C and D lie on the improper line. Hence A and B are proper points. The intersection P = AC ∩ BD exists.
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•
Moreover, point P is a proper point. Indeed if P , C would be two improper points, then point A would be improper. If P , D would be two improper points, then point B would be improper. • Points A , P are distinct, otherwise A, B, D would lie on a line. Points B , P are distinct, otherwise A, B, C would lie on a line. • Finally, point P cannot lie on the line AB. Otherwise all points A, B, P, C, D would lie on this single line. Hence we have found three proper points A, B and P not lying on a line, thus confirming axiom (A.3). Proof of item (b). The set of points P of the original projective plane is cut down to the set of points A = P \ l∞ of the affine plane. All the parallel lines in one pencil, extended, have the same intersection point with the improper line l∞ . There is a bijection between these pencils and the improper points on l∞ . Hence the reconstruction of the improper points via pencils brings back the original projective plane. Definition 3.19 (Dual projective plane). The dual of a projective plane is the projective plane where the points are the lines of the original one, and vice versa. The "points" of the dual projective plane are the lines of the primal projective plane. The "lines" of the dual projective plane are the points of the primal projective plane. 3.7.
The Fano Plane
Problem 3.24. The smallest projective plane is obtained by projective completion of the affine plane of order 2. Find and describe this projective plane. How many points, and how many lines does it have. How are they situated.
Figure 3.13. Fano’s seven-point projective plane
Answer. We begin with the affine plane with the four points A, B, D and E and six lines. The pencil of parallels AB k DE produces the improper point C. Similarly, the pencil of parallels AE k DB produces the improper point F and the pencil AD k BE produces the improper point G. Finally, we draw the improper line through the points C, F and G. We get a projective plane with seven points and seven lines. Each line has exactly three points. In each point intersect exact three lines.
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Problem 3.25. Give a highly symmetric illustration for the Fano plane based on an equilateral triangle. Your symmetric illustration is really isomorphic to the projective plane from page 55 above, which was obtained in problem 3.24 by completion. Denote the seven points with the same names in both drawings, consistently in a way to show the isomorphism. After you have obtained the isomorphism, color the lines with seven different colors. Give the corresponding lines in the other model the same colors.
Figure 3.14. The symmetric drawing of the Fano-plane is really isomorphic to the projective completion of the affine plane of order 2.
Answer. The figure on page 56 give an illustration based on an equilateral triangle. To check that this symmetric illustration is isomorphic to the illustration I have given on the left side, one needs to names the points in both illustrations in a way that the incidence relations hold for the same names. Thus the isomorphism is given by the correspondence of names. To find such an isomorphism, the key observation is that a triangle can be mapped to any triangle, but afterwards the correspondence of the remaining points is uniquely determined. Problem 3.26 (Self-duality of the Fano geometry). The dual of a projective plane is obtained by making the originally given lines the new points, and the originally given points the new lines (recall definition 3.19). Find and describe the dual of the Fano plane. Answer. For the Fano plane, one can name lines in a way such that the incidence relations of the dual are exactly the incidence relations of the primal with capital and small letters switched. It take a while to rename lines in a way to show the self-duality directly in this manner. This way of choosing names gives an isomorphism of the primal and dual, and hence confirms self-duality. In the figure on page 57, the names have already been chosen to confirm self-duality. For example, the three points A, B, D lie on line c— and the lines a, d, b intersect at the point C.
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Figure 3.15. Fano’s seven-point incidence geometry is selfdual
Figure 3.16. The points and lines of the Fano-plane named in a way to confirm self-duality.
Problem 3.27. Here is, once more, a highly symmetric illustration for the Fano plane based on an equilateral triangle. Find the 7 × 7 incidence-matrix giving the incidence between the lines and points. What does it means geometrically that the matrix is symmetric? Would the matrix be always symmetric for any arbitrary choice of the names of points and lines?
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Answer. Here is the incidence matrix: a b c d e f g
A B C 0 1 1 1 0 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0
D E 0 1 0 0 1 0 1 0 0 1 0 0 1 1
F G 0 0 1 0 0 0 0 1 0 1 1 1 1 0
Since the incidence matrix is symmetric, we see once more that the Fano plane is isomorphic to its dual. For any arbitrary choice of the names of points and lines, a symmetric matrix would only arise after a suitable permutation of the rows. 3.8.
Projective plane with coordinates
Definition 3.20 (The projective plane over a field F). The "points" of the projective plane PF2 are the sets of equivalent triples (xλ, yλ, zλ), where x, y, z ∈ F and λ runs over the nonzero elements of F. The "lines" of the projective plane are equations ax + by + cz = 0 with coefficients a, b, c ∈ F, not all three equal to zero. A "point lies on a line" if and only if the coordinate triple (x, y, z) satisfies the equation of the line. The triples (x, y, z) ∈ F3 \ (0, 0, 0) are the homogeneous coordinates for the points of the projective plane. Similarly, the triples (a, b, c) ∈ F3 \ (0, 0, 0) are the homogeneous coordinates for the lines of the projective plane. Again, it is easy to check, but important to confirm: Main Theorem 1. The projective plane over any field is a projective plane. Proof. One needs to check the incidence axioms (P.1), (P.2) and (P.3) from the definition (3.20) of a projective plane—which is not standard College algebra, but almost as easy: P.1 We confirm that every two points lie on exactly one line. Take two points with homogeneous coordinates (x1 , y1 , z1 ) , (0, 0, 0) and (x2 , y2 , z2 ). Because these points are different, one gets (x2 , y2 , z2 ) , (x1 λ, y1 λ, z1 λ) for all λ ∈ F. The homogeneous coordinates (a, b, c) of a line through the two points are a solution of the system ax1 + by1 + cz1 = 0 (3.4) ax2 + by2 + cz2 = 0 These are two equations with three unknowns, hence there exists a nontrivial solution (a, b, c) , (0, 0, 0). The system (3.4) has rank 2, because (x1 , y1 , z1 ) and (x2 , y2 , z2 ) are linearly independent. Hence the solution space is one dimensional, and hence (a, b, c) are the homogeneous coordinates for a unique line.
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P.2 We confirm that every two lines intersect at exactly one point. In the proof, the roles of points and lines interchanged: Take two lines with homogeneous coordinates (a1 , b1 , c1 ) , (0, 0, 0) and (a2 , b2 , c2 ). Because they are different, one gets (a2 , b2 , c2 ) , (λa1 , λb1 , λc1 ) for all λ ∈ F. The homogeneous coordinates (x, y, z) of the intersection point are a solution of the system a1 x + b1 y + c1 z = 0 a2 x + b2 y + c2 z = 0
(3.5)
Again, the solution space of system (3.5) is one dimensional, and hence (x, y, z) are the homogeneous coordinates for a unique point. P.3 Four points of which no three lie on a line are given by the homogeneous coordinates (1, 0, 0), (0, 1, 0), (0, 0, 1) and (1, 1, 1). Question. Definition 3.20 tells that the points of the projective plane l ∈ PF2 are defined to be the lines through the origin in the three dimension space F3 . On the other hand, one obtains a visual picture of the projective plane by adjoining to the set of points (x, y) of the affine plane the set of equivalence classes of parallel lines in this plane. How can both points of view be valid? Answer. To obtain the isomorphism between these two models of the projective plane, we take intersections with the plane P = {(x, y, 1)} : x, y ∈ F}. Let be given any point l ∈ PF2 . In coordinates this means l = {(aλ, bλ, cλ) : λ ∈ F} We take the intersection l ∩ P and distinguish two cases: a b In case that c , 0, we get a unique intersection point ( , , 1) which determines a proper c c a b point corresponding to the point ( , ) of the affine plane. c c • In case that c = 0, we get no intersection point. But we see that the given line l is parallel to the line {(aλ, bλ, 1) : λ ∈ F} lying in the intersection plane and determining the equivalence class of parallel lines for an improper point. These two cases yield the proper and improper points of the projective plane. The isomorphism between the abstract PF2 definition, and the visual definition of the projective plane using improper points is provided. •
The points of the affine plane are the proper points of the projective plane, whereas the equivalence classes of parallel lines in the affine plane are the improper points of the projective plane. The set of the points of the projective plane with the homogeneous coordinates (x, y, 0). is the improper line. From the Cartesian plane over a field, the projective plane is obtained by adjoining the improper elements. One gets back the Cartesian plane after deleting the improper elements. Problem 3.28 (Three points on a line). In a projective coordinate plane three points with the homogeneous coordinates (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ) are given. Check that the three points lie on a line if and only if the determinant x1 x2 x3 y1 y2 y3 = 0 z1 z2 z3
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Simple solution. The three points of the projective plane with the homogeneous coordinates (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ) lie on a line if and only if the three lines through the origin {(λx1 , λy1 , λz1 ) : λ ∈ F}
{(λx2 , λy2 , λz2 ) : λ ∈ F}
{(λx3 , λy3 , λz3 ) : λ ∈ F}
lie in a plane in F3 . It is know that this happens if and only if the determinant of their coordinates from above is zero.
Figure 3.17. The complete quadrilateral.
Theorem 3.3. Given is any projective coordinate plane PF2 . A quadrangle ABCD has its three intersections X = AD ∩ BC, Y = AC ∩ BD and Z = AB ∩ CD lying on a line if and only if field F has characteristic two. Proof. For any two given quadrangles x1 , x2 , x3 , x4 ∈ PF2 and x10 , x20 , x30 , x40 ∈ PF2 , there exists a projective mapping which takes xi 7→ xi0 for i = 1, 2, 3, 4. This is shown in Theorem 33. Hence it is enough to prove the theorem above for just one special quadrangle. Conveniently, we choose A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1), D = (1, 1, 1) In this special case, the points X, Y, Z turn out to have the homogeneous coordinates X = (0, 1, 1), Y = (1, 0, 1), Z = (1, 1, 0) By the previous problem, they lie on a line if and only if 0 1 1 1 0 1 1 1 0 is zero. Elementary arithmetic yields the value 2 for the determinant. Hence the points X, Y and Z lie on a line if and only if 2 = 0 which is true if and only if the field F has characteristic two. Proposition 3.3. The dual of a projective coordinate plane of a field is isomorphic to the plane.
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Proof. Once commutativity holds, the statement is easy to check since both points and lines use the same triples of homogeneous coordinates. A point P with homogeneous coordinates (x, y, z) lies on a line l with homogeneous coordinates (a, b, c) if and only if ax + by + cz = 0 The points are left vector spaces identifying (x, y, z) ∼ (xλ, yλ, zλ) with arbitrary λ , 0. But the lines are right vector spaces, in which one has to identify, (a, b, c) ∼ (µa, µb, µc) with arbitrary µ , 0 as homogeneous coordinates of the same projective line. Without assuming commutativity, a difference arises. For the dual geometry, exchange of points with lines means that the dual points become right vector spaces, whereas the dual lines become left vector spaces. The involution mapping σ σ : Points 7→ Lines between left and right vector spaces does not exist in the non commutative case. Only after having assumed commutativity, this mapping is provided simply by the identity. We get an incidence preserving bijection, and hence are able confirm that the geometry is self dual. Point P lies on line l if and only if line σ(P) goes through point σ−1 (l). Remark. For a non-Desargues projective plane, the claim of Proposition 3.3 need not to be true. Problem 3.29 (Self-duality of the Fano plane by means of coordinates). Explain a systematic procedure to get the dual of the Fano plane, using homogeneous coordinates. Give an illustration based on the equilateral triangle. Answer. The order of the Fano geometry is two. The homogeneous coordinates are the seven nonzero elements of Z2 × Z2 × Z2 . Three points lie on a line if and only if the determinant of their homogeneous coordinates is zero modulo two. Corresponding to the symmetric illustration, one can set: A : = (1, 0, 0), B := (0, 1, 0), C := (0, 0, 1), D : = (1, 1, 0), E := (1, 0, 1), F := (0, 1, 1), G := (1, 1, 1) In homogeneous coordinates, the dual of a line is the point with homogeneous coordinates orthogonal (dot product zero) to the coordinates of all the points on the line. For all lines, I have calculated their dual point, and have chosen names for the lines according to homogeneous coordinates of the dual point. I have indicated this bijection by choosing the same capital and small letters for points and lines. The preservation of the incidence relations is now easy to check directly. Problem 3.30 (The group of symmetries of the Fano plane). The permutations of the points that carry collinear points to collinear points are called collineations, automorphisms or symmetries of the plane. Determine the order and describe the group of the automorphisms for the Fano plane.
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Figure 3.18. Both illustrations confirm the self-duality
Answer. The group of automorphisms of the Fano plane is a simple group of order 168. To find and count the automorphisms, fix one triangle. As already mentioned in problem 3.24, a triangle can be mapped to any triangle, but afterwards the correspondence of the remaining points is uniquely determined. One has 7 choices to map the first vertex, 6 choices to map the second, and only 4 choices to map the third vertex, since the third vertex cannot be mapped onto the line given by the images of the first two vertices. Hence there are 7 × 6 × 4 = 168 automorphisms. There are subgroups which fix all three points of a given lines. They have order 4 and consist of the identity and pairs of two transpositions of the remaining four points. Furthermore, there are larger subgroups which fix a given line 123 as one object. They can be understood by the action on the four remaining points 4, 5, 6, 7. Since all permutations of these four points turn out to be poosible, they have order 24 and consist of the identity, pairs of two transpositions of the remaining points 4, 5, 6, 7, and permutations (123)(456), and of type (12)(4567). These four cases given permutations belonging to different conjugacy classes. Finally, a permutation of one 7-cycle can be an automorphism. If the permutation (1234567) is an automorphism, either point 4 or point 6 is the third point on the line 12. Both cases do occur. They belong to different conjugacy classes. From the above, we get 6 conjugacy classes: the identity 21 permutation of type (45)(67), with the three points 1, 2, 3 lying on one line 56 permutations of type (123)(456) with the three points 1, 2, 3 lying on one line, but the three points 4, 5, 6 not on a line 42 permutations of type (12)(4567) with points 1, 2, 3 collinear
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24 permutations of type (1234567) with 1, 2, 4 collinear 24 permutations of type (1234567) with 1, 2, 6 collinear Since these numbers add up to 168, we have found all conjugacy classes. Since 1 and 168 are the only divisors of 168 which are sums of sizes of conjugacy classes including the identity, the automorphism group has no normal subgroups. I leave it to reader to find all other subgroups. Problem 3.31 (Character table). Find the character table for the group of the automorphisms of the Fano plane.
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3.9.
Finite affine and projective incidence planes
Proposition 3.4 (The finite affine plane of order n). For every affine plane (a) Each line has the same number n of points. (b) The lines can be partitioned into n + 1 classes, each containing n parallel lines. (c) In every point intersect n + 1 different lines. (d) There are n2 points and n2 + n lines.
Figure 3.19. n + 1 = 5 pairwise intersecting lines in an affine plane of order n = 4.
Proof. Let n be the number of points on any line. Using the unique parallels, one shows that every other line (intersecting or parallel) has the same number of points. Take any two lines l and m intersecting in point L1 . No third line can be parallel to both of them. Line m, and the parallels to line m through the n − 1 points L2 , . . . , Ln on line l yield a pencil of n parallel lines. There cannot exist more parallels, because that would give more intersection points on line l. Hence the lines occur in pencils of n parallels. Take a second point M2 on line m. One gets the n + 1 lines l, m = L1 M2 , L2 M2 , L3 M2 , . . . , Ln M2
(3.6)
which are pairwise intersecting. If a line intersects all these n + 1 lines, it is equal to one of them. Hence any line is parallel to one of the lines in the list (3.6). Thus we get n + 1 pencils of n parallel lines, which are n2 + n lines altogether. Because of the n + 1 pairwise intersecting lines, in every point there intersect n + 1 lines, each one of which is either one of the lines (3.6) or a parallel to one of them. The intersection points of the parallels to the intersecting lines l and m yield n2 points. There can be no other point that this square grid, because there would be more than n + 1 lines through any outside hypothetical point.
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Definition 3.21 (Order of an affine plane). The number of points on any line is called the order of the affine plane. Proposition 3.5 (The projective completion of a finite affine plane). From an affine plane of order n, one gets a projective plane with n2 + n + 1 points and the same number of lines. Each line has the same number n + 1 of points. Through each point go n + 1 lines. Conversely, after delating any line from any given projective plane, one is left with an affine plane. In the case above with are left with n2 points. Reason. Discarting any line and its points as improper leave an affine plane. By problem (3.4), we are left with n2 proper points and n2 + n proper lines. Each line has n points. By adjoining the improper elements, we get n + 1 extra improper points, which together yields n2 + n + 1 points for the original projective plane. Too, we get one extra line, and hence n2 + n + 1 lines in the projective plane. Each proper line gets one extra improper point—in its direction at infinity—and thus has n + 1 points. The same number of points lie on the improper line. Through each proper point go n + 1 lines, as before. Through each improper point go the n parallel proper lines and the improper line, hence n + 1 lines together. Definition 3.22 (Order of a projective plane). The number of points on any line minus one, or, equivalently, the number of lines through a given point minus one, is called the order of the projective plane. Thus the affine plane, and the projective plane corresponding to it by concatenating and deleting improper elements, have the same order. Problem 3.32. Use homogeneous coordinates and the shorthand formula PF2 =
F3 \ {(0, 0, 0)} F \ {0}
to count the points of a projective plane over a finite field with n points. Convince yourself that you get the same result as explained above. Answer. Since (F3 \ {(0, 0, 0)}) has n3 − 1 points, and we take equivalence classes over (F \ {0})— which has n − 1 points—we can count the points of the projective plane |PF2 | =
n3 − 1 = n2 + n + 1 n−1
This is the same result as obtained earlier. We need just one extra line—the improper line— additionally to the n2 + n lines from the affine plane of order n. 3.10. Elementary propositions for three-dimensional incidence spaces Among the consequences of the axioms of incidence, Hilbert spells out two propositions. Proposition 3.6 (Hilbert’s Proposition 1).. (a) Any two different lines have either no or exactly one point in common. (b) Any two different planes have either no point in common, or they have one line and no further points in common. (c) A plane and a line not lying in this plane have either no or exactly one point in common.
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Problem 3.33. Prove statement (b). Use definite names for the objects, explain the steps, and state which axioms are involved. Prove statement (c). Write a short clear proof. Answer. (b) Take any two different planes α , β. If they do not intersect, we are ready. Assume that the two planes have point of intersection A in common. By axiom (I.7), the two planes have a further point B , A in common. By axioms (I.1) and (I.2), there exists a unique line a through points A and B. By axiom (I.6), every point of line a lies in the plane α. Reasoning about the same points A and B and line a, axiom (I.6) yields that every point of line a lies in the plane β, too. In other words, we get the inclusion a⊆α∩β We need still to check that the two planes α and β do not have any point outside the line a in common. Take any point C lying in both planes α and β. We assume towards a contradiction that point C does not lie on the line a. Both planes α and β contain the three points A, B and C. By axiom (I.5), there exists at most one plane through the three points A, B and C. Hence these two planes α and β would be equal, contradicting the assumption α , β. Hence the two planes α and β have no point outside the line a in common. Together we get the equality a=α∩β We have shown that two planes intersecting at one point have a line through this point as their intersection. (c) Take a plane α and a line a not lying in α. We assume towards a contradiction they intersect at two different points A , B. By axiom (I.6), every point of a lies in the plane α, contradicting the assumption that the line a is not lying in α. Hence the plane α and line a have at most one point in common. Proposition 3.7 (Hilbert’s Proposition 2).. (a) Given a line and a point not lying on it, there exists a unique plane containing the line and the point. (b) Given two different intersecting lines, there exists a unique plane containing these two lines. As far as two dimensional geometry is concerned, Proposition 3.6 reduces to the one simple statement that any two different lines either intersect at one point, or are parallel. Proof. (a) Given is the line a and a point C not lying on it. By axiom (I.3a), there exist two points A , B on the line a. The three points A, B, C do not lie on a line. Hence by axioms (I.4) and (I.5), there exists a unique plane α through these three points. By axiom (I.6), all points of line a lie in the plane α. Hence this plane contains both the given line and the given point. Uniqueness of the plane follows by axiom (I.5).
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(b) Given are two different lines a and b, intersecting in point C. By axiom (I.3a), there exists a second point A , C on line a, and a point B , C on the line b. The three points A, B and C do not lie on one line, since the given lines a , b are different. Hence, again by axioms (I.4) and (I.5), there exists a unique plane α through these three points. By axiom (I.6), all points of line a, as well as of line b lie in the plane α. Hence this plane contains both given lines. Uniqueness of the plane follows by axiom (I.5). Lemma 3.4. Any three lines in three-dimensional incidence space that go through one point but do not lie in a plane are different from each other. Proof. Let the three lines g, h, l all go through point P but not lie in a plane. The assumption that two of them are equal leads to a contradiction. Indeed, under the assumption that two of them would be the same line g = h, Hilbert’s proposition 3.7 part (b) would imply existence of a plane through line g and the third line l, since they intersect at point P. Hence all three lines would lie in a plane, a contradiction. Because of this contradiction, we see all three lines are different from each other. Proposition 3.8 (Three lines through any point). Through any point of a three-dimensional incidence space, there exist three lines not lying in a plane. Especially, these three lines are different from each other. A second variant. We use a modification of the proof for lemma 3.2. Let any point P be given. By axiom (I.8), there exist at least four points not lying in a plane. We call them A, B, C and D. In the case that point P is among the four points A, B, C, D, we may assume P = A. We draw the three lines AB, AC and AD. We check that these three lines do not lie in a plane,—otherwise the four points A, B, C, D would lie in a plane, contrary to the assumption. In the case that point P is different from all four points A, B, C, D, we draw all four lines PA, PB, PC and PD. • •
In the case that either lines PA, PB, PC or lines PA, PB, PD do not lie in a plane, we are ready. It may happen that both lines PA, PB, PC lie in the plane γ, and lines PA, PB, PC lie in a plane δ. But it cannot happen that any three lines, say e.g. PA, PB, PC, are equal. Otherwise Hilbert’s proposition 3.7 part (a) would imply existence of a plane through line l = PA = PB = PC and the fourth point D. Thus the four points A, B, C, D would lie in a plane, contrary to the assumption. Hence lines PA, PB, PC determine the plane γ uniquely. The lines PA, PB, PD determine the plane δ uniquely. Moreover, these are two different planes,—otherwise all four lines PA, PB, PC, PD and all the four points A, B, C, D would lie in a plane, contrary to the assumption. Now Hilbert’s proposition 3.6 part (b) implies that the intersection γ ∩ δ is a unique line, which is line PA = PB. We conclude that the lines PB, PC, PD cannot lie in a plane,—otherwise all four lines PA, PB, PC, PD and all the four points A, B, C, D would lie in a plane, contrary to the assumption. Hence we have obtained three lines PB, PC and PD not lying in a plane.
In all cases we have obtained three different lines through the arbitrary point P.
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Proposition 3.9 (Every plane is spanned by three points). In every plane, there exist three points not lying on a line. Hence any plane can be uniquely specified by three points lying on it. Proof. By axiom (I.4a), there exists a point P in the given plane π. By proposition 3.8, there exist three lines g, h and l through point P, which do not lie in a plane. By Hilbert’s proposition 3.7 part (b), there exists a plane spanned by the lines g and h, which we denote by α := span(P, g, h). Similarly, there exist the planes β := span(P, g, l) and γ := span(P, h, l). All three planes are different from each other,—otherwise the three lines g, h, l would lie in a plane. By Hilbert’s Proposition 3.6 part (b), one gets the intersection lines α ∩ π, β ∩ π and γ ∩ π. Two of them may be equal. Nevertheless, it is impossible that they are all three equal,—otherwise all three line g, h, and l would be equal. We may assume the two lines α ∩ π , β ∩ π to be different. By axiom (I.3a), a second point Q , P on the line α ∩ π exists. Similarly, a point R , P exists on the second line β ∩ π, again by axiom (I.3a). All three points P, Q, R lie in the given plane π, but not on one line. By axiom (I.5), they determine the plane π uniquely. Hence we say they span the plane π. 3.11. Three-dimensional Euclidean incidence geometry For three dimensional incidence geometry, but no congruence assumed, one needs a strict version of the axiom of parallelism, including existence and uniqueness. (IV*) Strict axiom of parallelism Given is a line a and a point P not lying on a. In the plane determined by the point P and the line a, there exists a unique parallel m to a through point P Definition 3.23 (Parallel lines and planes in three dimensions).. (a) Two different lines which lie in one plane and do not intersect are called parallel. (b) Two different planes which do not intersect are called parallel. (c) A line and a plane that do not intersect are called parallel. In the following, we assume at first the axioms of incidence (I.1) through (I.8), and Hilbert’s parallel axiom (IV) in order to assure the uniqueness of a parallel to a given line through a given point. The strict axiom of parallelism (IV*) is required later at the point where the existence of the parallel lines and planes is needed. By generalizing Proclus’ Lemma 3.3 and problem 3.3 to three dimensions, we are led to the following results. Proposition 3.10.. Given is a plane and two parallel lines. We assume that the plane intersects one of them. Then either one of the three cases occurs: (i) both lines lie in the plane; (ii) one of the lines lies in the plane, and the other one is parallel to the plane; (iii) both lines intersect the plane at a single point.
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Proof. Let the plane α and the line a intersect at point A. The given lines a and b are parallel, and both lie in the plane β. The two planes α and β can be either equal or different. The special case α = β leads to alternative (i). In the other case α , β, the two planes intersect at point A. By Hilbert’s Proposition 1(b) (see Proposition 3.6), they intersect in a line c := α ∩ β. Either one of the following cases occurs: • c = a. Hence line a lies in the plane α and line b is parallel to the plane α, leading to case (ii); • c , a. Now we use Proclus’ Lemma in the plane β, and conclude that line c intersects the parallel b k a, too. The intersection point B := b ∩ c lies in the plane α, too. Hence the second parallel b intersects the given plane, as to be shown. This leads to case (iii). Proposition 3.11.. (a) a plane intersecting one of two parallel lines at only one point intersects the other line, too. (b) a line intersecting one of two parallel planes and not lying entirely in the first plane, intersects the other plane, too. (c) a third plane intersecting one of two parallel planes intersects the other plane, too. Proof of item (b). Let the line c and the plane α intersect at point A. Let β k α be second parallel plane. We choose any point Q in the plane β, which is possible by axiom (I.4a). Let γ := span(c, Q) be the plane containing line c and point Q, which exists because of Proposition 3.7 part (a). Let p := α ∩ γ and q := β ∩ γ be the respective intersection lines, which exist according to Proposition 3.7 part (b). The lines p k q are parallel since they lie in the parallel planes α k β. Now we use Proclus’ Lemma for these two parallels in the plane γ. The given line c intersects the first parallel p in point A, hence it intersects the second parallel q in a point B := q ∩ c ⊂ β ∩ a. This is the intersection of the second parallel plane β with line a to be obtained. Proof of item (c). Let the third plane γ , α, β and the plane α intersect, say in the line a. Let β k α be second parallel plane. In the plane γ, we choose a second line c which intersects line a, say in a point A. The line a lies entirely in the plane α, but the line c does not lie entirely in the plane α. We can now apply part (b) and get that the line c intersects the plane β in some point B = c ∩ β ⊂ γ ∩ β. Hence the given plane γ intersects the second parallel plane β k α, as to be shown. The contrapositive of Proposition 3.11 part (a) and (b) can be formulated as follows: Proposition 3.12.. (a’) If one of two parallel lines is parallel to a plane, then the second line either lies entire in this plane, or is parallel to the plane, too. (b’) a line which is parallel to one of two parallel planes, either lies entirely in the second plane or is parallel to the other plane, too. Proposition 3.11 part (c) has further easy consequences:
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Lemma 3.5 (Uniqueness of a parallel plane). For any plane α and point P not lying on it, there exists at most one parallel plane through this point. Proof. Assume towards a contradiction that β , γ are two distinct planes through point P, and both β k α and γ k α. Now plane γ is the third plane, intersecting one of two parallel planes β k α, namely plane β in the point P. Hence by part (c) of Proposition 3.11, the plane γ intersects the other plane α, too. This contradicts the assumption γ k α. From this contradiction, we conclude there cannot exist two distinct planes through point P which are both parallel to plane α. Thus the uniqueness of the parallel plane; parallel to the plane α through point P not lying on α; has been confirmed. Proposition 3.13.. Two planes which are both parallel to a third plane are either equal or parallel planes. Consequently, being equal or parallel is an equivalence relation among planes. Proof. We assume β k α and γ k α: both planes β and γ are parallel to plane α. If β , γ, the planes β and γ cannot intersect;—otherwise there would be two distinct planes through the intersection point P, and both are parallel to plane α. By Lemma 3.5 the parallel plane is unique, and consequently β = γ contrary to the assumption. Hence either β = γ, or β k γ. More important, being equal or parallel defines an equivalence relation among lines, too. This follows from the following a bid more difficult result. Effectively, we have to construct the third plane of a prism with the three lines as edges. Proposition 3.14.. Two lines which are both parallel to a third line are either equal or parallel lines. Consequently, being equal or parallel is an equivalence relation among lines. Proof. Let the third line b , a, c be parallel to both lines a k b and c k b. Let α be the plane in which a and b lie and γ be the plane in which c and b lie. In the case that these two planes are equal, we are back to the two dimensional version of problem 3.3. We conclude that the lines a and c are either parallel or equal. We now assume that the planes α , γ are different. We choose any point C on the line c and let β = span(a, C) be the plane in which line a and point C lie. The planes β , γ are different,— otherwise all three planes α, β and γ would be equal. We define the intersection line d = β ∩ γ and have to confirm that c = d. Thus we get the third plane at the surface of a prism with the three lines a, b and c as edges. The lines b, c, d all lie in the plane γ. Lines b k c are parallel by assumption. Since b ∩ d = (α ∩ γ) ∩ (β ∩ γ) = (α ∩ β) ∩ (α ∩ γ) = a ∩ b = ∅ we see that the lines b k d are parallel, too. We now use Proclus’ Lemma in the plane β. The line d intersects the parallel c k b in point C. If lines d and c would be different, the line d would intersect the parallel b, too. That is a contradiction to the last formula above. Hence the lines d and c are equal. Thus we conclude that both lines: line a and line c = d = β∩γ lie in the plane β = span(a, C). Moreover, the lines a and c do not intersect since a ∩ c = a ∩ d = (α ∩ β) ∩ (β ∩ γ) = ∅ Hence the lines a and c are parallel, as to be shown.
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As a consequence of Proposition 3.12, part (a’) and (b’), the two equivalence relations for parallelism of lines and planes are compatible. In the following, we shall use existence of the parallel lines and planes. Because there is no direct construction of parallels, we need not only Hilbert’s parallel axiom, but the strict axiom of parallelism (IV*). Proposition 3.15 (Existence and uniqueness of a parallel plane). We assume the axioms of incidence (I.1) through (I.8) and the Euclidean parallel postulate (IV*). For any plane α and point P not lying on it, there exists a unique parallel plane through this point.
Figure 3.20. Construction of the plane through a given point parallel to a given plane.
Proof. The easier proof of uniqueness is left to the reader. To show existence, we use two intersecting lines in the given plane, and their parallels to the given point. As shown in Proposition 3.9, we can choose three points A, B and C spanning the given plane α. In the plane spanned by points A, B and the given point P, we get the parallel m through P to the line AB. In the plane spanned by points A, C and the given point P, we get the parallel l through P to the line AC. By Hilbert’s Proposition 2, there exists a unique plane π containing the intersecting lines m and l. Clearly this plane contains point P. We need still to check that the planes α and π are parallel. Assume towards a contradiction that the planes α , π intersect, say at point X. By Hilbert’s Proposition 1b, they would intersect in a line x through point X. All three lines m, l and x lie in the plane π. The lines m and x do not intersect because of Lemma ??, hence lines m k x are parallel. Similarly, we see that lines l k x are parallel. The line x would have two different intersecting parallel m and l, contradicting uniqueness of the parallel. The only way to avoid the contradiction is planes α and π being parallel.
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Figure 4.1. Desargues’ configuration in parallel setting.
4. The Theorems of Desargues and Pappus 4.1. Desargues’ Theorem For a triangle 4ABC, it is assumed that the three vertices A,B and C do not lie on a line. The order of the three vertices A, B and C matters. The lines a = BC, b = AC, and c = AB are the sides of the triangle. See also definition 14.3 above. 1 Proposition 4.1 (Theorem of Desargues in parallel setting).. (i) If two triangles in a plane have three pairs of parallel corresponding sides, then the three lines through their corresponding vertices either meet in one point, or all three are parallel. (ii) If the three lines through the corresponding vertices of two triangles in a plane either meet in one point, or all three are parallel, and moreover, two pairs of corresponding sides are parallel, then the third pair of corresponding sides are parallel, too. The following definition strictly speaking refers to the projective completion of the Euclidean space or plane. Definition 4.1 (Triangles in perspective). The vertices of two triangles 4ABC and 4A0 B0C 0 lie in perspective from a point O iff the three lines AA0 , BB0 and CC 0 through corresponding vertices go through one point O, or are all three parallel. In either case, we may either say simply that the two triangles lie in perspective. Definition 4.2. The sides of two triangles lie in perspective iff the three intersection points a ∩ a0 , b∩b0 and c∩c0 of corresponding sides lie on a line. Again, we have to assert that these intersections exist either as proper or improper points. Remark. The colorings of the first and the second part of the figure on page 73 are chosen in a way that color red always refers to the conclusion. In the upper figure, red colored parts stress the conclusion of converse Lemma 4.2. In the lower figure red colored parts stress the conclusion of the Lemma 4.1. 1
We cannot and need not at this point use the axioms of order.
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Figure 4.2. The Theorem of Desargues in parallel setting.
Lemma 4.1 (First part of Desargues’ Theorem). If two triangles are in perspective, and, furthermore, two pairs of corresponding sides are parallel, then the third pair of sides are parallel, too. Lemma 4.2 (Converse of the first part of Desargues’ Theorem). If the sides of two triangles are pairwise parallel, then the two triangles are either in perspective from a point, or the three lines through pairs of corresponding vertices are parallel. Question. Convince yourself that the first part ?? of Desargues’ Theorem implies the converse 4.2. Answer. Question. Convince yourself that the converse 4.2 implies the first part ?? of Desargues Theorem. Answer. 4.2.
Theorem of Desargues and related theorems in projective setting
Proposition 4.2 (Theorem of Desargues in projective setting). We assume the axioms of incidence (I.1) through (I.8) and the strict axiom of parallelism (IV*). The vertices of two triangles lie in perspective if and only if its sides lie in perspective. For any two triangles, for the assertions occurring above, I use the short hands (Verts) "the vertices of two triangles lie in perspective" (Sides) "the sides of two triangles lie in perspective"
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Figure 4.3. Desargues’ Theorem.
Problem 4.1. Assume that the triangles lie in different planes. Convince yourself that (Verts) ⇒ (Sides). Problem 4.2. Assume that the triangles lie in different planes. Convince yourself that (Sides) ⇒ (Verts). Lemma 4.3. If (Verts) ⇒ (Sides) holds for any two triangles lying in different planes, it holds for any two different triangles in the same plane, too. Reason. Given are two triangles 4ABC and 4A0 B0C 0 in the same plane π and its vertices lie in perspective, say at center Z. At least one pair of corresponding sides are different. We can assume that the sides AC and A0C 0 are different and intersect at point Q. Let X be a point not lying in the plane π. The plane spanned by points Z, B and X contains at least three points on every line. 1 Hence there exists a further point D , X, D , B on the line XB. Let D0 be the intersection of lines ZD and XB0 . Indeed, if B , B0 , all seven points X, Z, D, D0 , B, B0 are different (why?). We use point X as center of projection to produce a new three dimensional Desargues configuration, consisting of the triangles 4ADC and 4A0 D0C 0 . They lie in different planes (why?). By construction, their vertices lie in perspective. Hence, by the three-dimensional Desargues theorem, their sides lie in perspective. The three intersection points Q = AC ∩ A0C 0 , S := AD ∩ A0 D0 , T := CD ∩ C 0 D0 exist and lie on a line. A central projection with center X onto the plane π maps Q 7→ Q, S 7→ P, T 7→ R where P := AB ∩ A0 B0 , Q = AC ∩ A0C 0 , R := BC ∩ B0C 0 Hence these three points lie on a line, as to be shown. 1
At this point, we have used axiom [P.3] "There exist four points of which no three lie on a line", from the definition 3.20 of a projective plane.
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Figure 4.4. Projecting the three-dimensional Desargues to the two-dimensional one.
Remark. Work in the completion of the Euclidean space. Convince yourself that including cases were several of the vertices or sides turn out to be equal do not lead to contradictions. Lemma 4.4. If (Verts) ⇒ (Sides) holds for any two triangles lying in the same plane, the converse (Sides) ⇒ (Verts) holds for any two triangles in the same plane, too. Reason. Given are two triangles 4ABC and 4A0 B0C 0 in the same plane, the sides of which lie in perspective. The three points P := AB ∩ A0 B0 , Q = AC ∩ A0C 0 , R := BC ∩ B0C 0 lie on a line. Let Z = BB0 ∩ CC 0 be the intersection of the lines through two pairs of corresponding vertices. We apply the first part of Desargues theorem to the triangles 4PBB0 and 4QCC 0 . Their vertices lie in perspective with center R. Hence their sides lie in perspective. Thus the three intersections points A = PB ∩ QC , A0 := PB0 ∩ QC 0 , Z := BB0 ∩ CC 0 lie on a line. Hence the vertices of triangles 4ABC and 4A0 B0C 0 lie in perspective with center Z, as to be shown. Remark. Convince yourself that including cases were several of the vertices or sides turn out to be equal do not lead to contradictions. Problem 4.3. Convince yourself: If (Verts) ⇒ (Sides) holds for any two triangles lying in different planes, the converse (Sides) ⇒ (Verts) holds for any two triangles in different planes, too.
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Figure 4.5. Desargues implies the converse Desargues.
Proposition 4.3. If an affine plane is embedded into a three dimensional incidence geometry, where the axioms of incidence (I.1) through (I.8) and the Euclidean parallel postulate (IV*) hold, then the Theorem of Desargues holds in this affine plane, respectively its completion. Proof. This proof puts together the lemmas above and is left as an exercise for the reader.
The Theorem of Desargues 4.1 in parallel setting is a special case of the general theorem of Desargues 4.2 in projective setting, This special case is produced by the extra assumption that the line witnessing the sides of the triangles are in perspective, is an improper line. A different special case may be produced by the assumption that the point witnessing the vertices of the triangles are lying in perspective, is an improper point. In that manner, we get the following less common case worth to be stated: Lemma 4.5. Given is an affine plane having a projective completion for which Desargues’ Theorem holds. Assume the sides of the triangles 4ABC and 4A0 B0C 0 lie in perspective, and the lines AA0 and BB0 are parallel. Then the line CC 0 is parallel to these two lines. Problem 4.4. Given is an affine plane of order n ≥ 4 having a projective completion for which Desargues’ Theorem holds. Given are two different parallel lines AA0 k BB0 and the point C. Use Lemma 4.5 to construct the parallel to line AA0 through the point C. The main outline for the procedure. In the figure on page 77, points and lines are constructed in the order 1 2 R 3 P 4 5 Q 6 7 C0 Get the intersection point R = AB ∩ A0 B0 . We choose any point P on the line BC.
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Figure 4.6. Constructing the parallel to two given parallel lines.
Get the intersection point Q = CA ∩ PR. Get the intersection point C 0 = A0 Q ∩ PB0 . The line CC 0 k AA0 is the parallel to be constructed. Complete proof including considerations for the exceptional cases. In an affine plane of order n ≥ 4 are given two different parallel lines a = AA0 k b = BB0 and the point C, not lying on either line. In the special case that points A, B, C lie on a line, we replace A by a third point on the line a, different from A and A0 . Thus we may assume that the triangle 4ABC exists. On the line AB, we choose any point R , A, B different from A and B. One see that R , B0 since R lies neither on the line A nor the line b. The line RB0 exists and intersects the line b. By Proclus’ lemma, the line RB0 intersects the line a, too. We call the intersection point A0 , deleting the former point with this name. Since B , B0 , we see that A , A0 , too. Let Q0 be the intersection point of line CA with the parallel to line BC through point R. (It does not matter if Q0 is improper.) On the line CA, there exists a (proper) point Q lying neither on line a nor line b and different from Q0 . The points Q , R are different,—otherwise the three points A, B, C would lie on a line, contradicting the choice of A, B, C made above. By the choice of point Q, the lines QR and BC intersect. We call the intersection point P. Lemma 4.6 (Claim). I now claim that • P , B0 , and P does not lie on line b; • Q , A0 and Q does not lie on line a; • the lines PB0 and QA0 are not parallel. Reason for the lemma. • P = B0 would imply that point B0 would lie on the line BC and hence 0 B = B , which is impossible. P does not lie on line b,—otherwise P = B0 ; • Q = A0 is impossible since Q does not lie on the line a. • Assume towards a contradiction that lines PB0 k QA0 are parallel. We use the projective completion and let C 0 = PB0 ∩ QA0 be the improper intersection point in the completion. Now C , C 0 since C is a proper point. The three lines PB0 k QA0 k CC 0 are all parallel to each other.
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By the second part of the Theorem of Desargues 4.2 in projective setting, the lines PB0 k QA0 k CC 0 k AA0 k BB0 are all parallel to each other. Hence points P, B, B0 lie on a line, which is line b. This is impossible as pointed out above. The points Q, A, A0 lie on a line, which is line a. This is impossible as pointed out above. By the lemma, the lines PB0 and QA0 are not parallel. Instead they have a proper intersection point C 0 . By the second part of the Theorem of Desargues 4.2 in projective setting, any two triangles with their sides lying in perspective, have their vertices lying in perspective. The sides of the two triangles 4ABC and 4A0 B0C 0 lie in perspective, as the line PRQ witnesses. Hence the vertices of these two triangles lie in perspective, too. Since the lines AA0 k BB0 are different and parallel, the center Z of perspective is an improper point. The conclusion that points C, C 0 and the improper point Z lie on a line yields AA0 k BB0 k CC 0 . Thus we have obtained the requested parallel. Theorem 4.1 (The Little Theorem of Desargues). Given is an affine plane, respectively its projective completion. We assume that • the vertices of two triangles lie in perspective; • the center of projection and intersections points on two pairs of corresponding sides lie on a line L. Then the intersection point of the third pair of corresponding sides lies on the line L, too. Thus the sides of the triangles lie in perspective. Proposition 4.4 (Converse of the Little Theorem of Desargues). We assume that • two triangles lie in the same plane and their sides are in perspective at the line L; • the intersection point of two lines through corresponding vertices lies on the line L. These assumptions imply that the vertices of the triangles lie in perspective. Remark. In an affine plane, we have to assert that the relevant intersections exist either as proper or improper points. So these theorems really refers to the projective completion of the affine plane. Problem 4.5. Convince yourself that in any affine plane, the Little Theorem of Desargues implies its converse. Proposition 4.5. In any affine plane, the Little Theorem of Desargues implies its converse. The Theorem of Desargues implies the Little Theorem of Desargues. Definition 4.3 (Configuration of scissors). A scissor is called the system of two distinct intersecting or parallel lines, with two points on each one of them, and the quadrilateral zig-zaging between these doublets. A configuration of scissors consists of two such scissors lying between the same pair of lines, together with consistent bijections between the points, and the sides of the two scissors. An example is shown in the figure on page ??. Theorem 4.2 (Theorem of scissors). If the two scissors from a configuration of scissors have three pairs of corresponding parallel sides, their fourth pair of corresponding sides are parallels, too. An illustration is given in the figure on page 79.
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Figure 4.7. The Theorem of scissors.
Theorem 4.3 (Little Theorem of scissors). If a configuration of scissors lies between two parallel lines, and the two scissors have three pairs of corresponding parallel sides, their fourth pair of corresponding sides are parallels, too. Problem 4.6. Draw an illustration for the Little Theorem of scissors.
Figure 4.8. The Little Theorem of scissors.
Answer. An illustration is given in the figure on page 79. Theorem 4.4 ("Scissors Theorem"). For any affine plane,
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Figure 4.9. Prove the Theorem of scissors.
(a) validity of the Desargues Theorem implies validity of the Theorem of scissors. (b) validity of the Little Desargues Theorem implies validity of the Little Theorem of scissors. Problem 4.7. Give a proof of part (a). You may assume the Desargues Theorem together with its converse. Into the figure on page 80, put the extra lines for the proof in green, and mark and name the extra points needed. Write a paragraph for the proof. Answer. Let X be the intersection point of lines AD and BC, and correspondingly let X 0 = A0 D0 ∩ B0C 0 . The triangles 4ABX and 4A0 B0 X 0 have their sides, and hence, by the converse Desargues Theorem, their vertices in perspective. Thus the three lines AA0 , BB0 and XX 0 intersect at one point. Hence the two triangles 4XCD and 4X 0C 0 D0 have their vertices in perspective. Hence, by the Desargues Theorem, the two triangles have their sides in perspective. Thus the lines CD and C 0 D0 are parallel, as claimed. Remark. In a more comprehensive treatment of the Theorems of Desargues and Pappus, and the corresponding Little Theorems, it is customary to use the setting of a projective plane. Neither the Theorem of Desargues, nor the Theorem of Pappus, nor the corresponding Little Theorems hold automatically in every projective plane. In the affine setting, assuming only the axioms of incidence (I.1) through (I.3) and the Euclidean parallel postulate (IV*), is not enough to deduct the Little Theorem of Desargues, or the Theorem of Desargues. The first example of a non-Desarguesian plane is due to Frederic Moulton (1902). A detailed exposition is available in Hilbert’s Foundations, and in the recent book by Stillwell [34]. In the Moulton plane, it is easy to get an instance, where even the Little Theorem of Desargues is violated. Remark. The embedding into a three dimensional geometry is the most natural assumption leading to the Theorem of Desargues. For Hilbert and his followers this is a major motivation why to include three dimensions into the axiomatic setting. Remark. It is an rather easy but important fact that the Theorem of Desargues holds for all incidence geometries obtained from coordinates with values in a field or skew field.
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Figure 4.10. How to construct a tiling.
4.3. Tiling in perspective view Given is a parallelogram, or even simply a square or rectangle. Clearly it is possible to tile the Euclidean plane with shifts of the parallelogram. How can one draw the perspective view of such a tiling? From the given first quadrangle, we can find the horizon since two pairs of opposite sides intersect in two points on the horizon. To construct shifts of the first quadrangle, we use the fact that the shifted quadrangles have parallel diagonals. We need to use just one diagonal of the first quadrangle, and get in a unique way three adjacent ones. The drawing on page 81 shows the process in perspective, as well as simply from the view of the Euclidean plane. In both figures is done the construction with the following steps: • Construct parallel p1 to the diagonal, • find intersection point S 2, • draw parallel p3 through S 2 to the horizontal side, • intersection point S 4 with the diagonal, • draw the parallel p5 through S 4 to the vertical side. The other intersection points which complete the four adjacent parallelograms are obviously given at that point. A deeper mathematical question is the following. The process can obviously be done in any affine plane. Do there exist affine planes for which this process leads to contradictions. Especially, is the third diagonal automatically parallel. And how about the three other diagonals. Are they parallel? Are there indeed three points on the other long diagonal? Proposition 4.6. Assume the Little Theorem of Desargues holds. Then the construction from above of the four translates of a quadrangle can be done such that the parallel sides and parallel diagonals of the translated quadrangles intersect on four points on the horizon. In other words, the obvious incidences asked for above do hold. The figures on page 82 show how to complete the proof in five steps. The first three steps depend on the Little Theorem of Desargues; the last two steps on the converse Little Theorem of Desargues.
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Figure 4.11. How to prove the expected incidences;—step 1.
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Figure 4.12. How to prove the expected incidences;—step 2.
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Figure 4.13. How to prove the expected incidences;—step 3.
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Figure 4.14. How to prove the expected incidences;—all steps.
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4.4. The Prime Power Conjecture about Non-Desarguesian Planes From analytic geometry we are so much used to coordinates that one could easily overlook the main difficult and important questions arising now. Problem I. From which properties can one recognize that a given projective plane is constructed as a plane over some field? As Hilbert has shown, to answer this question, one needs two major theorems from projective geometry: the Theorems of Pappus and Desargues. We give only the main result and postpone a detailed proof to later sections. (see "Arithmetic of Segments—Hilbert’s Road from Geometry to Algebra".) Main Theorem 2. A projective plane is isomorphic to a plane obtained from a skew field if and only if the Theorem of Desargues holds in this plane. Main Theorem 3. A projective plane is isomorphic to a plane obtained from a field if and only if the Theorem of Pappus holds in this plane. Since the projective completion is possible for an arbitrary affine plane without the use of coordinates, the Main Theorem 3 implies the converse of the elementary Theorem 3.2. Hence we can state: Theorem 4.5. Any affine plane is isomorphic to a plane over a field if and only if the Theorem of Pappus holds in its projective completion. Problem II. Do there exist projective planes which are not coordinate planes over some field. Indeed, not all incidence geometries can be obtained from coordinates. The first example of a non-Desarguesian plane is due to Frederic Moulton (1902), and is available in Hilbert’s Foundations, and in the recent book by Stillwell [34]. Problem III. Do there exist finite projective planes which are not coordinate planes over some field. The first example of a finite non-Desarguesian plane was published in 1907 by O. Veblen and J.H.M. Wedderburn [39]. Indeed, these authors have obtained four non-isomorphic projective planes of order 9. They have 81 points! In the section on latin squares we given an independent account of Veblen’s arguments. 1 Open Problem IV. How many non-isomorphic planes do exist of order any prime power? Open Problem V, the Prime Power Conjecture. There do not exist projective planes of any order different from a prime power. This is a long-standing open conjecture. 1
Not all of Veblen’s results are reproduced.
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4.5. Hilbert’s investigation about the Theorem of Desargues For Hilbert’s detailed investigations about the Theorem of Desargues, it is enough to use the following formulation "in parallel setting". This is not any real restriction. By means of the projective completion, one can get back to the general Theorem of Desargues, spelled out in Proposition 4.2 above. Recall definition 3.12 of the affine plane: only the axioms of incidence (I.1),(I.2),(I.3) and the Euclidean Parallel Postulate 3.3 are assumed. We begin with an overview of Hilbert’s results, detailed proofs follow afterwards. Proposition 4.7. If an affine plane is embedded into a three-dimensional incidence geometry, where the all axioms of incidence (I.1) through (I.8) and the strict parallel axiom (IV*) hold, then the Theorem of Desargues is valid in this affine plane. Proposition 4.8. In an affine plane, where the Theorem of Desargues in parallel setting Proposition 4.1 is valid, too, one constructs the Desarguan arithmetic of segments. In this way, one does not always obtain a field, but at least a skew field. If additionally, the axioms of order (II.1) through (II.4) are valid, we get an ordered skew field. Proposition 4.9. From a given skew field D, one constructs a coordinate plane. The steps have already been explained above in Theorem 3.2 about the Cartesian plane over an arbitrary field. But now, the terms in the equation for a line ax + by + c = 0
(3.2)
need to be kept in the order as written down. For this coordinate plane, the axioms of incidence (I.1),(I.2),(I.3) and the strict axiom of parallelism (IV*), and the Theorem of Desargues in parallel setting Proposition 4.1 are valid. The skew field D to start with is isomorphic to the Desarguan arithmetic of segments one can construct, in a further step, from the coordinate plane. Moreover, if the skew field is ordered, the axioms of order (II.1) through (II.4) are valid in the coordinate plane, too. Proposition 4.10. From a given skew field D, one constructs even a three dimensional coordinate space. The steps are well-known from three dimensional analytic geometry. In the equations for lines and planes, the terms need to be kept in the order analogous to equation (3.2). In this geometry, all incidence axioms (I.1) through (I.8) and the strict axiom of parallelism (IV*) are satisfied. Hence the Theorem of Desargues is valid, both the version in parallel setting Proposition 4.1, as well as the general version given by Proposition 4.2. Moreover, if the skew field is ordered, the axioms of order (II.1) through (II.4) are valid, too. We see from these investigations: Main Theorem 4. Any projective plane is isomorphic to a plane obtained from a skew field if and only if the Theorem of Desargues holds in this plane. Remark. The Theorem of Desargues can only be proved using either all incidence axioms (I.1) through (I.8), including the three-dimensional ones. As a second road, the Theorem of Desargues can be proved in a Hilbert plane, using all congruence axioms, including the SAS axiom and the Euclidean parallel postulate. Indeed, by Theorem 18.1 from the section "Arithmetic of Segments— Hilbert’s Road from Geometry to Algebra", we know that any Pythagorean plane is isomorphic to the Cartesian plane F2 over its field F of segment lengths. By Proposition 4.9, in a coordinate plane of a field or even skew field, the Theorem of Desargues in parallel setting Proposition 4.1 is valid.
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4.6. The Mouton plane We proceed to the definition of the simplified Moulton plane. Over any ordered field F, the Cartesian plane is constructed in the usual way. We know that the axioms of incidence (I.1)(I.2)(I.3a)(I.3b), and order (II.1) through (II.4) are valid,—in their usual interpretation. The complete explanation and justification are given by Definition 3.14 and Theorem 3.2. With real coordinates, the axioms of continuity (V.1)(V.2) are valid, too. This ordered affine plane is now used as the underlying reality, and a new model is constructed by redefining the notion of "line". From the coordinate system, one obtains the x-axis, and a positive direction on it since the field is assumed to be ordered. We mark the left half-plane of line x as the upper half-plane {(x, y) : x, y ∈ F and y > 0}. The right half-plane is the lower half-plane {(x, y) : x, y ∈ F and y < 0}. Let any line l which is not parallel nor equal to x, be given. Let x+ be the ray from the intersection L := x ∩ l in the positive x-direction. Let k be the ray on the line l in the lower halfplane, and h be the ray on l in the upper half-plane, both with the same vertex L. For a line l with positive slope m > 0, let h0 the ray with vertex L and slope m/2.
Figure 4.15. Lines in the Moulton plane.
Definition 4.4 (Simplified Moulton plane). A "line" of the new geometry is now each one of the following: the x-axis; any line parallel or perpendicular to the x-axis; any line of negative slope; for any line with positive slope the union of the two rays k ∪ h0 . The "order" of the points on the new lines is defined in the obvious way. Problem 4.8. Convince yourself that from the notions above, one has obtained a planar geometry for which the axioms of incidence (I.1),(I.2),(I.3), all axioms of order (II), and the Euclidean Parallel Postulate 3.3 are all valid. Remark. This non-Desarguesian plane is due to Frederic Moulton (1902), and is available in Hilbert’s Foundations. The simplified version given in Definition 4.4 is used in the recent book by Stillwell [34].
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Proposition 4.11. There exists an order affine plane where the following violations occur: Neither the Theorem of Desargues, not the Little Theorem of Desargues, nor the Theorem of Pappus are valid. It is not possible to embed the planar geometry in a three-dimensional incidence geometry. Proof. By the figures below on page 89, page 90, and page 95, one sees easily that neither the converse part of the Theorem of Desargues, nor the Little Theorem of Desargues, nor the Little Pappus Theorem are valid in the new affine plane. To get such easy counterexamples, it is convenient to use a case in which only very few "lines" of the new geometry are affected by their intersection with the x-axis. The impossibility of an embedding into a three-dimensional incidence geometry follows via Proposition 4.7.
Figure 4.16. In the Moulton plane, the Desargues Theorem is not valid.
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Figure 4.17. In the Moulton plane, the Little Desargues Theorem is not valid.
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4.7.
Theorem of Pappus and related theorems
Definition 4.5 (Pappus’ configuration). We call the system of two distinct intersecting or parallel lines, with three points on each one of them, and the M-W hexagon zig-zaging between those triplets the configuration of Pappus.
Figure 4.18. For Pappus’ configuration: If BC 0 k B0 C and AC 0 k A0 C, then AB0 k A0 B.
In the present figure, we have points A, B, C on the horizontal line, three points A0 , B0 , C 0 on the tilted line. In the hexagon AB0CA0 BC 0 there are three pairs of opposite sides: the three pairs of lines A0 B and AB0 B0C and BC 0 C 0 A and CA0
the parallelism of which has to be investigated. Thus we are lead to conjecture that the hexagon of Pappus’ configuration containing two pairs of opposite parallel sides shall have a third pair of opposite parallel sides, too. A formulation with definitely specified variables is given here. Theorem 4.6 (Pappus Theorem in parallel setting). Let A, B, C and A0 , B0 , C 0 be both three points on two intersecting lines, all different from the intersection point. If the lines BC 0 and B0C are parallel, and the lines AC 0 and A0C are parallel, then the lines AB0 and A0 B and parallel, too. Theorem 4.7 (Pappus’ Theorem in projective setting). Let A, B, C and A0 , B0 , C 0 be three points on two intersecting lines, respectively.The three intersection points X = BC 0 ∩ B0C, Y = AC 0 ∩ A0C and Z = AB0 ∩ A0 B of three pairs of opposite sides of hexagon AB0CA0 BC 0 lie on a line. 4.8. Theorem of Hessenberg There are remarkably different routes to a proof of the Pappus’ Theorem! Theorem 4.8 (Theorem of Hessenberg). For an affine plane, validity of Pappus’s Theorem implies Desargues’ Theorem.
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Figure 4.19. The Theorem of Pappus in the projective setting.
Figure 4.20. Pappus’s Theorem implies Desargues’ Theorem
Proof. We shall proof that the second part of Desargues’ Theorem holds under the given assumptions. Furthermore, we shall assume that the two triangles 4ABC and 4A0 B0C 0 are in perspective from point O, and that AB k A0 B0 and AC k A0C 0 . We give the proof under the following simplifying assumption: OB0 k A0 C 0
(4.1)
Draw the parallel to line OB through point A. Let L be the intersection point of this parallel with line A0C 0 . Let M be the intersection point of the parallel with line OC. Let N be the intersection point of lines LB0 and AB.
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Figure 4.21. The case with OABC a parallelogram can be handled, too
Question. Why do points L, M and N exist? Answer. We now use Pappus’ Theorem for three different configurations: Step 1: Use Pappus’ Theorem in configuration ONALA0 B0 . Because AB = NA k A0 B0 and AL k B0 O, we conclude ON k LA0 . Now ON k LA0 = A0C 0 and A0C 0 k AC imply ON k AC. Step 2: Use Pappus’ Theorem in configuration ON MACB. Because ON k AC and MA k BO, we conclude N M k CB. Step 3: Use Pappus’ Theorem in configuration ON MLC 0 B0 . Because ON k LC 0 = LA0 and ML k B0 O, we conclude N M k C 0 B0 . Finally, N M k CB and N M k C 0 B0 imply CB k C 0 B0 , as to be shown.
Figure 4.22. The Little Pappus Theorem asserts for a hexagon AC 0 BA0 CB0 that has its vertices alternating on two parallel lines: if BC 0 k B0 C and AC 0 k A0 C, then AB0 k A0 B.
4.9.
Relations to the "Little Theorems"
Theorem 4.9 (Little Pappus Theorem). Let A, B, C and A0 , B0 , C 0 be both three points on two parallel lines. If the lines BC 0 and B0C are parallel, and the lines AC 0 and A0C are parallel, then the lines AB0 and A0 B and parallel, too.
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Figure 4.23.
The Little Desargues Theorem asserts for two triangles with vertices on three parallel lines: if AB k A0 B0 and BC k B0 C 0 , then AC k A0 C 0 .
Theorem 4.10 (Little Desargues Theorem). If corresponding vertices of two triangles lie on three parallel lines, and, furthermore two pairs of corresponding sides are parallel, then the third pair of sides are parallel, too. Theorem 4.11 ("Little Hessenberg Theorem"). For an affine plane, validity of the little Desargues Theorem implies validity of the little Pappus Theorem.
Figure 4.24. The Little Desargues Theorem implies the Little Pappus Theorem.
Proof. Given are points A, B, C and A0 , B0 , C 0 on two parallel lines such that BC 0 k B0C and AC 0 k A0C. We draw the parallel to line AC 0 through point B0 , and the parallel to line BC 0 through point A0 . These two lines intersect in a point D. The little Desargues Theorem is now applied to the two triangles 4ACB0 and 4C 0 A0 D. Corresponding vertices are indeed joint by three parallel lines. Too, there are two pairs of parallel sides: AC k C 0 A0 and CB0 k A0 D. Hence the little Desargues Theorem assures that AB0 k C 0 D Secondly, one applies the little Desargues Theorem to triangles 4CBA0 and 4B0C 0 D. Again, corresponding vertices are indeed joint by three parallel lines. Two pairs of parallel sides are
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CB k B0C 0 and CA0 k B0 D. Hence the little Desargues Theorem assures that BA0 k C 0 D Both instances of the little Desargues theorem together imply AB0 k BA0 , as to be shown.
Figure 4.25. In the Moulton plane, the Little Pappus Theorem is not valid.
Remark. For simplicity all four configurations— Pappus, Desargues and Little Pappus, Little Desargues— were given above in the affine version. The affine version deals with parallel lines. Hilbert’s foundations [22] use the affine version. As explained in definition 3.17, improper elements can be adjoined to produce a projective plane from a given affine plane. The pencils of parallel lines are denoted as improper points for the different directions of these pencils. The line through all improper points is called the improper line. All four configurations and theorems— Pappus, Desargues and Little Pappus, Little Desargues— have a corresponding version in the projective plane. In the projective version, the statement that any lines are parallel has to be replaced by the statement that they intersect on the improper line. 1 Theorem 4.12 (The logical relations between Pappus, Desargues and scissors). All four configurations and theorems— Pappus, Desargues and Little Pappus, Little Desargues—are now taken in the projective plane. The logical relations between them are summed up in the following diagram:
1
Hessenberg Th. 4.8
Pappus y
−−−−−−−−−−−−−→
Little Pappus
←−−−−−−−−−
Theorem 4.11
Desargues y
Scissors Theorem 4.4
−−−−−−−−−−−−−−→
Scissors Theorem 4.4
scissors y
Little Desargues −−−−−−−−−−−−−−→ Little scissors
Stillwell’s book [34] states only the projective versions.
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5. The Axioms of Order and Their Consequences Definition 5.1 (Segment). Let A and B be two distinct points. The segment AB is the set consisting of the points A and B and all points lying between A and B. The points A and B are called the endpoints of the segment, the points between A and B are called the interior points, and the remaining points on the line AB are called the exterior points of the segment. Definition 5.2 (Triangle). We define a triangle to be the union of the three segments AB, BC and CA. The three points A, B and C are assumed not to lie on a line. These three points are the vertices, and the segments BC, AC, and AB are the sides of the triangle. For a segment, it is assumed that the two endpoints A and B are different. For a triangle 4ABC, it is assumed that the three vertices do not lie on a line. Remark. For the correct statement of the Theorems about congruence, similarity of two triangles, as well as for the Theorem of Desargues, and for the purpose of fixing an orientation, we assign a definite order to the vertices and sides of the relevant triangles. 5.1.
Order of points on a line
Proposition 5.1 (Hilbert’s Proposition 3). For any two points A and C, there exists at least one point B on the line AC lying between A and C.
Figure 5.1. How to get a point inside the given segment AC.
Proof. By the axiom of incidence (I.3), there exists a point E not lying on the line AC, and by axiom of order (II.2), there exists a point F such that E is a point inside the segment AF. By the same axiom, there exists a point G such that C is a point inside segment FG. Now we use Pasch’s axiom (II.4) for the triangle 4ACF and line EG. Question. Why is 4ACF a triangle? Answer. Points A and C are two different points by assumption. The third vertex F does not lie on line AC— pother point E would lie on that line, too, contrary to the construction. Question. Why are E and G two different points?
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Answer. Assume towards a contradiction E = G. In that case, lines AF and CF would intersect in that point, hence F = E = G, contradicting the definition of point F. For the triangle 4ACF and line EG, Pasch’s axiom yields that the line intersects a second side of the triangle besides side AF. But line EG does not intersect segment FC, because the intersection point of the lines EG and FC is G, which lies outside the segment FC. Hence line EG intersects the third side of triangle 4ACF, which is segment AC. The intersection point B is a point between A and C, existence of which was to shown. −−→ Definition 5.3 (Ray). Given two distinct points A and B, the ray AB is the set consisting of the points A and B, the points inside the segment AB, and all points P on the line AB such that the given point B lies between A and P. The point A is called the vertex of the ray. −−→ Problem 5.1. Explain why the ray AB contains more points than the segment AB. −−→ Answer. The axiom of order (II.2) implies that the ray AB contains a point C such that A ∗ B ∗ C. The axiom of order (II.1) implies that A, B, C are three distinct points lying on line AB. The axiom of order (II.3) implies that point C does not satisfy A ∗ C ∗ B and hence does not lie in the segment AB. Proposition 5.2 (Hilbert’s Proposition 4, also called "Three-point Theorem"). Among any three different points A, B and C lying on a line, there exists exactly one lying between the two other points. Remark. In the first edition of the foundations of geometry, proposition 5.2 had been postulated as an axiom. The proof given below was published in 1902 by E. H. Moore [13]. Proof. The axiom or order (II.3) states that at most one of the three points lies between the two others. We need to prove that actually one of the three points does lie between the two others. Assume that neither A not C lies between the two other points. The construction shows that B does lie between A and C. One chooses a point D not lying on line AC. Next we choose a point G on the −−→ ray BD such that D lies between B and G. Being used repeatedly, we have to mention the follow consequence of axioms (II.3) and (II.4): A consequence of Pasch’s axiom. If a line cuts one side of a triangle, and the extension of a second side, then the line cuts the third side. Now we use Pasch’s axiom for the triangle 4BCG and the line AD. Because this line intersects side BG, but not side BC, the line intersects the third side CG, say in point E. A similar application of Pasch’s axiom, now for triangle 4ABG and line CD yields existence of a point F between A and G. Now use Pasch’s axiom a third time—for triangle 4AEG and the line CF. Because this line intersects side AG, but not side GE, the line intersects the third side, which is the segment AE. But this intersection point has to be the intersection of line CF with line AE, which his point D. Hence point D lies between A and E. Finally use Pasch’s axiom a fourth time—for triangle 4AEC and the line GB. Because this line intersects side AE, but not side CE, the line intersects the third side AC. But this intersection point has to be the intersection of lines GB and AC, which his point B. Hence point B lies between A and C.
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Figure 5.2. Assume that neither A nor C lie between the two other of the three points A, B and C. The construction shows that B does lie between A and C.
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Figure 5.3. Two intersecting segments.
Problem 5.2. We assume A ∗ B ∗ C and A ∗ D ∗ E, and points A, C, E do not lie on a line. Prove that the segments BE and CD intersect each other. Provide a drawing. (i) Why are the two lines BE and CD different? (ii) Why does the segment BE intersect the line CD? (iii) Why does the segment CD intersect the line BE? (iv) Why are these two intersection points equal? Why do the segments BE and CD intersect each other? Answer. (i) The two lines CD and BE are distinct, otherwise all points A, B, C, D, E would lie on a line. (ii) We use Pasch’s axiom for triangle 4ABE and line CD. This line intersects side AE, but not side AB. Hence line CD intersects the side BE, say at point X. (iii) We use Pasch’s axiom for triangle 4ACD and line BE. This line intersects side AC, but not side AD. Hence line BE intersects the side CD, say at point Y. (iv) By item (i), the lines CD and BE are different. Hence they have at most one intersection point. By items (ii) and (iii), they indeed intersect, and their unique intersection point is X = Y. Hence this point X is the intersection of the segments BE and CD. 5.2. Bernays’ Lemma One can say that Pasch’s axiom tells: a line entering a triangle has to leave it, too. It can be proved that a line cannot intersect all three sides of a triangle. Proposition 5.3 (Bernays’ Lemma). A line cannot cut all three sides of a triangle. Proof. 1 We assume that line l intersects all three sides of triangle 4ABC and derive a contradiction. If the sides BC, CA and AB would intersect line l in the points D, E and F, these three points would be all different. 1
Hilbert credits this proof to his collaborator Paul Bernays.
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Figure 5.4. Bernays’ Lemma would not hold— and line A, E 0 , E, C would be equal to line E 0 , C, D, B—all points collapse onto one line.
By Proposition 4, one of the three points lies between the two others. Without loss of generality, we can assume that point D lies between E and F. We now apply Pasch’s axiom to the triangle 4AEF 2 and the line BC. This line cut the side EF in point D. Hence it intersects a second side, say side AE in point E 0 or side AF in a point F 0 . In both cases we get a contradiction. Here are the details for the first case: E 0 , C, because E 0 lies inside the segment AE, but C lies −−→ on the ray AE outside that segment. The four points A, E 0 , E, C lie on one line. Too, the four points E 0 , C, D, B lie on one line. But these two lines have the two points E 0 and C in common. Hence they are identical. Thus all three points A, B, C lie on one line, contradicting the assumption that 4ABC is a triangle. In the second case, one gets a similar contradiction: F 0 , B, because F 0 lies inside the segment −−→ AF, but B lies on the ray AF outside that segment. The four points A, F 0 , F, B lie on one line. Too, the four points F 0 , B, D, C lie on one line. But these two lines have the two points F 0 and B in common. Hence they are identical. Thus all three points A, B, C lie on one line, contradicting the assumption that 4ABC is a triangle. Hence a line cannot intersect all three sides of a triangle. 5.3.
Plane separation
Definition 5.4. Given is plane α and a line a lying in this plane. We say that the points A and A0 lie on the same side of line a—or in the same half plane— in case that segment AA0 does not intersect the line a. The points A and B lie on different sides of line a in case that the segment AB does 2
The figure does not show this triangle correctly
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intersect the line a. The two open regions one gets in this way are called the half-planes bounded by the line a. 1 The next theorem shows that any line lying in plane separates this plane into exactly two halfplanes. Proposition 5.4 (Hilbert’s Proposition 8, the "Plane separation Theorem"). Each line a, lying in a plane α, separates the points of this plane which do not lie on the line, into two nonempty regions R and S called half planes which have the following properties: (i) every point lying in the plane α lies either on the line a, or in the half plane R, or in the half-plane S, but not in any two of these three sets; (ii) for any two points A in the half-plane R and B in the half-plane S, the segment AB intersects line a; (iii) for any two points A and A0 in the half-plane R, the segment AA0 does not intersect the line a; (iv) for any two points B and B0 in the half plane S, the segment BB0 does not intersect the line a. The following proof of the plane separation theorem uses only the axioms of order and Bernay’s Lemma.
Figure 5.5. Generic cases in the proof of the plane separation theorem.
A simple proof of the plane separation theorem. By axiom (I.3a), there exist two points P , Q on the line a. By Proposition 3.9, there exist three points in the plane α, not lying on a line. 1 . Thus there exists a third point R in the plane α, not lying on the line a. The three points P, Q and R span the plane α. 1 1
More exactly: half-planes of the plane α bounded by the line a. For purely two-dimensional geometry, we get this statement directly from axiom (I.3b)
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Figure 5.6. Special cases in the proof of the plane separation theorem.
We can now separate the points of the plane α into those lying on the line a and, additionally, the following two sets: R : = {A ∈ α : A < a, the segment RA does not intersect the line a} S : = {B ∈ α : B < a, the segment RB intersects the line a} By axiom (II.2), there exist a point S such that R ∗ P ∗ S . In other words, the point S lies on the −−→ extension of the ray RP beyond point P. Similarly, we get point T such that R ∗ Q ∗ T . Since the points P, Q and R do not lie on a line, neither the three points R, S and T lie on a line. Since R ∈ R, and S , T ∈ S, both sets are nonempty. From the definition, it is clear that each point of the plane α lies in exactly one of the three sets R, a or S. Thus item (i) has been checked. In the special case that A = R or A0 = R, the statements (ii) and (iii) follow from the definition. Moreover, we use repeatedly the following statement: Lemma 5.1. Assume that three points A ∈ R and B ∈ S and R as above lie on a line r, but point S does not lie on this line. Then line a intersects segment AS , but does not intersect segment BS . Reason for the Lemma. In the triangle 4AS R, the line a intersects the side S R by the construction of point S , but does not intersect side AR. By Pasch’s axiom the line a intersects the third side AS , as to be shown. In the triangle 4BS R, the line a intersects both sides S R and BR. By Bernay’s Lemma, the line a cannot intersect all three sides of a triangle. Hence the line a does not intersect the third side BS , as to be shown. We now assume that A , R and A0 , R, and check statements (ii),(iii) and (iv). (ii) Given are a point A in the region R and a point B in the region S. We distinguish two cases:
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•
•
The three points R, A and B do not lie on a line. We apply Pasch’s axiom to the triangle 4RAB and line a. The assumptions A ∈ R and B ∈ S mean that line a does not intersect the side AR, but does intersect the side BR of the triangle. By Pasch’s axiom the line a intersects a second side which can only be side AB, as to be shown. We consider now the special case that points R, A and B lie on a line. Either the three points S , A and B or the three points T, A and B do not lie on a line. Since the roles of S and T can be switched, I need only to consider the first possibility. We use Pasch’s axiom for triangle 4ABS . Since line a intersects the side AS but not the side S B, the line a intersects the side AB, as to be shown.
(iii) Given are two points A and A0 in the region R. We distinguish two cases: •
•
The three points R, A and A0 do not lie on a line. We apply Pasch’s axiom to the triangle 4RAA0 and line a. The assumptions A ∈ R and A0 ∈ R mean that line a does not intersect neither the side AR nor the side A0 R. By Pasch’s axiom the line a cannot intersect the third side AA0 , as to be shown. We consider now the special case that points R, A and A0 lie on a line. Either the point S or point T does not lie on this line. Since the roles of S and T can be switched, I need only to consider the first possibility. We use Bernay’s Lemma for triangle 4AA0 S . Since line a intersects both sides AS and A0 S , the line a does not intersect the side AA0 , as to be shown.
(iv) Given are two points B and B0 in the region S. We distinguish two cases: •
•
The three points R, B and B0 do not lie on a line. We apply Pasch’s axiom to the triangle 4RBB0 and line a. The assumptions B ∈ S and B0 ∈ S mean that line a does not intersect both sides BR and B0 R. By Bernay’s Lemma, the line a cannot intersect the third side BB0 , as to be shown. We consider now the special case that points R, B and B0 lie on a line. Either the point S or point T does not lie on this line. Since the roles of S and T can be switched, I need only to consider the first possibility. We use Pasch’s axiom for triangle 4BB0 S . Since line a does not intersect neither side BS nor side B0 S , the line a does not intersect the side BB0 , as to be shown.
Question. Why is a plane mentioned in the plane separation theorem? • • •
These are just the points of any given plane that are separated into the points on the line and in the two half planes. Too, to keep all rigor, one needs the plane just to tell which points are separated by the theorem. The theorem is not true for a line in the three dimensional space.
Problem 5.3 (Almost no problem). Given is a plane α, any line a lying in this plane, and any point A on the line a. For any ray r with vertex A, either one of the following possibilities occurs: • the entire ray lies on the line a; • except for its vertex, the ray lies in one of the two half-planes of line a; • only the vertex of the ray lies in the plane α. Except for its vertex, the ray lies in one of the two half-spaces of plane α
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Remark. In the second case, we simply say "the ray r lies in the respective half-plane";—in the third case "the ray r lies in the respective half-space". −−→ Answer. Let B a second point on the ray and assume r = AB. By definition, the further points of the ray are the points P such that A ∗ P ∗ B, as well as the points Q such that A ∗ B∗ Q. The following cases are possible: • In the case that point B lies on the line a, we see by axiom (II.1) that the points P and B, respectively the points B and Q, lie on the line a, too. • In the case that point B does not lie on the line a, but in the plane α, we see by the plane separation Theorem 5.4 that point B lies in either one of the two half-planes. By the same Theorem 5.4 the points P and B, respectively the points B and Q lie in the same half-plane as B. Thus the entire ray,—except its vertex A,—lies in this half-plane, too. • In the case that point B does not lie in the plane α, we see use Proposition 3.6, part (c). By this proposition, the plane α and the line AB have exactly the point A in common. Hence only the vertex of the ray lies in the plane α. We see by the space separation Theorem 5.11 that point B lies in either one of the two halfspaces of plane α. By the same Theorem the points P and B, respectively the points B and Q lie in the same half-space as B. Thus the entire ray,—except its vertex A,—lies in this half-space, too. Problem 5.4. In an ordered incidence plane α, any line a is given. Prove that both half planes of the line a contain at least three points which do not lie on a line. Proof. We begin as in the proof of the plane-separation theorem. By axiom (I.3a), there exist two points P , Q on the line a. By axiom (I.3b), there exists a third point R in the plane α, not lying on the line l. In the case that this point lies in the given half-plane H, we put A := R. Otherwise, we proceed as in the proof of the plane-separation theorem. By axiom (II.2), there exist a point S such that R ∗ P ∗ S . In that case, we put A := S . In any case, we have obtained one point A in the given half-plane. Now we twice use proposition 5.1. There exists a point B lying between A and P. Since the line AB intersects the line a outside the segment AB, the plane separation theorem implies that points A and B lie in the same half-plane H of line a. Too, there exists a point C lying between A and Q. All three points A, B, C lie in the same halfplane H. Moreover points A, B, C cannot lie on a line,—otherwise all points A, B, C, P, Q would lie on a line. Hence we have obtained three points A, B, C as required. 5.4.
Four-point and n-point Theorems
Definition 5.5. Given any n points lying on a line. We say that the list [A1 , A2 , . . . , An ] or A1 ∗ A2 ∗ · · · ∗ An represents the order of the points A1 , A2 , . . . , An iff the order relations Ai ∗ A j ∗ Ak hold for all 1 ≤ i < j < k ≤ n. Proposition 5.5 (Four-point Theorem). Any four points on a line can be put in exactly two ways into a list that represents their order. These two lists are just reversed to each other, as for example the lists A ∗ B ∗ C ∗ D and D ∗ C ∗ B ∗ A are. Reason for uniqueness. Assume we have the list A ∗ B ∗ C ∗ D showing the order, as well as a second list. Since points B and C both lie between two other points among A, B, C, D, they cannot appear in the second list as neither the first nor last item. Neither can the second list be A∗C ∗ B∗ D, nor D ∗ B ∗ C ∗ A. Hence the reversed list D ∗ C ∗ B ∗ A is the only possible second list.
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Problem 5.5. Prove the Four-point Theorem 5.5 directly from the Three-point Theorem 5.2 and the Plane separation Theorem 5.4. Answer. Given are four different points on a line l. Choose any three among them. By the Threepoint Theorem, one of them lies between two others, and is now renamed C. We draw a line c through C different from l and use plane separation. In one half-plane of c lie two of the remaining points, in the other half-plane only the last remaining point, which we call D. The two points remaining points in the same half plane are named A and B. The three-point theorem holds for points A, B, C. Point C does not lie between A and B, as one sees from the plane separation theorem. We name B the point lying between C and the other one, finally named A. We have thus obtained the order relations A∗ B∗C,
A ∗C ∗ D,
B∗C ∗D
To confirm the fourth order relation, we draw a third line b through B different from l, and use plane separation once more. Point C and D lie in the same half-plane of b whereas A lies in the opposite half-plane from C. Hence A and D lie in opposite half-planes, confirming the order relation A∗B∗D. Definition 5.6. Let A, B and C be three points on the given line with B lying between A and C. −−→ −−→ The two rays BA and BC are called the opposite rays if their common vertex B lies between A and C. We say that two points in the same ray lie on the same side of the vertex. Two points in the opposite rays lie on different sides of the vertex. Proposition 5.6 ("Line separation Theorem"). Given is a line and a vertex lying on it. Each point of the line except the vertex is contained in exactly one of the opposite rays originating from the vertex. Proof. Given is a line and a point B on this line. By axiom (I.3a) there exists a second point on the line, which we name A By axiom (II.2), there exists a third point C on the line such that A ∗ B ∗ C. −−→ We want to check that an arbitrary point P , B is contained in exactly one of the rays BA and −−→ BC. If P = A or P = C, the assertion follows from the three-point Theorem. Otherwise, P , A, P , B and P , C, and the assertion follows from the four-point Theorem, and its Corollary. Indeed, the four points {A, B, C, P} can be put into an order representing list, with A preceding B. In that list, B precedes C, since A ∗ B∗C. For the order representing lists [P, A, B, C] −−→ or [A, P, B, C], the point P is contained in the ray BA. For the order representing lists [A, B, P, C] or −−→ [A, B, C, P], the point P is contained in the ray BC. Question. There are three pairs of order relations among four points A, B, C, D on a line that imply all four alphabetic order relations. The order representing list is A ∗ B ∗ C ∗ D. Which are these three pairs? Answer. If either A ∗ B ∗ C and B ∗ C ∗ D
or
(1)
A ∗ B ∗ C and A ∗ C ∗ D
or
(2)
A ∗ B ∗ D and B ∗ C ∗ D then all four alphabetic order relations follow.
(3)
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Remark. The Line separation Theorem does not follow neither simply from the axioms (II.1)(II.2) and (II.3), nor from the Three-point Theorem. For the beginner, that may not be obvious. Indeed, the very first edition of Hilbert’s foundations makes the four-point theorem an axiom. The following counterexample has been constructed by Hartshorne. The "points" are the integers modulo 5. Hence there exist just five points. There is one "line" through all five points. The "order relation" is defined by requiring a ∗ b ∗ c if and only if
2b ≡ a + c (mod 5)
(5.1)
Question. Convince yourself that axioms (II.1) and (II.2) are valid for this new ordering; of any three distinct points 0, a, b exactly one lies between the two others. There are six cases to be checked: (0, a, b) = (0, 1, 4), (0, 2, 3), (0, 1, 2), (0, 2, 4), (0, 1, 3), (0, 3, 4) In each case, find the middle point and write down the valid order relation. Show that of any three distinct points, exactly one lies between the other two. Answer. Axiom (II.1) is valid for this new ordering. Indeed, a ∗ b ∗ c implies c ∗ b ∗ a since a + c = c + a. Axiom (II.2) is valid. Given two points a and c, let b :≡ 2c − a (mod 5). Since a + b ≡ 2c (mod 5), we get a point b (beyond the segment) such that point c lies between a and b. Of any three distinct points 0, a, b exactly one lies between the two others. Indeed, for the six possible cases, the order turns out to be: 1∗0∗4, 2∗0∗3, 0∗1∗2, 0∗2∗4, 0∗3∗1, 0∗4∗3 Given any three distinct points x, y, z, let a := y − x and b := z − x. We order the three points 0, a, b and finally get for the three points x, y, z the corresponding order relation. Problem 5.6. To see that the line separation Theorem does not follow neither simply from the axioms (II.1)(II.2) and (II.3), nor from the Three-point Theorem, we proceed as follows. Convince yourself that for the order defined above a∗x∗c a∗c∗y
if and only if if and only if
x ≡ 3(a + c) (mod 5) y ≡ 2c − a (mod 5)
For better reading I denote the five points now by P0 , P1 , P2 , P3 , P4 . • Of which three points consists the segment P0 P2 . −−−−→ • Of which four points consists the ray P0 P2 . • Of which points consists the segment P0 P3 . −−−−→ • Of which points consists the ray P0 P3 . −−−−→ −−−−→ • Why are P0 P2 and P0 P3 two opposite rays. −−−−→ −−−−→ • Which points are common to the opposite rays P0 P2 and P0 P3 . • Why is the line separation Theorem not valid.
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The above order provides a counterexample to the Four-point Theorem, too: 1 1
∗2 ∗3 2 ∗0 ∗3 ∗3 ∗0
and are both valid; nevertheless
But not 1 ∗ 0 ∗ 3 as naturally expected, in case that the Four-point Theorem is valid. Problem 5.7. In the affine plane Z5 × Z5 we introduce the following "order relation". For any three points with coordinates A = (a1 , a2 ), B = (b1 , b2 ) and C = (c1 , c2 ), we require A ∗ B ∗ C if and only if 2b1 ≡ a1 + c1
(mod 5) and 2b2 ≡ a2 + c2
(mod 5)
(5.2)
Convince yourself that the axioms (II.1)(II.2) and (II.3) are valid, but Pasch’s axiom (II.4) is not valid. Answer. In the affine plane Z5 × Z5 , we may denote the points by (i, k) with i = 0, 1, 2, 3, 4 and k = 0, 1, 2, 3, 4. We leave the first part of the problem to the reader. To show that Pasch’s axiom is not valid, take the triangle with vertices A = (0, 0), B = (4, 0) and C = (2, 2). The only interior points of the sides are the midpoints Mc = (2, 0), Ma = (3, 1) and Mb = (1, 1). From Problem 3.19 and the figure on page 51, we check which are the five points on each line. There exist 6 lines through point Mc : • the horizontal one is AB, the vertical one is McC, which both go through a vertex and are excluded by the assumptions of Pasch’s axiom; • the lines Mc Ma and Mc Mb for which the assertion of Pasch’s axiom hold; • finally the lines Mc P and Mc Q where P = (3, 2) and Q = (1, 2). These two line meet only one side of the triangle 4ABC. Hence Pasch’s axiom is not valid. Recall definition 5.5 for a list [A1 , A2 , . . . , An ] or A1 ∗ A2 ∗ · · · ∗ An that represents the order of n points on a line. Lemma 5.2. There are exactly two lists which both represent the order of n points A1 , A2 , . . . , An lying on a line. The two lists are reversed of each other. Reason for uniqueness. In the case of 3 points, uniqueness follows directly from axioms (II.1) and (II.3). Here is the induction step "n 7→ n + 1": Assume the uniqueness lemma holds for orderings of up to n points on a line. Let two ordering, say [A1 ∗ A2 ∗ · · · ∗ An+1 ] = O, and a second list L, of n + 1 points on the line a be given. The points A2 . . . An all lie between A1 and An+1 . Hence they may not occur as first nor last item of the second list L. Thus list L has either A1 or An+1 as its first item. In the second case, we reverse the list, and get a second list L0 starting with A1 and ending with An+1 , too. We delete the last item An+1 from both lists to be compared, and get two lists of n points which represent the same ordering, both beginnig with point A1 . By the induction assumption, these two (truncated) lists are equal. Hence the two given lists O and L are either equal or reversed to each other. Proposition 5.7 (The n-point Theorem). Any n points on a line can be put in exactly two ways into a list that represents their order. These two lists are just reversed to each other. Problem 5.8. Prove proposition 5.7 by induction from the three-point theorem 5.2 and the plane separation theorem 5.4.
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Induction start for n = 3: The proposition is just the three-point theorem, proposition 5.2.
Induction step "n 7→ n + 1": Assume that proposition 5.7 holds for ordering up to n points on a line. On the line a, any n + 1 points A1 , A2 , . . . , An and P are given. By the induction assumption, the points A1 , A2 , . . . , An may have been ordered. Thus we may assume the order relations A1 ∗ A2 ∗ · · · ∗ An
(5.3)
to hold. We draw a line p , a through the point P and use plane separation. By possibly reversing the order list (5.3), we may assume that point A1 lies in the half plane H which contains more or at least equally many points as the opposite half plane. Let A1 and A j be the points lying in the half plane H for which the integer j ∈ {2 . . . n} is maximal. A special case occurs if all points A1 . . . An lie in the half plane H. By possibly reversing the order list, we may assume additionally the order relation A1 ∗ An ∗ P to holds After these agreements, we are left with two cases (i) and (ii), for which we now finish the argument. (i) The points Ak for k ∈ { j + 1 . . . n} lie in the opposite half plane. We draw a further line a j , a through point A j . Using plane separation we see that Ai ∗A j ∗ Ak and A j ∗P∗Ak imply Ai ∗A j ∗P for all i ∈ {1 . . . j − 1}. Once more, we use plane separation with the line p through the point P to see that Ai ∗A j ∗P and A j ∗P∗Ak imply Ai ∗ P∗Ak for all i ∈ {1 . . . j − 1} and all k ∈ { j + 1 . . . n}. Finally, we use plane separation with the line p through the point P to see that A j ∗ Ak ∗Al and A j ∗P∗Ak imply P∗Ak ∗Al for all k ∈ { j + 1 . . . n} and l ∈ {k + 1 . . . n}. (ii) A special case occurs if all points A1 . . . An lie in one half plane. Moreover A1 ∗ An ∗ P. We draw a further line an , a through point An and use plane separation to see that A1 ∗ An ∗P and A1 ∗A j ∗An imply A1 ∗A j ∗ P for all j ∈ {2 . . . n}. Next we use plane separation with the line a j , a through the point A j to see that A1 ∗ A j ∗P and A1 ∗Ai ∗A j imply Ai ∗A j ∗P for all i ∈ {1 . . . j − 1} and j ∈ {2 . . . n}.
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In both cases, all instances of three point orders contained in the order A1 ∗ A2 ∗ · · · ∗ A j ∗ P ∗ A j+1 ∗ · · · ∗ An have been confirmed. Thus the inductive proof of proposition 5.7 is completed.
Proposition 5.8. For any different points A , B, and for all natural numbers n, there exist at least n points between A and B. Proof. We construct a sequence of points A1 , A2 , . . . , all of which are different and lie between the two given points A and B. The existence of the point A1 between A and B was proved in Proposition 5.1. Assume that n points A1 , A2 , . . . , An between A and B have been constructed, and there order is represented by the list [A, A1 , A2 , . . . , An , B] We choose any point An+1 between An and B. As shown in Proposition 5.7, the order of the n + 1 points is now represented by the list [A, A1 , A2 , . . . , An , An+1 , B] Hence one has obtained n + 1 different points between A and B.
Corollary 4. For any finite incidence plane, it is impossible to define an order relation for which all the axioms of order (II.1),(II.2),(II.3),(II.4) hold. Remark. Hilbert’s Proposition 7 simply states: "there exist infinitely many points lying between any two different points." Such an assertion has not been exactly proved. We have only shown that "for any natural number n, there exist more than n different points on the given line." Since the axiomatic set theory of Zermelo Fraenkel has become common knowledge, it is clear that existence of any infinite set needs an extra axiom. Such an axiom of infinity is part of Zermelo Fraenkel set theory, but not of Peano arithmetic. Indeed, no proof by induction yields existence of any infinite set. A similar case occurs with respect to the common-place statement "there exist infinitely many primes." Any of the known proofs only verify the assertion "For any natural number n, there exist more than n prime numbers." Remark. Of course, one wants to know more about the structure of a line than the assertion made by proposition 5.8. To this end, we need to assume Peano arithmetic is true; and moreover use the important results obtained at several places below in these notes. Thus one is lead to the following results. We fix a unit segment CD. There exist segments having the length given by any positive rational number p/q ∈ Q, under either one of the following assumptions: • for any affine plane coordinate plane over an ordered field; • for any affine ordered Pappian plane; • for any Hilbert plane where both axioms of continuity hold,—the Archimedean axiom (V.1) as well as some completeness axiom are both required; • for any Pythagorean plane. In the latter case, a segment having length p/q is constructible for any positive rational number p/q. Problem 5.9. State in detail which theorems are needed to confirm the assertions of this remark.
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5.5.
Angles
Definition 5.7 (Angle). An angle is the union of two rays with common vertex, and not lying on one line. The common vertex of these rays is called the vertex of the angle. The rays are called the sides of the angle. For an angle, we may use the two ray notation ∠(h, k);— the two rays h and k are supposed to have common vertex and no lie on a line. As an alternative, we use the three letter notation. Here the middle letter denotes the vertex. −−→ −−→ Thus the angle ∠BAC has the vertex A, and the sides AB and AC. Remark. In the section on congruence, we introduce by Definition 7.14 the notion of oriented angle. In the section about similar triangles, we introduce by definition 21.3 the group of unwound angles.
Figure 5.7. Interior and exterior of an angle are intersection and union of two half planes.
Definition 5.8 (Interior and exterior domain of an angle). The interior domain of an angle lying in a plane A is the intersection of two corresponding half planes—bordered by the sides of the angle, and containing points on the other side of the angle, respectively. The exterior of an angle is the union of two opposite half planes—-bordered by the sides of the angle, and not containing the points neither in the interior nor on the legs of the angle. Half planes, interior and exterior of an angle all do not include the lines or rays on their boundary. Thus the interior of ∠BAD is the intersection of the half plane of AB in which D lies, and the half plane of AD in which B lies. The exterior of ∠BAD is the union of the half plane of AB opposite to D, and the half plane of AD opposite to B. Remark. Note the "exterior of an angle" is defined differently from "exterior angle of a triangle". The latter is any supplementary angle to an interior angle of the triangle. Problem 5.10. Given an angle ∠(h, k) lying in a plane A. Let K be the intersection of the half plane of h in which k does not lie with the half plane of k in which h does not lie. Explain what the set K is like and provide a drawing. Give a short exact description for this set. Answer. From the drawing on page 111, one sees that the set K is the interior of an angle. Indeed, it is the interior of the vertical angle ∠(h0 , k0 ) corresponding to the given angle ∠(h, k). Problem 5.11. Given an angle ∠(h, k) lying in a plane A.
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Figure 5.8. Interior of the vertical angle.
(a) Let points H ∈ h and K ∈ k be given, both different from the vertex of the angle. Explain why all interior points of segment HK lie in the interior of the angle ∠(h, k). (b) Let A = h ∩ k be the vertex of the angle ∠(h, k) and D be any point in the interior of the angle. −−→ Explain why all points of the ray AD except A lie in the interior of the angle ∠(h, k). Answer. (a) Take any point P interior to segment HK, thus H ∗ P ∗ K. Point P lies in the same half plane of h in which point K, and hence the ray k lies. Similarly, point P lies in the same half plane of k in which point H, and hence the ray h lies. Hence point P lies in the interior of the angle ∠(h, k). Indeed, all interior points of segment HK lie in the interior of the angle ∠(h, k). −−→ (b) Take any point P , A on the ray AD. Hence either P = D, or A ∗ P ∗ D, or A ∗ D ∗ P. In each case, point P lies in the same half plane of h in which point D lies. But point D lies in the interior of the angle. Hence both point P and point D lie in the same half plane of h in which the ray k lies. Similarly, point P lies in the same half plane of k in which point D and hence the ray h lies. Hence point P lies in the interior of the angle ∠(h, k). Indeed, all −−→ points of the ray AD except the vertex A lie in the interior of the angle ∠(h, k). A second way to answer. (a) Given are points H ∈ h and K ∈ k, both different from the vertex of −−→ the angle. By the previous problem 5.3, the entire ray HK lies in the same half-plane of h as point K. This is the half-plane of h in which ray k lies, too. We call it K. −−→ The entire ray KH lies in the same half-plane of k as point H. This is the half-plane of k in which ray h lies, too. We call it H. The interior of the angle ∠(h, k) is the intersection of half-planes K ∩ H. The intersection −−→ −−→ HK ∩ KH is the interior of the segment HK. Since −−→ −−→ HK ∩ KH ⊆ K ∩ H we see that the interior of the segment HK lies in the interior of the angle ∠(h, k).
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(b) Let A = h ∩ k be the vertex of the angle ∠(h, k), and D be any point in the interior of the angle. −−→ By the previous problem 5.3, the entire ray AD \ {A} lies in the same half-plane of h as point D. Since point D is in the interior of the angle, this is half-plane K of h in which ray k lies. −−→ The entire ray AD \ {A} lies in the same half-plane of k as point D. Since point D is in the interior of the angle, this is half-plane H of k in which ray h lies. −−→ Hence the ray AD \ {A} lies in the intersection K ∩ H, which is the interior of the angle ∠(h, k).
Figure 5.9. A ray interior of an angle intersects a segment from side to side.
Proposition 5.9 (The Crossbar Theorem). A segment with endpoints on the two sides of an angle and a ray emanating from its vertex into the interior of the angle intersect. −−→ Proof. Let ray r = AC lie in the interior of the given angle ∠BAD, and let B and D be arbitrary −−→ points on the two sides of this angle. We have to show that the ray AC intersects the segment BD. −−→ Let F be any point on the ray opposite to AB. We apply Pasch’s axiom to triangle 4FBD and line l = AC. The line intersects the side FB of the triangle in point A, and does not pass through neither one of the vertices F, B, D. We check that side FD does not intersect line AC. −−→ Indeed, the points inside segment FD and the points inside ray AC lie on different sides of line AD. But the points inside segment FD and inside the opposite ray lie on different sides of line AB. (Points A, D, F are exceptions, but neither can segment FD and line AC intersect in any of these points.) Hence—by Pasch’s axiom—the third side BD intersects line AC, say at point Q. Segment BD, −−→ and hence the intersection point Q are in the interior of ∠BAD. Since only the ray AC, but not its −−→ opposite ray, lies in the interior of ∠BAD, the intersection point Q lies on the ray AC. Here is a drawing, to show how Pasch’s axiom is applied. Warning. A point in the interior of an angle may not lie on any segment going from side to side of the angle. To spell it out precisely: Given an angle ∠BAD and a point Q in its interior, there does not always −−→ −−→ exist any two points G and H on the rays AB and AD such that Q lies on the segment GH. Indeed, in hyperbolic geometry, the union of all segments from side to side of an angle span only the interior of an asymptotic triangle—and this is a proper subset of the interior of the angle.
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Figure 5.10. The Crossbar Theorem is proved using Pasch’s axiom.
Lemma 5.3. Given are two opposite rays r and s with common vertex O, one half-plane H of line r ∪ s, and two distinct rays a and b from vertex O lying in this half-plane H. There the two following cases to occur: (i) either the ray a lies in the interior of angle ∠(r, b) and the ray b lies in the interior of angle ∠(s, a); (ii) or the ray b lies in the interior of angle ∠(r, a) and the ray a lies in the interior of the angle ∠(s, b).
Figure 5.11. About any two angles.
Proof. The entire figure lies in the half-plane H of line r ∪ s. We may choose points R and S such −−→ −−→ −−→ that r = OR and s = OS . We choose any point B on the ray b such that b = OB. We use the plane separation theorem with the line of ray b. Asking about the ray a, either one of the two following cases occur. Both are shown in the figure on page 113.
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(i) The rays r and a lie in the same half plane of ray b. Hence the ray a lies in the interior of angle ∠(r, b). The Crossbar Theorem implies that the ray a intersects the segment BR, say in point A. From the order relation B ∗ A ∗ R, we see that points B and R lie on different sides −−→ of ray a. Hence points B and S lie on the same side of ray a. Hence ray b = OB lies in the interior of angle ∠(s, a). (ii) The rays r and a lie in opposite half planes of ray b. Hence the ray a lies in the interior of angle ∠(s, b). The Crossbar Theorem implies that the ray a intersects the segment BS , say in point A. From the order relation B ∗ A ∗ S , we see that points B and S lie on different sides of ray a. Hence points B and R lie on the same side of ray a. Hence ray b lies in the interior of angle ∠(s, a). One may at this point already introduce the notions of supplementary angles and vertical angles. Definition 5.9 (Supplementary Angles). Two angles are called supplementary angles or a linear pair, if and only if they have a common vertex, they both have one side on a common ray, and the two other sides are the opposite rays on a line. Definition 5.10 (Vertical Angles). Two angles are called vertical angles, iff they have a common vertex, and their sides are two pairs of opposite rays on two lines. Proposition 5.10 (About n points in a plane). Among any n ≥ 3 points lying in an ordered incidence plane but not on a line, there exist three points P, Q, R such that all n points lie in the interior or on the boundary of the angle ∠PQR.
Figure 5.12. Any n points lie in the interior or on the boundary of an angle.
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Induction start for n = 3: The three given points P, Q, R do not lie on a line, hence the angle ∠PQR exists. Induction step "n 7→ n + 1": Assume that proposition holds for any n points. Given are now these points, and an extra point Pn+1 . We distinguish these cases: (i) The points P1 . . . Pn lie on a line l. By the n-point proposition, they can be put into an ordered list P1 ∗ P2 ∗ · · · ∗ Pn . By assumption point Pn+1 does not lie on the line l. We put P := P1 , the first item in the list; R := Pn , the last item in the list; and choose Q := Pn+1 as vertex of the angle. Thus all n + 1 points lie inside or on the boundary of the angle ∠PQR. (ii) The points P1 . . . Pn do not lie on a line l. By the induction assumption, there are points A, B, C among them such that points P1 . . . Pn lie in the interior or on the boundary of the angle ∠ABC. • •
•
In case that the extra point Pn+1 lies in the interior or on the boundary of the angle ∠ABC, we are ready. In case that the extra point Pn+1 lies in the interior of a supplementary angle of the angle ∠ABC, all points P1 . . . Pn+1 lie in the interior or on the boundary of either angle ∠ABPn+1 or angle ∠Pn+1 BC. We choose any points A ∗ B ∗ A0 and C ∗ B ∗ C 0 . Assume that the extra point Q := Pn+1 −−→ lies in the interior of the vertical angle ∠A0 BC 0 or on the ray BA0 . This case in shown in the figure on page 114. We draw the segments Pi Pn+1 for i = 1 . . . n, obviously including the segments APn+1 , BPn+1 and CPn+1 . Since the points P1 . . . Pn either lie on the line AB, or the opposite side of AB as point Pn+1 , all these segments intersect the line AB . By the n-point theorem we may order these intersection points into a list S 1 ∗ · · · ∗ S n . −−−−−→ Let P be a point among P1 . . . Pn on the first ray Pn+1 S 1 , and R be a point among −−−−−→ P1 . . . Pn on the last ray Pn+1 S n . Now all points P1 . . . Pn+1 lie in the interior or on the boundary of the angle ∠PQR.
5.6.
Space separation
Proposition 5.11 (Hilbert’s Proposition 10, "Space separation Theorem"). Each plane α separates the remaining points of the space into two nonempty regions R and S with the following properties: • Every point not lying in the plane α lies either in region R, or in region S, but not in both regions. • for any points A in region R and B in region S, segment AB intersects plane α; • for two points A and A0 in the same region, segment AA0 does not intersect α. Definition 5.11. In case that segment AA0 does not intersect the plane α, we say that the points A and A0 lie on the same side of plane α—or in the same half space. The points A and B lie on different sides of plane α in case that the segment AB does intersect this plane. The two open regions one gets in this way are called the half spaces bounded by the plane α.
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Proof. Take any point A not lying in the plane α. Since by axiom I.8, there exist at least four points which do not lie in a plane, such a point A does exist. We can now separate the points not lying in the plane α into the following two sets: R : = {B : B < α, the segment AB does not intersect the plane α} S : = {C : C < α, the segment AC intersects the plane α} Each point lies in exactly one of the three sets R, α or S. Since A ∈ R this set is nonempty. Question. Explain why the set S is nonempty. Answer. Take a point P ∈ α. By axioms (I.1) and (II.2), there exists a point Q on the line AP such that A ∗ P ∗ Q. Since the segment AQ intersects the plane α, the point Q lies in the set S. We check the following facts: (a) Take any two points B and B0 in the domain R. Then the segment BB0 does not intersect α. (b) Take any two points C and C 0 in the domain S. Then the segment CC 0 does not intersect α. (c) Take points B in domain R and C in domain S. Then the segment BC intersects the plane α. Consider assertion (a). The three points A, B and B0 lie either a line l or span a plane β. If this line or plane does not intersect the plane α, we are ready. Take the case that points A, B and B0 lie on a line l intersecting plane α in a point D. Assumption (a) implies that A and B lie in the same ray of line l from vertex D. Similarly, points A and B0 lie in the same ray from D. By line separation (Proposition 5.6), we conclude that points B and B0 , too, lie in the same ray from D. Hence segment BB0 does not intersect the plane α. Take the case that points A, B and B0 do not lie on a line. By the axioms I.4 and I.5, they determine a unique plane β. Take the case that the planes β and α intersect. By Hilbert’s Proposition 1 (Proposition 3.6), the intersection α ∩ β is a line k. The assumption (a) implies that points A and B lie on the same side of k in the plane β. Similarly A and B0 lie in the same half-plane. We use plane separation (Proposition 5.4) for the plane β, and conclude that points B and B0 lie in the same half-plane. Hence segment BB0 does not intersect the plane α. Consider assertion (c). Take points B ∈ R and C ∈ S. The segment AC intersects the plane α in a point D. If A = B, we are ready. The points A, B and C lie either on a line l or span a plane β. Take the case that points A, B and C lie on a line. This line l intersects plane α in point D. Assumption (c) implies that A and B lie in the same ray of line l from vertex D. But points A and C lie on opposite rays from D. By line separation (Proposition 5.6), we conclude that points B and C, too, lie on opposite rays from D. Hence segment BC intersects the plane α in point D. Take the case that points A, B and C do not lie on a line. By the axioms I.4 and I.5, they determine a unique plane β. The planes β and α intersect in point D. By Hilbert’s Proposition 1 (Proposition 3.6), the intersection α ∩ β is a line k. The assumption (c) implies that points A and B lie on the same side of k in the plane β, but A and C lie in opposite half-planes. We use plane separation (Proposition 5.4) for the plane β, and conclude that points B and C lie opposite half-planes. Hence segment BC intersects the line k, and hence the plane α in a point E. We leave it to the reader to check case (b).
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Figure 5.13. The segment between two points on different sides of a plane intersects the plane.
5.7.
Interior and exterior of a triangle
Definition 5.12 (Interior and exterior of a triangle). The interior domain of a triangle is the intersection of the three half-planes of the sides in which the respectively opposite vertices lie. By Int4 ABC we denote the set of interior points of triangle 4ABC. The exterior domain of a triangle consists of the points of the plane which do not lie neither in the interior domain, nor on the sides, nor at the vertices of the triangle. Problem 5.12. Show that the interior domain of a triangle is nonempty.
Figure 5.14. The interior of a triangle is the intersection of three half planes.
Reason why the interior domain of a triangle is nonempty. Take a triangle 4ABC. Let H be the half plane of line AB in which point C lies, K be the half plane of line BC in which point A lies,
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and L be the half plane of line CA in which point B lies. By definition, the interior domain of the triangle 4ABC is the intersection H ∩ K ∩ L. By definition, the interior domain of the triangle 4ABC is the intersection H ∩ K ∩ L. By Proposition 5.1 (Hilbert’s Proposition 3), there exists a there exists a point D lying between A and B. By the same reasoning, there exists a point E lying between C and D. I claim that E ∈ H ∩ K ∩ L, which is the interior of triangle 4ABC. Indeed, the segment EC lies entire in H since D ∗ E ∗ C and point E lies in the same half plane H of line AB as point C. Because of A ∗ D ∗ B and D ∗ E ∗ C, segments DB and DE lie entirely in L. Hence points B, D and E lie in the same half plane L of line CA. Similarly, since segments AD and DE lie in the half plane K, we see that points A, D and E lie in the same half plane K of line BC. Problem 5.13. By choosing an appropriate point on a side of the triangle, and an appropriate point on the segment from this point to the opposite vertex, one can obtain any point in the interior of a triangle. Give a reason. Proof. We use a reasoning reversing the steps of problem 5.12. Let E be any point in the interior of −−→ triangle 4ABC. The ray r = CE lies in the interior of the angle ∠ACB. By the Crossbar Theorem, the ray r intersects the side AB. Let the intersection point be D. Since the given point E lies in the interior of the triangle, points E and C lie on the same side of AB, and hence C ∗ E ∗ D. Thus point E has been obtained as described in the problem: by choosing point D on a side AB of the triangle, and then point E on the segment DC to the opposite vertex. The exterior domain of a triangle consists of the points of the plane which do not lie neither in the interior domain, nor on the sides, nor at the vertices of the triangle. It is equal to the union H 0 ∪ K 0 ∪ L0 of the three opposite half planes. Problem 5.14. Show that the half-planes of the three sides of a triangle, which are opposite to the third vertex, respectively, do not intersect.
Figure 5.15. The three half planes H 0 , K 0 , L0 opposite to the interior ones have empty intersection.
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Answer. The three half planes H 0 , K 0 , L0 opposite to H, K, L have empty triple intersection. Indeed, the intersection H 0 ∩ K 0 is the interior of angle ∠C 0 BA0 . This is the vertical angle of ∠ABC— I have chosen any points A ∗ B ∗ A0 and C ∗ B ∗ C 0 . −−→ −−→ The rays BA0 and BC 0 lie entirely inside the half plane L. Hence the interior of the angle ∠A0 BC 0 which is the intersection H 0 ∩ K 0 is a subset of L and does not intersect L0 . The purpose of the following exercises is to check the Jordan curve Theorem in the case that the closed curve is a triangle. Problem 5.15. Given is a triangle and a line through an interior point of the triangle. Show that the line either intersects two sides of the triangle, or it intersects only one side and goes through the opposite vertex.
Figure 5.16. A line through an interior point of a triangle intersects either two sides, or goes through a vertex and intersects the opposite side.
Answer. Let the line l go through the point P in the interior of triangle 4ABC. We distinguish two cases: (a) The line goes through a vertex of the triangle. Say the line goes through vertex A. Because −−→ point P lies in the interior of angle ∠BAC, the Crossbar theorem shows that the ray AP intersects the opposite side BC at some point, say Q. −−→ (b) The line does not go through any vertex of the triangle. We draw the ray AP. Because −−→ point P lies in the interior of angle ∠BAC, the Crossbar theorem shows that the ray AP intersects the opposite side BC at some point, say Q. We now apply Pasch’s axiom to the triangle 4ABQ and the line l. This line intersects side AQ at point P. Hence it intersects a second side, either BQ ⊂ BC or AB, say at point D. In both cases, the line l intersects one side of the given triangle 4ABC. By Pasch’s axiom it intersects a second side, say at point E. By Bernays’ lemma, the line l intersects exactly two sides of the given triangle. Problem 5.16 ("non-Pasch"). Give a reason why a line which intersects the extensions of two different sides of a triangle lies entirely in the exterior domain of the triangle.
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Answer. Since the given line l is different from the lines on all three sides, it can only intersect a given side of the triangle or its extension, but not both. Assume line l intersects both the extensions of sides AB and BC. Could line l still intersect the third side CA? No, in this case, by Pasch’s axiom, it would intersect a second side, either AB or BC, too, and hence not its extension. Problem 5.17. Given is a triangle. (a) Prove that any segment PR from an interior point P to an exterior point R on the extension of a side of the triangle intersects a second side of the triangle. (b) Prove that any segment PQ from an interior point P to an exterior point Q intersects either a vertex or a side of the triangle. Answer. (a) Given is a segment PR from an interior point P of triangle 4ABC to an exterior point R on the extension of a side. We may assume that R ∗ A ∗ B, the other cases being similar. Points R and P lie on different sides of line AC. Hence segment RP intersects line AC, say in point S . Why does point S lie on segment AC? By the previous problem, the line PQ cannot intersect the extensions of two different sides of the triangle. Hence point S lies on the side AC. (b) Given is a segment PQ from an interior point P to an exterior point Q of the triangle. The two points P and Q lie either in different half planes of one side of the triangle, say AB, or point Q lies on the extension of a side AB. In the second case we have shown in part (a) that segment PQ intersects a side of the triangle. In the first case, segment PQ intersects line AB, by plane separation. Let the intersection point be R, hence P ∗ R ∗ Q. There are two cases: If point R lies on the side AB, we are ready. If point R lies on the extension of side AB, we go back to part (a) and conclude that segment RP intersects a second side of the triangle. Remark. In hyperbolic geometry, it is possible that a line passes through the extension of one side, and is parallel (even asymptotically parallel!) to the two other sides of a triangle. Problem 5.18. Given are any three points on a line, where B lies between A and C. Assume point D does not lies on this line. Show that the interior of triangle 4ACD is the union of the interiors of the triangles 4ABD and 4BCD and the interior of the segment BD. 5.8.
Convexity
Definition 5.13 (Convex set). A set of points is called convex, if and only if with any two points P and Q, the segment PQ is contained in the set. Lemma 5.4. Any segment is a convex set. Reason. This statement is a consequence of the four- and five-point theorems. Given is any segment AB, and two points P and Q from this segment. Let R be any point from the segment PQ. We need to consider two cases, the remaining ones are similar. (i) P = A is an endpoint, and Q an interior point of the segment. Thus the order relations A ∗ Q ∗ B and P = A ∗ R ∗ Q hold. The four-point theorem implies the ordered list A ∗ R ∗ Q ∗ B. Hence we conclude A ∗ R ∗ B, telling the point R lies in the segment AB, as to be shown.
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(ii) P and Q are both interior points of the segment. Thus the order relations A ∗ P ∗ B, A ∗ Q ∗ B, and P ∗ R ∗ Q hold. The five points can appear in either of two ways in an ordered list: one obtains either A ∗ P ∗ R ∗ Q ∗ B or A ∗ Q ∗ R ∗ P ∗ B. In both cases we conclude A ∗ R ∗ B, telling the point R lies in the segment AB, as to be shown. Lemma 5.5. Any ray is a convex set. Lemma 5.6. Any half-plane is a convex set. Reason. Given is any plane α, a line l lying in this plane, and a half-plane H ⊂ α bounded by this line. Let P and Q be any two points from the half-plane H. By the plane separation theorem, the segment PQ does not intersect the line l. Let R be any point from the segment PQ. The plane separation theorem implies that point R does not lie on the line l. By the previous lemma 5.4, any segment is a convex set, and hence PR ⊆ PQ. Hence the line l does not intersect the (shorter) segment PR, neither. By the plane separation theorem, we conclude that points P and R lie in the same half-plane. Thus we have checked that R ∈ H, as claimed. Lemma 5.7. The intersection of any convex sets is convex. Hence the interior of a triangle is a convex set. Definition 5.14. A polygon or polygonal curve is called strictly convex, iff for any segment AB, BC, CD, . . . , KL the remaining vertices lie in one half plane of the respective segment. A polygon or polygonal curve is called convex, iff for any segment AB, BC, CD, . . . , KL the remaining vertices lie in one half plane or on the line of the respective segment. It is easy to see that any closed and strictly convex polygon is simple. Too, any closed and convex polygon is simple, provided its vertices which happen to lie on one line appear in their natural order, such that the corresponding segments intersect only at their endpoints. Lemma 5.8. Assume no three of the four points A, B, C, D lie on a line. For the quadrilateral ABCD the following statements are equivalent: (a) The quadrilateral ABCD is strictly convex. (b) Any three of the following statements are true: (i) Vertices C and D lie in one half plane of side AB; (ii) vertices D and A lie in one half plane of side BC; (iii) vertices A and B lie in one half plane of side CD; (iv) vertices B and C lie in one half plane of side DA. (c) The diagonals AC and BD intersect at an interior point Q. (d) The triangles 4ABC and 4ADC lie on different sides of line AC; and the triangles 4BCD and 4BAD lie on different sides of line BD.
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Proof that (i)(ii)(iv) imply (c). Items (i) and (ii) imply that point D lies in the interior of the angle −−→ ∠ABC. By the Crossbar Theorem, segment AC intersects the ray BD at some point Q. Items (i) and (iv) imply that point C lies in the interior of the angle ∠BAD. By the Crossbar −−→ Theorem, segment BD intersects the ray AC at some point R. Since no three of the four points A, B, C, D lie on a line, the lines AC and BD are distinct, and hence have a unique intersection point. Hence Q = R, and this point is the intersection of the diagonals AC and BD. Proof that (c) implies (a). By the assumptions A ∗ Q ∗ C and B ∗ Q ∗ D, the three points C, D and Q lie in one half plane of line AB, confirming (i). We get three similar valid reasonings by the cyclic permutation A 7→ B 7→ C 7→ D 7→ A. Thus we confirm items (ii),(iii) and (iv). The four statements (i) through (iv) mean that the quadrilateral is strictly convex. It can now be left to the reader to complete the entire proof. 5.9.
Topology of the ordered incidence plane
Problem 5.19. In an ordered incidence plane α, any point P is given. Prove there exists three rays a, b, c with vertex P, lying in the plane α such that • a and b lie in different half planes of c; • b and c lie in different half planes of a; • c and a lie in different half planes of b. I say these three rays make a "star shape". Provide a drawing to illustrate your proof.1 Answer. A drawing is given on page 122. By axiom (I.3b), there exist three points in the plane α,
Figure 5.17. How to get three rays from given vertex P in a star-shape.
not lying on a line. As explained in Lemma 3.2, we deduct that there exist two points A and B, −−→ −−→ different from P, for which the lines PA and PB are different. Hence the rays a = PA and b = PB are different and lie on different lines. 1
I find it rather impossible to understand the reasoning without a drawing,—hence without a drawing the problem cannot be counted.
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By axiom (II.2), there exists a point A0 such that A ∗ P ∗ A0 . Similarly, there exists a point B0 such that B ∗ P ∗ B0 . Now we use Hilbert’s proposition 3. By this proposition, there exists a point −−→ C lying between A0 and B0 . Define the ray c = PC. We now check that these three rays make a "star shape". Use plane separation by line of c. Points A0 and B0 lie in different half-planes of c; points A and A0 lie in different half-planes of c; points B and B0 lie in different half-planes of c. Hence points A and B lie in different half-planes of c, and hence rays a and b lie in different half-planes of c. Use plane separation by line of a. Points B and B0 lie in different half-planes of a; points B0 and C lie in the same half-plane of a. Hence points B and C lie in different half-planes of a, and hence rays b and c lie in different half-planes of a. Use plane separation by line of b. Points A and A0 lie in different half-planes of b; points A0 and C lie in the same half-plane of b. Hence points A and C lie in different half-planes of b, and hence rays a and c lie in different half-planes of b.
Figure 5.18. The three ray star makes a triangle with common vertex in the interior, and conversely.
−−→ −−→ −−→ Lemma 5.9. Assume the three rays a := PA, b := PB and c := PC are a star shape. Then any point X not lying on the three rays lies in the interior of exactly one of the angles ∠APB, ∠BPC, or ∠CPA. Proof. The reader may check that plane separation leave no other possibilities.
Lemma 5.10. Given are a point P and three rays a, b, c with vertex P lying in an ordered incidence plane α. We assume these rays to make a star shape, as explained in the previous problem 5.19. Let A, B and C be any three points lying on the three rays a, b, and c, respectively, and different from vertex P. Then the vertex P lies in the interior of triangle 4ABC. −−→ Proof. By axiom (II.2) there exists a point A0 such that A ∗ P ∗ A0 . The ray a0 := PA0 is the ray −−→ opposite to ray a := PA. −−→ −−→ Because of the star shape, rays PA and PB lie on different sides of line PC. By construction, −−→ −−→ the opposite rays PA and PA0 lie on different sides of line PC. Hence plane separation implies that −−→ −−→ the rays PA0 and PB lie on the same side of line PC.
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−−→ The next argument is similar, just exchanging B and C. Because of the star shape, rays PA and − − → −−→ −−→ PC lie on different sides of line PB. By construction, the opposite rays PA and PA0 lie on different −−→0 −−→ sides of line PB. Hence plane separation implies that the rays PA and PC lie on the same side of line PB. −−→ Both arguments together show that the ray PA0 lies in the interior of angle ∠BPC. Because of the star shape the rays b and c do not lie on a line and this angle exists. −−→ By the Crossbar Theorem, the segment BC intersects the ray PA0 . We call the intersection −−→ point X and see that B ∗ X ∗ C. Moreover X ∗ P ∗ A since X lies on the ray PA0 . Thus points A and P lie on the same side of line BC. By switching A, B, C cyclically, we see that Points A and P lie on the same side of line BC. Points B and P lie on the same side of line CA. Points C and P lie on the same side of line AB. All three statements together imply that point P lies in the interior of triangle 4ABC.
Lemma 5.11. Given is a triangle 4ABC and any interior point P of the triangle. Then the three −−→ −−→ −−→ rays a := PA, b := PB and c := PC are a star shape. Proof. No two of the three rays a, b, c lie on a line,—otherwise point P would lie on a side of the triangle. The point P and hence the ray a lie in the interior of angle ∠BAC. The Crossbar Theorem −−→ yields that the (opposite) ray AP intersects the side BC, say at point X. Hence the rays b and c lie on opposite sides of the line PA. By switching a, b, c cyclically, we see that Rays a and b lie on opposite sides of line PC. Rays b and c lie on opposite sides of line PA. Rays c and a lie on opposite sides of line PB. −−→ −−→ −−→ All three statements together imply that the three rays a := PA, b := PB and c := PC are a star shape. Lemma 5.12. Given is a triangle and a ray r emanating from an interior point P of the triangle. The ray r either intersects exactly intersects exactly one of the vertices or one side of the triangle. Let this intersection point be called X. The interior points of segment PX lie in the interior domain of the given triangle. Proof. Let the ray r have vertex P in the interior of triangle 4ABC. Note that the ray cannot intersect a vertex and still another side of the triangle,—otherwise point P would lie on a side of the triangle. In the case that the ray goes through a vertex of the triangle, we are ready. Now assume that the ray r does not go through any vertex of the triangle. Take any point R , P on the ray. By Lemma 5.9 the point R, and hence the ray r lies in the interior of exactly one of the angles ∠APB, ∠BPC, or ∠CPA. We may assume the ray r lies in the interior of angle ∠APB. Hence the Crossbar theorem shows that the ray intersects the side BC at some point, say X. I leave it to the reader to check that any point between P and X lies in the interior of triangle 4ABC. Lemma 5.13. Given are three points A, B, C in the interior of a triangle 4EFG. Then the sides and vertices from triangle 4ABC, as well as the interior of 4ABC lie in the interior of triangle 4EFG.
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Proof. The interior of any triangle is the intersection of three half-planes which are known to be convex. As already seen by Lemma 5.7, this intersection is convex, too. Hence all three segments AB, BC and CA lie in the interior of triangle 4EFG. In the special case that points A, B, C lie on a line we are ready. Now we assume that triangle 4ABC exists. We take any point P in the interior of triangle 4ABC, and any line l through this point. By the result of problem 5.15, the line either intersects two sides of the triangle, or it intersects only one side and goes through the opposite vertex. Let X1 and X2 be the two intersection points of the line l with the triangle. All three segments AB, BC and CA and hence the points Xi lie in the interior of triangle 4EFG. Since the interior of a triangle is convex, the entire segment X1 X2 lies in the interior of triangle 4EFG. By Int4 ABC we denote the set of interior points of triangle 4ABC. Definition 5.15 (The topology of an ordered incidence plane). Any subset O of an ordered incidence plane α is called open, if and only if for every point P ∈ α, there exists a triangle 4ABC such that P lies in the interior of this triangle and Int4 ABC ⊆ O. Problem 5.20. Check for the topology from definition 5.15 the postulates for the open sets are valid: (O1) ∅ and X are open sets. (O2) If U, V are open, their intersection U ∩ V is open. (O3) If Uα is a family of open sets, with indexes α ∈ I, their union
S
α∈I
Uα is open.
Answer. (O1) ∅ open is vacuously true. The entire plane α is open. Indeed given any point P ∈ α, by problem 5.19 there exist three rays a, b, c with vertex P making a star shape. As in Lemma 5.10, we choose any three points A, B and C lying on the three rays a, b, and c, respectively, and different from vertex P. By Lemma 5.10 the vertex P lies in the interior of triangle 4ABC. On the other hand, Int4 ABC ⊆ α holds. (O2) Take any open sets U and V. To check whether their intersection is open, take any point P ∈ U ∩ V. There exist triangles 4EFG and 4XYZ such that Int4 EFG ⊆ U and Int4 XYZ ⊆ V, and point P lies in the interior of both triangles. By problem 5.19 there exist three rays a, b, c with vertex P making a star shape. By Lemma 5.12, the ray a intersects the (union of the sides from) triangle 4EFG at one point Ua , and the triangle 4XYZ at one point Va . By the Three-point Theorem 5.2, either P ∗ Ua ∗ Va or P ∗ Va ∗ Ua or Ua = Va , exclusively. Take the first case. By Proposition 5.1, there exists a point A such that P ∗ A ∗ Ua . Now the Four-point Theorem 5.5 implies the both order relations P ∗ A ∗ Ua and P ∗ A ∗ Va to hold. Moreover, by Lemma 5.12 the point A lies in the interior of both triangles 4EFG and triangle 4XYZ. Similarly, there exists a point B such that both P ∗ B ∗ Ub and P ∗ B ∗ Vb hold, and B lies in the interior of both triangles 4EFG and triangle 4XYZ. Finally, there exists a a point C such that both P ∗ C ∗ Uc and P ∗ C ∗ Vc hold, and C lies in the interior of both triangles 4EFG and triangle 4XYZ. By Lemma 5.13, the sides and vertices from triangle 4ABC, as well as the interior of 4ABC lie in the interior of triangle 4EFG, as well as of triangle 4XYZ.
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Figure 5.19. The nonempty intersection of interiors of two triangles contains a triangle.
By Lemma 5.10, the vertex P lies in the interior of triangle 4ABC. We already know that 4EFG ⊆ U and 4XYZ ⊆ V. Altogether, we have obtained P ∈ Int4 ABC ⊆ Int4 EFG ∩ Int4 XYZ ⊆ U ∩ V We see the intersection U ∩ V contain a triangle 4ABC around each one of its points, and hence is open, as to be checked. (O3) Let Uα be a nonempty family of open sets, with indexes α ∈ I. To check whether their union S U := α∈I Uα is open, take any point x ∈ U. There exist β ∈ I and a triangle such that P ∈ Int4 ABC ⊆ Uβ . Hence [ Uα P ∈ Int4 ABC ⊆ Uβ ⊆ α∈I
We see the union U contains a triangle around each one of its points, and hence is open, as to be checked. 5.10.
Left and right, orientation
Definition 5.16 (Coterminal rays). Two rays are called coterminal if one of them can be obtained from the other one by extension or deletion of a segment. Problem 5.21. It is easy to see that being coterminal defines an equivalence relation among rays. Explain the details.
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Problem 5.22. The opposite rays of two coterminal rays are coterminal, too. Explain the details. Definition 5.17 (Oriented line). An oriented line is a line AB together with the direction specified −−→ by the ray AB, where A and B are any points lying on the line. In other words, an oriented line is a line together with an equivalence class of coterminal rays lying on it. Problem 5.23. Convince yourself that there are exactly two equivalence classes of coterminal rays on a given line. Hence there are two possibilities to assign an orientation to a line. Theorem 5.1 (Orientation Theorem). In any given plane, one can assign a left and a right half plane to every oriented line in a consistent way such that • reversing the orientation of a line switches its left and right half plane; • any point in the interior of an angle lies in the left half plane of on of its sides and the right half plane of the other side of the angle. Hence there exist exactly two possibilities to assign a left and a right half plane to all oriented lines in a plane. −−→ For one ray OA, we define—rather arbitrarily or really determined by physics— one of the half-planes of line OA to be the left half-plane, and the other one the right half-plane. Typically, −−→ we take as ray OA the positive x-axis, and agree the half-plane {(x, y) : y > 0} to be lying left to the positive x-axis. Once the left and right half planes have been agreed on for one ray, the left and −−→ a right half-plane are uniquely determined for any ray AB, too. Justification. We begin by an agreement what are to be the left and right half-plane for one fixed −−→ −−→ ray OA. After that agreement, for any second ray OB with the same vertex, the left and right half−−→ planes are determined. Indeed, if point B is left of ray OA, then point A lies in the right half plane −−→ of ray OB. One checks that for the opposite rays, the half-planes get exchanged. −−→ Problem 5.24. Convince yourself that the statement of Theorem 5.1 holds for any two rays OB and −−→ −−→ OC from vertex O. Begin with the case that B and C are two points in opposite half planes of OA. Answer. It is enough to confirm the statement: −−→ −−→ −−→ If B ∈ left [OA] and C ∈ right [OA] and B ∈ left [OC] −−→ then C ∈ right [OB] Point B is neither in the interior of angle ∠AOC nor in the interior of the vertical angle. Hence point B lies in the interior of one of the supplementary angles of ∠AOC, and points A and C lie on the −−→ −−→ same side of line OB. Furthermore A is right of OB. Hence C is right of OB. −−→ Problem 5.25. Convince yourself that the statement of Theorem 5.1 holds for any two rays OB and −−→ −−→ OC from vertex O for which points B and C lie in the same half plane of OA. −−→ Answer. I need to consider the cases that B and C are two points in the same half plane of OA. It is enough to confirm the statement: −−→ −−→ If B, C ∈ left [OA] and B ∈ left [OC] −−→ then C ∈ right [OB]
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Point B is neither in the interior of angle ∠AOC nor in the interior of the vertical angle. Hence point B lies in the interior of one of the supplementary angles of ∠AOC, and points A and C lie on the same side of line OB. Because of the position of B, points A and C are on different sides of line −−→ −−→ OB. Furthermore A is right of OB. Hence C is right of OB. −−→ In the next step, take any oriented line AB, on which point O does not lie. Now left- and right half-plane have already determined by the requirement of the orientation theorem. Indeed −−→ • if point B happens to lie in the right half plane of ray OA, then B lies in the left half plane of −−→ −−→ the opposite ray AO and hence point O lies in the right half plane of ray AB; −−→ • in the opposite case, it turns out that point O lies in the left half plane of ray AB provided that −−→ point B lies in the left half plane of ray OA. We still need to verify that no contradiction can arise. Problem 5.26. Convince yourself that the above definition gives the left and right planes in the −−−→ −−→ same way, if the ray AB is replaced by any coterminal ray A0 B0 . Is the requirement of the orientation theorem true for any given angle ∠CAB? Does a point in the interior of the angle lie in the left half plane of on of its sides and the right half plane of the other side of the angle? Problem 5.27. Assume that point O lies in the interior of angle ∠CAB. −−→ Convince yourself that point C lies in the left half plane of ray AB if and only if point B lies in −−→ the right half plane of ray AC. Hence the requirement of the orientation theorem that any point in the interior of an angle lies in the left half plane of on of its sides and the right half plane of the other side of the angle is satisfied. −−→ Solution. By the Crossbar Theorem, the segment BC and the ray AO intersect. Hence −−→ −−→ B ∈ left [OA] ⇔ C ∈ right [OA] −−→ By the definition above, the point O lies in the left half plane of ray AB if and only if point B lies −−→ in the left half plane of ray OA. Too, the agreement states that the point O lies in the right half −−→ −−→ plane of ray AC if and only if point C lies in the right half plane of ray OA. In the end, we get the equivalences: −−→ −−→ −−→ C ∈ left [AB] ⇔ O ∈ left [AB] ⇔ B ∈ left [OA] −−→ −−→ −−→ ⇔ C ∈ right [OA] ⇔ O ∈ right [AC] ⇔ B ∈ right [AC] The remaining cases correspond to point O lying in the vertical or one of the supplementary angles of the given angle ∠CAB. By applying the reasoning above to the requested vertical or one of the supplementary angle, we can verify the requirement of the orientation theorem for each of these cases. We have confirmed there exist exactly two ways to assign a left and a right half plane to every oriented line, and have checked that the assignment is consistent. Indeed, if any point C lies in the −−→ −−→ left half-plane of any ray AB, then point B lies in the right half-plane of ray AC.
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Theorem 5.2. The interior domain of a triangle 4ABC is either the intersection of the left half−−→ −−→ −−→ planes of the three rays AB, BC, CA, or the intersection of the three right half-planes of these rays. In either case, the intersection of the three opposite half-planes is empty. Definition 5.18 (Orientation of a triangle). If the interior domain of a triangle 4ABC is the −−→ −−→ −−→ intersection of the left half-planes of the three rays AB, BC, CA, we say that the triangle is oriented counterclockwise or positive. If the interior domain of a triangle 4ABC is the intersection of the right half-planes of the three −−→ −−→ −−→ rays AB, BC, CA, we say the triangle is oriented clockwise or negative. Proposition 5.12. Assume the points A , B and P , Q lie on the same line and the points C and D do not lie on this line. The orientation of the triangles 4ABC and 4ABD are equal if C and D lie in the same halfplane of AB, and opposite if C and D lie in opposite half-planes. −−→ −−→ The orientation of the triangles 4ABC and 4PQC are equal if the rays AB and PQ point in the same direction, and opposite if the rays point in opposite directions. Indication of reason. If A ∗ B ∗ C, and point P does not lie on line ABC, then the triangles 4PAB, 4PAC and 4PBC all have the same orientation. Problem 5.28. Given is a triangle 4ABC and three rays a with vertex A, ray b with vertex B, and ray c with vertex C, pointing into the interior of the triangle. (a) Given a reason why any two of these rays, say rays a and b, intersect in the interior of the triangle. Define the intersection points Z := a ∩ b , X := b ∩ c , Y := c ∩ a (b) If two of the points X, Y, Z are equal, then all three are equal. Explain why. Provide a drawing for this case. (c) If the points X, Y, Z are distinct, they do not lie on a line. Explain why. Answer. By the Crossbar Theorem, the ray a intersects the opposite side BC, say in a point P. Similarly the ray b intersects the opposite side CA, say in a point Q. It is convenient to introduce a notation for proofs using the plane separation theorem repeatedly. Point C and Q lie on the same side of line a, but points C and B lie on different sides of line a. We summarize these two facts by writing B |a C, Q By the plane separation theorem, we conclude that points B and Q lie on on different sides of line a. Hence the segment QB intersects the line a. The intersection point R even lies on the ray a. This is confirmed in the same way as in the proof of the Crossbar Theorem. Hence point R lies the interior of angle ∠BAC. Switching the roles of rays a and b, we see that point R lies in the interior of angle ∠ABC, too. Hence point R lies in the interior of triangle 4ABC. If two of the points X, Y, Z are equal, this point is the intersection of all three rays a, b, c. Hence X = Y = Z confirming item (b). To check item (c), assume towards a contradiction that X , Y and the three points X, Y, Z lie on a line. On this line would lie all three rays a, b, c and all three vertices A, B, C of triangle 4ABC, contradicting the definition of a triangle.
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Figure 5.20. Three rays pointing from the vertices into the interior of a triangle.
We continue the previous problem and now assume that the triangle 4XYZ exists. Two cases can occur. Either (i) A ∗ Z ∗ Y or (ii) A ∗ Y ∗ Z. Problem 5.29. Begin with the case (i). The order of the points C, X, Y and the points B, X, Z is determined. Explain how this happens. Answer. By the assumption (i) A ∗ Z ∗ Y and point Z lies on line b. Hence points A and Y lie on different sides of line b. Too, the points A and C lie on different sides of line b. As a shorthand we write A |b Y, C We use the plane separation theorem with line b and conclude that the points Y and C lie on the same side of line b. Now we determine the order of the points C, X, Y. Since point X lies on line b, and points Y and C lie on the same side of this line, we conclude that either C ∗ Y ∗ X or Y ∗ C ∗ X. The last possibility is excluded since the points X, Y lie on the ray c emanating from point C. We conclude that C ∗ Y ∗ X. To get the order of points B, X, Z, we use the half planes of line a. Points C and X are in different half planes since C ∗ Y ∗ X and Y ∈ a. Too, points B and C are in different half planes. Indeed C |a X, B Hence points X and B are in the same half plane of line a. Since point Z is on this line, one concludes B ∗ X ∗ Z. Problem 5.30. Take now the case (ii). Determine the order of the point B, X, Z and the points C, X, Y. Answer. You can confirm A |c Z, B and B |a X, C By the assumption (ii) A ∗ Y ∗ Z. We use the plane separation theorem with line c. Since point Y ∈ c, points A and Z are on different sides of line c. Too, the points A and B lie on different sides of line c. Hence we conclude A |c Z, B Hence points B, Z in the same half plane on c. Thus we determine the order of the points B, X, Z. The points X, Z lie on the ray b emanating from point B, and point X lies on line c. We conclude that X ∗ Z ∗ B.
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To get the order of points C, X, Y, we use the half planes of line a. The points B and C lie on different sides of line a. Moreover point Z lies on line a and points X and B are in different half planes of line a. Hence B |a X, C and points X and C are in the same half plane of line a. Since point Y is on this line, one concludes C ∗ X ∗ Y. Problem 5.31. We assume that triangle 4ABC is oriented counterclockwise. Convince yourself that each one of the two cases (i)(ii) considered above, the triangle 4XYZ is oriented counterclockwise, too −−→ −−→ −−→ Answer. (i) A ∗ Z ∗ Y and A ∈ left [ BC] imply A ∈ left [XY], hence Z ∈ left [XY]. Hence the triangle 4XYZ is oriented counterclockwise. −−→ −−→ −−→ (ii) A ∗ Y ∗ Z and A ∈ right [CB] imply A ∈ right [XY], hence Z ∈ left [XY]. Hence the triangle 4XYZ is oriented counterclockwise. 5.11.
The restricted Jordan Curve Theorem
Definition 5.19 (Open and closed polygon). A list of segments AB, BC, CD, . . . , KL is called a polygon or polygonal curve. The points inside the segments and their endpoints are called the points of polygon. If the endpoint L is equal to the beginning point A = L, the polygon is called closed; if A , L the polygon is open. The points A, B, C, . . . , K are called the vertices, and the segments AB, BC, CD, . . . , KL are called the sides of the polygon. Definition 5.20. A polygon or polygonal curve is called simple, if the vertices—with possible exception of the first and last—are all different, and any two sides do not intersect except at their common endpoint when they follow each other in the list AB, BC, CD, . . . , KL.
Figure 5.21. The restricted Jordan Curve Theorem.
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Proposition 5.13 (Restricted Jordan Curve Theorem). A simple closed polygonal curve C partitions the points of the plane not lying on the polygon into two regions, called the interior and exterior of the polygon, which have the following properties: • If A is a point of the interior, and B a point of the exterior region, then each polygonal curve connecting A to B has at least one point in common with the curve C. • If A and A0 are two points of the interior region, then there exists a polygonal curve connecting A to A0 which has no point in common with the curve C. • Similarly, if B and B0 are two points of the exterior region, there exists a polygonal curve connecting B to B0 which has no point in common with the curve C. • There exists a line which does not cut the interior region. • Every line cuts the exterior region. Sketch of a proof for the Euclidean plane. 1 The set R2 \C consists of finitely many components. As one follows the polygon C, the points near to the right, are all in the same component as some reference point R. Similarly, the points near to the left are all in the same component as some reference point L. Furthermore, one can choose points R and L near to the same side of the polygon C and to each other. Any point X in R2 \ C can be connected by a polygonal curve inside this set R2 \ C to one of these these two points R and L. Hence R2 \ C consists of at most two connected components. To prove that the points on the left and the right lie in different components, consider rays. We call a ray exceptional, if it contains a vertex of the curve C. For a given vertex P, there exist at most finitely many directions of exceptional rays. Each ray that is not exceptional, has only finitely many intersection points with the curve C. As one turns the ray around a fixed vertex P, the number of intersection points changes only at the exceptional directions. Nearby, before and after passing through an exceptional direction, the parity of the number of intersection points is the same. Hence we can assign a even parity or odd parity to the vertex P, depending of the number of intersection points being even or odd. Next we show that the parity is constant inside each component of the set R2 \ C. Let X and Y be two points in the same component. There exists a polygonal curve P connecting X and Y inside this component. We can assume that the rays along the sides of curve P are not exceptional. This can be achieved by an arbitrarily small adjustment of the curve P. After this adjustment, the two endpoints of each segment of P have the same parity, since the intersection set of the ray with C changes continuously as one moves the vertex of the ray along one side of P, keeping the ray parallel. Hence any two points in the same component of R2 \ C have the same parity. The endpoints of a small segment crossing the polygon C have opposite parities. Points far away from the polygon have even parity, because there exists a ray not intersecting the curve C. Hence the unbounded outside component of R2 \ C has even parity, and the inside bounded component has odd parity. Remark. A complete proof in the present context is given by G. Feigl [14]. The really interesting point is that none of the further axioms of congruence, continuity or parallels are needed. Especially, the Jordan Curve Theorem holds for the hyperbolic plane, too. On the other hand, the Restricted Jordan Curve Theorem 5.13 is not valid for the projective plane, and hence neither in double elliptic geometry. The reason is that the improper points and line of the projective plane change the topological structure. 1
This proof follows H. Tverberg [37].
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Remark. The more general Jordan curve theorem says that any closed simple continuous curve separates the Euclidean plane into an interior and exterior domain. It is credited to Camille Jordan (1836-1922), but Jordan’s original argument was in fact inadequate. The first correct proof is credited to Oswald Veblen (1880-1960).
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6. Finite Affine and Projective Incidence Planes and Latin Squares 6.1.
Latin squares
Definition 6.1. A n × n square matrix where each row and column is a permutation of the numbers 1, 2, . . . , n is called a Latin square. Definition 6.2. Two Latin squares ai j and bi j are called orthogonal if the pairs (ai j , bi j ) are different at all n2 positions (i, j). Equivalently, two Latin squares are orthogonal, if the mapping (i, j) ∈ {1, 2, . . . , n} × {1, 2, . . . , n} 7→ (ai j , bi j ) ∈ {1, 2, . . . , n} × {1, 2, . . . , n}
(6.1)
is a bijection. The search for orthogonal Latin squares goes back to the following problem posed by Leonhard Euler in 1782: A very curious question, which has exercised for some time the ingenuity of many people, has involved me in the following studies, which seem to open a new field of analysis, in particular in the study of combinatorics. The question revolves around arranging 36 officers to be drawn from 6 different ranks, and at the same time from 6 different regiments, so that they are also arranged in a square so that in each line—both horizontal and vertical—there are 6 officers of different ranks and and regiments. Euler’s problem is just equivalent to finding two orthogonal Latin squares of side 6—one specifying the rank, the second specifying the regiment for each officer. It is easy to find two orthogonal 3 × 3 Latin squares: 1 2 3 2 3 1 3 1 2
and
1 3 2 2 1 3 3 2 1
(6.2)
Problem 6.1. Solve Euler’s problem for a 4 × 4 square: Take 16 cards with numbers 1, 2, 3, 4 and suits ♦, ♥, ♠, ♣ and put them into a 4 × 4 square such that each row and column contains exactly one number and one suit. Answer. Here is the solution I got in five minutes: 1 2 3 4
2 1 4 3
3 4 1 2
4 3 2 1
and
1 2 3 4
3 4 1 2
4 3 2 1
2 1 4 3
The cards are:
1♦ 2♥ 3♠ 4♣
2♠ 1♣ 4♦ 3♥
3♣ 4♠ 1♥ 2♦
4♥ 3♦ 2♣ 1♠
(6.3)
Euler correctly proved that two orthogonal squares exist for every odd side n ≥ 3 and for every side divisible by four. Furthermore, he wrote: I have examined a very great number of tables . . . and I do not hesitate to conclude that one cannot produce an orthogonal pair of order 6, and that the same impossibility extends to n = 6, 10, 14, . . . and in general to all orders which are unevenly even. Thus, Euler conjectured that no solution exist any n ≡ 2 mod 4.
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Problem 6.2 (An attempt of solve Euler’s officer problem). Put 36 officers in a 6 × 6 square with places (i, j), were i, j = 1, . . . , 6. We (and the officers) calculate modulo six. Regiment and rank numbers are given by if i is even i + j regi j = 2i + j − 1 if i is odd if i is even 2i + j − 2 ranki j = i + j − 2 if i is odd Find out which combinations of regiment- and rank-numbers come up more than once. Which ones are missing. It is simplest to make up a 6 × 6 matrix indexed by the regiment- and rank-numbers and tally. Answer. The matrices of the regiment- and rank-numbers are 2 3 6 regi j = 5 4 1
3 4 1 6 5 2
4 5 2 1 6 3
5 6 3 2 1 4
6 1 4 3 2 5
1 2 5 4 3 6
6 3 2 ranki j = 1 4 5
and
1 4 3 2 5 6
2 5 4 3 6 1
3 6 5 4 1 2
4 1 6 5 2 3
5 2 1 6 3 4
To find out how often each combination of regiment and rank number comes up, and which do not come up at all, one has to tally these combinations. The result is: 2 0 2 number of occurrences of (reg, rank) = 0 2 0
0 2 0 2 0 2
2 0 2 0 2 0
0 2 0 2 0 2
2 0 2 0 2 0
0 2 0 2 0 2
It is easy to check directly that regi j + ranki j is always even. From the tally, we see that all the combinations with regi j + ranki j = 2 modulo 6 come up exactly twice. About 177 years later, it was discovered that Euler’s conjecture is right for order n = 6, but wrong for all larger orders. Main Theorem 5 (Existence of orthogonal Latin squares). Nonexistence of orthogonal Latin squares is a rare exception: G.Tarry, 1900 There do not exist two orthogonal Latin squares of side 6. R.C.Bose, E.T.Parker, S.S. Shrikhande, 1959 Two orthogonal Latin squares exist for all sides n , 2 and n , 6.
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Here are two orthogonal Latin squares of side ten: 0 8 9 5 7 6 3 1 2 4
4 1 8 9 6 7 0 2 3 5
1 5 2 8 9 0 7 3 4 6
7 2 6 3 8 9 1 4 5 0
2 7 3 0 4 8 9 5 6 1
9 3 7 4 1 5 8 6 0 2
8 9 4 7 5 2 6 0 1 3
3 4 5 6 0 1 2 7 8 9
6 0 1 2 3 4 5 8 9 7
5 6 0 1 2 3 4 9 7 8
and
0 6 5 9 3 8 7 4 1 2
7 1 0 6 9 4 8 5 2 3
8 7 2 1 0 9 5 6 3 4
6 8 7 3 2 1 9 0 4 5
9 0 8 7 4 3 2 1 5 6
3 9 1 8 7 5 4 2 6 0
5 4 9 2 8 7 6 3 0 1
4 5 6 0 1 2 3 7 9 8
1 2 3 4 5 6 0 8 7 9
2 3 4 5 6 0 1 9 8 7
(6.4)
The exact maximum number N(n) of mutually orthogonal squares of side n is still not known in all cases. Especially, it is not known whether there are three mutually orthogonal squares of side 10. In 1988, after having completed the computer proof that N(10) ≤ 8, Lam from the Concordia group writes: I suspect that finding three mutually orthogonal Latin squares of order 10 is even more work than what we have done. Besides the obvious N(3) = 2, this would be the only case with N(n) = 2, possibly. 1 The following basic fact gives the connection with finite incidence geometry. Proposition 6.1 (The basic correspondence of affine planes and Latin squares). From an affine plane of order n can be constructed n − 1 pairwise orthogonal Latin squares of side n. Conversely, these n − 1 orthogonal n × n Latin squares define an affine plane of order n. Reason. Take any two intersecting lines l and m. The n2 intersection points (i, j) of these two lines and their parallels build a square grid of size {1, 2, . . . , n} × {1, 2, . . . , n}. Take now any pencil of n parallel lines, not parallel to neither l nor m— and number them in any way by {1, 2, . . . , n}. A Latin square is defined by placing at grid point (i, j) the number of the line in the pencil passing through the grid point. Two Latin squares ai j and bi j derived from two different pencils a and b cannot have the equal pairs (ai j , bi j ) at two positions (i, j) and (i0 , j0 ). Indeed ai j = ai0 j0 and bi j = bi0 j0 would imply that the line through points (i, j) and (i0 , j0 ) is both in the pencil a and the pencil b. Conversely, these n − 1 orthogonal Latin squares define n − 1 pencils of parallel lines in n − 1 different directions— all different from the horizontal and vertical lines of the grid. Together with the latter lines, one gets n + 1 pencils of n parallel lines in n + 1 different directions. These are all n2 + n lines of an affine plane with n2 points. The order of the affine plane is n—equal to the side of the given Latin squares. 6.2. Latin squares from finite fields As explained in proposition 6.1, there is a one-to-one correspondence between and finite affine plane of order n and a maximal set of n − 1 orthogonal Latin squares of side n. We now construct the Latin squares corresponding to the Cartesian plane over the finite Galois field GF(pr ). The indices i and j of the rows and columns, and the matrix elements ai j of the Latin squares are now elements of GF(pr ). 1
This is my information from [9] and [30], p.783.
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Corollary 5 (Latin squares from finite fields). Let the side n = pr be a prime power. For this side, there exist n − 1 mutually orthogonal Latin squares. With addition and multiplication from the finite Galois field GF(n), and slopes m running over all pr − 1 non zero values m ∈ GF(pr ) \ {0}, their elements are ai j = i + m j (6.5) for all i, j ∈ GF(pr ). Problem 6.3. Explain once more what are the points and lines of the affine plane given by these Latin squares. Confirm the basic counting from proposition 3.4. Answer. The points are ordered pairs (i, j) of elements i, j ∈ GF(n). The lines are the sets of points, specified as follows: The horizontal lines are
{(i, j) : j ∈ GF(n)} for all i ∈ GF(n) {(i, j) : i ∈ GF(n)} for all j ∈ GF(n)
The vertical lines are
For each slope m ∈ GF(n) \ {0}, a Latin square defines a set of parallel lines {(i, j) : i + m j = i0 , i, j ∈ GF(n)} for all intercepts i0 ∈ GF(n). (a) Each line has n points. (b) The lines can be partitioned into n + 1 classes— horizontal, vertical, and n − 1 sloped pencils, each containing n parallel lines. (c) In every point intersect n + 1 different lines. (d) There are n2 = p2r points and n(n + 1) lines. Problem 6.4. Find four mutually orthogonal Latin squares of side 5. Answer. Since 5 is a prime, the relevant Galois field is GF(5) = Z5 . Thus the arithmetic is done modulo 5. With m = 1, −1, equation (6.5) yields the tables 0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 2 3
and
0 1 2 3 4
4 0 1 2 3
3 4 0 1 2
2 3 4 0 1
1 2 3 4 0
1 2 3 4 0
4 0 1 2 3
2 3 4 0 1
With m = 2, −2, equation (6.5) yields the tables 0 1 2 3 4
2 3 4 0 1
4 0 1 2 3
1 2 3 4 0
3 4 0 1 2
and
0 1 2 3 4
3 4 0 1 2
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Remark (Construction of the Galois field GF(9)). One needs to get a generator of the multiplicative group GF(9) \ {0}, which is known to be cyclic. The field GF(9) is a 2-dimensional vector space over Z3 . Hence the generator α is a zero of a quadratic irreducible polynomial P(x) ∈ Z3 [x]. All elements of GF(9) are zeros of the polynomial x9 − x. The factoring x9 − x = x(x4 + 1)(x2 + 1)(x + 1)(x − 1) implies that P(x) has to be a divisor of x4 + 1— otherwise the order of α would not be 8, as required for the generator. Since in Z3 [x] one can factor x4 + 1 = x4 − 3x2 + 1 = (x2 − x − 1)(x2 + x − 1) there are two choices for the irreducible factor of which α is a zero. I choose P(x) = x2 − x − 1, and require α2 = α + 1. Because there exists only one unique finite field for an given order, there are always some arbitrary choices for the corresponding irreducible polynomial P(x). Next we express the powers αi in the basis 1, α: α α2 α3 α4 α5 α6 α7 α8
α 1+α 1 + 2α 2 2α 2 + 2α 2+α 1
Problem 6.5. Take addition and multiplication as in Z2 . The Galois field GF(4) is {0, 1, α, 1 + α}, where α is a generator of the multiplicative group. It satisfies α2 + α + 1 = 0. Find three mutually orthogonal Latin squares of side 4. Answer. The equation (6.5), with m = 1, α, 1 + α yields the three tables below. Because of their unusual names, I have repeated the indices i and j in an extra column left and row on top. To get Latin squares in ordinary fashion, one can substitute α := 2 and 1 + α := 3, and add one to all entries. In that way, we arrives once more at the solution given for problem 6.1. m=1 0 1 α 1+α
0 0 1 α 1+α
m=1+α 0 1 α 1+α
1 1 0 1+α α
0 0 1 α 1+α
α α 1+α 0 1
1 1+α α 1 0
1+α 1+α α 1 0
α 1 0 1+α α
m=α 0 1 α 1+α
0 0 1 α 1+α
1 α 1+α 0 1
α 1+α α 1 0
1+α 1 0 1+α α
1+α α 1+α 0 1
Problem 6.6. Find eight mutually orthogonal Latin squares of side 9. Put as indices of the rows and columns the field elements 0, α1 , α2 , . . . α8 . Calculate with slopes m = αµ . Get the symmetric square for µ = 8, and explain how the other ones for µ = 1, . . . 7 are obtained from it.
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Answer. The elements of the Latin squares become a00 = 0 ai0 = αi a0 j = α
for i = 1, . . . , 8
µ+ j
for j = 1, . . . , 8 µ+ j
ai j = α + α i
for i, j = 1, . . . , 8
For µ = 8 one gets a symmetric square 0 α α2 α3 α4 α5 α6 α7 α8
α α5 α3 α8 α7 0 α4 α6 α2
α2 α3 α6 α4 α α8 0 α5 α7
α3 α8 α4 α7 α5 α2 α 0 α6
α4 α7 α α5 α8 α6 α3 α2 0
α5 0 α8 α2 α6 α α7 α4 α3
α6 α4 0 α α3 α7 α2 α8 α5
α7 α6 α5 0 α2 α4 α8 α3 α
α8 α2 α7 α6 0 α3 α5 α α4
The other seven square are obtained by cyclic permutation of the last seven columns, leaving the first column fixed. Problem 6.7. Follow the description from problem (6.6) and really write out the eight Desarguesian Latin squares of side 9. Answer. For µ = 8 one gets a symmetric square. And for µ = 1 one gets 0 1 2 3 4 5 6 7 8
1 5 3 8 7 0 4 6 2
2 3 6 4 1 8 0 5 7
3 8 4 7 5 2 1 0 6
4 7 1 5 8 6 3 2 0
5 0 8 2 6 1 7 4 3
6 4 0 1 3 7 2 8 5
7 6 5 0 2 4 8 3 1
8 2 7 6 0 3 5 1 4
0 1 2 3 4 5 6 7 8
2 3 6 4 1 8 0 5 7
3 8 4 7 5 2 1 0 6
4 7 1 5 8 6 3 2 0
5 0 8 2 6 1 7 4 3
6 4 0 1 3 7 2 8 5
7 6 5 0 2 4 8 3 1
8 2 7 6 0 3 5 1 4
1 5 3 8 7 0 4 6 2
8 2 7 6 0 3 5 1 4
1 5 3 8 7 0 4 6 2
2 3 6 4 1 8 0 5 7
0 1 2 3 4 5 6 7 8
4 7 1 5 8 6 3 2 0
5 0 8 2 6 1 7 4 3
6 4 0 1 3 7 2 8 5
7 6 5 0 2 4 8 3 1
8 2 7 6 0 3 5 1 4
1 5 3 8 7 0 4 6 2
2 3 6 4 1 8 0 5 7
3 8 4 7 5 2 1 0 6
For µ = 2 and µ = 3 one gets 0 1 2 3 4 5 6 7 8
3 8 4 7 5 2 1 0 6
4 7 1 5 8 6 3 2 0
5 0 8 2 6 1 7 4 3
6 4 0 1 3 7 2 8 5
7 6 5 0 2 4 8 3 1
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For µ = 4 and µ = 5 one gets 0 1 2 3 4 5 6 7 8
5 0 8 2 6 1 7 4 3
6 4 0 1 3 7 2 8 5
7 6 5 0 2 4 8 3 1
8 2 7 6 0 3 5 1 4
1 5 3 8 7 0 4 6 2
2 3 6 4 1 8 0 5 7
3 8 4 7 5 2 1 0 6
4 7 1 5 8 6 3 2 0
0 1 2 3 4 5 6 7 8
6 4 0 1 3 7 2 8 5
7 6 5 0 2 4 8 3 1
8 2 7 6 0 3 5 1 4
1 5 3 8 7 0 4 6 2
2 3 6 4 1 8 0 5 7
3 8 4 7 5 2 1 0 6
4 7 1 5 8 6 3 2 0
5 0 8 2 6 1 7 4 3
4 7 1 5 8 6 3 2 0
5 0 8 2 6 1 7 4 3
6 4 0 1 3 7 2 8 5
0 1 2 3 4 5 6 7 8
8 2 7 6 0 3 5 1 4
1 5 3 8 7 0 4 6 2
2 3 6 4 1 8 0 5 7
3 8 4 7 5 2 1 0 6
4 7 1 5 8 6 3 2 0
5 0 8 2 6 1 7 4 3
6 4 0 1 3 7 2 8 5
7 6 5 0 2 4 8 3 1
For µ = 6 and µ = 7 one gets 0 1 2 3 4 5 6 7 8
7 6 5 0 2 4 8 3 1
8 2 7 6 0 3 5 1 4
1 5 3 8 7 0 4 6 2
2 3 6 4 1 8 0 5 7
3 8 4 7 5 2 1 0 6
6.3. Finite Non-Desarguesian planes The first example of a finite non-Desarguesian plane was published in 1907 by O. Veblen and J.H.M. Wedderburn [39]. Indeed, these authors have obtained four non-isomorphic projective planes of order 9. In the following pages, I give an independent access to part of Veblen’s arguments. Remark (Conjugations in the Galois field GF(9)). A conjugation is defined as an automorphism of a field, which is its own inverse, but not the identity. The complex conjugation in the complex number field is the most common example. The Galois field GF(9) has indeed a conjugation. The multiplicative group GF(9) \ {0} is known to be cyclic. As above, we use a generator α. The mapping defined by αi := α si for all i = 1, . . . , 8 and 0 := 0 (6.6) is an automorphism of the multiplicative group, for s = 3, 5, 7, since these numbers are relatively prime to the order 8 of the multiplicative group. Since one calculates modulo 3 in the field GF(9) (αi + α j )3 = α3i + α3 j
for all i, j = 1, . . . , 8
follows directly from the binomial theorem. Hence the mapping (6.6) is a field automorphism. It is indeed the well-known Frobenius automorphism. Now one can check that 1 + α = −(1 + α), which makes 1 + α take the role of the imaginary unit for the present conjugation. Problem 6.8. Check that neither the choice s = 5 nor s = 7 leads to a field automorphism.
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Remark (The perfect squares in the Galois field GF(pr )). Given is any a ∈ GF(pr )\{0}. A necessary condition for solvability of the equation x2 = a (6.7) follows from the fact that pr − 1 is the order of the multiplicative group GF(pr ) \ {0}: a
pr −1 2
=1
(6.8)
is necessary for equation (6.7) to have a solution. Since the equation (6.8) holds for exactly half of the members of the multiplicative group GF(pr ) \ {0}, and, too, the equation (6.7) is solvable for right-hands sides a in exactly half of the members of this group, one concludes that a is a perfect square if and only if relation (6.8) holds. Remark (A second algebraic structure in the Galois field GF(9)). We define the addition as before, and the new pseudo-multiplication by setting a · b, if a is a perfect square (6.9) a♦b := a · b, if a is not a perfect square With that new pseudo-multiplication, GF(pr ) \ {0} is a non-commutative group. But only the righthanded distributive law a♦(b + c) = a♦b + a♦c (6.10) holds, whereas the left-handed distributive law is violated. Question. Find a formula for the inverse of the pseudo-multiplication. Question. Check that the pseudo-multiplication ♦ makes the set GF(pr ) \ {0} a non-commutative group. Question. Give the reason for the right-handed distributive law. Question. Give an example where the left-handed distributive law does not hold. Answer. Since for example α2 + α4 = α, the sum of two perfect squares need not be a perfect square. Especially (α2 + α4 )♦α , α2 ♦α + α4 ♦α Problem 6.9. As above, we use pairs of the field elements 0, α1 , α2 , . . . α8 as coordinates, and as indices for the matrix elements of the Latin squares, too. Find eight Latin squares of side 9, which are essentially different from (6.2). Construct the sloped lines by means of the pseudomultiplication. Answer. The elements of the of Latin squares become a00 = 0 ai0 = αi µ
a0 j = α ♦α
for i = 1, . . . , 8 j
ai j = αi + αµ ♦α j
for j = 1, . . . , 8
(left-slope)
for i, j = 1, . . . , 8
For even µ, one gets the same Latin squares as in the Desarguesian case given above in (6.2). The odd µ, the new j-th column is the old (3 j mod 8)-th column. Thus columns 1 and 3, columns 2 and 6, as well as columns 5 and 7 are exchanged. Problem 6.10. The following basic facts were checked in problem (6.3) for the Desarguesian case.
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(a) The lines are the sets of n = pr points each. (b) There are n + 1 pencils of n parallel lines. (c) In each point, n + 1 lines intersect. (d) Together, there are n2 = p2r points and n2 + n lines. Convince yourself that these facts are still valid for non-Desarguesian Latin squares. Answer. Are the items (a) through (d) strong enough to get an affine plane? Clearly not, indeed, we need check axioms (A.1), (A.2) and (A.3). Problem 6.11. Check axiom (A.1)—existence and uniqueness of the line between any two points (i, j) and (i0 , j0 ). Answer. The case that either i = i0 or j = j0 leads to a unique horizontal or vertical line. We now assume that i , i0 and j , j0 . If these two points lie on a line, it has to have some slope m , 0 such that i + m♦ j = i0 + m♦ j0 Because of the right-handed distributive law (6.10), we conclude that i − i0 = m♦( j − j0 ) Because the pseudo-multiplication ♦ makes the set GF(n) \ {0} a multiplicative group, and both i − i0 , 0, j − j0 , 0, one can solve this equation uniquely for the slope m. Now it is easy to get the unique line connecting the two given points (i, j) and (i0 , j0 ). Axiom (A.3) is indeed even easier to check. Problem 6.12. Check axiom (A.3). Answer. The three points with coordinates (0, 0), (0, n), (n, 0) do not lie on a line. This is a good step forward. Still, we are left with the problem to check axiom (A.2)—existence and uniqueness of the parallel. Axiom (A.1) clearly implies that two different lines can intersect in at most one point. The existence of a parallel follows to any given line through any given point is stated in item (c). The hard part is to confirm uniqueness of the parallel. As Veblen has observed, here simple counting helps better than calculation. Problem 6.13. Show that any finite system of points and lines, which satisfies items (a), (d) and axioms (A.1) and (A.3), satisfies axiom (A.2), too. Answer. Axiom (A.1) clearly implies that two different lines can intersect in at most one point. The existence of a parallel to any given line through any given point is stated in item (b). The hard part is to confirm uniqueness of the parallel. Towards a contradiction, we assume that there exists a line l and a point P not on line l, and two parallels p and q to line l pass through point P. A contradiction arises because so many lines would lead to more points than there exist. Here are the details: By item (a), there are n points L1 , . . . , Ln on line l. On each of the n different lines L1 P, . . . , Ln P there exist n−2 points different from L1 , . . . , Ln , P. All these points are different, yielding n(n−1)+1 points. Too, there are n−1 points on the parallel p, which are different from P and the points above. Altogether, we have found n(n − 1) + 1 + n − 1 = n2 points. But by item (d), these are all the points that exist. Hence point P is the only point on the second parallel q, contradicting item (a).
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Here is an alternative to Veblen’s argument to confirm axiom (A.2). Problem 6.14. Why does a direct calculation fail to yield uniqueness of parallels? Answer. One needs to show that two lines of different slopes m , n always intersect. The case involving a horizontal or vertical line is easy. Let these two lines be given by equations i + m♦ j = i0
and
i + n♦ j = i1
One concludes that m♦ j − n♦ j = i0 − i1
(6.11)
But then one is stuck. Because there is no left-hand distributive law, one cannot solve for j. One cannot tell whether the two lines are parallel or do intersect. Problem 6.15. How can one get around the problem, deal with equation (6.11), and confirm axiom (A.2) nevertheless? Answer. Again, we assume that m , n. One needs to show that the mapping j ∈ GF(n) 7→ m♦ j − n♦ j ∈ GF(n)
(6.12)
is a bijection. Because the field is finite, it is enough to show that this mapping is injective. Hence we assume m♦ j − n♦ j = m♦ j0 − n♦ j0 and check whether j and j0 are equal. One gets immediately m♦ j − m♦ j0 = n♦ j − m♦ j0 But now the right-hand distributive law can be used, again. Hence m♦( j − j0 ) = n♦( j − j0 ) If j , j0 , the difference j − j0 would have a multiplicative inverse, and m = n would follow, contradicting the assumption m , n. Hence one concludes that j = j0 , which confirms that the mapping defined by (6.12) is injective. As a map between finite sets of the same size, it is a bijection, too. Hence equation (6.11) can be uniquely solved for j. One has proved this fact even without being able to write a formula for the solution. Thus two lines with different slopes always intersect, confirming axiom (A.2) Problem 6.16. Follow the description from problem (6.9) and really write out the eight nonDesarguesian Latin squares of side 9. Answer. For even µ, one gets the same Latin squares as in the Desarguesian case given above in (6.2). The odd µ, the new j-th column is the old 3 j mod 8-th column. Thus columns 1 and 3, columns 2 and 6, as well as columns 5 and 7 are exchanged. Veblen [39] even constructs two further non-Desarguesian affine planes of order 9. Finally, let’s make the point! Veblen’s ingenious example now begs the question: Open Problem III. For which orders do there exist affine planes at all? It is a long-standing conjecture that affine planes exist only if the order is a prime power. The best result toward a solution of this conjecture is the Bruck-Ryser Theorem:
144
Main Theorem 6 (Nonexistence of projective planes (R.H. Bruck and H.J.Ryser), 1949). No projective plane with order M does exist, if the number number satisfies the following three conditions: (a) M is not a prime power. (b) M ≡ 1 mod 4 or M ≡ 2 mod 4. (c) M , a2 + b2 for all integers a and b. Problem 6.17. Use the Bruck-Ryser Theorem to show that no projective planes of order M ≡ 6 mod 8 can exist. Answer. One needs to check the three conditions: (a) Indeed M is not a prime power, since M = 2m and m ≡ 3 mod 4 is odd. (b) M ≡ 2 mod 4 does hold. (c) Assume towards a contradiction that M = A2 + B2 . Both A and B would be odd, and A−B m = M2 = a2 + b2 would hold with a = A+B 2 and b = 2 . Since m ≡ 3 mod 4 is odd, one number among a and b is even, and the other one is odd. But this implies a2 + b2 ≡ 1 mod 4 contradicting m ≡ 3 mod 4. The smallest orders for which the existence of a projective plane remains open at this point are 10, 12, 15, 18, 20, . . . Only the smallest number in this list has been settled: Main Theorem 7 (Lam, Thiel and Swiercz, 1989). There exist no projective plane of order 10. The computer proof of this theorem was started by Lam and McKay at Concordia in 1979. The final run on a Cray supercomputer was running from September 1986 until November 1988. Lam estimates that the proof involved about two orders of magnitude more computing power than the proof of the famous Four Color Theorem by Wolfgang Haken and Kenneth Appel in 1976. The Concordia group investigated different ways of attacking the problem. Lam writes The final program we wrote is based on a lot of experimentation. If we had just written a program without doing the estimation, we’d probably have a program that was too long to be run anywhere. 6.4. A note on projective spaces The n-dimensional projective space over a field F is defined quite similar to a projective plane (see Definition 3.20). The "points" of the projective space PFn are the sets of equivalent n + 1-tuples (x1 λ, x2 λ, . . . xn+1 λ), where (x1 , x2 , . . . xn+1 ) ∈ Fn+1 and λ runs over the nonzero elements of F. In this manner, one uses homogeneous coordinates and an extra n + 1-th dimension. As a shorthand one writes PFn =
Fn+1 \ {0} F \ {0}
The k-dimensional subspaces of PFn are given in homogeneous coordinates by the k + 1dimensional subspaces through the origin in Fn+1 . This interpretation works for k = 0, 1, . . . n. The n-dimensional projective space over some finite Galois field GF(q) is denoted by PG(n, q).
145
Problem 6.18. Count the points of the projective space PG(n, q). Answer. The n-dimensional projective space over the Galois field GF(q) has Galois field has q points.
qn+1 −1 q−1
points since the
In general, the number k-dimensional subspaces of PG(n, q) is given by the product ! k Y n+1 qn+1−i − 1 (qn+1 − 1)(qn − 1) · · · (qn+1−k − 1) = = k + 1 q i=0 qi+1 − 1 (q − 1)(q2 − 1) · · · (qk+1 − 1) This is a Gaussian binomial coefficient, a q-analogue of a binomial coefficient. In the limit q → 1, one gets back to the ordinary binomial coefficient. Main Theorem 8 (Veblen-Young Theorem). Every finite projective space of geometric dimension n ≥ 3 is isomorphic to a n-dimensional projective space over some finite Galois field GF(q); thus isomorphic to one to the spaces PG(n, q). Remark. Note that this result holds only for dimension at least 3. In sharp contrast, the projective planes are much harder to classify, as not all of them are isomorphic with one of the PG(2, q). The Desarguesian planes are those which are isomorphic with a PG(2, q). These are exactly those planes satisfying Desargues’s theorem. But there exist non-Desargesian planes—and even the prime power conjecture about their order is still open. The smallest 3-dimensional projective space is built over the field GF(2) = Z2 and is denoted by PG(3, 2). It has 15 points, 35 lines, and 15 planes. Each of the 15 planes contains 7 points and 7 lines. As geometries, these planes are isomorphic to the Fano plane. Every point of PG(3, 2) is contained in 7 lines and every line contains three points. In addition, two distinct points are contained in exactly one line and two planes intersect in exactly one line. In 1892, Gino Fano was
Figure 6.1. PG(3, 2) but not all the lines are drawn.
˘ S¸ a three dimensional geometry containing 15 points, the first to consider such a finite geometry âA 35 lines, and 15 planes, with each plane containing 7 points and 7 lines. Problem 6.19 (Kirkman’s Schoolgirl Problem 1850). Kirkman’s schoolgirl problem is a problem in combinatorics proposed by Rev. Thomas Penyngton Kirkman in 1850 as Query VI in The Lady’s and Gentleman’s Diary (pg.48). The problem states:
146
Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily so that no two shall walk twice abreast. Answer. There are 7 days of the week, and 3 girls in each of 5 group. There are 35 = 7 × 5 different combinations for three girls to walk together. Two of the seven non-isomorphic solutions of Kirkman’s schoolgirl problem provide a visual representation of the Fano 3-space PG(3, 2). Some diagrams for this problem can be found at: Beutelspacher, Albrecht; Rosenbaum, Ute (1998), Projective geometry: from foundations to applications, Cambridge University Press, ISBN 978-0-521-48277-6; 978-0-521-48364-3, MR1629468 Each color represents the day of the week (seven colors, blue, green, yellow, purple, red, black, and orange). The definition of a Fano space says that each line is on three points. The figure represents this showing that there are 3 points for every line. This is the basis for the answer to the schoolgirl problem. This figure is then rotated 7 times. There are 5 different lines for each day, multiplied by 7 (days) and the result is 35. Then, there are 15 points, and there are also 7 starting lines on each point. This then gives a representation of the Fano 3-space, PG(3,2). http://mathworld.wolfram.com/KirkmansSchoolgirlProblem.html
gives the solution
Sun Mon Tue Wed Thu Fri Sat
ABC ADH AEM AFI AGL AJN AKO
DEF BEK BHN BLO BDJ BIM BFG
GHI CIO CGK CHJ CFM CEL CDN
JKL FLN DIL DKM EHO DOG EIJ
MNO GJM FJO EGN IKN FHK HLM
http://en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem#Notes http://en.wikipedia.org/wiki/Thomas_Kirkman http://en.wikipedia.org/wiki/Finite_geometry Remark. The problem can be generalized to n girls walking in triplets for (n − 1)/2 days, with the requirement, again, that no pair of girls walk in the same row twice. The number n has to be an odd multiple of 3: thus n ≡ 3 mod 6. It is this generalization of the problem that Kirkman discussed first, while the famous special case n = 15 was only proposed later. A complete solution to the general case was given by D. K. Ray-Chaudhuri and R. M. Wilson [15] in 1968, but had already been settled by Lu Jiaxi [16] in 1965.
147
7. Congruence of Segments, Angles and Triangles This section deals with David Hilbert’s axiomatization of neutral geometry, following Hilbert’s foundations of geometry with some modifications. 7.1.
Congruence of segments
Proposition 7.1 (Congruence is an equivalence relation). Congruence is an equivalence relation on the class of line segments. Question. Which three properties do we need to check for a congruence relation? Answer. For an arbitrary relation to be a congruence relation, we need to check (a) reflexivity: Each segment is congruent to itself, written AB AB. (b) symmetry: If AB A0 B0 , then A0 B0 AB. (c) transitivity: If AB CD, and CD EF, then AB EF. Proof of reflexivity, for a bottle of wine from Hilbert personally. 1 We need to show that each segment is congruent to itself. Take any given segment AB. We transfer the segment to a ray starting from any point C. By Hilbert’s axiom (III.1) , there exists a point D on that ray such that AB CD. Now use Hilbert’s axiom (III.2) : "If A0 B0 AB and A00 B00 AB, then A0 B0 A00 B00 ." Hence, for a case with other notation, AB CD and AB CD imply AB AB. Proof of symmetry. Assume that AB CD. Because of reflexivity CD CD. Once more, we use Hilbert’s axiom (III.2) : "If A0 B0 AB and A00 B00 AB, then A0 B0 A00 B00 ." Hence CD CD and AB CD imply CD AB. Proof of transitivity. Assume AB CD and CD EF. Because of symmetry EF CD. Now we use Hilbert’s axiom (III.2) : "If A0 B0 AB and A00 B00 AB, then A0 B0 A00 B00 ." For a case with other notation, this means that AB CD and EF CD imply AB EF, as to be shown.
1
which one can take gracefully as a thank-you—for translating from page 15 of the millenium edition of "Grundlagen der Geometrie"–etc.
148
Proposition 7.2 (Existence and uniqueness of segment transfer). Given a segment AB and given a ray r originating at point A0 , there exists a unique point B0 on the ray r such that AB A0 B0 . Question. Which part of this statement is among Hilbert’s axioms? Which axiom is used? Answer. The existence of the segment A0 B0 is postulated in Hilbert’s axiom of congruence (III.1). Question. How does the uniqueness of segment transfer follow? Which axioms are needed for that part? Answer. The uniqueness of segment transfer follows from the uniqueness of angle transfer, stated in (III.4), and the SAS Axiom (III.5).
Figure 7.1. Uniqueness of segment transfer
Proof of uniqueness. Assume the segment AB can be transferred to the ray r from A0 in two ways, such that both AB A0 B0 and AB A0 B00 . We choose a point C 0 not on the line A0 B0 . One obtains the congruences A0 B0 A0 B00 , A0C 0 A0C 0 , ∠B0 A0C 0 ∠B00 A0C 0 We are using axiom (III.2), the fact that a segment is congruent to itself, and an angle is congruent to itself by axiom (III.4c). By the axiom (III.5), this implies ∠A0C 0 B0 ∠A0C 0 B00 . By the uniqueness −−−→ −−−−→ of angle transfer, as stated in axiom (III.4b), this implies that the rays C 0 B0 = C 0 B00 are equal. −−−→ −−−→ Hence B0 = B00 is the unique intersection point of the two different rays r = A0 B0 and C 0 B0 . We have shown that AB A0 B0 and AB A0 B00 imply B0 = B00 . Thus segment transfer is unique. Proposition 7.3 (Subtraction of segments). Given are three points on a line such that A ∗ B ∗ C, −−−→ and two points B0 and C 0 on a ray A0 B0 emanating from A0 . Suppose that AB A0 B0 , AC A0C 0 then BC B0C 0 and A0 ∗ B0 ∗ C 0 follow. −−−→ Proof. On the ray originating from B0 , opposite to B0 A0 , we transfer segment BC. We get the seventh point S such that BC B0 S , and A0 ∗ B0 ∗ S . By Hilbert’s axiom (III.3) on segment addition, AB A0 B0 and BC B0 S imply now AC A0 S . On the other hand, it is assumed that AC A0C 0 . Both points S and C 0 lie on the same ray −−0−→0 A B . As explained in Proposition 7.2, the transfer of segment AC is unique. Hence we get C 0 = S . Finally we substitute equals. Thus BC B0 S and A0 ∗ B0 ∗ S imply BC B0C 0 and A0 ∗ B0 ∗ C 0 , as to be shown.
149
Figure 7.2. Segment subtraction
Definition 7.1 (Segment comparison). For any two given segments AB and CD we say that AB is less than CD, iff there exists a point E between C and D such that AB CE and C ∗ E ∗ D. In this case, also say that CD is greater than AB. We write CD > AB or AB < CD. equivalently. Proposition 7.4 (Segment comparison holds for congruence classes). Assuming AB A0 B0 and CD C 0 D0 , we get: AB < CD if and only if A0 B0 < C 0 D0 .
Figure 7.3. Segment comparison for congruence classes
−−→ Proof. Transfer segment AB onto ray CD, and get AB CE. The assumption AB < CD implies −−−→ that C ∗ E ∗ D. Too, we transfer segment A0 B0 onto ray C 0 D0 , and get A0 B0 C 0 E 0 . We now use segment subtraction for points C, E, D and C 0 , E 0 , D0 . Since CD C 0 D0 by assumption, and CE AB A0 B0 C 0 E 0 —by assumption and construction and transitivity—, and C ∗ E ∗ D; segment subtraction yields ED E 0 D0 and C 0 ∗ E 0 ∗ D0 . Thus E 0 lies between C 0 and D0 , from which we conclude A0 B0 < E 0 D0 . Proposition 7.5 (Transitivity of segment comparison). If AB < CD and CD < EF, then AB < EF. −−→ Proof. I can assume that all three segments lie on the same ray AB, and A = C = E. (Some transferring of segments will produce segments congruent with the given ones which satisfy these requirements.) Now, having done this, we get AB < AD and AD < AF. By definition this means A ∗ B ∗ D and A ∗ D ∗ F. Question. Express these order relations in words. Answer. Point B lies between A and D, and point D lies between A and F. By Theorem 5 from Hilbert’s foundations, any four points on a line can be notated in a way that all four alphabetic order relations hold. (I shall that statement the "Four point Theorem"). Now the four points A, B, D, F satisfy the order relations A ∗ B ∗ D and A ∗ D ∗ F, which implies they are already notated in alphabetic order. Hence the two other order relations follow: A ∗ B ∗ F and B ∗ D ∗ F. But A ∗ B ∗ F means by definition that AB < AF, as to be shown. Corollary 6. If AB < AD and AD < AF, then BD < BF.
150
Figure 7.4. Transitivity of segment comparison
Proposition 7.6 (Any two segments are comparable). For any two segments AB and CD, one and only one of the three cases (i)(ii)(iii) occurs: Either (i) AB < CD or (ii) AB CD or (iii) CD < AB. −−→ Proof. Transfer segment AB onto the ray CD. We get AB CB0 . Now the two segments CB0 and CD start at the same vertex and lie on the same ray. By Theorem 4 from Hilbert’s foundations, for any given three points on a line, exactly one lies between the other two. Hence the three points C, B0 , D can satisfy either (i) C ∗ B0 ∗ D. or (ii) B0 = D. or (iii) C ∗ D ∗ B0 . These cases correspond to the three cases as claimed. Indeed, in case (i), C ∗ B0 ∗ D implies by definition AB < CD. Indeed, in case (ii), B0 = D implies AB CB0 = CD by construction. In case (iii), we get CD < CB0 . Since AB CB0 , Proposition 7.4 implies CD < AB, as to be shown.
Figure 7.5. To extend a line—.
−−→ Lemma 7.1 ("simple fact"). Given a segment AB and a ray CD, there exists a point R on this ray such that the segment DR is longer than the segment AB. Proof. By the axiom of order (II.2), there exists a point P such that C ∗ D ∗ P. By the axiom of congruence (III.1), there exists a point Q on the line CD, lying on the same side of D as point P, such that AB DQ. By the axiom of order (II.2), there exists a point R such that P ∗ Q ∗ R. −−→ In the drawing on page 150, the given ray CD is extended to obtain a point P outside the −−→ segment CD. Next we transfer the given segment AB onto the extension ray DP. Finally, the ray −−→ PQ can be extended even more to get a point R outside the segment PQ. The construction has produced points C, D, Q, R on a line satisfying the first two order relations below. We invoke the Four point Theorem, Proposition 5.5. C∗D∗Q and D∗Q∗R imply C∗D∗ R
151
−−→ The third order relation confirms that point R lies on the ray CD. The order relation D∗ Q∗R means that the segment DR is longer than the segment DQ. By construction, the segments AB DQ are congruent. We see that DR > DQ AB By Proposition 7.4, segment comparison holds for congruence classes. We conclude DR > AB, as required. Remark. What does Euclid’s Second Postulate "To extend a line" really mean? A bid of thought shows that this postulate means more than just Hilbert’s axiom of order (II.2). Even extenbding as many times as one wants needs not lead very far. Remember Achilles and the tortoise! But clearly we often need to extend a given line beyond a given segment. This process involves a combination of the axiom of order (II.2) with the axiom of congruence (III.1) for the segment transfer. Another situation occurs when one knows, for any other reason, some point E to exist beyond which one has to extend the line. Even in that situation, we need to invoke the Four-point Theorem (Hilbert’s Proposition 5) for the possibility of further extension beyond E. So we end up with different possible interpretations of Euclid’s postulate! Definition 7.2 (Sum of segments). The sum of any two segments AB and EF is defined to be the segment AC were C is the point such that A ∗ B ∗ C and EF BC. −−→ In other words, the sum segment AB + EF is obtained by extending the ray AB to the point C by a segment BC congruent to the second summand EF. Proposition 7.7 (Congruence classes of segments are an ordered Abelian group). The sum of segments is defined on equivalence classes of congruent segments because of axiom (III.3) and the existence and uniqueness of segment transfer, as shown in Proposition 7.2. These equivalence classes have the following properties: Commutativity a + b = b + a. Associativity (a + b) + c = a + (b + c) Comparison Any two segments a and c satisfy either a < c, or a = c or a > c. Difference a < c if and only if there exists b such that a + b = c. Comparison of Sums If a < b and c < d, then a + c < b + d. Put together with their negatives −a for each congruence class a, and finally the neutral element 0, these equivalence classes are an ordered Abelian group. Proof. By Hilbert’s axiom of congruence (III.3), the sum is defined on equivalence classes of congruent segments. Now we check the items stated: Commutativity: Let segment AB represent the equivalence class a. By axiom (III.1), we can choose point C such that A ∗ B ∗ C, and segment BC represents the equivalence class b. Since BC CB, and AB BA, these latter two segments represent the classes b and a. By the definition of segment addition CA = CB + BA, and the segment CA represents b + a. Since AC CA, we conclude a + b = b + a.
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Associativity: We construct (a + b) + c. To this end, choose a segment AB of congruence class a, in short AB ∈ a. Again, choose point C such that A ∗ B ∗ C, and segment BC ∈ b. Furthermore, choose point D such that A ∗ C ∗ D, and segment CD ∈ c. We now know that AD ∈ (a + b) + c On the other hand, we begin by constructing at first b+c. By the four-point theorem, A∗B∗C and A ∗ C ∗ D imply B ∗ C ∗ D. Hence BD ∈ b + c. The construction of a + (b + c) is finished by finding point H such that A ∗ B ∗ H, and segment BH ∈ b + c. We now know that AH ∈ a + (b + c) The uniqueness of segment transfer implies H = D. Finally, we see that AD = AH, and hence (a + b) + c = a + (b + c). Comparison: Let any segments a and c be given. These equivalence classes can be represented −−→ −−→ by segments AB ∈ a and AC ∈ c on the same ray AB = AC. The three point theorem implies that either B = C, or A ∗ B ∗ C, or A ∗ C ∗ B. Hence either a = c, or a < c, or c < a. In the last case a > c as shown above. Difference: Assume a < c. These equivalence classes can be represented by segments AB ∈ a and AC ∈ c such that A ∗ B ∗ C. Now BC represents the class b such that a + b = c. The converse is as obvious. Comparison of Sums The proof is left to the reader. Proposition 7.8 (Comparison of supplements). Let a segment PQ and two points A, B in it be given. Assume that PA < PB. Then BQ < AQ.
Proof. The reader has to try himself. 7.2.
Some elementary triangle congruences
Definition 7.3 (Triangle, Euler’s notation). For a triangle 4ABC, it is assumed that the three vertices do not lie on a line. I follow Euler’s conventional notation for vertices, sides and angles: in triangle 4ABC, let A, B and C be the vertices, let the segments a = BC, b = AC, and c = AB be the sides and the angles α := ∠BAC, β := ∠ABC, and γ := ∠ACB be the angles. For a segment AB, it is assumed that the two endpoints A and B are different. For a triangle 4ABC, it is assumed that the three vertices do not lie on a line. Definition 7.4 (Congruence of triangles). Two triangles are called congruent iff all three pairs of corresponding sides, and all three pairs of corresponding angles are congruent. Thus the congruence of triangles 4ABC 4A0 B0C 0 means the six congruences AB A0 B0 , BC B0C 0 , CA C 0 A0 and ∠ABC ∠A0 B0C 0 , ∠BCA ∠B0C 0 A0 , ∠CAB ∠C 0 A0 B0
(7.1)
Note that one has to put the vertices, sides and angles of the two triangles into a definite order.
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Definition 7.5 (Isosceles triangle, equilateral triangle, right triangle). A triangle with two congruent sides is called isosceles. A triangle for which all three sides are congruent, is called equilateral. A triangle with at least one right angle is called a right triangle. Problem 7.1. For which geometries does exist an isosceles right triangle? Answer. An isosceles right triangle exists in any Hilbert plane, especially in Euclidean and hyperbolic geometry. In spherical geometry, too, exists an isosceles right triangle. Proposition 7.9 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert’s theorem 11] In a triangle with two congruent sides, the opposite angles are congruent. An isosceles triangle has congruent base angles. Question. Formulate the theorem with specific quantities from a triangle 4ABC like drawn in the figure on page 153. Answer. If b c, then β γ.
Figure 7.6. An isosceles triangle
Proof. Assume that the sides AB AC of triangle 4ABC are congruent. We need to show that the base angles β = ∠ABC and γ = ∠BCA are congruent. Define a second triangle 4A0 B0C 0 by setting 1
A0 := A , B0 := C , C 0 := B To apply SAS congruence, we match corresponding pieces: (1) ∠BAC ∠CAB since by axiom (III.4d), an angle is congruent to the angle with the sides reversed. As the points are defined, we know ∠CAB = ∠B0 A0C 0 . Hence ∠BAC ∠B0 A0C 0 . (2) AB A0 B0 . Question. Explain why this holds. Answer. AB AC because we have assumed the triangle to be isosceles, and AC = A0 B0 by construction. Hence AB AC. 1
It does not matter that the second triangle is just "on top" of the first one.
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(3) Similarly, we show AC A0C 0 : Indeed, AC AB because we have assumed the triangle to be isosceles, and congruence is a symmetric relation, and AB = A0C 0 by construction. Hence AC A0C 0 . Finally, we use axiom (III.5). Items (1)(2)(3) imply ∠ABC ∠A0 B0C 0 = ∠ACB. But this is just the claimed congruence of base angles. Remark. This proof is far simpler than Euclid’s proof for the congruence of the base angles of an isosceles triangle. It is attributed to Pappus of Alexandria, A.D. 300. Hilbert only gives the hint that Axiom (III.4d) and (III.5) are used for the proof.
Figure 7.7. How to simply get ASA congruence
Proposition 7.10 (ASA Congruence). [Theorem 13 in Hilbert] Two triangles with a pair of congruent sides, and pairwise congruent adjacent angles are congruent. Independent proof of ASA congruence. Given are the triangles 4ABC and 4A0 B0C 0 , with congruent sides AB A0 B0 and two pairs of congruent adjacent angles ∠BAC ∠B0 A0C 0 and −−−→ ∠ABC ∠A0 B0C 0 . The segment AC is transferred onto the ray A0C 0 . On this ray, one gets a point X such that AC A0 X (7.2) Now axiom (III.5) is applied to the triangles 4ABC and 4A0 B0 X. In the figure on page 154, the three pairs of matching pieces used to prove the congruence are stressed in matching colors. From axiom (III.5) one concludes ∠ABC ∠A0 B0 X (7.3) We now have obtained both ∠ABC ∠A0 B0 X, and ∠ABC ∠A0 B0C 0 as assumed, too. Hence uniqueness of angle transfer, postulated by axiom (III.4b), implies the rays to be equal: −−0→ −−0−→0 BX=BC −−→ −−−→ Thus the point X lies both on this ray, and by construction on the ray A0 X = A0C 0 , too. By axiom (I.2), these two rays have a unique intersection point, since they do not lie on the same line, and hence X = C 0 . By substitution of equals, equations (7.2) and (40.21) imply AC A0C 0 and ∠ABC ∠A0 B0C 0
(7.4)
Finally we show BC B0C 0 by a similar argument with vertices A and B exchanged. Hence the two triangles are congruent, as to be shown.
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Proposition 7.11 (SAS Congruence). [Theorem 12 in Hilbert] Given are two triangles. We assume that two sides and the angle between these sides are congruent to the corresponding pieces of the second triangle. Then the two triangles are congruent. Short proof of SAS-congruence Proposition 7.11. Given are two triangles 4ABC and 4A0 B0C 0 . We assume that the angles at A and A0 as well as two pairs of adjacent sides are matched: AB A0 B0 , AC A0C 0 , ∠BAC ∠B0 A0C 0 By axiom (III.5) we conclude ∠ABC ∠A0 B0C 0 Now the ASA-congruence stated in Proposition 7.10 above yields the triangle congruence 4ABC 4A0 B0C 0 to be shown. Question. We see from Proposition 7.11 that Hilbert’s SAS-axiom (III.5) implies the full SAS congruence theorem. Why is Hilbert’s Axiom (III.5) weaker than the SAS congruence theorem? Answer. In Hilbert’s Axiom, only congruence of a further pair of angles is postulated. In the SAS congruence theorem, all pieces of the two triangles are stated to be pairwise congruent.
Figure 7.8. Transfer of a triangle.
Proposition 7.12 (Transfer of a triangle). Any given triangle 4ABC can be transferred into any given half-plane H 0 of any given ray r, such that one obtains a congruent triangle 4A0 B0C 0 lying −−−→ in the prescribed half-plane, and the given ray is r = A0 B0 , emanates from vertex A0 and lies on the side A0 B0 . Problem 7.2. Provide a drawing for Proposition 7.12 and prove the Proposition, starting from the axioms and ASA-congruence. State clearly which axioms you use, and where one has to use the ASA-congruence. Proof. Using axiom (III.1), we transfer the segment AB onto the ray r to obtain the congruent segment AB A0 B0 . Next we use axiom (III.4), and transfer the angle ∠BAC into the given halfplane H 0 to obtain the congruent angle ∠BAC ∠(r, k). Thus the newly produced angle has the given ray r as one side. Onto the other side k of the newly produced angle, we transfer the segment −−−→ −−−→ AC and obtain the congruent segment AC A0C 0 . Since by construction r = A0 B0 and k = A0C 0 , we see that ∠BAC ∠B0 A0C 0 . From axiom (III.5)—which we now use even literally!—we conclude the congruence ∠ABC 0 0 0 ∠A B C . We are now in the position to use the ASA-congruence, stated in Proposition 7.10 for the triangles 4ABC and 4A0 B0C 0 . Hence we conclude that the triangle congruence 4ABC 4A0 B0C 0 , as to be shown.
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Figure 7.9. Extended ASA congruence
Proposition 7.13 (Extended ASA-Congruence). Given is a triangle, a segment congruent to one of its sides, and a half-plane H 0 bounded by this segment . The two angles at the vertices of this side are transferred to the endpoints of the segment, and reproduced in the same half plane. Then the newly constructed rays intersect, and one gets a second congruent triangle. Short proof of the extended ASA-congruence Proposition 7.13. Given are the triangle 4ABC −−−→ and the segment A0 B0 AB. We define the ray r = A0 B0 . By Proposition 7.12 about the transfer of a triangle, we transfer the triangle 4ABC into the given half-plane H 0 . We obtain a congruent triangle 4A0 B0C 0 with vertex C 0 lying in the prescribed half-plane. Question. In which point does the extended ASA congruence theorem extend the usual ASA congruence theorem? Answer. The extended ASA theorem differs from the usual ASA-congruence theorem, because existence of a second triangle is not assumed—besides the first triangle, only a second segment is given. It is proved that the two newly produced rays do intersect. Proposition 7.14 (Preliminary Converse Isosceles Triangle Proposition). If the two base angles of a triangle are congruent to each other, the triangle is isosceles. Problem 7.3. • Formulate the proposition 7.14 above with specific quantities from a triangle 4ABC. • Provide a drawing. • Prove the theorem as stated. Proof. I use the formulation: "If β γ and γ β, then b c." I may still use the drawing from page 153. We use ASA-congruence to prove the proposition. Assume that the angles β = ∠ABC and γ = ∠ACB are congruent in 4ABC. We need to show that the two sides b = AC and c = AB are congruent. Define a second triangle 4A0 B0C 0 by setting once more A0 := A , B0 := C , C 0 := B To apply ASA congruence, we match corresponding pieces: (1) BC B0C 0 : Indeed BC = C 0 B0 = B0C 0 , because a segment is congruent to itself and the order of the endpoints of a segment is arbitrary. Hence BC B0C 0 .
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(2) β β0 : Indeed, β γ by assumption, and γ = ∠ACB = ∠A0 B0C 0 = β0 by construction. Hence β β0 . (3) Similarly, one shows that γ γ0 : γ β by assumption, and β = ∠ABC = ∠A0C 0 B0 = γ0 by construction. Hence γ γ0 . Via ASA congruence, items (1)(2)(3) imply that 4ABC 4A0 B0C 0 , and hence especially AB A0 B0 = AC as to be shown. Remark. Later on in Proposition 7.31, we shall avoid the awkward assumption: "If the two base angles of a triangle are congruent to each other". Hilbert proves the converse isosceles triangle theorem as his Theorem 24, after having the exterior angle theorem at his disposal. 7.3. Congruence of angles Recall definition 5.9 of supplementary angles, and definition 5.10 of vertical angles. Remark. After having proved the results below, we may also correctly say that a pair of supplementary angles adds up to two right angles, or 180◦ . But this statement is not an acceptable definition for supplementary angles. Indeed one has produced a bad example for circular reasoning. The notion of right angle can only be defined,—by definition 7.8 below,—to be an angle congruent to its supplementary angle. Such a definition requires that one already knows what supplementary angles are. Remark. Nevertheless, it has become common usage to talk about angles adding to 180◦ as supplementary angles in a relaxed sense. In order to distinguish the relaxed notion from the strict definition 5.9, the term "linear pair" has been suggested for the primary basic notion. That practice is only followed by some authors, but neither by Hilbert nor by Hartshorn. Proposition 7.15 (Congruence of Supplementary Angles). [Theorem 14 of Hilbert] If an angle ∠ABC is congruent to another angle ∠A0 B0C 0 , then its supplementary angle ∠CBD is congruent to the supplementary angle ∠C 0 B0 D0 of the second angle. Proof. The three steps of the proof each identify a new pair of congruent triangles. Step 1: One can choose the points A0 , C 0 and D0 on the given rays from B such that AB A0 B0 , CB C 0 B0 , DB D0 B0 Because of the assumption ∠ABC ∠A0 B0C 0 , SAS congruence (Theorem 12 of Hilbert, see Proposition 7.11 above) now implies that 4ABC 4A0 B0C 0 . In the drawing, the three pairs of matching pieces used to prove the congruence are stressed in matching colors. Congruence of the
Figure 7.10. Congruence of supplementary angles, the first pair of congruent triangles
two triangles implies AC A0C 0 and
∠BAC ∠B0 A0C 0
(1)
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Step 2: By axiom (III.3), adding congruent segments yields congruent segments. Hence the segments AD and A0 D0 are congruent. Now SAS congruence (Theorem 12 of Hilbert, see Proposition 7.11 above) implies that the (greater) triangles, too, are congruent in the two figures below. In the drawing, the three pairs of matching pieces used to prove the congruence are stressed in matching colors. The congruence 4CAD 4C 0 A0 D0 implies
Figure 7.11. Congruence of supplementary angles, the second pair of congruent triangles
CD C 0 D0 and
∠ADC ∠A0 D0C 0
(2)
Step 3: At last we consider the two triangles 4BCD and 4B0C 0 D0 on the right side. In the drawing, the three pairs of matching pieces used to prove the congruence are stressed in matching colors. Again by using SAS congruence (Theorem 12 of Hilbert, see Proposition 7.11 above), we see the
Figure 7.12. Congruence of supplementary angles, the third pair of congruent triangles
two triangles are congruent. Finally ∠CBD ∠C 0 B0 D0 , as to be shown.
Corollary 7. Adjacent angles congruent to supplementary angles are supplementary, too.
Figure 7.13. Supplementary angles yield points on a line
Proof. Given are supplementary angles ∠ABC and ∠DBC, a further congruent angle ∠ABC ∠A0 B0C 0 , and a point D0 with A0 and D0 lying on different sides of line B0C 0 . We shown that the angles ∠A0 B0C 0 and ∠D0 B0C 0 are supplementary if and only if ∠CBD 0 0 0 ∠C B D . Above, we have already shown one direction: If the angles ∠A0 B0C 0 and ∠D0 B0C 0 are supplementary, then ∠CBD ∠C 0 B0 D0 . Now we show the converse. Assume that ∠CBD ∠C 0 B0 D0 . We have to check whether −−−→ point B0 lies between A0 and D0 . Choose any point D00 on the ray opposite to B0 A0 . Congruence of supplementary angles (Hilbert’s Theorem 14, see Proposition 7.15 above) implies ∠CBD
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∠C 0 B0 D00 . On the other hand, ∠CBD ∠C 0 B0 D0 is assumed. Angle ∠CBD is transferred uniquely −−→ along ray B0C into the half plane opposite to A0 . Indeed, by axiom III.4, angle transfer produces a −−−−→ −−−→ unique new ray. Hence B0 D00 = B0 D0 . and the four points A0 , B0 , D0 , D00 lie on one line, with point B0 between A0 and D0 . Hence angles ∠A0 B0C 0 and ∠D0 B0C 0 are supplementary, too. Proposition 7.16 (Congruence of Vertical Angles). [Euclid I.15] Vertical angles are congruent. Proof. This is an easy consequence of Hilbert’s Theorem 14 about supplementary angles (see Proposition 7.15 above). Take any two vertical angles ∠ABC and ∠A0 BC 0 . We assume that vertex B lies between points A and A0 on one line, as well as between the two points C and C 0 on a second line. As shown in the figures, angle ∠ABC 0 has angle (i) ∠ABC as supplementary angle. Secondly,
Figure 7.14. Two pairs of supplementary angles yield vertical angles
angle ∠ABC 0 has (ii) ∠A0 BC 0 as supplementary angle, too. An angle is congruent to itself, as stated in Axiom (III.4). Hence especially ∠ABC 0 ∠ABC 0 . By Theorem 14 of Hilbert (see Proposition 7.15 above), angles supplementary to congruent angles are congruent, too. Hence we conclude congruence of the two vertical angles: ∠ABC ∠A0 BC 0 , as to be shown. −−→ Proposition 7.17 (The main case of angle-addition). Given is an angle ∠ABC and a ray BG in 0 0 0 0 0 its interior, as well as a second angle ∠A B C . We assume that points A and C lie in different half −−−→ planes of ray B0G0 . Furthermore, we assume that ∠CBG ∠C 0 B0G0 and ∠ABG ∠A0 B0G0 . −−−→ Then the ray B0G0 lies in the interior of angle ∠A0 B0C 0 , and the two angles ∠ABC ∠A0 B0C 0 are congruent. Proof. By the Crossbar Theorem, a segment going from one side of an angle to the other, and a −−→ −−→ ray in the interior of that angle always intersect. Hence there exists a point H such that BG = BH and A ? H ? C. For simplicity, we choose G = H from the beginning. Too, we may assume that A0 , C 0 and G0 are chosen such that BA B0 A0 , BC B0C 0 and BG B0G0 . The proof uses three pairs of congruent triangles, in step (1)(2)(4), respectively. Step (3) is needed to show A0 ∗ G0 ∗ C 0 −−−→ and the ray B0G0 lies in the interior of angle ∠A0 B0C 0 . Step (1): The SAS-congruence axiom implies 4BGC 4B0G0C 0 because of ∠GBC ∠G0 B0C 0 , GB G0 B0 and BC B0C 0 which hold by assumption and the remarks above. In the drawing, the three pairs of matching pieces used to prove the congruence are stressed in matching colors. From the congruence of these two triangles, we conclude that
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Figure 7.15. Angle addition
Figure 7.16. the first pair of congruent triangles
∠BCG ∠B0C 0G0 ∠BGC ∠B0G0C 0
(1) (1b)
Step (2): As a second step, the SAS-congruence axiom implies 4BGA 4B0G0 A0 because of ∠GBA ∠G0 B0 A0 , GB G0 B0 and BA B0 A0 which hold by assumption and the remarks above. Again, in the drawing, the three pairs of matching pieces used to prove the congruence are stressed in matching colors. −−−→ Step (3): It is assumed that points A0 and C 0 lie in different half planes of ray B0G0 . Hence the angles ∠B0G0 A0 and ∠B0G0C 0 are adjacent angles. They are congruent to the supplementary angles ∠BGA and ∠BGC. The congruence ∠BGA ∠B0G0 A0 follows from step (2), and ∠BGC ∠B0G0C 0 follows from step (1). As derived in our Corollary 7, adjacent angles congruent to supplementary angles are supplementary, too. Hence the two angles ∠B0G0 A0 and ∠B0G0C 0 are supplementary. Hence point G0 lies −−−→ between points A0 and C 0 , the three points A0 , G0 and C 0 lie on a straight line, and the ray B0G0 lies in the interior of angle ∠A0 B0C 0 . Step (4): By segment addition (Hilbert’s axiom (III.3)), AG A0G0 and GC G0C 0 , and A ∗ G ∗ C, A0 ∗ G0 ∗ C 0 imply AC A0C 0 (2) To set up a third pair of congruent triangles, we use (1) (2) and the assumption BC B0C 0
(3)
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Figure 7.17. the second pair of congruent triangles
The SAS axiom shows that 4ABC 4A0 B0C 0 because of (1)(2)(3). I have stressed these three pairs of matching pieces in matching colors. Finally, ∠ABC ∠A0 B0C 0 follows from 4ABC 4A0 B0C 0 ,
Figure 7.18. the third pair of congruent triangles
−−→ Proposition 7.18 (The main case of angle subtraction). Given is an angle ∠ABC and a ray BG − − − → in its interior, as well as a second angle ∠A0 B0C 0 and a ray B0G0 in its interior. If ∠CBG ∠C 0 B0G0 and ∠ABC ∠A0 B0C 0 then ∠ABG ∠A0 B0G0 .
as to be shown.
Proposition 7.19 (Angle-Addition and Subtraction). [Theorem 15 in Hilbert] Given are three rays h, k, l with the common vertex B lying in a plane a, and three rays h0 , k0 , l0 with the common vertex B0 lying in a plane a0 . There are the cases of angle subtraction or angle addition. angle addition: in this case the rays h and k lie in the different half-planes of l, and the rays h0 and k0 lie in different half-planes of l0 . angle subtraction: in this case the rays h and k lie in the same half-plane of l, and the rays h0 and k0 lie in the same half-plane of l0 ; Assume either angle addition or angle subtraction occurs and ∠(h, l) ∠(h0 , l0 )
and
∠(l, k) ∠(l0 , k0 ),
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Then, either ∠(h, k) ∠(h0 , k0 ) or both h and k are opposite rays, and h0 and k0 , too, are opposite rays.
Figure 7.19. Three subcases for angle addition.
Proof. We explain only the case of angle addition. As shown in the figure on page 162, we led to three different subcases. (i) the two rays k and l lie in one half-plane of ray h; (ii) the two rays k and l lie in different half-planes of ray h; (iii) the rays h and k are opposite rays. In the special case (iii), the angles ∠(h, l) and ∠(l, k) are adjacent supplements. Hence by Corollary 7, the congruent angles ∠(h0 , l0 ) and ∠(l0 , k0 ) are supplementary, too. In other words we get opposite rays h0 and k0 . We consider the two main cases. In the first case (i), we use Proposition 7.17 about the main −−−→ case of angle-addition. As see in this Proposition, A0 ∗ G0 ∗ C 0 and the ray l0 = B0G0 lies in the −−−→ −−−→ interior of angle ∠A0 B0C 0 . Hence the two rays l0 = B0G0 and k0 = B0C 0 lie in one half-plane of ray h0 . The claim ∠(h, k) ∠(h0 , k0 ) follows by angle addition as in Proposition 7.17. In the second case (ii), we use the opposition rays l− of l and l−0 from l0 . Now Proposition 7.15 about the congruence of supplementary angles implies ∠(h, l− ) ∠(h0 , l−0 )
and
∠(l− , k) ∠(l−0 , k0 )
Since we are back to case (i), we can conclude the claim ∠(h, k) ∠(h0 , k0 ). The case of angle subtraction does not lead to subcases, and can be left to the reader.
Proposition 7.20 (Transfer of three rays). Given are three different rays h, k, l with the common vertex B lying in a plane a, and two rays h0 , k0 with the common vertex B0 lying in a plane a0 . We assume that the three rays h, k and l lie on three different lines. Finally, we assume the angles ∠(h, k) ∠(h0 , k0 ) to be congruent. Then there exists a unique ray l0 lying in the plane a0 , emanating from the vertex of ∠(h0 , k0 ) such that ∠(h, l) ∠(h0 , l0 ) and ∠(l, k) ∠(l0 , k0 ) (7.5) This ray l0 lies in the interior of angle ∠(h0 , k0 ) if and only if the ray l lies in the interior of angle ∠(h, k).
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Proof. We begin with the case that the ray l lies in the interior of angle ∠(h, k). The common vertex of the rays h, k, l is the point B. We choose any points A on the ray h and C on the ray k. By the Crossbar Theorem, there exists the intersection point G of segment AC with the ray of l. −−−→ We transfer the triangle 4ABC onto the ray B0 A0 = k0 , into the half plane containing ray k0 . By the assumption ∠(h, k) ∠(h0 , k0 ) and the uniqueness of angle transfer, the point C 0 lies on the ray k0 . −−−→ We transfer the triangle 4ABG onto the ray B0 A0 = k0 , into the half plane containing ray k0 to −−−→ obtain the congruent triangle such that 4ABG 4A0 B0G0 . Thus we obtain the new ray l0 := B0G0 . −−→ −−−→ The uniqueness of angle transfer implies A0G = A0C 0 . Moreover, A ∗ G ∗ C implies A0 ∗ G0 ∗ C 0 . We see that the new ray l0 lies in the interior of angle ∠(h0 , k0 ), and the claimed congruences ∠(h, l) ∠(h0 , l0 ) and ∠(l, k) ∠(l0 , k0 ) indeed hold. It is left to the reader to check the uniqueness of the ray l0 . In the case that the ray l does not lies in the interior of angle ∠(h, k), it lies in the interior of either the vertical angle of one of the two supplementary angles of angle ∠(h, k). Here we have used the assumption that the three rays h, k, and l lie on three different lines. By the argument above, we obtain the ray l0 , and we see that it lies in in the interior of either the vertical angle of one of the two supplementary angles of angle ∠(h0 , k0 ). The congruences (7.5) can now be obtained by means of Proposition 7.15 about the congruence of supplementary angles. 7.4. SSS congruence The "Theorem about the symmetric kite" is needed as a preparation to get SSS congruence. Definition 7.6 (The Symmetric Kite). A symmetric kite or simply kite is a quadrilateral XZ1 YZ2 , where Z1 and Z2 are two points on different sides of diagonal XY, and XZ1 XZ2 and YZ1 YZ2 . Proposition 7.21 (The Symmetric Kite). [Theorem 17 of Hilbert] Let Z1 and Z2 be two points on different sides of line XY, and assume that XZ1 XZ2 and YZ1 YZ2 . Under these assumptions (i) the angles ∠XZ1 Y ∠XZ2 Y are congruent; (ii) the angles ∠XYZ1 ∠XYZ2 are congruent; (iii) the triangles 4XYZ1 4XYZ2 are congruent. Proof. The congruence of the base angles of isosceles 4XZ1 Z2 yields ∠XZ1 Z2 ∠XZ2 Z1 . Similarly, one gets ∠YZ1 Z2 ∠YZ2 Z1 . By plane separation, the segment Z1 Z2 intersects the line XY. Let F be the point of intersection. According to the three-point theorem, we distinguish the cases (a) X ∗ F ∗ Y (b) X ∗ Y ∗ F (c) Y = F (d) F ∗ X ∗ Y (e) X = F By angle addition in case (a) or substraction in cases (b) and (d), respectively, we conclude that ∠XZ1 Z2 ∠XZ2 Z1 and ∠YZ1 Z2 ∠YZ2 Z1 imply ∠XZ1 Y ∠XZ2 Y
(*)
−−−→ Angle addition, as proved in Proposition 7.17 is needed in case ray Z1 Z2 lies both inside the angle ∠XZ1 Y and the angle ∠XZ2 Y. Angle subtraction from Proposition 7.18 is used for the case that ray −−−→ Z1 Z2 lies both outside the ∠XZ1 Y and the ∠XZ2 Y.
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Figure 7.20. The symmetric kite
In the special cases (c) and (e) that either point X or Y lies on the line Z1 Z2 , one gets the same conclusion even easier. Here are drawings for the three cases (a),(b) and (c). One now applies SAS congruence to the triangles 4XZ1 Y and 4XZ2 Y. Indeed, the angles at Z1 and Z2 and the adjacent sides are pairwise congruent. Hence the assertion ∠XYZ1 ∠XYZ2 follows from axiom (III.5). Moreover, SAS congruence theorem 7.11 yields the triangle congruence 4XYZ1 4XYZ2 . Before getting the general SSS-congruence, I consider one further special case. Lemma 7.2 (Lemma for SSS Congruence). [Euclid I.7] Assume that the two triangles 4ABC and 4AB0C have a common side AC, and all three corresponding sides are congruent, and the two vertices B and B0 lie on the same side of line AC. Then the two triangles are identical. −−→ Proof. We transfer the angle ∠BAC onto the ray AC, on the side of line AC opposite to B and B0 . On the newly produced ray, we transfer segment AB, starting at vertex A. Thus we get point B00 , and segment AB00 AB. From SAS-congruence (Theorem 12 of Hilbert, see Proposition 7.11 above), one concludes that BC B00C. As stressed in the first drawing, one has constructed a symmetric kite with the four points X := A, Y := C, Z1 := B, Z2 := B00 Hence by Theorem 17 (see Proposition 7.21 above), ∠B00 AC ∠BAC
Figure 7.21. Which kite is symmetric?
But wait! Another choice of four points to get a kite is X := A, Y := C, Z1 := B0 , Z2 := B00
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—replacing B by B0 . Because of the assumptions, and the construction of the first kite, AB0 AB AB00 and B0C BC B00C. We see that the second kite, stressed in the second drawing, satisfies the assumptions of Hilbert’s Theorem 17 (see Proposition 7.21 above), too. Now we conclude from Hilbert’s Theorem 17 (see Proposition 7.21 above) that ∠B00 AC ∠B0 AC. Finally, −−→ −−→ the uniqueness of angle transfer implies that AB = AB0 . Since AB AB0 , and segment transfer was shown to produce a unique point, we have confirmed that B = B0 . Thus the two triangles 4ABC and 4AB0C are identical, as to be shown. Remark. We do not need to assume that angle congruence is an equivalence relation. But therefore we have to consider two kites, by choosing two of the three triangles, as shown in two drawings. Then we can use that transferring an angle gives a unique ray. From Hilbert’s Theorem 17 (see Proposition 7.21 above), and the Lemma 7.2 above, we see that SSS-congruence holds for any two triangles with a common side. Now we can easily get the general case of Proposition 7.22 (SSS Congruence). [Theorem 18 in Hilbert’s Foundations] Two triangles with three pairs of congruent sides are congruent. End of the proof of SSS-congruence. Assume the two triangles 4ABC and 4A0 B0C 0 have corre−−−→ sponding sides which are congruent. We transfer the angle ∠BAC onto the ray A0C 0 , at vertex A0 , to the same side of A0C 0 as B0 . On the newly produced ray, we transfer segment AB, starting at vertex A0 . Thus we get point B0 , such that A0 B0 AB. Because of SAS-congruence (Theorem 12 of Hilbert, see Proposition 7.11 above), we get 4ABC 4A0 B0C 0
(*)
Hence especially, AB A0 B0 A0 B0 and BC B0C 0 B0C 0 . We can now apply the Lemma‘7.2 to the two triangles 4A0 B0C 0 and 4A0 B0C 0 . Hence B0 = B0 , and the assertion follows from (*). Proposition 7.23 (The diagonals of the rhombus bisect each other perpendicularly). Given are four different points A, B, C, D lying in one plane such that the segments AB BC CD DA are congruent. Then the segments AC and BD bisect each other perpendicularly at their common midpoint. Answer. Points A and C lie on different sides of line BD. Otherwise the Lemma for SSS-congruence would imply A = C, contrary to the assumption that four different points A, B, C, D are given. Similarly, we see that points B and D lie on different sides of the other diagonal AC. Hence the two segments AC and BD intersect. Let point M be their intersection point. We draw the horizontal diagonal AC. The upper- and lower triangles have two pairs of congruent bases angles (see second figure). Angle addition implies ∠ABD ∠CBD. By SAS congruence, the left- and right triangles 4ABD 4CBD are congruent (see third figure). Furthermore, the latter two triangles are both isosceles. Hence 4ABD 4CDB holds, too, and ∠ABD ∠CDB (see first figure in the second row). Similarly, we prove ∠CAB ∠ACD. (Why?) By ASA congruence, we get 4MAB 4MCD (see second figure in the second row). Hence the diagonals AC and BD bisect each other (see last figure in the second row). Since the isosceles triangle 4ABC has congruent base angles, SAS congruence yields 4MAB 4MCB. Hence the angles ∠AMB ∠CMB are congruent supplements, and hence they are right angles. Hence the diagonals AC and BD are perpendicular to each other.
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Figure 7.22. The diagonals of the rhombus bisect each other.
Figure 7.23. Completing Euclid’s proof of (I.7).
Problem 7.4. Provide the missing parts to complete Euclid’d proof of (I.7), in the case he −−→ considers. As shown in the drawing on page 166, we assume that the ray AD is in the interior of angle ∠CAB, and point D is outside the triangle 4ABC. We need to confirm the following items: −−→ (i) the ray DA lies in the interior of angle ∠CDB; −−→ (ii) the ray CB lies in the interior of angle ∠ACD. Use the axioms of order, the Crossbar Theorem, and so on, to provide a detailed justification.
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−−→ Answer. By the Crossbar Theorem, the ray AD intersects the segment BC, say in point Q. Hence −−→ B ∗ Q ∗ C and A ∗ Q ∗ D. Since B ∗ Q ∗ C and point Q lies on the ray DA, this ray lies in the interior −−→ of angle ∠CDB. Since A ∗ Q ∗ D and point Q lies on the ray CB, this ray lies in the interior of angle ∠ACD.
Figure 7.24. Completing Euclid’s proof of (I.7).
Problem 7.5. We now finish Euclid’d proof of (I.7), in the case he considers. The triangles 4ABC and 4ABD are assumed to lie on the same side of line AB. Moreover, as shown in the drawing on −−→ page 166, we assume that the ray AD is in the interior of angle ∠CAB, and point D is outside the triangle 4ABC. Show that the congruences AC AD and BC BD cannot both hold. Show that these congruences would yield two isosceles triangles. Get from the previous problem 7.4 their base angles would satisfy both α < β and β < α,—which is impossible. Nevertheless, provide a drawing to explain this impossible case. Answer. The assumed congruences yields two isosceles triangles: 4ACD with congruent base angles α, and 4BCD with congruent base angles β. Angle comparison at vertex D yields α < β −−→ since by item (i), the ray DA lies in the interior of angle ∠CDB β. −−→ Angle comparison at vertex C yields β < α since by item (ii), the ray CB lies in the interior of angle ∠ACD α. 7.5.
Right, acute and obtuse angles
Proposition 7.24 (Theorem 19 in Hilbert’s Foundations). If two angles ∠(h0 , k0 ) and ∠(h00 , k00 ) are congruent to a third angle ∠(h, k), then the two angles ∠(h0 , k0 ) and ∠(h00 , k00 ) are congruent, too. Proof for a bottle of wine from Hilbert to A. Rosenthal. 1 Let the vertices of the angles be O, O0 , O00 . Choose points A, A0 , A00 on one side of the three angles, respectively, such that O0 A0 OA and O00 A00 OA. Similarly choose points B, B0 , B00 on the three remaining sides, respectively, such that O0 B0 OB and O00 B00 OB. Now the assumption of SAS congruence (Theorem 12 of Hilbert, see Proposition 7.11 above) are met for both 4A0 O0 B0 and 4AOB, as well as 4A00 O00 B00 and 4AOB. Hence A0 B0 AB , A00 B00 AB 1
Hilbert gives credit for this proof to A. Rosenthal (Math. Ann. Band 71)
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We use axiom (III.2) : "two segments congruent to a third one are congruent to each other". Hence 4A0 B0 O0 and 4A00 B00 O00 have three pairs of congruent sides. By the SSS-congruence Theorem 18, we conclude that ∠(h0 , k0 ) ∠(h00 , k00 ), as to be shown. Proposition 7.25 (Congruence is an equivalence relation). Congruence is an equivalence relation on the class of angles. Question. Which three properties do we need to check for a congruence relation? Answer. For an arbitrary relation to be a congruence relation, we need to check (a) reflexivity: Each angle is congruent to itself, written α α. (b) symmetry: If α β, then β α. (c) transitivity: If α β, and β γ, then α γ. Question. How can we claim reflexivity? Answer. Reflexivity is given by the last part of axiom (III.4) Proof of symmetry, for a bottle of wine from A. Rosenthal to Hilbert. 1 By Theorem 19 (see Proposition 7.24 above), ∠(h0 , k0 ) ∠(h, k) and ∠(h00 , k00 ) ∠(h, k) imply ∠(h0 , k0 ) ∠(h00 , k00 ). Hence, with just other notation, we see that β β and α β imply β α. Proof of transitivity. Assume α β and β γ. Because of symmetry γ β. Now we use Theorem 19 (see Proposition 7.24 above). Hence, just with other notation, α β and γ β imply α γ. Definition 7.7 (Angle comparison). Given are two angles ∠BAC and ∠B0 A0C 0 . We say that ∠BAC −−→ is less than ∠B0 A0C 0 , iff there exists a ray A0G in the interior of ∠B0 A0C 0 such that ∠BAC ∠B0 A0G. In this case, we also say that ∠B0 A0C 0 is greater than ∠BAC. We write ∠B0 A0C 0 > ∠BAC and ∠BAC < ∠B0 A0C 0 , equivalently. Given is a ray r with vertex O and a half-plane H of this ray. We may compare any given angles α and β, by transferring them onto the ray r, with their second side in the half plane H. We obtain the congruent angles ∠(r, a) α and ∠(r, b) β. Because of the postulate (III.4), the relative position of the rays a and b can be used for comparison of the angles α and β. Lemma 7.3. In the above situation, no matter how the ray r and the half plane H are chosen, the following equivalence holds: α < β holds if and only if the ray a lies in the interior of the angle ∠(r, b). As a consequence, one obtains the following propositions. Proposition 7.26 (Angle comparison holds for congruence classes). Assume that α α0 and β β0 . α < β if and only if α0 < β0 . Proof. The reader should do it on her own. 1
Hilbert gave credit and got back a good bottle
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Proposition 7.27 (Transitivity of angle comparison). If α < β and β < γ, then α < γ. Proof. After having done some transfer of angles, I can assume that all three angles α, β, γ have −−→ the common side AB, and lie on the same side of AB. Choose any point E on the second side of γ = ∠BAE. Because β < γ, the second side of β is in the interior of the largest angle γ. Hence, by the Crossbar Theorem the segment BC intersects that second side of β, say at point D, and β = ∠BAD as well as B ∗ D ∗ E. Because α < β, the second side of α is in the interior of β . Hence, by the Crossbar Theorem the segment BD intersects that second side of α, say at point E, and α = ∠BAE as well as B ∗ C ∗ D. Any four points on a line can be ordered in a way that all four alphabetic order relations hold. (see
Figure 7.25. Transitivity of comparison of angles
Theorem 5 in Hilbert, which I call the four-point Theorem 5.5). Now the four points B, C, D, E satisfy the order relations B ∗ C ∗ D and B ∗ D ∗ E. Therefore they are already put in alphabetic order. Hence B ∗ C ∗ E . This shows by definition that α = ∠BAC < ∠BAE = γ.
Figure 7.26. The case (i) where α < β.
Let s be the ray opposite to ray r. As already explained in lemma 5.3, plane separation by the ray b leads to the following cases to be distinguished: (i) either the ray a lies in the interior of angle ∠(r, b) and the ray b lies in the interior of angle ∠(s, a), as shown in the figure on page 169;
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(ii) or the ray b lies in the interior of angle ∠(r, a) and the ray a lies in the interior of the angle ∠(s, b), as shown in the figure on page 170; (iii) or the two rays a and b are equal.
Figure 7.27. The case (ii) where β < α.
We have obtained the following result: Proposition 7.28 (All angles are comparable). For any two angles α and β, one and only one of the three cases (i)(ii)(iii) occurs: Either (i) α < β or (ii) β < α or (iii) α β. With r and s opposite rays, the above figures contain the supplementary pairs α ∠(r, a) and S (α) ∠(s, a) β ∠(r, b) and S (β) ∠(s, b) Hence lemma 5.3 yields the result Proposition 7.29 (Comparison of supplements). If α < β, then their supplements S (α) and S (β) satisfy S (α) > S (β). Definition 7.8 (right angle). A right angle is an angle congruent to its supplementary angle. Proposition 7.30 (All right angles are congruent (Proclus 410-485 A.D.)). All right angles are congruent Question. How is a right angle defined? Answer. A right angle is an angle congruent to its supplementary angle. Proof in the conventional style. Let α be a right angle, and β = S (α) α be its congruent supplement. Similarly, we consider a second pair of right angles α0 and β0 = S (α0 ) α0 being its congruent supplement. The question is whether both right angles α and α0 are congruent to each other. By Proposition 7.28 all angles are comparable. Hence the two angles α and α0 satisfy just one of the following relations.
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Either (i) α < α0
or (ii) α0 < α
or (iii) α α0 .
We need to rule out cases (i) and (ii) by deriving a contradiction. I only need to explain case (i), because (ii) is similar. We now assume α < α0 towards a contradiction. By the comparison of supplements from Proposition 7.29, we get S (α) > S (α0 ). Because a right angle is congruent to its supplement we get α S (α) > S (α0 ) α0 . As explained in Proposition 7.26, angle comparison holds for congruence classes. Hence we get α > α0 . Thus, from the assumption α < α0 , we have derive α > α0 . Transitivity holds by proposition 7.27 and implies α < α. But axiom (III.4c) states that an angle is congruent to itself, and hence α α. By proposition 7.28, all angles are comparable. It cannot happen that both α < α and α α would hold. Thus case (i) leads to a contradiction. Similarly, case (ii) leads to a contradiction. The only remaining possibility is case (iii): Any two given right angles α and α0 are congruent. Definition 7.9 (acute, right and obtuse angles). A right angle is an angle congruent to its supplementary angle. An acute angle is an angle less than a right angle. An obtuse angle is an angle greater than a right angle. Remark. If an angle α is acute, its supplement S (α) is obtuse. Furthermore, the angle is less than its supplement. If an angle β is obtuse, its supplement S (β) is acute. Furthermore, the angle is less than its supplement. Reason. Assume α < R, where R denotes a right angle. From the comparison of supplements done in Proposition 7.29, we conclude S (α) > S (R). But by the definition of a right angle R S (R). Hence S (α) > S (R) R, and because comparison is for congruence classes, we get S (α) > R. Hence α < R < S (α), and by transitivity (Problem 10.2), we conclude α < S (α). Similarly, we explain that β > R implies S (β) < R and S (β) < β. Proposition 7.31 (Converse Isosceles Triangle Proposition). [Euclid I.6, Theorem 24 of Hilbert] A triangle with two congruent angles is isosceles. Question. Formulate the theorem with specific quantities from a triangle 4ABC. Answer. If α β, then a b. Proof. Given is a triangle 4ABC with α β. By Hilbert’s Theorem 19 (see Proposition 7.24 above) and proposition 7.25, congruence of angles is an equivalence relation. Hence α β implies β α. In the preliminary version given as Proposition 7.14, we have shown that α β and β α together imply a b. Hence the 4ABC is isosceles, as to be shown. Proposition 7.32 (The Hypothenuse Leg Theorem). Two right triangles for which the two hypothenuse, and one pair of legs are congruent, are congruent. Question. Of which one of SAS, SSA, SAA, ASA, SSS congruence contains the hypothenuse-leg theorem as a special case? Is there a corresponding unrestricted congruence theorem? Answer. The hypothenuse-leg theorem is a special case of SSA congruence. There is no unrestricted SSA congruence theorem.
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Figure 7.28. First proof of the hypothenuse leg theorem
First proof of the hypothenuse-leg theorem 7.32 . Given are two right triangles 4ABC and 4A0 B0C 0 . As usual, we put the right angles at vertices C and C 0 . We assume congruence of the hypothenuses AB A0 B0 and of one pair of legs AC A0C 0 . We build a kite out of two copies of triangle 4ABC and two copies of triangle 4A0 B0C 0 . To this −−→ end, one transfers angle ∠C 0 A0 B0 onto the ray AC into the opposite half plane and gets by transfer of triangle (see Proposition 7.12) the congruence 4C 0 A0 B0 4CAB00 . Especially B0C 0 B00C
(7.6)
Since all right angles are congruent (see Proclus’s Proposition 7.30), the two supplementary right angles at vertex C imply that points B, C and B00 lie on a line. −−→ Next we transfer segment CA onto the ray opposite to CA and get point D such that CA CD. We produce the two triangles to the right, as shown in the second row of the figure on page 172. Using the right angle, SAS congruences imply 4ABC 4DBC and 4AB00C 4DB00C. Since AB A0 B0 AB00 and DB AB AB00 DB00 , one has constructed a symmetric kite with the four points X := A, Y := D, Z1 := B, Z2 := B00 We can apply Hilbert’s kite-theorem 7.21 and conclude ∠DAB = ∠XYZ1 ∠XYZ2 = ∠DAB00 The congruence of the same angles is ∠CAB = ∠CAB00 . Hence SAS congruence implies now 4CAB 4CAB00 and BC B00C (7.7) Together, the formulas (7.6) and (7.7) imply B0C 0 BC. Since all right angles are congruent, a final SAS congruence using the right angles implies congruence 4ABC 4A0 B0C 0 of the originally given triangles, as to be shown.
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7.6.
Constructions with Hilbert tools
Definition 7.10 (Hilbert tools). By incidence axiom (I.1) and (I.2), we can draw a unique line between any two given points. By axiom (III.1) and Proposition 7.2 from the very beginning, we transfer a given segment uniquely to a given ray. Finally by axiom III.4, we transfer a given angle uniquely on a given ray into the specified half plane. Constructions done using only these means are called constructions by Hilbert tools. Now we show that a right angle actually exists, and do some further basic constructions with the Hilbert tools. The only means of construction are those granted by Hilbert’s axioms. Problem 7.6 (Drop a Perpendicular). Given is a line OA and a point B not on this line. We have to drop the perpendicular from point B onto line OA. −−→ Construction 7.1 (Dropping a Perpendicular). Draw ray OB. Transfer angle ∠AOB, into the −−→ half plane opposite to B, with ray OA as one side. On the newly produced ray, transfer segment OB to produce a new segment OC OB. The line BC is the perpendicular, dropped from point B onto line OA.
Figure 7.29. Drop the perpendicular, the two cases
Proof of validity. Question. Why do the line OA and the segment BC intersect.
Answer. Line OA and the segment BC intersect because points B and C lie on different sides of line OA. I call the intersection point M.
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Question. It can happen that O = M. How does one proceed in this special case. Answer. In the special case that O = M, we know that A , M and ∠AMB ∠AMC
(**)
But these is a pair of congruent supplementary angles, because M lies between B and C. By definition, an angle congruent to its supplementary angle is a right angle. Hence ∠AMB is a right angle. In the generic situation O , M, we distinguish two cases: (i) points M and A lie on the same ray with vertex O; (ii) point O lies between M and A. In both cases we show the two triangles 4OMB and 4OMC are congruent. Indeed, the angles ∠MOB ∠MOC at vertex O are congruent, both in case (i) and (ii). Question. In case (i), one gets this congruence directly from the construction since the rays −−→ −−→ OA = OM are equal. How does one proceed in case (ii).
Figure 7.30. Congruences needed in the two cases.
−−→ −−→ Answer. In case (ii), the rays OA and OM are opposite. Thus ∠MOB and ∠AOB, as well as ∠MOC and ∠AOC are supplementary angles. Again ∠AOB ∠AOC because of the angle transfer done in the construction. Now Theorem 14 of Hilbert (see Proposition 7.15 above) tells that supplements of congruent angles are congruent. Hence we get ∠MOB ∠MOC once again. Question. Finish the argument showing the triangle congruence 4MOB 4MOC. Answer. The adjacent sides OB OC are congruent by construction, too. Finally the common sides OM is congruent to itself. Hence the congruence follows by SAS congruence. The drawing stresses the pieces matched to prove the congruence. Because the two triangles are congruent, the corresponding angles at vertex M are congruent, too. Hence ∠OMB ∠OMC (*) which is a pair of congruent supplementary angles.
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Question. How is a right angle defined. How do we see that ∠OMB is a right angle. Answer. A right angle is, by definition, an angle congruent to its supplementary angle . Hence ∠OMB is a right angle because of formula (*). Corollary 8. A right angle exists and is unique. Hence it deserve a special name. Proof. By axiom (I.3b) there exist three points A, B, C not lying on a line. We use the construction 7.1 to drop the perpendicular from point C onto the line AB. At the foot point of the perpendicular, we obtain a right angle. It has been proved in lemma 7.30 that all right angles are congruent (Proclus 410-485 A.D.) Remark. Traditionally, we put a right angle equal to 90◦ . But note that such a convention is only allowable after one has proved that a right angle exists and is unique. Proposition 7.33 (Existence of an Isosceles Triangle). For any given segment AB, on a given side of line AB, there exists an isosceles triangle. Remark. Of course, the isosceles triangle one has obtained, is not unique. Construction 7.2 (Construction of an isosceles triangle). Given is a segment AB. Choose any point P not on line AB, in the half plane specified. Compare the two angles ∠BAP and ∠ABP. In −−→ the drawing, ∠BAP is the smaller angle. Transfer the smaller angle, with ray BA as one side, and the newly produced ray rB in the same half plane as P. Ray rB and segment AP do intersect. The intersection point C lies in the half plane as required, and 4ABC is isosceles.
Figure 7.31. Construction of an isosceles triangle
Proof of validity. By Proposition 7.28, all angles are comparable. Comparison of the two angles ∠BAP and ∠ABP leads to one of the three possibilities: Either (i) The two angles are congruent
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or (ii) α = ∠BAP < ∠ABP = β or (iii) α = ∠BAP > ∠ABP = β. It is enough to consider cases (i) and (ii). In case (i) let C = P. In case (ii) we have transferred −−→ the smaller angle ∠BAP, onto ray BA, and the newly produced ray rB in the same half plane as P. By the Crossbar Theorem, a segment going from one side of an angle to the other, and a ray in the interior of an angle always intersect. By the Crossbar Theorem, ray rB and segment AP intersect. The intersection point C lies between A and P, hence both points C and P lie in the same half plane of line AB. Hence point C lies in the half plane as required. By construction, the base angles of triangle 4ABC are congruent. Hence the triangle is isosceles by Euclid I.6 (Converse Isosceles Triangles). Remark. One may suggest to make it part of the construction how to compare the two angles ∠BAP and ∠ABP. One possibility is to transfer both angles to the other vertex, B or A, respectively. Only the newly produced ray from the smaller angle does intersect the opposite side of 4ABP. Problem 7.7 (Erect a Perpendicular). Given is a line l and a point R on this line. We have to erect the perpendicular at point R onto line l. Construction 7.3. Choose any point A , R on line l. Transfer segment AR onto the ray opposite to −−→ RA to get a segment RB AR. As explained in Construction 7.2, construct any isosceles triangle 4ABC. The line CR is the perpendicular to the given line l through point R.
Figure 7.32. Erect a perpendicular
Reason for validity. The two triangles 4RAC and 4RBC are congruent by SAS congruence. Indeed, ∠RAC ∠RBC by the construction of the isosceles triangle. Furthermore, we have a pair of congruent adjacent sides: AR BR by the construction above, and AC BC because 4ABC is isosceles. Now the triangle congruence 4RAC 4RBC implies that ∠ARC ∠BRC. But these two angles are supplementary angles. Because congruent supplementary angles are right angles, we have confirmed that ∠ARC is a right angle, as to be shown.
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Definition 7.11 (The perpendicular bisector). The line through the midpoint of a segment and perpendicular to the segment is called the perpendicular bisector. Construction 7.4 (Construction I of the perpendicular bisector). Construct two different isosceles triangles over segment AB. The line connecting the third vertices of the two isosceles triangles is the perpendicular bisector.
Figure 7.33. The perpendicular bisector via the kite
Proof of validity. Let 4ABC be the first isosceles triangle, and 4ABD be the second one. Question. Why do the points A and B lie on different sides of CD? We need just to use the Lemma 7.2 to SSS-congruence! Answer. If points A and B would lie on the same side of line CD, the congruence 4ACD 4BCD would imply A = B, contradicting to the endpoints of a segment being different. Because A and B lie on different sides of line CD, the segment AB intersects the line CD, say at point M. Now we have the symmetric kite ACBD with two congruent triangles (left and right in the figure). By the kite Theorem (Proposition 7.21 above), we conclude the congruence 4ACD 4BCD. Of the three points M, C, D exactly one lies between the two others. We can assume that C does not lie between D and M, this assumption can be achieved by possibly interchanging the names C and D. Question. Give a detailed reason for the triangle congruence 4ACM 4BCM. Answer. Indeed, the triangles 4ACM and 4BCM have congruent angles at the common vertex C, and two pairs of congruent adjacent sides AC BC and MC MC. Now we get by SAS congruence that 4ACM 4BCM. Matching pieces are stressed in the drawing.
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Figure 7.34. Get the perpendicular bisector via: (a) a convex kite, (b) a nonconvex kite
Question. Use the triangle congruence 4ACM 4BCM to confirm that line CD is the perpendicular bisector of the given segment AB. Answer. From the last triangle congruence we get ∠AMC ∠BMC. Because M lies between A and B, these are two congruent supplementary angles. Hence they are right angles. Too, the triangle congruence implies AM BM. Hence M is the midpoint of segment AB. We have shown that the segment AB and the line CD intersect perpendicularly at the midpoint of segment AB. Hence CD is the perpendicular bisector. Remark. There are the possibilities of an (a) convex kite, or (b) a non convex kite. (a): If the two isosceles 4ABC and 4ABD lie on different sides of AB, points C and D are on different sides of AB, and it is clear that the segment CD intersects the line AB. (b): Too, it is possible to use two different isosceles triangles on the same side of AB. To get the second isosceles triangle 4ABD, one transfers as base angles any two congruent angles which are less than the base angles of the first triangle 4ABC. The figure ACBD is still a symmetric—but non convex—kite. As proved in Hilbert’s kite Proposition 7.21, the two triangles left and right in the figure are still congruent: 4ACD 4BCD. The line CD still intersects the segment AB, because points A and B are on different sides of CD, but the two segments AB and CD do not intersect each other. Corollary 9. The perpendicular bisector of any segment is unique. A point lies on the perpendicular bisector of segment AB if and only if its distances AP and BP to the endpoints of the segment are congruent. Proof. Since the midpoint of a segment is unique, and all right angles are congruent by Proclus’ proposition 7.30, the uniqueness of angle transfer by axiom (III.4) implies uniqueness of the perpendicular bisector. Given any point P for which AP BP. If P = M turns out to be the midpoint of segment AB, the distances AP BP are congruent. Assume that M , P. In that case, points A, B and M do not lie on a line since the midpoint is unique. One can use the isosceles triangle 4ABP for the construction of the unique perpendicular bisector p. Hence point P lies on the bisector. Conversely given any point P , M on the perpendicular bisector p. We may use SAS congruence to show the triangle congruence 4AMP 4BMP. Indeed, they both have a right
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angles at vertex M, and the adjacent sides MP MP and MA MB are pairwise congruent. Hence AP congBP, as claimed. Problem 7.8. Given any plane α and two points A and B in it. Give an independent proof showing that any point of the plane with congruent distances to these two points lies on the perpendicular bisector of segment AB.
Figure 7.35. A point with congruent distances to both endpoints lies on the perpendicular bisector.
Proof. Assume point P lies in the plane α and has congruent distances AP BP to the endpoints of the segment. Let M be the midpoint of segment AB. If P = M, we are ready since M is a point on the perpendicular bisector. Hence we may assume P , M. The congruence 4AMP 4BMP follows from SSS congruence . Indeed, PM PM since a segment is congruent to itself, AP BP by assumption, and AM BM holds since M is the midpoint of AB. Hence ∠AMP ∠BMP are congruent supplementary angles, and hence right angles. Since the right angle is unique by Proposition 7.30, and angle transfer produces a unique ray by axiom (III.4), the point P lies on the perpendicular bisector, as to be shown. Construction 7.5 (Construction II of the perpendicular bisector— the way one really wants it). Construct any isosceles triangle over the given segment. Drop the perpendicular from its third vertex. Problem 7.9. Give a short proof of validity for construction II of the perpendicular bisector. Use the construction done above and the hypothenuse leg Theorem 7.32 Short proof of validity of construction II. Let 4ABC be the isosceles triangle constructed at the first step. Next we drop the perpendicular from point C onto line AB. As explained in construction 7.1, we get the reflected point D and the perpendicular CD. The foot point M is the intersection of segment CD and line AB. Question. Use the hypothenuse leg Theorem to confirm the triangle congruence 4AMC 4BMC.
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Figure 7.36. Construction of the perpendicular bisector—the natural way
Answer. We know from construction 7.1 above that the angle ∠AMC is a right angle, and hence ∠AMC ∠BMC. We know from construction 7.2 of an isosceles triangle that AC BC. Too, the segment MC MC is congruent to itself. Hence the hypothenuse leg Theorem 7.32 yields the triangle congruence 4AMC 4BMC. This triangle congruence implies that the segments AM BM are congruent. This is impossible for a point on the line AB outside the segment AB. Hence point M lies between A and B, and is the midpoint of segment AB. Thus we have confirmed that CD is the perpendicular bisector of segment AB, as to be shown. Definition 7.12 (The angular bisector). The ray in the interior of an angle which bisects the angle into two congruent angles is called the angular bisector. Construction 7.6 (Construction of the angular bisector). One transfers two congruent segments AB and AC onto the two sides of the angle, both starting from the vertex A of the angle. The perpendicular, dropped from the vertex A onto the segment BC, is the angular bisector.
Figure 7.37. The angular bisector
Question. Reformulate the description of this construction, including the details for all steps.
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Answer. We transfer congruent segments AB AC onto the sides of the given angle. Draw the −−→ line BC, and transfer the base angle ∠ABC to the ray BC, on the side of line BC opposite to vertex −−→ A. On the new ray, we transfer segment AB to get the new segment BD BA. The ray AD is the bisector of the given angle ∠BAC. Short proof of validity. By assumption, the three points A, B, C do not lie on a line. Let M be the foot point of the perpendicular dropped from point A onto line BC. Question. Confirm the triangle congruence 4AMB 4AMC.
Figure 7.38. The essential pair of congruent triangles
Answer. Indeed ∠AMB ∠AMC R, because a right angle is congruent to its supplement. It does −−→ not matter whether M lies on the ray BC or the opposite ray. Too, we have by construction the pair of congruent sides AB AC. Finally AM AM. The hypothenuse leg Theorem 7.32 yields the triangle congruence 4AMB 4AMC. −−→ Question. Explain why the ray AM lies inside the given ∠BAC and bisects this angle. Answer. From the triangle congruence 4AMB 4AMC, we get ∠MAB ∠MAC, and MB MC. Since point M lies on the line BC, the last congruence shows that M lies between B and C, too. −−→ Hence ray AM lies inside the given angle ∠BAC and bisects this angle. −−→ Proposition 7.34 (Existence of the angular bisector). For any angle ∠BAC, there exists a ray AD inside the given angle such that ∠DAB ∠DAC. 7.7. Remarks about angles We have already defined supplementary angles (see Definition 5.9.) Moreover, a right angle is defined to be an angle congruent to its supplement (see Definition 7.8.) Because we have a valid construction 7.1 to drop a perpendicular from a point onto a line, we know a right angle really exists. Moreover, we have proved that all right angles are congruent (see Proposition 7.30.) Hence it is legitimate to introduce a special notation. I use either the capital letter R, or the conventional 90◦ . Such a convention can neither replace the geometric Definition 7.8, nor the proofs of existence and uniqueness! −−→ Definition 7.13 (The sum of two angles). Let an angle ∠ABC and a ray BG in its interior. Angle ∠ABC is called the sum of the angles ∠ABG and ∠GBC. One writes ∠ABC = ∠ABG + ∠GBC.
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Figure 7.39. Three cases for adding angles: they (i) are supplementary added
(ii) can be added
(iii) or cannot be
Lemma 7.4. Given are two angles ∠ABG and ∠GBC with same vertex B lying on different sides of −−→ a common ray BG. Exactly one of three possibilities occur about the angles ∠ABG and ∠GBC: (i) They are supplementary. The three points A, B and C lie on a line. (ii) They can be added. Their sum is ∠ABG + ∠GBC = ∠ABC. Points A and G lie on the same side of line BC. Points C and G lie on the same side of line AB. (iii) They cannot be added. Points A and G lie on different sides of line BC. Points C and G lie on different sides of line AB. (The sum would be an over-obtuse angle.)
Figure 7.40. If A and G lie on different sides of BC, then C and G lie on different sides of AB.
Proof. Suppose that neither case (i) nor (ii) occurs. Under that assumption, either (a) points A and G lie on different sides of line BC—or (b) points C and G lie on different sides of line AB.
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Suppose case (a) occurs. Segment AG intersects line BC, say in point H. We use Pasch’s axiom for triangle 4CHG and line AB. This line does not intersect side GH, but intersects side CH. Indeed, point B lies between C and H, since C and A, and hence C and H are on different sides of line BG. Now Pasch’s axiom—for triangle 4CHG and line AB—implies that this line intersects side CG, say at point Pasch. Thus points C and G lie on different sides of line AB. We have shown case (iii) to occur. Suppose case (b) occurs. The same argument—with A and C exchanged— shows that case (iii) occurs once more. Remark. Because of these theorems, angle addition and subtraction are defined for congruence classes of angles. Proposition 7.35 (About sums of angles). The sum of angles is defined on equivalence classes of congruent angles. It satisfies the following properties: Commutativity If α + β exists, then β + α exists and β + α α + β. Associativity (α + β) + γ α + (β + γ) and existence of one side implies existence of the other one. Difference α < γ if and only if there exists β such that α + β γ. Comparison of Sums 1 If α β and γ < δ and β + δ exists, then α + γ exists and α + γ < β + γ. Comparison of Sums 2 If α < β and γ < δ and β + δ exists, then α + γ exists and α + γ < β + γ. Proof. By Hilbert’s Theorem 15 and Proposition 7.25, the sum of angles is defined on equivalence classes of congruent angles. Now we check the items stated: Commutativity: Let α ∠ABG and β ∠GBC where points G and C lie both on the same side of line AB. With that setup, α + β ∠ABC, which is assumed to exist. Since the order of the two rays of an angle is defined to be chosen arbitrarily, ∠ABC = ∠CBA ∠CBG + ∠GBA β + α hence the latter angle exists and β + α α + β. Associativity: The proof is left to the reader. Difference: Let α ∠ABG and γ ∠ABC where points G and C lie both on the same side of −−→ line AB. With that setup, α < γ iff the ray BG lies inside the angle ∠ABC iff α + β γ with β ∠GBC. Comparison of Sums 1: Assuming that α β and γ < δ and β + δ exists, we know there exists ε such that γ + ε δ. Hence (α + γ) + ε α + (γ + ε) β + δ and α + γ < β + γ where the former is shown to exist.
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Comparison of Sums 2: Assuming α < β and γ < δ, we know there exist angles η and ε such that α + η β and γ + ε δ. Hence (α + γ) + η + ε (α + η) + (γ + ε) β + δ and α + γ < β + γ where the former is shown to exist. Problem 7.10. Explain why the angular bisectors of two supplementary angles are perpendicular to each other.
Figure 7.41. Bisecting a pair of supplementary angles yields perpendicular rays.
Answer. Take any pair of supplementary angles, for instance α = ∠ABC 0 and β = ∠ABC in the first figure on page 184. −−→ −−→ We bisect these two angles, and draw the vertical angle ∠A0 BC. Opposite rays BE and BE 0 are −−→ the bisectors of the vertical angles α = ∠ABC 0 and angle ∠A0 BC. Let BD be the bisector of the supplement β = ∠ABC. Both angles ∠EBD and its supplement ∠DBE 0 are the sum of angles α/2 and β/2. Hence they are congruent. Thus they are a pair of congruent supplementary, and hence right angles, as to be shown. Proposition 7.36 (Supplements of acute and obtuse angles). Let R denote a right angle. For any angle γ and its supplement S (γ) exactly one of the following three cases occurs: Either (1) or (2) or (3). (1) γ < R, S (γ) > R and γ < S (γ) (2) γ R, S (γ) R and γ S (γ) (3) γ > R, S (γ) < R and γ > S (γ). Proof. All angles are comparable (see Proposition 7.28). For angle γ and the right angle R, exactly one of the three cases holds: Either (i) γ < R or (ii) γ R or (iii) γ > R. From SAS congruence (Theorem 12 of Hilbert, see Proposition 7.11 above), we know that α β implies S (α) S (β). From Proposition 7.29 about comparison of supplements, we know that α < β implies S (α) > S (β). Hence with the help of Propositions 7.25 and 7.26 we get:
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in case (i), one concludes S (γ) > S (R) R > γ, and hence (a) γ < S (γ), in case (ii), one concludes S (γ) S (R) R γ, and hence (b) γ S (γ), in case (iii), one concludes S (γ) < S (R) R < γ, and hence (c) γ > S (γ). By Proposition 7.28 above, all angles are comparable. Hence the two angles γ and its supplement S (γ) satisfy either (a) or (b) or (c). This observation allows one to get the converse of the conclusions shown above. For example, (a) excludes both (ii) and (iii), and hence implies (i). Thus we get (a) implies (i) , (b) implies (ii) , (c) implies (iii); and finally (a) if and only if (i) , (b) if and only if (ii) , (c) if and only if (iii). This leads to the mutually exclusive cases (1) (2) (3), as originally stated. Corollary 10 (All right angles are congruent). Proof. The proposition 7.36 about supplements of acute and obtuse angles yields an easy proof that all right angles are congruent: Let R be the right angle as has been constructed in Construction 7.1. Now suppose γ = R0 is another right angle. By definition of a right angle, this means that γ S (γ). Hence case (b) above occurs, which implies (ii): γ R. You see that existence of a right angle make proving its uniqueness a bit easier. Proposition 7.37 (Uniqueness of angular bisector). The angular bisector is unique. −−→ −−→ Proof. Suppose both rays BG and BG0 bisect angle ∠BAC. The angles α := ∠ABG and α0 := ∠ABG0 are comparable, because all angles are comparable by Proposition 7.28. Suppose towards a contradiction that α < α0 . As specified in Proposition 7.35, one can add inequalities of angles. Hence we conclude that ∠ABC = ∠ABG + ∠GBC α + α < α0 + α0 ∠ABG0 + ∠G0 BC ∠ABC
(?)
which is impossible. Similarly, the case α > α0 can be ruled out. Hence α α0 , and hence by −−→ −−→ uniqueness of angle transfer BG = BG0 , as to be shown. 7.8.
Orientated angles
Definition 7.14 (Orientated angle). An orientated angle is an ordered pair (h, k) of two rays h and k with a common vertex. It is allowed that h and k are equal or opposite rays. For emphasis, I denote the orientated angle by x (h, k). Let an orientated angle between two lines p and q be an ordered pair (p, q). Definition 7.15 (Degenerate angle). An orientated angle x (h, k) is called degenerate if h and k are either equal or opposite rays. These degenerate angles are the zero and the straight angle. An orientated angle x (h, k) is called non-degenerate if the two rays h and k do not lie on the same line. We agree to allow degenerate angles only for orientated and unwound angles, but not for common angles without orientation. As shown by the Orientation Theorem 5.1, in every ordered incidence plane, it is possible to define and fix an orientation. In every ordered incidence plane, it is possible to define an orientation. The orientation is fixed, as soon as one has agreed which are the left and right half plane for one fixed ray.
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Definition 7.16 (Two orientations for angles). Two orientated non-degenerate angles x (h, k) and x (h0 , k0 ) have the same orientation if and only if either • ray k lies in the left half-plane of ray h, and ray k0 lies in the left half-plane of ray h0 (they are both counter clockwise), or • ray k lies in the right half-plane of ray h, and ray k0 lies in the right half-plane of ray h0 (they are both clockwise). Two orientated non-degenerate angles x (h, k) and x (h0 , k0 ) have the opposite orientation if and only if either • ray k lies in the left half-plane of ray h, and ray k0 lies in the right half-plane of ray h0 , or • ray k lies in the right half-plane of ray h, and ray k0 lies in the left half-plane of ray h0 . Definition 7.17 (Congruence of orientated angles). Two orientated angles x (h, k) and x (h0 , k0 ) are called congruent in either one of the following cases • h = k and h0 = k0 , or • h and k as well as h0 and k0 are opposite rays, or • they are congruent in the usual sense ∠(h, k) ∠(h0 , k0 ) and the orientated angles x (h, k) and x (h0 , k0 ) have the same orientation. Problem 7.11. We have seen in Proposition 7.25 that the congruence of ordinary angles is an equivalence relation. Convince yourself that the congruence of orientated angles in an equivalence relation, too. We denote the congruence class of right angles by R. The positively orientated angles take values between 0 and 2R. The straight angle x (k, k0 ) where k and k0 are opposite rays takes by convention the value 2R. The negatively orientated angles take values between −2R and 0. We have seen in Proposition 7.26 that the comparison of ordinary angles is well defined for congruence classes. Problem 7.12. Convince yourself of the following facts: • The congruence of orientated angles is an equivalence relation. • All zero angles x (h, h) are congruent. • All straight angles x (h, k) with two opposite rays h and k are congruent. • The orientated angles x (h, k) and x (k, h) are congruent if and only if the two rays h and k are either equal or opposite. • There exist right angles in two different orientations. • Only orientated right angles with the same orientation are congruent. The sum of two orientated angles is defined simpler than the sum of common angle from definition 7.13. Definition 7.18 (The sum of orientated angles). Any two orientated angles x (h, k) and x (k, l) have the sum x (h, k)+ x (k, l) = x (h, l) The operation of addition is compatible with congruence. Proposition 7.38. The addition of congruence classes of orientated angles is well-defined. With this addition, the congruence classes of orientated angles are an Abelian group. This is not a free group since 4R = 0. Hence it is neither an ordered group. Proposition 7.39. The group of orientated angles is isomorphic to the group of rotations around a fixed point.
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7.9.
The exterior angle theorem
Definition 7.19 (Exterior angle of a triangle). The exterior angle of a triangle is the supplement of an interior angle. Proposition 7.40 (The Exterior Angle Theorem). [Euclid I.16. Theorem 22 in Hilbert] The exterior angle of a triangle is greater than both nonadjacent interior angles. −−→ Proof. For the given 4ABC, we can choose point D on the ray opposite to AB, such that AD CB. We compare the two nonadjacent interior angles γ = ∠ACB and β = ∠ABC to the exterior angle δ = ∠CAD. As a first step, we show Lemma 7.5. The exterior angle δ = ∠CAD is not congruent to the interior angle ∠ACB = γ. Proof of Lemma‘7.5. Assume towards a contradiction that δ γ. The supplementary angle of δ is α = ∠CAB. The supplementary angle of γ is ∠ACE, with −−→ a point E on the ray opposite to CB. Supplementary angles of congruent angles are congruent (Theorem 14 in Hilbert, see Proposition 7.15 above). Hence ∠CAB ∠ACE
(?1)
On the other hand, we use SAS-congruence for the two triangles 4ABC and 4A0 B0C 0 with
Figure 7.42. The impossible situation of a congruent exterior angle
A0 := C , B0 := D , C 0 := A Those two triangles would be congruent by SAS congruence. Indeed the angles at C and C 0 are congruent by our assumption γ = ∠ACB ∠CAD = δ. Too, the adjacent sides are pairwise congruent because of CB AD = C 0 B0 by construction, and CA AC = C 0 A0 by axiom (III.1b). 1 From Axiom (III.5), we conclude that the corresponding angles at A and A0 = C are congruent: ∠CAB ∠C 0 A0 B0 = ∠ACD We have shown that angle ∠CAB is congruent to both (?1) ∠ACE and (?2) ∠ACD. 1
I would prefer to add the item (III.1b) AB BA to Hilbert’s axioms.
(?2)
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−−→ Now we use the uniqueness of angle transfer. Transferring angle ∠CAB with one side CA, into −−→ the half plane not containing B, yields as second side of the angle once the ray CE and the second −−→ time CD. Hence these two rays are the same. Hence points E and D both lie on the line la := BC. In other words, the four points B, C, D, E all lie on one line. Because B and D lie on opposite rays with vertex A, these two points are different: B , D. 2 Because a line is uniquely specified by two of its points, this implies la = BC = BD. We have just seen that point C lies on this line. By construction, point A lies on the line BD, too. Hence all the points A, B, C, D, E lie on the same line. This contradicts the definition of a triangle. By definition of a triangle, its three vertices do not lie on one line. This contradiction confirms the original claim δ γ. Remark. Given are any three different points A, B, C not lying on a line. We choose point D on −−→ −−→ the ray opposite to AB, such that AD CB. We choose the point E on the ray opposite to CB for which AB CE. The assumption that the exterior angle δ = ∠CAD is congruent to the interior angle ∠ACB = γ has the following consequences: • one gets congruent triangles 4ABC 4CDA; • D=E • the points B, A, D lie on a line; • the points B, C, D lie on a line. Remark. No contradiction arises from the assumption δ γ in the cases that either (a) or (b) hold: (a) D , B, but the line DB is not unique; (b) D = B. Remark (Cutting a slice of orange in a strange way). In spherical geometry, the case (a) is possible. The figure constructed does exist. We do not get a contradiction, the conclusion is just that the two points B and D are antipodes, but still D , B. The two congruent triangles
Figure 7.43. Two ways from B to D
4ABC 4A0 B0C 0 = 4CDA are obtained from each other by a rotation of 180◦ around the midpoint M of segment AC. The vertices A and C lie on two different lines BAD and BCD through these two antipodes. Too, the sum of the segments is congruent going both ways from B to D. Hence BC DA and CD AB, and hence the sums BC + CD DA + AB are congruent. Both turn out to be 180◦ . The union of the triangles 4ABC 4A0 B0C 0 = 4CDA has the shape as the peel of a slice of orange, which is cut along a line through M. This line need not be perpendicular to neither one of the great circles BAD nor BCD. 2
Note that just this simple remark is not true in elliptic geometry!
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Remark. In double elliptic geometry, the case (b) is possible. A model of double elliptic geometry can be obtained by identifying antipodes on the sphere. The next Lemma rules out the possibility δ < γ. Lemma 7.6. The exterior angle δ = ∠CAD is not smaller that the interior angle ∠ACB = γ. Proof of Lemma 7.6. Suppose towards a contradiction that δ < γ. We transfer the exterior angle
Figure 7.44. The impossible situation of a smaller exterior angle
−−→ δ = ∠CAD with one side CA, into the half plane containing B. The second side of the angle we get −−→ is a ray CB0 inside the angle ∠ACB. By the Crossbar Theorem, it meets the segment AB in a point B0 . We can now apply the first Lemma 7.5 to 4AB0C. This smaller triangle would have exterior angle δ congruent to the interior angle γ0 = ∠ACB0 . This is impossible by Lemma 7.5. Thus we have both ruled out the possibility that δ = γ, or that δ < γ. By Proposition 7.28, all angles are comparable. Hence there remains only the possibility that δ > γ. Lemma 7.7. The exterior angle δ β = ∠ABC.
=
∠CAD is greater than the interior angle
Proof of Lemma 7.7. We compare angle β = ∠ABC to the exterior angle δ = ∠CAD. Choose any
Figure 7.45. Comparing the other nonadjacent angle
−−→ point F on the ray opposite to AC. The vertical angles δ = ∠CAD ∠FAB = δ0 are congruent by Euclid I.15. Now we are back to the case already covered, because the interior angle β = ∠ABC and exterior angle ∠FAB lie on opposite sides of triangle side AB, which is also part of one side of each of these angles. By Lemma 7.5 and 7.6, we conclude ∠FAB > ∠ABC and hence δ > β.
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Remark. Here is an other way to prove Lemma 7.7: Define 4A3 B3C3 by setting A3 := A, B3 := C, C3 := B. Now we can apply Lemma7.5 and 7.6 to this new triangle and get δ0 = δ3 > γ3 = β. By congruence of the vertical angles δ = ∠CAD ∠FAB = δ0 we get δ > β. Thus the proof of the exterior angle theorem is finished.
Below I give a second proof of the exterior angle Theorem, following Euclid’s ideas. This proof is presupposing a valid construction to bisect a segment. To avoid a circular reasoning, there are two possibilities: • Either one is following Hilbert’s line of reasoning. He proves at first the exterior angle Theorem by the arguments on Lemmas 7.5, 7.6, 7.7 above, and afterwards gives the construction 7.7 below for the bisection of a segment. The justification of his method of bisection presupposes the exterior angle theorem. • One may at first give valid constructions for an isosceles triangle, and the perpendicular bisector of a segment. This line of reasoning is based on the converse isosceles triangle proposition 7.31, already proved above, in a way different from Hilbert’s proof. Moreover, one needs the construction 7.2 of an isosceles triangle as well as the construction 7.5 of the perpendicular bisector. After the validity of these steps is provided, one may uses Euclid’s idea explained below to prove the exterior angle theorem.
Figure 7.46. A proof of the exterior angle theorem presupposing bisection of a segment.
Proof of the exterior angle Theorem following Euclid’s ideas. In the given triangle 4ABC, we −−→ bisect the side AC and let M be the midpoint. We extend the ray BM and construct on the extension a point F such that MB MF. As one easily checks via SAS congruence, one obtains congruent triangles 4MCB 4MAF (7.8) Indeed, these triangles have congruent vertical angles at vertex M , and by construction two pairs of congruent sides MC MA and MB MF. From the triangle congruence (7.8) we obtain congruent angles γ := ∠ACB ∠CAF. −−→ We choose any point D on the ray opposite to AB. Since by construction A ∗ M ∗ C and −−→ B ∗ M ∗ F, all three points C, M and F lie on the same side of line AB. Hence the ray AF
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lies in the interior of the angle δ := ∠CAD, which is an exterior angle of triangle 4ABC. One obtains γ ∠CAF < δ ∠CAD. As to be shown, the exterior angle δ is indeed greater than the nonadjacent interior angle γ. Proposition 7.41. [Euclid I.17] The sum of any two interior angles of a triangle is less than two right angles. 7.10.
Congruence of z-angles
Definition 7.20 (Alternate interior angles or z-angles). Let a transversal t intersect two lines a and b at the points A and B. A pair of alternate interior angles or simply z-angles are two angles −−→ −−→ α := ∠(AB, rA ) and β := ∠( BA, rB ) They have vertices A and B. One of their respective sides are the rays rA ⊆ a and rB ⊆ b lying on the lines a respectively b, and lying in opposite half-planes of the transversal t. Their other sides −−→ −−→ are the rays AB and BA, which both contain the segment AB. Proposition 7.42 (Congruent z-angles imply parallels). [Euclid I.27] If two lines form congruent z-angles with a transversal, they are parallel. Proof. This is an immediate consequence of the exterior angle theorem. We argue by contradiction. Suppose the two lines would intersect in point C. One of the z-angles is an exterior angle of triangle 4ABC, the other one is a nonadjacent interior angle. Lemma 7.5 does imply that the two z-angles are not congruent— contradicting the assumption from above. Hence the two lines a and b cannot intersect.
Figure 7.47. A pair of z-angles.
Remark. Referring to the pair of z-angles shown in the figure on page 191 the Proposition 7.42 tells that αβ⇒akb This is equivalent to its contrapositive a∦b⇒α β which relates even more directly to the exterior angle theorem.
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As a consequence of Euclid I.27, the existence of a parallel can be proved in neutral geometry. One parallel to l through point P is conveniently constructed as "double perpendicular". Proposition 7.43 (Existence of a parallel). For every line l and for every point P lying not on l, there exists at least one parallel m to l through point P. Proof. Given is line l and a point P not on l. One drops the perpendicular from point P onto line l and denotes the foot point by F. Next, one erects at point P the perpendicular to line PF. Thus, one gets a line, which we call m. Because the two lines l and m form congruent z-angles with the ← → transversal PF, Euclid I.27 or I.28 imply that l and m are parallel. This leaves open the question whether or not m is the unique parallel to line l through point P. Remark. Similar to the exterior angle theorem, these are facts of neutral geometry. They are is valid both in Euclidean and hyperbolic geometry. Note that Euclid I.2 through I.28 are theorems that hold for every Hilbert plane. Indeed Euclid I.29 is the first theorem in Euclid’s elements that uses the Euclidean parallel postulate. Warning. The converse of Euclid I.27 is does not hold in hyperbolic geometry. Indeed, in hyperbolic geometry, parallels can form non-congruent z-angles with any transversal. Recall that the Euclidean Parallel Postulate 3.3 postulates both existence and uniqueness of the parallel to a given line through a given point. On the other hand, Hilbert’s Parallel Postulate for plane geometry 3.4 postulates only the uniqueness of the parallel to a given line through a given point. Because of Proposition 7.43, Euclid’s and Hilbert’s parallel postulate turn out to be equivalent for any Hilbert plane. In other word, we can state the definition equivalent to definition 2.2: Definition 7.21 (Pythagorean plane). A Pythagorean plane is a Hilbert plane for which the Euclidean parallel postulate holds. Proposition 7.44 (In a Pythagorean plane, parallels imply congruent z-angles). [Euclid I.29] In a Pythagorean plane, any two parallel lines form congruent z-angles with any transversal. Referring to the pair of z-angles shown in the figure on page 191 akb⇒αβ −−→ −−→ Proof. We transfer the angle α := ∠(AB, rA ) onto the ray BA into the same half plane as the ray rB −−→ −−→ from angle β := ∠( BA, rB ). We obtain a new ray s such that α ∠( BA, s). Now the line a and the line on ray s form congruent z-angles with the transversal AB. According to proposition 7.42 they are parallel. On the other hand, the lines a and b are parallel by assumption. Finally, the line a has a unique parallel through point B since the plane is assumed to be Pythagorean. Hence we conclude that the line b and the line on ray s are equal. Since the rays rB and s lie both on the line b, have vertex B and lie in the same half-plane of AB, they are equal. We conclude that as to be shown −−→ −−→ α ∠( BA, s) = ∠( BA, rB ) β Proposition 7.45 (A Hilbert plane where parallel lines are a means to transport of an angle, is Pythagorean). Assume that in the Hilbert plane to be considered, any two parallel lines form congruent z-angles with any transversal. Then for every line l and for every point P lying not on l, there exists a unique one parallel m to l through point P. In other words, the plane is Pythagorean.
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Remark. By the force of proposition 7.44 and 7.45, we get the following insight: Only in a Pythagorean plane can parallel lines be used as a means for the transport of an angle. Corollary 11. In any Hilbert plane are equivalent (i) Hilbert’s Parallel Postulate 3.4; (ii) the Euclidean Parallel Postulate] 3.3. 7.11.
Consequences of the exterior angle theorem
Proposition 7.46 (Immediate consequences of the exterior angle theorem). (i) Every triangle can have at most one right or obtuse angle. (ii) The base angles of an isosceles triangle are acute. (iii) The foot point of a perpendicular is unique. (iv) Given a line l and a point O not on l. At most two points of l have the same distance from O. (v) A circle and a line can intersect in at most two points. Proof. (i): Suppose angle α of 4ABC is right or obtuse: α ≥ R. By Proposition 7.36, its supplement is right or acute: S (α) ≤ R. But the supplement is an exterior angle: δ = S (α). By the exterior angle theorem, the two other nonadjacent interior angles of the triangle are less than that exterior angle. Hence they are acute: β, γ < δ = S (α) ≤ R, and hence β, γ < R. (ii): The two base angles being congruent, by (i), they cannot be both right or obtuse. (iii): Given a line l and a point O not on l. Suppose there are two perpendiculars, with foot points F and G. The 4OFG would have two right angles, contradicting (i). (iv): Given a line l and a point O not on l. Suppose towards a contradiction there are three points
Figure 7.48. No three points have the same distance from a line
A, B, C on the line l such that OA OB OC. Of any three points, one lies between the two others (Theorem 4 of Hilbert). We may suppose by renaming that A ∗ B ∗ C.
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The base angle α of triangle 4OBC is an exterior angle of the other triangle 4OAB. Hence the exterior angle theorem yields γ > α. But the role of the two triangles can be switched: The base angle γ of triangle 4OAB is an exterior angle of the other triangle 4OBC. Hence the exterior angle theorem yields α < γ. Both inequalities together are impossible. This contradiction implies there can exist at most two points on a line which have congruent distances from point O. (v) For the case that the center O of the circle does not lie on the line, the statement is confirmed by item (iv) above. For the case that the center O of the circle lies on the line, the statement follows from line separation and the uniqueness of segment transfer stated in Proposition 7.2.
Figure 7.49. Assumed is ε ≤ β. Why does the ray r intersect the side AC.
Problem 7.13. As shown in the figure on page 194, given is a triangle 4ABC, a point D on the side AB, and a ray r emanating from point D into the interior of the triangle. We assume that this −−→ ray forms the angle ε ≤ β with the ray DA. Use the exterior angle theorem and the crossbar theorem to show, in neutral geometry, that the ray intersects the side AC.
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Figure 7.50. Since ε ≤ β < δ the ray r intersects the side AC.
Answer. We draw the segment DC and use the exterior angle theorem for triangle 4DBC. In this −−→ triangle the angle δ = ∠(DA, r) is an exterior angle and β = ∠ABC is a nonadjacent interior angle. One concludes β < δ. Hence the assumptions imply ε ≤ β < δ, which means that the ray r lies in the interior of the angle δ. We use the crossbar theorem for the triangle 4ADC, and conclude that the ray r intersects the opposite side AC, as to be shown. Problem 7.14. Given is a strictly convex quadrilateral with two pairs of congruent opposite sides. As shown in Lemma 5.8, each diagonal separates the quadrilateral into two triangles in opposite half planes of their common side. Prove in neutral geometry that the two pairs of opposite sides are parallel. Provide a drawing.
Figure 7.51. A parallelogram in neutral geometry.
Answer. Given is the quadrilateral ACBD with two pairs of congruent opposite sides AC BD and BC AD. We draw the diagonal AB and get the congruent triangles 4ABC 4BAD, as one confirms by SSS congruence. Hence the angles β = ∠ABC and α = ∠BAD are congruent. We see that the diagonal AB transverses the lines of the opposite sides DA and BC with congruent z-angles. By Euclid I.27, we conclude that the opposite sides are parallel.
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Figure 7.52. Addendum to the extended ASA congruence.
Proposition 7.47 (Addendum to the extended ASA-Congruence). Given is a triangle 4ABC, a shorter segment A0 B0 < AB and a half-plane H 0 bounded by this segment. The two angles at the vertices A and B are transferred to the endpoints of the segment A0 B0 , and reproduced in the same half plane H 0 . Then the newly obtained rays rA0 and r0B intersect, say at point C 0 . One obtains a triangle 4A0 B0C 0 , lying in the half plane H 0 , and having at vertices A0 and B0 the angles ∠B0 A0C 0 ∠BAC and ∠A0 B0C 0 ∠ABC. Problem 7.15. Prove proposition 7.47, using extended ASA-congruence, Euclid I.27 (see proposition 7.42), and Pasch’s axiom. −−−→ Proof. One transfers segment AB to the ray A0 B0 and gets a segment A0 B2 AB with A0 ∗ B0 ∗ B2 . Question. Which assumption has been used to get A0 ∗ B0 ∗ B2 ? Answer. Since A0 B0 < AB, we get A0 ∗ B0 ∗ B2 . The two angles at the vertices A and B have been transferred to the endpoints of the segment A0 B0 , and reproduced in the same half plane H 0 . We transfer the segment AC onto the newly obtained ray rA0 and obtain a congruent segment A0C2 AC. Thus we get a triangle 4A0 B2C2 and check by SAS congruence that 4A0 B2C2 4ABC. Now we apply Pasch’s axiom to triangle 4A0 B2C2 and line l0 on the ray r0B . This line intersects the triangle side A0 B2 at point B0 , by construction. Line l0 does not intersect the segment B2C2 . Question. Why are the lines l0 and B2C2 parallel? Answer. The lines l2 = B2C2 and l0 form congruent angles β with the base A0 B0 . By Euclid I.27, the two lines are parallel. Hence Pasch’s axiom implies that line l0 intersects the third side A0C2 of 4A0 B2C2 . We call the intersection point C 0 . Thus we have produced a 4A0 B0C 0 , the angles of which at vertices A0 and B0 are congruent to the corresponding angles of 4ABC. Remark. In Euclidean geometry, the two rays rA0 and r0B always intersect, no matter whether A0 B0 < AB, A0 B0 AB, or A0 B0 > AB. In all three cases, the 4A0 B0C 0 is similar to 4ABC. Remark. Here is what happens in hyperbolic geometry: In hyperbolic geometry, similar triangles are always congruent, this implies that 4A0 B0C 0 and 4ABC are not similar if either A0 B0 < AB or A0 B0 > AB!
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In the case just considered, A0 B0 < AB and γ0 > γ. In the opposite case that A0 B0 > AB, the second triangle 4A0 B0C 0 has either a different angle γ0 < γ, or does not exist at all. The last possibilities occurs because the two rays rA0 and r0B do not intersect at all. This happens always, once the segment A0 B0 is long enough. Proposition 7.48 (Comparison of sides implies comparison of angles). [Euclid I.18, Theorem 23 of Hilbert] In any triangle, across the longer side lies the greater angle. Proof. In 4ABC, we assume for sides AB and BC that c = AB > BC = a. The issue is to compare the angles α = ∠CAB and γ = ∠ACB across these two sides. We transfer the shorter side BC at the common vertex B onto the longer side. Thus one gets a segment BD BC, with point D between B and A. Because the 4BCD is isosceles, it has two congruent base angles δ = ∠CDB ∠DCB Because B ∗ D ∗ A, we get by angle comparison at vertex C δ = ∠DCB < γ = ∠ACB Now we use the exterior angle theorem for 4ACD. Hence α = ∠CAB < δ = ∠CDB By transitivity, these three equations together imply that α < γ. Hence the angle α across the smaller side CB is smaller than the angle γ lying across the greater side AB. In short, we have shown that c > a ⇒ γ > α.
Figure 7.53. Across the longer side lies the greater angle
Proposition 7.49 (Comparison of angles implies comparison of sides). [Euclid I.19] In any triangle, across the greater angle lies the longer side. Proof. In 4ABC, we assume for two angles that α = ∠CAB < γ = ∠ACB. The issue is to compare the two sides BC and AB lying across the angles and show a = BC < AB = c. As shown in Proposition 7.6, any two segments are comparable. Question. Please repeat the reasoning for convenience. −−→ Answer. We transfer segment BC along the ray BA, and get a segment BD BC. By Theorem 4 in Hilbert, of the three points A, B, D on a line AB, exactly one lies between the two others. This leads to the following three cases: Either (i) A ∗ D ∗ B and a < c. or (ii) A = D and a = c. or (i) D ∗ A ∗ B and a > c.
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We can now rule out cases (ii) and (iii). In case (ii), the 4ABC is isosceles, and Euclid I.5 would imply α = γ, contrary to the hypothesis about these two angles. In case (iii), Euclid I.18 or Theorem 23 of Hilbert, would imply α > γ, contrary to the hypothesis. Hence only case (i) is left, and a < c, as to be shown. In short, we have shown that γ > α ⇒ c > a. Question. Explain how Euclid I.18 and Euclid I.19 are logically related. Answer.
Euclid I.18 in shorthand: c > a ⇒ γ > α.
Euclid I.19 in shorthand: γ > α ⇒ c > a. Euclid I.19 is the converse of Euclid I.18. Question. Does Euclid I.19 follow from Euclid I.18 by pure logic? Why not? Answer. No, the converse does not follow purely by logic. Question. Which fact does the proof above work nevertheless? Answer. Because any two segments are comparable, we get the converse, nevertheless. Corollary 12. A triangle with two congruent angles is isosceles. Proof. Assume α γ for the given triangle. Since c > a ⇒ γ > α and c < a ⇒ γ < α, and any two segments are comparable, only a c is possible. Proposition 7.50. The hypothenuse is the longest side of a right triangle. Proof. By Proposition 7.46(i), a triangle can have at most one right or obtuse angle. Hence a right triangle 4ABC with the right angle at vertex C has acute angles at the vertices A and B. As shown in Proposition 7.49 (Euclid I.19), comparison of angles of a triangle implies comparison of its sides. Hence the two legs across of the acute angles A or B are shorter than the hypothenuse across the right angle. Proposition 7.51 (The foot point has the shortest distance). Given is a line and a point O not on the line. The foot point F of a perpendicular from O to the line is the unique among all points of the line, which has the shortest distance. Proof. Call the given line l and let O be the given point not on l. Let T be the foot point of the perpendicular dropped from O onto the line l. Take any different point A , T on the line l. The 4AT O has a right angle at vertex T , which is its largest angle by item (i) from proposition 7.46. The hypothenuse AO lies across to T , hence by Euclid I.19 it is the longest side. Remark. Because all other point on l except the foot point have strictly greater distance from O, we can once more conclude that the foot point of the perpendicular is unique. Proposition 7.52 (The Triangle Inequality). [Euclid I.20] In any triangle, the sum of two sides is greater than the third side.
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Figure 7.54. The foot point has the shortest distance
Figure 7.55. The triangle inequality
Proof. In 4ABC, we compare the sum AB + BC to the third side AC. According to definition 7.2, −−→ the sum AB + BC is obtained by transfer of the segment BC onto the ray opposite to BA. Thus we get segment BD BC and the sum AB + BC = AD. Too, the construction yields an additional triangle 4DBC, which turns out to be isosceles. Its two congruent base angles are ∠CDB and ∠DCB. In the figure on page 199, both are notated by δ. Because point B lies between A and D, angle comparison at vertex C yields δ ∠DCB < ∠DCA = η Euclid I.19 tells that in a triangle, across a larger angle lies a longer side. We use Euclid I.19 for the larger triangle 4ACD. Hence the side AD lying across the greater angle η is longer than the side AC, which lies across the smaller angle δ. We have obtained AC < AD. By construction, the latter segment is AD = AB + BD = AB + BC. This confirms that the sum of two sides AB + BC of the given triangle is larger than the third side AC. Corollary 13. For any three points A, B, C hold the inequalities AC ≤ AB + BC
and |AC| ≤ |AB| + |BC|
for the segments, and their lengths, respectively. Moreover, AC AB + BC respectively |AC| = |AB| + |BC| holds if and only if point B lies between A and C or is equal to one of them.
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Proposition 7.53. (i) The sum of the segments of any polygon is equal to or longer than the distance of its first and last vertex. (ii) The sum of the segments A0 A1 , A1 A2 , A2 A3 , . . . , An−1 An is equal to the segment from its first vertex A0 to its last vertex An if and only if all vertices lie between them and occur in the order A0 ∗ A1 ∗ A2 ∗ A3 ∗ · · · ∗ An . Proof of item (i). Recall definition 5.19 of a polygon. We proceed by induction on the number of its segments A0 A1 , A1 A2 , A2 A3 , . . . , An−1 An . The assertion holds for a polygon A0 A1 , A1 A2 with two segments since A0 A2 ≤ A0 A1 + A1 A2 according to the triangle inequality, and its corollary 13. For the induction step, we proceed from n to n + 1 segments. We know that A0 An+1 ≤ A0 An + An An+1 by the triangle inequality and A0 An ≤ A0 A1 + A1 A2 + · · · + An−1 An by the induction assumption. The associativity of the segment addition has already been stated as Proposition 7.7. Hence A0 An+1 ≤ A0 An + An An+1 ≤ (A0 A1 + A1 A2 + · · · + An−1 An ) + An An+1 A0 A1 + A1 A2 + · · · + An−1 An + An An+1
as to be shown.
Proof of item (ii). We proceed by induction on the number of the polygon’s segments. For simplicity, I may assume that all these points are different. The assertion holds for a polygon A0 A1 , A1 A2 with two segments since A0 A2 = A0 A1 + A1 A2 ⇔ A0 ∗ A1 ∗ A2 or A1 = A0 or A1 = A2 according to corollary 13. For the induction step, we proceed from n to n + 1 segments. Assume the equality A0 A1 + A1 A2 + · · · + An An+1 = A0 An+1
(7.9)
to hold. In that case A0 An+1 ≤ A0 An + An An+1 ≤ (A0 A1 + A1 A2 + · · · + An−1 An ) + An An+1 = A0 An+1
(7.10)
and hence equality occurs everywhere and A0 An = A0 A1 + A1 A2 + · · · + An−1 An Now the induction assumption implies that the first n vertices lie in the order A0 ∗ A1 ∗ A2 ∗ A3 ∗ · · · ∗ An Moreover the statement (7.10) implies A0 An+1 = A0 An + An An+1 and hence according to corollary 13, we get the order relation A0 ∗ An ∗ An+1 Via the n-point Theorem 5.7 we get the order relation A0 ∗ A1 ∗ A2 ∗ A3 ∗ · · · ∗ An ∗ An+1
(7.11)
for all n + 1 segment as claimed. The converse tells that statement 7.11 implies equality (7.9), and is even easier to check.
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Figure 7.56. SAA congruence
Proposition 7.54 (SAA-Congruence Theorem). [Theorem 25 in Hilbert] Assume two triangles have a pair of congruent sides, one pair of congruent angles across these sides, and a second pair of congruent angles adjacent to these sides. Then the two triangles are congruent. Proof. Let the triangles be 4ABC and 4A0 B0C 0 , and assume that AB A0 B0 , ∠BAC ∠B0 A0C 0 −−−→ and ∠ACB ∠A0C 0 B0 . We choose a point C 00 on the ray A0C 0 such that AC A0C 00 . By axiom (III.5) γ = ∠ACB ∠A0C 00 B0 Indeed, the SAS congruence even implies 4ABC 4A0 B0C 00
(*)
On the other hand, by assumption γ = ∠ACB ∠A0C 0 B0 If C 0 , C 00 , an exterior angle of 4C 0C 00 B0 would be congruent to a nonadjacent interior angle, which is impossible by the exterior angle theorem. Hence we can conclude that C 0 = C 00 . Because of (*), this implies 4ABC 4A0 B0C 0 , as to be shown. Question. How does this theorem differ from the ASA congruence (Hilbert’s Theorem 13, and Proposition 7.10 above)? Answer. Of the two pairs of angles that are given (or compared), one pair lies across the pair of given sides. Second proof of the hypothenuse-leg theorem 7.32. Take two right 4ABC and 4A0 B0C 0 with b = b0 , c = c0 and γ = γ0 = R. One transfers segment CB and segment C 0 B0 , both on the ray opposite to −−− → C 0 B0 , and get new segments C 0 D0 CB as well as C 0 E 0 C 0 B0 . From the construction and SAS congruence, we conclude 4ABC 4A0 D0C ,
4A0 B0C 0 4A0 E 0C 0
(1)
Hence especially A0 B0 A0 D0 A0 E 0
(2)
By Proposition 7.46(iv), at most two points can have the same distance from a line. Hence not all three points B0 , D0 , E 0 can be different. The only possibility left is D0 = B0 , because they both lie on the opposite side of A0C 0 than B0 .
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Figure 7.57. The hypothenuse leg theorem
Question. For completeness, explain once more. Assume that D0 , E 0 towards a contradiction. Take the case that B0 ∗ D0 ∗ E 0 shown in the drawing. (The other cases can be dealt with similarly.) Independent answer. There are two isosceles 4B0 A0 D0 and 4DA0 E 0 . Question. What would happen with their base angles at vertex D0 ? Answer. The two base angles would be supplementary. Question. Why is this impossible? Answer. This is impossible, because the exterior angle theorem would imply that each one of them is larger than the other one. This contradiction leave only the possibility that D0 = E 0 . Now the required congruence 4ABC 4A0 B0C 0 follows from D0 = E 0 and (1).
Proposition 7.55 (The Hinge Theorem). [Euclid I.24] Increasing the angle between two constant sides increases the opposite side of a triangle. Corollary 14 states the equivalence we get by taking from [Euclid I.24] and [Euclid I.25] together: Corollary 14. Given are two triangles with two pairs of congruent sides. In the first triangle, the angle between them is smaller, congruent, or greater than in the second one if and only if the opposite side is smaller, congruent, or greater in the first triangle. Proof. Given are 4ABC and 4A0 B0C 0 with a = a0 and c = c0 . Assuming β < β0 , we have to check whether b < b0 . One can assume that A = A0 , B = B0 , and put the two points C and C 0 lie on the same side of line AB. By the hypothesis β < β0 , the two points A and C 0 lie on different sides of line BC. Hence this line intersects the segment AC 0 . −−→ Indeed, as stated by the Crossbar theorem, the ray BC intersects the segment AC 0 . Let D be the intersection point. We now distinguish three cases: (a) B ∗ D ∗ C. This case occurs if the triangle side BC to be rotated is long enough. (b) C = D.
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Figure 7.58. The Hinge Theorem
Figure 7.59. The Hinge Theorem: turning a long side
(c) B ∗ C ∗ D. In this case the triangle side BC to be rotated is rather short. In the border line case (b), we see directly that β < β0 implies A ∗ C ∗ C 0 and hence AC < AC 0 , as to be shown. In the two other cases, we apply Euclid I.19 to triangle 4ACC 0 . Thus it is enough to show that this triangle has a larger angle ε = ∠ACC 0 at vertex C than the angle ε0 = ∠AC 0C at vertex C 0 . The 4BCC 0 is isosceles. Hence, by Euclid I.5, it has two congruent base angles, which we denote by ϕ. In case (a), we proceed as follows: Because B and C lie on opposite sides of AC 0 , comparison of angles at vertex C 0 yields ε0 < ϕ (1) And because A and C 0 lie on opposite sides of line BC, comparison of angles at vertex C yields ϕ<ε Because of transitivity, (1)(2) together imply ε0 < ε and hence AC 0 > AC.
(2)
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Figure 7.60. The Hinge Theorem: the borderline case
Figure 7.61. The Hinge Theorem: turning a short side
In case (c), we show that ε is an obtuse, and ε0 is an acute angle. Because A and C 0 lie on opposite sides of line BCD, we get ∠C 0CD < ∠C 0CA = ε (3) but ∠C 0CD is an exterior angle of the isosceles 4BCC 0 . Because the base angles of an isosceles triangle are acute, its supplement ∠C 0CD is obtuse. Hence by (3), ε is obtuse, too. Since the triangle 4ACC 0 can have at most one right or obtuse angle, the angle ε0 is acute. Now ε0 < R < ε implies again ε0 < ε, and hence Euclid I.19 yields AC 0 > AC. Proposition 7.56 (The Midpoint of a Segment). [Theorem 26 in Hilbert] Any segment has a midpoint. Construction 7.7 (Construction of the Midpoint of a Segment with Hilbert Tools). Given is a segment AB. Choose any point C not on the line AB. Transfer angle ∠CAB into the half-plane of −−→ AB opposite to point C, with the second endpoint B as vertex and ray BA as one side. Next we transfer congruent segments AC BD onto the newly produced legs of these two angles. Finally, ← → line AB and segment CD intersect at the midpoint M.
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Figure 7.62. Hilbert’s construction of the midpoint
(a) Here is a drawing. (b) Explain why lines AB and CD intersect. Answer. By construction, points C and D lie on different sides of line AB. Hence, by the plane separation theorem, line AB and segment CD intersect. Let M be the intersection point. From the plane separation theorem, too, it follows that the intersection point M lies between C and D. But it turns out to be harder to see why M lies between A and B! (c) Show that M = A is impossible.
Figure 7.63. M = A is impossible
−−→ Answer. In that case line l = AC would go through point D. The ray AC would be an extension of side AD of the triangle 4ABD. This triangle would have the interior angle
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∠ABD congruent to the exterior angle ∠CAB, contradicting the exterior angle theorem. (Remember: an exterior angle is always greater than a nonadjacent interior angle.) (d) Show that M ∗ A ∗ B is impossible.
Figure 7.64. Point A lying between M and B is impossible
Answer 1. We use Pasch’s axiom for 4MBD and line l = CA. Which conclusion do you get? Answer. The line CA enters 4MBD on the side MB. By Pasch’s axiom, this line either (i) intersects side DB, or (ii) goes through point D, or (iii) intersects side MD. In all three cases, we derive a contradiction: Case (i): suppose line l intersects segment DB, say at point G. The 4ABG would have the interior ∠ABG congruent to the exterior ∠CAB. This contradicts the exterior angle theorem, which tells an exterior angle is always greater than a nonadjacent interior angle. Case (ii): suppose line l goes through point D. The 4ABD would have the interior ∠ABD congruent to the exterior ∠CAB. This contradicts the exterior angle theorem, which tells an exterior angle is always greater than a nonadjacent interior angle. Case (iii): suppose line l intersects segment MD, say at point F. Points C , F are different, because they lie on different sides of AB. The lines AC and MD intersect both in point C and in point F,. Because this are two different points, they determine a line uniquely. Hence all five points C, M, D, A, F lie on one line. Hence we are back to the case M = A, ruled out earlier.
Answer following Hilbert. We apply the exterior angle theorem twice. Here is a sketchy drawing for that impossibility. 1 In triangle 4ABD, the interior angle at vertex B is β = ∠DBM, which is smaller than the exterior angle at vertex ε = ∠BMC. In triangle 1
The drawing, too, occurs in the millenium edition of "Grundlagen der Geometrie", page 26.
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Figure 7.65. Point A lying between M and B is impossible
4AMC, the angle ε from above is interior angle at vertex M, and hence smaller than the exterior angle α = ∠BAC at vertex A. Now transitivity yields β < ε < α. On the other hand, the angles ∠DBA = β and α = ∠CAB are congruent by construction. This contradiction rules out the case M ∗ A ∗ B. (e) Now we know that M lies between A and B, we finally can prove that M is the midpoint. Question. Which congruence theorem is used for which triangles? Answer. One uses SAA congruence for 4AMC and 4BMD. Indeed, the angles at A and B
Figure 7.66. Apply the SAA congruence
are congruent by construction, and the angles at vertex M are congruent vertical angles. (Both statements are only true because M lies between A and B!) The sides AC and BD opposite to those angles are congruent by construction. Hence the two triangles are congruent, and especially AM MB
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Figure 7.67. The generic situation for Proposition 7.57, for which we prove: Two segments CX and DY on different sides of XY are congruent if and only if midpoint M of segment CD lies on line l. The third figure shows the case with both conditions true.
Problem 7.16. Given is a convex quadrilateral with two pairs of congruent opposite sides. Give a short proof that the diagonals bisect each other. You may use the result of problem 7.14 and assume Hilbert’s construction of the midpoint of a segment is valid. Provide a drawing.
Figure 7.68. The diagonals of a parallelogram bisect each other.
Answer. Since the quadrilateral is convex, the opposite vertices C and D lie on different sides of the diagonal AB. We have proved in problem 7.14 that the angles β = ∠ABC and α = ∠BAD are congruent. Hence the figure repeats exactly Hilbert’s construction of the midpoint of segment AB. Hence the diagonal segments intersect in the midpoint M of diagonal AB. Similarly, we prove that M is the midpoint of the other diagonal CD, too. Proposition 7.57. The midpoint of a segment lies on a given line l if and only if the two endpoints of a segment have the same distance to the line, and lie on the opposite sides of it. Proof. Assume that the two endpoints C and D of the given segment lie on opposite sides of l. Furthermore, assume CX DY are the congruent segments to the foot points X and Y. Let Q be the intersection point of line l and segment CD, which exists by plane separation. In the special case X = Y, we are ready immediately. Otherwise, we need to see why Q lies between
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X and Y! This is done is the same way as in Hilbert’s construction 7.11 of the midpoint, which is now applied to the segment AB = XY. Finally, one obtains the triangle congruence 4CQX 4DQY
(oneflier)
via SAA congruence, using the right angles at X and Y, vertical angles at vertex M, and the congruent segments CX DY. Hence CQ DQ, confirming that Q = M is the midpoint of segment CD. Conversely, assume that the midpoint M of segment CD lies on the line l. In the special case X = Y we are ready immediately. Otherwise, we need to confirm, once more, why M lies between X and Y! Again one needs to use the exterior angle theorem. Finally, one obtains the triangle congruence 4CMX 4DMY (oneflier) via SAA congruence, using the right angles at X and Y, vertical angles at vertex M, and the congruent segments CX DY, as to be shown. 7.12. SSA congruence Next we study the possibilities and difficulties with SSA congruence. Thus the matching pieces of the two given triangles are two pairs of sides, and one pair of angles opposite to one of these sides. I follow Euler’s conventional notation: in triangle 4ABC, let a = BC, b = AC, and c = AB be the sides and α := ∠BAC, β := ∠ABC, and γ := ∠ACB be the angles. Problem 7.17. Investigate an example Use straightedge and compass to construct in Euclidian geometry all triangles (up to congruence) with γ = ∠ACB = 30◦ , side AC = 10 units and side AB given as below. How many non-congruent solutions (none, one, two?) do you get in each case? How many of them are acute, right or obtuse triangles? Measure and report the angle β = ∠ABC for all your solutions. Make clear by the drawings what happens in all cases, especially how many acute, right, and obtuse triangles you get as solutions. Hint: It is convenient to construct all triangles with one common side AC. Check carefully to find the obtuse angles! (a) AB = 4 units. Answer (a). One has to begin the construction by drawing a segment AC of length 10, and a ray r ←−− with vertex C, forming an angle of 30◦ with CA. The point B has to lie on the ray r, as well as on a circle of radius AB around A. In case (a), the circle does not intersect this ray, hence there is no solution. (b) AB = 5 units. Answer. In case (b), the circle just touches the ray r, hence there is one solution, with a right angle at B. (c) AB = 5.5 units. Answer. In case (c), the circle intersect the ray r in two points, hence there are two non congruent solutions. At vertices B and B0 , I measure the angles about β = 67◦ and β0 = 113◦ . One solution is an acute, the other an obtuse triangle. (d) AB = 6 units. Answer. Again in case (d), the circle intersects the ray r in two points, and there are two non congruent solutions. At vertices B and B0 , I measure the angles about β = 57◦ and β0 = 123◦ . Both solutions are obtuse triangles. Indeed, the first solution has the obtuse angle α = 180◦ −β−γ = 93◦ .
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Figure 7.69. A Euclidean example for SSA triangle construction
Proposition 7.58 (SSA Matching Proposition). Given are two triangles. Assume that two sides of the first triangle are pairwise congruent to two sides of the second triangle, and that the angles across to one of these pairs are congruent. Then either the two triangles are congruent, or the angles across the other pair of congruent sides add up to two right angles. Proof. It is assumed that two sides and the angle across one of these sides of 4ABC can be matched to congruent pieces of 4A0 B0C 0 : c = AB A0 B0 = c0 , b = AC A0C 0 = b0 , γ = ∠ACB ∠A0C 0 B0 = γ0
(SSA)
We have to show that either (a) or (b) holds: (a) 4ABC 4A0 B0C 0 . (b) The two angles across the second matched side β = ∠ABC and β0 = ∠A0 B0C 0 are (congruent to) supplementary angles. One of them is acute and the other one is obtuse. We begin by reducing the problem to the special case that A = A0 , C = C 0 and that B and B0 = D lie on the same side of AC. We use the SAS congruence to construct a triangle 4ADC, such that 4A0 B0C 0 4ADC, with points B and D on the same side of AC. The drawing below shows that procedure. Question. Explain how you get the triangle 4ADC congruent to 4A0 B0C 0 . −−→ Answer. I just transfer segment C 0 B0 onto the the ray CB and get C 0 B0 CD. The congruence 4A0 B0C 0 4ADC, follows from SAS, given in Theorem 12 of Hilbert (see Proposition 7.11 above).
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Figure 7.70. Matching two triangles with SSA
Assume that congruence (a) does not hold. We have to prove that case (b) occurs. We know that α , α0 , since otherwise the SAS-congruence theorem implies 4ABC 4A0 B0C 0 , which again would be case (a) just ruled out. Without loss of generality, we can assume α0 < α. Since ∠DAC = α0 < α = ∠BAC, the point D lies between B and C, as shown in the drawing. Since AD A0 B0 AB by assumption, triangle 4ABD is isosceles, with baseline BD. By Euclid I.5, the two base angles of an isosceles triangle are congruent. One of them is the angle β = ∠ABD. By Proposition 7.46(ii), these base angles are always acute. The angle β0 = ∠A0 B0C 0 ∠ADC is the supplement to the second base angle ∠ADB ∠ABD = β. Hence β and β0 are congruent to supplementary angles. The first one is acute, the second one is obtuse, as claimed in (b). Corollary 15 (SSAA Congruence). Two triangles which have two pairs of congruent sides, and two pairs of congruent angles across to the latter, are congruent. Proposition 7.59 (Restricted SSA Congruence, case of a unique solution). As in the matching Proposition above, we assume that two sides and the angle across one of these sides of 4ABC can be matched to congruent pieces of 4A0 B0C 0 : AB A0 B0 , AC A0C 0 and ∠ACB ∠A0C 0 B0
(SSA)
Under the additional assumption that either (i) or (ii) hold, (i) The given angle γ lies across a side longer or equal to the other given side: c = AB ≥ AC = b. (ii) The given angle γ = ∠ACB is right or obtuse. congruence holds among all triangles matched by (SSA). In all these cases there do not exist any non congruent triangles with congruent sides b, c and angle γ. Actually assumption (ii) implies (i). Proof of Proposition 7.59. Assume that (ii) holds. Indeed, by Proposition 7.46(i), a triangle can have at most one angle which is right or obtuse. Hence assumption (ii) implies γ ≥ β. By Euclid I.19, across the greater angle of any triangle lies the longer side. Hence γ ≥ β implies c ≥ b, which is assumption (i). We now assume that c = AB ≥ AC = b holds, as stated by (i). By Euclid I.18, across the longer side of a triangle lies the greater angle. Hence c ≥ b implies γ ≥ β. A triangle cannot have two angles which are right or obtuse. Hence β and β0 are both acute or right. Now congruence follows because case (b) in the matching Proposition is ruled out.
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Corollary 16 (SSA Congruence for isosceles and right triangles). (i’) Two isosceles triangles with congruent legs and congruent base angles are congruent. (ii’) Any two right triangles with congruent hypothenuses and one pair of congruent legs are congruent. Proof. Assumption (i’) is a special case of (i), and (ii’) a special case of (ii).
Proposition 7.60 (Restricted SSA Congruence Theorem, case with non uniqueness). Again we assume that two sides and the angle across one of these sides of 4ABC can be matched to congruent pieces of 4A0 B0C 0 : AB A0 B0 , AC A0C 0 and
∠ACB ∠A0C 0 B0
(SSA)
Complementary to Proposition 7.59, we assume that neither (i) nor (ii) nor β R holds. In other words, we assume that the given angle γ = ∠ACB is acute and lies across the shorter given side: c = AB < AC = b, and angle β is not a right angle. Under these additional assumptions (a) There exist two non congruent triangles matched by (SSA). (b) Nevertheless, equivalent are (1) The two given triangles 4ABC and 4A0 B0C 0 are congruent. (2) (SSAA) The two triangles can be matched in two sides and the two angles across both given sides. (3) For both triangles, the angles β and β00 across the longer given side b are both acute or both right or both obtuse. (4) Both triangles are acute, or both are right, or both triangles are obtuse with the two obtuse angles at corresponding vertices. In this case just the two given triangles 4ABC and 4A0 B0C 0 —not all other triangles with those given angle and sides γ, c, b—are congruent.
Figure 7.71. Establish two solutions for SSA
Proof for part (a). We drop the perpendicular from vertex A onto the opposite side CB. The foot point F , B, because otherwise β would be a right angle. Transfer segment FB on the ray opposite
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−−→ to FB to produce segment FB00 FB. The two right triangles 4ABF and 4AB00 F are congruent by the Hypothenuse-Leg Theorem 15. Hence AB AB00 . Thus we got two non congruent triangles 4ABC and 4AB00C which nevertheless satisfy AB AB00 , AC AC and ∠ACB ∠ACB00
(SSA”)
For these non congruent triangles, the two angles β and β00 are supplementary. One of them is acute, the other one is obtuse. Proof for part (b). Clearly (1) implies (2) implies (3) implies (4). If (4) holds, the matching Proposition excludes non congruent solutions. Indeed, if both triangles are acute, both angles β and β0 are acute. Congruence follows by the SSA matching Proposition. If both triangles are right, both angles β and β0 are acute or right. In both cases, congruence follows by the SSA matching Proposition. If both triangles are obtuse, and obtuse angle occurs at corresponding vertices, then either both angles β and β0 are obtuse, or both angles α and α0 are obtuse. In the second case both angles both angles β and β0 are acute. In both cases, congruence follows by the SSA matching Proposition. Remark. Earlier on, we have carefully studied triangles with c = AB = 5 , 5.5 . 6
b = AC = 10
and γ = ∠ACB = 30◦
in Euclidean geometry. In the first case c = 5, one gets a unique right triangle as solution. In the second case c = 5.5 and c = 6, there are two non-congruent solutions, one of which is an acute triangle, the other one is an obtuse triangle with β0 > R. In the third case c = 6, there are two non-congruent solutions again. But both are obtuse triangles! One gets one triangle with α > R and β < R, as well as a second one with α0 < R and β0 > R. 7.13.
Reflection
Definition 7.22 (Reflection across a line). Given is a line l. The reflection across line l is the mapping r : P 7→ P0 which maps an arbitrary point P to a point P0 meeting the following requirements: • If P lies on the symmetry axis l, we put P0 := P. • If P does not lie on the symmetry axis, the point P0 is specified by requiring (i) P and P0 lie on different sides of l. (ii) the lines l and PP0 are perpendicular, intersecting at F. (iii) PF P0 F. −−→ Construction 7.8. Choose any two points A and B on the line l. Draw ray AP. Transfer angle −−→ ∠BAP, into the half plane opposite to P, with ray AP as one side. On the newly produced ray, transfer segment AP to produce a new segment AP0 AP. The point P0 is the reflection of the given point P across the line AB. Proof of validity. The line AB and the segment PP0 intersect, because P and P0 lie on different sides of the line. I call the intersection point F. We still have to distinguish several cases depending on the order of the three points A, B and F : −−→ • as drawn in the figure on page 214, the point F lies on the ray AB;
214
Figure 7.72. Reflect a point
−−→ • point F lies on the ray opposite to AB; • the special case that point F is equal to point A or B, In the first two cases (generic cases), we show by the SAS-congruence that the two triangles 4FAP and 4FAP0 are congruent. • Indeed, the angles at vertex A are congruent—either directly by construction, or in the second case by construction and because Proposition 7.15 tells that supplements of congruent angles are congruent. • The common side FA FA is congruent to itself, • and the other adjacent side AP AP0 by construction. From the congruence of the two triangles we conclude ∠AFP ∠AFP0 which is a pair of congruent supplementary, and hence right angles. Moreover FP FP0 . Hence all three requirements (i)(ii) and (iii) in the definition of the reflected point are satisfied. In the special case that point F turns out to be equal to point A, the result is proved even simpler. Proposition 7.61. Given are two points P, Q on the same side of l and their images P0 , Q0 . We show that (a) PQ P0 Q0 ← → (b) If the lines m := PQ and l intersect at M, then P0 , Q0 and M lie on a second line m0 . Question. For this exercise, you can if you like, do drawing and proof, in neutral geometry, on your own. Proof of (a). Let F and G be the foot points of the perpendiculars dropped from P and Q onto the symmetry axis l. We use SAS-congruence for the two triangles 4QGF and 4Q0GF.
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Figure 7.73. Reflection by a line l
Question. Indeed both triangles have a right angle at G—. Explain why. Answer. ∠QGF is right by definition of a reflection. Since a right angle is congruent to its supplementary angle, the supplementary angle ∠Q0GF is a right angle, too. —a common side GF and congruent sides GQ GQ0 , by definition of a reflection. Hence SAS congruence implies 4QGF 4Q0GF (1) Because of congruence (1), we get α = ∠QFG ∠Q0 FG , QF Q0 F
(2) (3)
Since ∠GFP ∠GFP0 is a right angle, angle subtraction yields β = ∠QFP ∠Q0 FP0
(4)
PF P0 F
(5)
By construction Using (3)(4) and (5), SAS-congruence implies that 4QFP 4Q0 FP0
(6)
Hence especially (a) holds.
Proof of (b). We use SAS-congruence for the two triangles 4QGM and 4Q0GM. Indeed both triangles have a right angle at G: ∠QGM is right by definition of a reflection. Since a right angle is congruent to its supplementary angle, the supplementary angle ∠Q0GM is a right angle, too. Furthermore, they have a common side GM and congruent sides GQ GQ0 , by definition of a reflection. Hence SAS congruence implies 4QGM 4Q0GM
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For the angle x between the line l of reflection and the given line m, one gets x = ∠QMG ∠Q0 MG
(8)
Now we argue similarly for the two triangles 4PF M and 4P0 F M. Thus we get y = ∠PMF ∠P0 MF = ∠P0 MG
(9)
Now by assumption, the three points P, Q and M lie on a line m. x = ∠QMG = ∠PMF = y
(10)
From (8)(9),(10) and transitivity of angle congruence, we get ∠Q0 MG = ∠P0 MG
(11)
−−−→ −−−→ Now by Hilbert’s axiom, the angle transfer produces a unique ray. Hence MQ0 = MP0 . Thus the uniqueness of angle transfer implies that the three points M, P0 and Q0 lie on one line. An alternative for the proof of (b) . (6) above implies γ = ∠QPF ∠Q0 P0 F
(7)
We apply the extended ASA congruence to triangle 4PF M and segment FP0 . The angles to be transferred to the endpoints of that segment are a right angle at F and γ at P0 . Transferring produces −−−→ as second sides of these angles part of the line l and the ray P0 Q0 . Hence those rays intersect, say at point M 0 . By ASA-congruence 4MFP 4M 0 FP0 Hence F M F M 0 . The points Q and Q0 and hence M and M 0 lie on the same side of PP0 . Hence uniqueness of segment transfer implies M = M 0 as to be shown. Theorem 7.1. The mapping of reflection preserves incidence and maps segments and angles to congruent ones. Hence it is an isometry. 7.14.
Applied problems
Figure 7.74. Where on the boundary can guys A and B meet the quickest?
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Problem 7.18. Two guys who are equally good runners stand at points A and B on the border of an L-shaped place shown in the figure on page 216. They are allowed to go across the place, but they want to meet at any point on the boundary. Which point do they need to choose to meet as quick as possible. Give a construction and explain your reasoning. Answer. The runners pass congruent segments AM and BM until they meet at some point M. Hence they meet where the perpendicular bisector of AB intersects the boundary. There are two such points, of which they choose the nearer one.
Figure 7.75. They meet where the perpendicular bisector of AB intersects the boundary.
Figure 7.76. Where on the boundary can guys A and B meet the quickest?
Problem 7.19. Repeat the problem 7.18 for the situation drawn in the two figures on page 217. Answer. Until they meet, the runners pass congruent distances. They need not pass any distance longer than their mutual distance AB. Hence their meeting point lies inside the two equilateral
218
Figure 7.77. Where on the boundary can guys A and B meet the quickest?
triangles erected on both sides of segment AB. They can meet either in point A, or point B, or a point where the perpendicular bisector of AB intersects the boundary. For the example drawn in the left part figure on page 218, these are the points N and F. But these points do not lies inside the two equilateral triangles erected on both sides of segment AB. Hence it is more advantageous for the runners to simply meet at point A or point B. For the example drawn in the right part figure on page 218, it is most advantageous for the runners to simply meet at the point of the perpendicular bisector denoted nearest. Problem 7.20. Given is a line l which is a plain mirror, a bundled light source at point P, and a point Q on the same side of the mirror, to which one wants the light ray to be reflected. Give a construction of the light ray from P to Q, reflected by the mirror. Provide a drawing and an explanation. Answer. We construct the reflected point Q0 of the point Q to which the light ray is to be directed. The light has to be directed along the segment PQ0 . The intersection of this segment with the (line of the) mirror gives the point R where the light ray is reflected. The reflected light goes from from point R to the given target point Q. Proof of validity. Let G be the foot-point of the perpendicular dropped from point Q onto the mirror. The angles ∠QRG ∠GRQ0 are congruent. The angle of the incident light ray with the mirror is a vertical angle of ∠GRQ0 , and the angle of the reflected light ray with the mirror is ∠QRG. Hence they are congruent, as required for the reflection of a light ray. Problem 7.21. In many technical and physical applications—for example for sensitive measurement of electrical current—one uses a small mirror attached to a twisting thin wire. A light beam shines onto the mirror and its reflected beam is depicted on a scale, which can be rather far away. Additionally it is assumed that the incoming light ray, and the perpendicular lines to the mirror in both positions lie in a plane. Thus we deal with the two-dimensional problem. Give the reason why turning the mirror by an angle θ results in turning the reflected beam by angle 2θ. Provide a drawing.
219
Figure 7.78. Reflect a light ray from P to a given point Q.
Figure 7.79. Turning the mirror by angle θ turns the reflected ray by the angle 2θ.
Answer. The problem is easiest under the assumption that the light ray hints the mirror exact at its point of rotation (respectively in three dimensions on its axis of rotation). By the law of reflection of light, the angle between the perpendicular to the mirror and the reflected ray is congruent to the angle between the incoming light ray and the perpendicular. Hence α = ∠LOP1 ∠P1 OR1 in the figure on page 219. The angle between the incoming light ray and its reflected ray is the sum ∠LOR1 = 2α of these two angles.
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As the mirror is turning by the angle θ, its perpendicular is turning by the congruent angle θ = ∠P1 OP2 between the perpendiculars; shown in blue. The angle between the incoming light ray and the perpendicular to the mirror in its second position is ∠LOP2 = α + θ, with plus sign for the situation as drawn. The angle between the perpendicular and the reflected ray increases by the same angle θ. After the turn of the mirror has occurred, the angle between the incoming light ray and its reflected ray is ∠LOR2 = 2(α + θ). The reflected light ray has been turned by the difference angle ∠R1 OR2 = ∠LOR2 − ∠LOR1 = 2θ. 7.15. Independence of the SAS-axiom Already in a commentary to Euclid, written about 1557 by Peletier, the author writes about the side-angle-side (SAS) congruence for triangles, Euclid I.4: The truth of this proposition belongs among the common notions, because to superimpose one figure to another is mechanics, not mathematics. The "principle of superposition" used by Euclid for the justification of side-angle-side (SAS) congruence of triangles, has been sharply criticised by Bertrand Russell (1872-1970). In his acerbic manner, he speaks of superposition—in no uncertain terms—as a tissue of nonsense. 1 Indeed, although the principle of superposition may seem reasonable enough in dealing with material triangles of small laboratory size, its legitimacy is questionable for working with conceptional entities. The assumption that moving a triangle is possible without alteration of its internal structure is just based on experience, not logic. We talk today of a physical thought experiment, instead of the historic term "mechanics" (still used by Gauss in such philosophical considerations). For the mathematician, the above discussion cannot be the endpoint. As already made clear in the preamble, the goals of Hilbert’s foundations of geometry are not only • to set up a complete and as simple as possible system of axioms; • to deduct from them the most important geometric theorems; • but finally, to clarify the meaning and importance of the different axioms and their consequences. Indeed, Hilbert proves the independence of the SAS axiom (III.5), and thus establishes the necessity of this axiom beyond any doubt. Theorem 7.2 (Hilbert). There exists a spatial geometry for which the axioms I, II, IV,V as well as (III.1),(III.2),(III.3),(III.4) are all satisfied, but the SAS axiom (III.5) is violated. Moreover, Euclid I.5 is violated. But the following Propositions are still valid in this model: Proposition 7.15 about the congruence of supplementary angles; Proposition 7.19 about angle-addition and subtraction; Proposition ?? about the transfer of an interior ray; Proposition 7.30 Euclid’s postulate that all right angles are congruent; Proposition 7.25 telling that congruence is an equivalence relation on the class of angles. 1
In his autobiography, Bertrand Russell penned this remarkable recollection: "At the age of eleven, I began Euclid, with my brother as tutor. This was one of the great events of my life, as dazzling as first love."
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We see that it is impossible to deduct the SAS axiom (III.5) from the remaining axioms I, II, (III.1),(III.2),(III.3),(III.4), IV,V, nor from the Propositions enumerated above. Proof. We define the "points", "lines", "planes" of our model to be the points, lines, planes of the common three-dimensional analytic geometry with coordinates from any real-closed field. 2 It is well-known that the axioms of incidence, order, parallelism and continuity are valid. The delicate point are of course the axioms of congruence. The transfer of angles is defined as in analytic geometry. Indeed, we take as usual Defintion 7.17 for the congruence of orientated angles and Definition 24.11 for congruence of angles. Thus, the transfer of angle is well defined and axiom (III.4) is valid. On the other hand, the length of the segment between the points A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ), is measured by a non-Pythagorean distance p dn(A, B) = (a1 − b1 + a2 − b2 )2 + (a2 − b2 )2 + (a3 − b3 )2 We define the congruence of any two segments AB and A0 B0 by postulating AB A0 B0 ⇔ dn2 (A, B) = dn2 (A0 , B0 ) Distance and congruence is thus defined differently from a Pythagorean plane. Nevertheless, one easily checks that congruence of segment is an equivalence relation and hence Hilbert’s axioms (III.1) and (III.2) hold. Moreover, one can still add two adjacent segments on a line and check that Hilbert’s axiom (III.3) is valid. Indeed, it is easy to use and modify the procedure from Problem 24.3 from the section "Coordinate Planes—Descartes’ Road from Algebra to Geometry", subsection about congruence in a Pythagorean plane. This—a bid tedious but not difficult exercise—is left to the reader. As the "coup de grâce", we now construct a counterexample showing that the axiom (III.5) is violated. Take the points √ 2 O = (0, 0, 0) , A = (1, 0, 0) , B = (−1, 0, 0) , C = (0, , 0) 2 In the new distance dn(O, A) = dn(O, B) = dn(O, C) = 1 We get two right triangles 4AOC and 4COB for which hold the congruences ∠AOC ∠COB ,
OA OC ,
OC OB
But, contrary to the conclusion from axiom (III.5), the angles ∠OAC and ∠OCB are not congruent. Moreover, the SAS congruence of the two triangle is violated for the sides opposite to the right angles since √ √ dn2 (A, C) = 2 − 2 and dn2 (B, C) = 2 + 2 are different. Finally the two triangles 4AOC and 4COB are isosceles, but do not have congruent base angles. Thus Euclid I.5 is not valid. Remark. A counterexample for two-dimensional geometry has been constructed by Hilbert as follows. In the three-dimensional Euclidean space, we take the planes α = {(x, y, z) : z = 0} and 2
Hilbert did simply use real coordinates.
β = {(x, y, z) : x = z}
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and project the plane β onto α by the parallel projection (x, y, x) ∈ β 7→ (x, y, 0) ∈ α In the plane α, we define as new "distance" by measuring the Euclidean distance of the preimage in the plane β. But we keep the Euclidean angle as "angle" in the plane α. We get in the plane α a new geometry, for which all axioms except (III.5) hold. 7.16.
The Moulton plane
Theorem 7.3 (Hilbert’s Proposition 54). There exists a planar geometry for which the axioms of incidence (I.1),(I.2),(I.3), all axioms of order (II.1),(II.2),(II.3),(II.4), the axioms of congruence (III.1),(III.2),(III.3),(III.4) but not the SAS-axiom (III.5), and the Euclidean Parallel Postulate 3.3 are all valid. But the following violations occur: The SAS axiom (III.5) is violated. Moreover, neither the Theorem of Desargues, not the Little Theorem of Desargues, nor the Theorem of Pappus are valid. It is not possible to embed the planar geometry in a three-dimensional incidence geometry. Remark. If one wishes, even the axioms of continuity (V) can be assured to be valid. We see once more that it is impossible to deduct the SAS axiom (III.5) from the remaining axioms I, II, (III.1),(III.2),(III.3),(III.4), IV,V. Remark. This non-Desarguesian plane is due to Frederic Moulton (1902), and is available in Hilbert’s Foundations. The simplified version given in Definition 4.4 is mentioned in the recent book by Stillwell [34]. Proof. We begin with any Pythagorean plane. Thus we have a Hilbert plane, for which the axioms of incidence (I.1)(I.2)(I.3a)(I.3b), order (II.1) through (II.4), and congruence (III.1) through (III.5) hold. Moreover, the axiom of parallelism (IV.1) is assumed to hold. By Proposition 7.43, the existence of a parallel is shown. Thus the Euclidean Parallel Postulate 3.3 is valid. This Pythagorean plane is now used as the underlying reality, and a new model is constructed by redefining the notions "lines", "congruence of segments", "congruence of angles". The part of the construction without congruence has already been explained in Definition 4.4, but is repeated for convenience. We choose one line x, a positive direction on it, and mark the left half-plane as the upper half-plane, the right half-plane as the lower half-plane. 1 Let any line l which is not parallel nor equal to x, be given. Let x+ be the ray from the intersection L := x ∩ l in the positive x-direction. Let k be the ray on the line l in the lower half-plane, and h be the ray on l in the upper half-plane. For a line l with positive slope, we choose any point P on the ray h and drop the perpendicular onto the x-axis. Let F be the foot point, M be the midpoint of the segment −−→ FP, and h0 = LM. In other words, this is the ray with the same vertex L as h but half its slope. A "line" of the new geometry is now each one of the following: the x-axis; any line parallel or perpendicular to the x-axis; 1
In case of a construction from a real coordinate plane, this line is the x-axis. With real coordinates, one is including the validity of axioms of continuity (V.1)(V.2).
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any line with obtuse angle ∠(h, x+ ); in case the angle ∠(h, x+ ) is acute, the "line" is the union of the two rays k ∪ h0 . The "order" of the points on the new lines is defined in the obvious way. The "distance" of any two points is defined to be the Euclidean distance, but as measured along the new lines. Two segments are congruent, if the distances between their endpoints are equal. Problem 7.22. Convince yourself that from the notions above, one has obtained a planar geometry for which the axioms of incidence (I.1),(I.2),(I.3), all axioms of order (II), the axioms of congruence (III.1),(III.2),(III.3), and the Euclidean Parallel Postulate 3.3are all valid. The angles which do not have their vertex on the x-axis are measured by constructing an angle with vertex on the x-axis and having its sides parallel to the given angle. To measure an angle ∠(m0 , l0 ) with vertex on the x-axis, we replace any of its side with positive slope by a ray from the same vertex with double its slope. Thus one goes make to the original unbroken lines. Problem 7.23. Convince yourself that from the notions above, all parts of axiom of congruence (III.4) are valid. By the figures below on page 89, page 90, and page 95, one sees easily that neither the converse part of the Theorem of Desargues, nor the Little Theorem of Desargues, nor the Little Pappus Theorem are valid in the new incidence plane. To get such easy counterexamples, it is convenient to use a case in which only very few "lines" of the new geometry are affected by their intersection with the x-axis. 7.17. Restriction of SAS congruence The SAS axiom restricted to triangles of the same orientation is discussed by Hilbert in an appendix to the Foundations of Geometry. Note that the triangles 4ABC and 4A0 B0C 0 have the same orientation if either both point C lies −−−→ −−→ in the left half-plane of ray AB and point C 0 lies in the left half-plane of ray A0 B0 ; or both point C −−−→ −−→ lies in the right half-plane of ray AB and point C 0 lies in the right half-plane of ray A0 B0 . Hilbert postulates SAS congruence restricted to triangles with the same orientation. and two additional axioms: III.5* Assume that the two triangles 4ABC and 4A0 B0C 0 have the same orientation and the congruences AB A0 B0 , AC A0C 0 , ∠BAC ∠B0 A0C 0 hold, then the congruence ∠ABC ∠A0 B0C 0 is also satisfied. III.6 If the angles ∠(h0 , k0 ) and ∠(h00 , k00 ) are both congruent to angle ∠(h, k), then the angles ∠(h0 , k0 ) and ∠(h00 , k00 ) are congruent to each other. III.7 If the rays c and d have the same vertex as the angle ∠(h, k) and lie in the interior of that angle, then the angles ∠(h, k) and ∠(c, d) are not congruent. Hilbert then announces the following result without a proof: Proposition 7.62 (The importance of Euclid I.5). We assume only the restricted SAS axiom (III.5*), axioms (III.6) and (III.7), and finally we assume Euclid (I.5) to be valid: the base angles of any isosceles triangle are congruent. These assumptions imply the unrestricted SAS axiom (III.5),— for any triangles independent of their orientation.
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Problem 7.24 (A hard problem, only for fans). Provide some drawings and prove this proposition. Definition 7.23 (Skew-Hilbert plane). A skew-Hilbert plane is any model for two-dimensional geometry where Hilbert’s axioms of incidence (I.1)(I.2)(I.3a)(I.3b), order (II.1) through (II.4), and congruence (III.1) through (III.4) , the restricted SAS-axiom (III.5*), and the axioms (III.6) and (III.7) hold. Lemma 7.8 (Transfer of a triangle with the same orientation). Any given triangle 4ABC can be transferred into either the left or the right half-plane of any given ray r, to obtain a congruent triangle with one vertex at the vertex of this ray, one side lying on the given ray, and having the same orientation as the original triangle. Lemma 7.9 (Preparations in a skew-Hilbert plane). The following Propositions are valid in any skew-Hilbert plane • The angle congruence is an equivalence relation. Too, Proposition 7.26 holds: angle comparison holds for congruence classes. • Proposition 7.11 about SAS congruence holds for two triangles with the same orientation. • As stated in Lemma 7.8, the transfer of a triangle is still possible under the restriction that the orientation is preserved. • Proposition 7.19 about angle-addition and subtraction (Theorem 15 in Hilbert) remains valid. • The Exterior Angle Theorem (Euclid (I.16). Theorem 22 in Hilbert, see Proposition 7.40) is valid. • The SAA-Congruence Theorem, Proposition 7.54 (Theorem 25 in Hilbert) is valid for triangles of the same orientation. • The midpoint of a segment can be obtained by the method used in Theorem 26 of Hilbert (see Proposition 7.56). Any segment has a midpoint. Lemma 7.10. The following Propositions are valid in a skew-Hilbert plane where additionally Euclid (I.5) is assumed. • From the Theorem 7.21 about the symmetric kite, part (i) holds. 1 • Euclid(I.18) (Theorem 23 of Hilbert, see Proposition 7.48 above) is valid: in any triangle, across the longer side lies the greater angle. • Euclid (I.19) (Proposition 7.49 above) is valid: in any triangle, across the greater angle lies the longer side. • We get Euclid (I.6), the converse of Euclid (I.5). Lemma 7.11. The diagonals of the rhombus bisect each other, they bisect the angles at the vertices of the rhombus, and they are perpendicularly to each other. For any isosceles triangle 4ABC, we can construct a rhombus ABCD which inherits three of its vertices from the triangle. Proof of the Lemma. The midpoint M of side AC can be obtained with Hilbert’s construction. From SAS axiom (III.5*) we derive the triangle congruences 4AMB 4CMB00
and
4CMB 4AMB00
Hence we have obtained a rhombus ABCB00 . Too, get four congruent angles at vertices A and C and see that the diagonal AC bisects the angles at these two vertices. Beginning with the isosceles 1
Part (ii) cannot be recovered at this point but only later.
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Figure 7.80. To get a rhombus without use of reflection.
triangle 4BAB00 we repeat a similar argument. Hence the diagonals of a rhombus bisect each other, and they bisect the angles at the vertices of the rhombus. It remains to check that the diagonals are perpendicular to each other. To this end, we transfer the hypothenuse BC of triangle 4BMC on the segment BA and obtain the congruent triangle 4BM 0 A 4MBC with the same orientation. One needs only axioms (III.1),(III.4) and the restricted SAS axiom (III.5*). From the first item (i) of the Theorem 7.21 about the symmetric kite, we conclude ∠BMA ∠BM 0 A. From the triangle congruence we conclude we conclude ∠BMC ∠BM 0 A. Hence ∠BMC ∠BMA are congruent supplementary, and hence right angles. Lemma 7.12. All points P of the perpendicular bisector of two points B and C have congruent distances BP CP from these two points. Proof. We may suppose that point P does not lie on the line BC. We take the smaller one of the two angles ∠CBP and ∠BCP and transfer it to the vertex and onto the side of the other one. −−→ Suppose to be definite that ∠BCP ≤ ∠CBP. We transfer ∠BCP onto the ray BC. By the Crossbar Theorem, the newly produced ray intersects the segment BP, say at point Q. The triangle 4CQB has congruent base angles, and hence from Euclid (I.6) we conclude BQ CQ. It remains to show that Q = P. Let M be the midpoint of segment BC. From the Proposition about the rhombus (Lemma 7.11) we know that line QM is the perpendicular bisector of segment BC. Both P and Q are the intersection point of the two different lines QM and BP, hence P = Q. Finally we have confirmed BP CP.
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Lemma 7.13. In a skew-Hilbert plane Euclid (I.5) has been assumed. Theorem 7.21 about the symmetric kite, parts (i) and (ii) both hold. Indeed, for a kite with congruent sides XZ1 XZ2 and YZ1 YZ2 , and Z1 and Z2 on different sides of XY, the triangles 4XZ1 Y 4XZ2 Y are congruent. End of the proof of the Theorem 7.21 about the symmetric kite. Let Z1 and Z2 be two points on different sides of line XY, and assume that XZ1 XZ2 and YZ1 YZ2 . As explained above, from the isosceles triangle 4Z1 XZ2 , we construct a rhombus Z1 XZ2 X 0 . Its diagonals XX 0 and Z1 Z2 intersect perpendicularly in the midpoint M of segment Z1 Z2 . We begin a similar procedure with the isosceles triangle 4Z1 YZ2 and obtain a second rhombus Z1 Y 0 Z2 Y. Its diagonals YY 0 and Z1 Z2 intersect perpendicularly in the midpoint M of segment Z1 Z2 . Hence all four points X, X 0 , Y, Y 0 lie on the perpendicular to segment Z1 Z2 , erected at its midpoint M,—which we recognize as the perpendicular bisector of Z1 Z2 . Since we have shown that the diagonals XX 0 and YY 0 bisect the angles at X and Y, we get the remaining angle congruences for triangles 4XZ1 Y 4XZ2 Y. Lemma 7.14. Assume that the triangles 4APB and 4APC lie on different sides of AP, have congruent sides AB AC and congruent angles ∠PAB ∠CAP between these sides and their common side. Then the two triangles are congruent. Proof. From the Proposition about the rhombus (Lemma 7.11) we know that line AP is the perpendicular bisector of segment BC. From Lemma 7.12, we know that BP CP. From the Theorem 7.21 about the symmetric kite we conclude that the triangles 4APB 4APC are congruent. End of the proof of the Proposition about importance of Euclid (I.5). Combine Lemma 7.14 with the Lemma 7.8 about the transfer of a triangle. Assume the two triangles 4ABC and 4A0 B0C 0 have different orientations, have corresponding congruent sides AB A0 B0 and AC A0C 0 , and congruent angles ∠BAC ∠B0 A0C 0 between them. −−−→ We transfer the angle ∠BAC onto the ray A0C 0 , at vertex A0 , to the same side of A0C 0 as B0 . On the newly produced ray, we transfer segment AB, starting at vertex A0 . Thus we get point B0 , such that A0 B0 AB. Because of Lemma ?? we get 4ABC 4A0 B0C 0 By Lemma 7.14 we conclude that 4A0 B0C 0 4A0 B0C 0 Hence 4ABC 4A0 B0C 0 as to be shown.
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8. Measurement and Continuity Hilbert postulates two axioms of continuity: the axiom of Archimedes, and the axiom of completeness. 8.1. The Archimedean axiom Given are any segments AB and CD. We now use the second one as a measure unit. The Archimedean axiom states existence of natural number n and as finite −−→ sequence of points A = A0 , A1 , A2 , . . . , An on the ray AB such that CD Ak−1 Ak for k = 1, 2, . . . n
and
A ∗ Ak−1 ∗ Ak for k = 2, 3 . . . n
(8.1)
and the point B either lies between An−1 and An , or B = An−1 . In Hilbert’s words, "n segments congruent to CD constructed contiguously from A, along a ray from A through B, will pass beyond B." As a shorthand notation, we shall write n · CD > AB
(8.2)
More colloquially, one can say that no segment is too long as to be measured in terms of a given unit segment. As a first consequence of this fact, we note that no segment is so short that it cannot
Figure 8.1. The Archimedean axiom. For the case drawn in the figure, it turns out that n = 7.
be measured in terms of a given unit segment. (Repeated bisection produces arbitrarily short segments). Given any segments PQ and EF, PQ there exists a natural number s such that s successive bisections produce a segment P s Q s = s 2 shorter that segment EF, written as a formula PQ < EF 2s Especially, not both points E and F can lie inside segment P s Q s or on its endpoints. Proof. We apply the Archimedean axiom to the segments AB := EF and CD := PQ. Hence there exists a natural number n such that n · EF > PQ (8.3) Let s be an integer such that 2 s ≥ n. We get 2 s · EF > PQ Bisection of all segments involved leads inductively to the inequalities 2 s−1 · EF > as to be shown.
PQ PQ PQ PQ , 2 s−2 · EF > , 2 s−3 · EF > , . . . , EF > s 2 4 8 2
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The Archimedean axiom allows the measurement of segments and angles using real numbers. These real numbers occur during the measurement process in the form of binary fractions. Since Hilbert, this axiom is also known as the axiom of measurement. To start the measuring process, one assigns the length one to some arbitrary—but convenient—segment. We shall call it the unit segment. Main Theorem 9 (Measurement of segments). Given is a unit segment OI. There exists a unique way of assigning a length to any segment. We denote the length of segment AB by |AB|. 1 We construct the length as an (finite or infinite) binary fraction. Too, this length is called the distance of points A and B. The length has the following properties: (1) Positivity If A , B, then |AB| is a positive real number, and |OI| = 1. (2) Congruence |AB| = |CD| if and only if AB CD. (3) Ordering |AB| < |CD| if and only if AB < CD. (4) Additivity The distances are additive for any three points on a line. With point B lying between A and C, we get |AC| = |AB| + |BC|. Proof. We begin by assuming existence of the measurement function, and really construct it. In a second step, we confirm that the construction has produced the result as claimed. Let M be the midpoint of the unit segment OI. Since OM MI, the additivity of segment lengths implies |OM| + |MI| = 1, and hence |OM| = |MI| = 12 . Successive bisections now produce segments of lengths 14 , 18 , . . . . A length is assigned to any given segment AB by the process indicated in the
Figure 8.2. This measurement yields |AB| = 1.1101 . . . as a binary fraction.
figure on page 228. As in the Archimedean axiom, one begins by constructing a finite sequence of −−→ points A0 = A, A1 , A2 , . . . , An on the ray AB such that −−→ −−→ OI Ak−1 Ak and AAk = AB for k = 1, 2, . . . n (8.4) This part of the process stops, and the integer part of the length |AB| is determined, as soon as point B either lies between An−1 and An , or B = An−1 . In the latter case, one gets the length |AB| = n − 1 exactly, and the measure process is finished. In general, the former case occurs. One concludes that n − 1 < |AB| < n, and needs to construct the digits d s of an finite or infinite binary fraction n − 1 . d1 d2 d3 . . . in order to achieve a measurement with more and more precision. 1 To begin the process, we set the lower bound L0 := An−1 , and the upper bound U0 := An . The measurement uses sequences for the lower bound, digit, and upper bound, which are constructed inductively by repeated bisections. 1
In projective geometry and other contexts, one uses a signed length, which I shall denote by AB. Of course, any real life measure has to stop such an infinite process, and hence cannot confirm an exact real number as result, but only a finite number digits, limited by the technical possibilities. 1
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Inductive measurement step to determine d s . For s = 1, 2, 3, . . . , let H s−1 be midpoint of segment L s−1 U s−1 . If point B lies between L s−1 and H s−1 , one puts new digit d s := 0 lower bound L s := L s−1 upper bound U s := H s−1 If point B lies between H s−1 and U s−1 , or B = H s , one puts new digit d s := 1 lower bound L s := H s−1 upper bound U s := U s−1 Hence the digit d s depends on whether point B lies left or right of H s−1 . In both cases, the approximation to |AB| obtained by the s-th step is |AL s | = n − 1 +
d1 d2 d3 ds + + + ··· + s 2 4 8 2
It can happen that B = H s−1 , in which case this approximation is exact, and the measure process is finished. In general, the measurement process does not stop, but produces an infinite fraction. By Cantor’s principal of boxed intervals, the infinite binary fraction is indeed a real number. Question. Explain once more, how the first digit d1 is determined. Answer. One puts the lower bound L0 := An−1 and the upper bound bound U0 := An , with An−1 and An obtained via the Archimedean axiom. Let H0 be the midpoint of segment L0 U0 . If point B lies between L0 and H0 , one puts new digit d1 := 0 current approximation |AL1 | = n − 1 lower bound L1 := L0 upper bound U1 := H0 If point B lies between H0 and U0 , or B = H0 , one puts new digit d1 := 1 current approximation |AL1 | = n − 1 + lower bound L1 := H0 upper bound U1 := U0
1 2
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Finally, one wants to be convinced that the congruence, ordering, and additivity properties stated as (2),(3) and (4) are really satisfied. We start by confirming (2a) If AB CD, then |AB| = |CD|. (3a) If AB < CD then |AB| < |CD|.
If AB > CD then |AB| > |CD|.
It is rather straightforward to deduct (2a) from Hilbert’s axiom III.3. Let any two segments AB < AC be given. Repeated bisection produces arbitrarily short segments. Hence there exists a number s such that OI (8.5) L s U s s < BC 2 Because we go through the measurement process explained above for segment AB, we know that B = L s or L s ∗ B ∗ U s . By the segment comparison above, not both points B and C can lie in the interval L s U s , but indeed L s ∗ U s ∗ C. Hence the binary fractions for points B and C differ in the sth place, and hence have less than s common places, furthermore |AB| , |AC|. Assume that t is the first place where the measurement fractions of points B and C are different: b1 = c1 , b2 = c2 , . . . , bt−1 = ct−1 , but bt , ct
(8.6)
It can happen by accident that t from equation (8.6) is much smaller than any number s for which equation (8.5) holds. The four-point theorem yields the natural order Lt−1 ∗ B ∗ C ∗ Ut−1 , and its generalization to five or more points leads to the natural order Lt−1 ∗ B∗ Ht−1 ∗C ∗ Ut−1 , or C = Ht−1 . Hence |AB| < |AC| follows by using the t-th approximations. The second part of (3a) is quite similar to check. Since all segments are comparable, items (2a) and (3a) implies the converse statements, and hence (2) and (3) are confirmed. The additivity (4) can at first be checked by induction to hold for all segments of integer lengths. Next, one can prove additivity for segments of which lengths are finite binary fractions, using induction on the number of digits occurring. I only explain the further case that the length of segment AB is given by a finite binary fraction, but segment BC by an infinite fraction. For any arbitrary lower and upper bound finite fractions BL < BC < BU, we get BL < BC < BU |BL| < |BC| < |BU| |AB| + |BL| < |AB| + |BC| < |AB| + |BU| |AL| < |AB| + |BC| < |AU| and on the other hand
Both statements
AB + BL < AB + BC < AB + BU AL < AB + BC < AU |AL| < |AB + BC| < |AU| |AL| < |AB| + |BC| < |AU| |AL| < |AB + BC| < |AU|
hold for all lower and upper bounds L and U. Hence the Archimedean axiom for the real numbers implies |AB| + |BC| = |AB + BC|.
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Corollary 17. Given any three points A, B and C, the equality |AB| + |BC| = |AC| holds if and only if the three points lie on a line and either two of them are equal or B lies between A and C. The measurement of angles is done quite similar to the measurement of segments. Contrary to the situation for segments, there exists a well defined and convenient unit of measurement, which is the right angle. The traditional measurement in degrees is obtained by assigning the value 90◦ to the right angle. Main Theorem 10 (Measurement of angles). There exists a unique way of assigning a degree measurement to any angle. We denote the value of angle ∠ABC by (∠ABC)◦ . 1 −−→ For additivity, we consider a ray BG in the interior angle ∠ABC. Following definition 7.13, the angle ∠ABC is the sum of the angles ∠ABG and ∠GBC, written as ∠ABC = ∠ABG + ∠GBC. The measurement has the following properties: (1a) Positivity For any angle ∠ABC, the degree measurement is a positive number and 0◦ < (∠ABC)◦ < 180◦ . (1b) Right angle The measurement of a right angle is 90◦ . (2) Congruence (∠ABC)◦ = (∠A0 B0C 0 )◦ if and only if ∠ABC ∠A0 B0C 0 . (3) Ordering (∠ABC)◦ < (∠A0 B0C 0 )◦ if and only if ∠ABC < ∠A0 B0C 0 . (4) Additivity If ∠ABC = ∠ABG + ∠GBC, then (∠ABC)◦ = (∠ABG)◦ + (∠GBC)◦ . Question. Why does bisecting any angle produce an acute angle?
Figure 8.3. The angular bisector
Answer. It is shown in the figure on page 231, how to bisect an angle. Let A be the vertex of the angle. One transfers two congruent segments AB and AC onto the two sides of the angle, both 1
In calculus and differential geometry, one uses often measurement using the arc length on the unit circle. In that case, the right angle is assigned the value π2 .
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starting from the vertex A. The perpendicular, dropped from the vertex A onto the segment BC, is the angular bisector. Let F be the foot point of the perpendicular. As a consequence of the exterior angle theorem, we have shown that the two further angles in a right triangle are acute. Hence the angle ∠BAF is acute, but this is just the bisected angle. Theorem 8.1 (The Archimedean property for angles, Hilbert’s Proposition 34). Assume that the Archimedean Axiom holds. For any two angles α and ε there exists a natural number r such that α <ε (8.7) 2r Proof of Proposition 8.1. The angle α2 is acute. If α2 ≤ ε, assertion (40.11) holds with r = 1, or r = 2 in case of equality. We need a construction for the remaining case α2 > ε. Choose any point C on one side of the angle α2 and drop the perpendicular on the other side of that angle. Let B be the foot point of that perpendicular, and let A be the vertex of α2 . Next we transfer angle ε at vertex A, with one side AB, and the other side inside the angle α2 = ∠CAB. By the Crossbar Theorem, there exists a point D where this other side intersects segment BC. By the Archimedean axiom, there exists a natural number n such that n · BD > BC
(8.8)
Now, one transfers the angle ε, repeating n times, always with vertex A, using common sides, and turning away from segment AB. Let C0 = B, C1 = D. Let the new sides of transferred angle ε −−→ intersect BC at points C1 = D, C2 , C3 , . . . . We distinguish two cases (8.9) or (8.10):
Figure 8.4. Many consecutive small angles surpass any angle.
−−→ (8.9) It can happen that one of the new sides do no longer intersect ray BC. Let m be the smallest −−→ number of angle transfers for which this happens. Since the second leg AC of angle α2 does −−→ intersect ray BC, we conclude that α m·ε> (8.9) 2 Otherwise the Crossbar Theorem would lead to a contradiction. (8.10) The other possible case is that the new sides produced by transferring angle ε all n times −−→ intersect ray BC, say at points C1 = D, C2 , C3 , . . . , Cn .
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−−→ Let E = Cn be the point where the new side of nth angle ε intersects BC. The segments C0C1 , C1C2 , C2C3 , . . . , Cn−1Cn cut on line BC by these angles ε = ∠Ck ACk−1 satisfy Ck−1Ck < Ck Ck+1 for k = 1, 2, 3, . . . , n − 1, as follows from Proposition 33 given below. Hence a simple induction argument yields Ck−1Ck ≥ BD and BCk > k · DB for k = 1, 2, 3, . . . , n − 1. Hence (46.7) implies BCn > n · BD > BC. Because ∠BACn = n · ε, we get α (8.10) n·ε> 2 Let r be an integer such that 2r−1 ≥ m or 2r−1 ≥ n, in case (a) or (b), respectively. Now (8.9) or (8.10) imply α 2r−1 · ε > 2 Bisection of those two angles leads inductively to the inequalities α α α 2r−2 · ε > , 2r−3 · ε > , . . . , ε > r 4 8 2 This finishes the proof of Proposition 34. Here is still the missing proposition, already used above. (Hilbert’s Proposition 33). Let triangle 4OPZ have a right angle at vertex P. Let X and Y be two points on segment PZ such that ∠XOY = ∠YOZ. Then XY < YZ.
Figure 8.5. Congruent angles cut longer and longer segments from a line.
−−→ Proof of Proposition 33. One transfers segment OX onto the ray OZ at vertex O. One gets the segment OX 0 OX. Because segment OZ is the side opposite to the obtuse angle in 4OXZ, and the obtuse angle is the largest angle of any triangle, we get OZ > OX from Euclid I.19. Hence the point X 0 lies between O and Z. From the exterior angle theorem (Euclid I.16) for 4OYZ one gets ∠OZY < ∠OY X. From the exterior angle theorem for 4OX 0 Y one gets ∠OY X 0 < ∠Y X 0 Z. Hence ∠X 0 ZY = ∠OZY < ∠OY X ∠OY X 0 < ∠Y X 0 Z
(8.11)
In figure 8.1, we see that α < β < γ. By Euclid I.19, the side opposite to a larger angle is larger. We use this theorem for 4X 0 YZ. Hence (8.11) implies X 0 Y < YZ and hence XY X 0 Y < YZ as to be shown.
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Sketch of the proof for the measurement of angles. Let the given angle be ∠BAC. We erect the −−→ perpendicular to the ray AB at vertex A. Depending on whether the perpendicular is inside the supplementary right angle, or inside the given angle, or the perpendicular coincide with the second −−→ side AC, the given angle is acute, obtuse or right. The measurement process is now most easily explained by taking the right angle as unit. Starting with that one of the two complementary right −−→ angles, inside of which the ray AC lies, one successively bisects angles—always keeping the ray −−→ AC inside of them. The left bounds of the bisected angles correspond to an infinite fraction 0 . d1 d2 d3 . . . for an acute, or 1 . d1 d2 d3 . . . for an obtuse angle. To obtain a measurement in the traditional degrees, these binary fractions have to be multiplied by 90 1 The further details are so similar to the case of segment measurement that we do not need to repeat. 8.2. Axioms related to completeness The clear-cut understanding of this material was achieved only in the late nineteenth century. There are several axioms for completeness, with very similar implications, having slight but deep differences. It is hard to say what is the most natural one of these axioms. Even Hilbert has suggested different axioms of continuity in different editions of his foundations of geometry. 8.2.1. Cantor’s axiom Following Baldus and Löbell [7], p.43, I state my favorite version of Cantor’s axiom. Cantor’s Axiom. There exists at least one segment A1 B1 with the following property: Given any sequence Ai Bi of boxed subsegments for i = 2, 3, . . . such that Ai−1 ∗ Ai ∗ Bi ∗ Bi−1 for all i = 2, 3, . . .
(8.12)
there exists a point X ∗ such that Ai ∗ X ∗ ∗ Bi for all i = 1, 2, 3, . . .
(8.13)
For any every Hilbert plane, the following similar, but a bid more general statement is an immediate consequence: Cantor’s principle of boxed intervals For every sequence of boxed segments Ai Bi such that either Ai−1 ∗ Ai ∗ Bi ∗ Bi−1 or Ai−1 = Ai , Ai−1 ∗ Bi ∗ Bi−1 or Ai−1 ∗ Ai ∗ Bi−1 , Bi−1 = Bi , (8.14) for all i = 2, 3, . . . , there exists a point X ∗ such that either Ai ∗ X ∗ ∗ Bi or X ∗ = Ai or X ∗ = Bi for all i = 1, 2, 3, . . .
(8.15)
Problem 8.1. Convince yourself, and explain that Cantor’s axiom implies Cantor’s principal of boxed intervals, just assuming the axioms of incidence, order and congruence. Main Theorem 11 (Completeness of measured segments and angles). Assume the axioms of incidence, order and congruence, as well as the Archimedean and Cantor’s axiom. 1
In binary notation 90 = 64 + 16 + 8 + 2 = 1011010
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Given is a unit segment OI of length |OI| = 1. For every real number r > 0, there exists a segment segment AB of length |AB| = r. For every real number 0 < d < 180, there exists an angle, which has the measurement ∠ABC = d, in traditional degrees. Remark. As far as the continuity axioms are concerned, one needs to assume the Archimedean axiom (V.1) and Cantor’s principle of boxed intervals (V.2). I have stated these axioms of continuity 14.1 among a simplified axiomatization of geometry 14.1. Indication of reason. The real number r can be given as an finite or infinite binary fraction r = n − 1 . d1 d2 d3 . . . The boxed intervals are constructed as explained in the measuring process. Let L0 U0 be a segment of unit length such that |AL0 | = n − 1 and |AU0 | = n. We define a sequence of boxed intervals L s U s by setting L s := L s−1 , U s := H s−1 if d s = 0 L s := H s−1 , U s := U s−1 if d s = 1 for s = 1, 2, 3, . . . . Here H s−1 is midpoint of segment L s−1 U s−1 . In both cases, the left lower approximation to |AB| obtained by the s-th step is |AL s | = n − 1 +
d1 d2 d3 ds + + + ··· + s 2 4 8 2
In the special case of a finite fraction, one gets X ∗ = H s−1 , in which case this approximation is exact, and the measure process is finished. For an infinite binary fraction, we need Cantor’s principal of boxed intervals: There exists a point X ∗ lying is in the infinite intersection of all intervals L s U s . The interval L0 X ∗ has length |L0 X ∗ | = r equal to the originally given number r. 8.2.2. Dedekind’s axiom This is now most popular axiom for continuity. Let us start with a definition. Definition 8.1 (Dedekind Cut). A Dedekind cut of the line l is a pair of sets {Σ1 , Σ2 } such that the set of points on a line l is the disjoint union of the two nonempty sets Σ1 and Σ2 , which have the following property: (P) If P1 and P3 are any two points in Σi , and the third point P2 lies between P1 and P3 , then P2 is in the same set Σi , for i = 1, 2. Dedekind’s Axiom. Assume that {Σ1 , Σ2 } is a Dedekind cut of the line l. Then there exists a cut point K ∗ on the line l such that Σ1 ∪ {K ∗ } and Σ2 ∪ {K ∗ } are the two opposite rays on the line l with vertex K ∗ . In his recent book [1], Greenberg writes on p.260 about this axiom: We bring in our deus ex machina, as classical Greek theater called it—a god comes down from heaven to save the day. This axiom was proposed by J.W.R. Dedekind in 1871. Here is what Dedekind says about continuity in "Stetigkeit und irrationale Zahlen" (1872):
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I find the essence of continuity in the following principle: "If all the points of a line fall into two classes in such a way that each point of the first class lies to the left of each point of the second class, then there exists one and only one point that gives rise to this division of all the points into two classes, cutting of the line into two pieces." As mentioned before, I believe I am not wrong if I assume that everyone will immediately admit truth of this assertion; most of my readers will be very disappointed to realize that by this triviality the mystery of continuity will be revealed. I am very glad if everyone finds the above principle so clear and so much in agreement with our own conception of a line; for I am not in a position to give any kind of proof of its correctness; nor is anyone else. The assumption of this property of a line is nothing else than an axiom by which we first recognize continuity of the line, through which we think continuity into the line. If space has any real existence at all, it does not necessary need to be continuous; countless properties would remain the same if it were discontinuous. And if we knew for certain that space was discontinuous, still nothing could hinder us, if we so desired, from making it continuous in our thought by filling up its gaps; this filling up would consist in the creation of new point-individuals, and would have to be carried out in accord with the above principle. Indeed, Dedekind’s axiom is a very strong axiom. It essentially introduces the real numbers into our geometry, which is not in the spirit of Euclid, but useful and even necessary from the engineering point of view. Dedekind’s axiom is by no means necessary to do interesting mathematics— on the contrary— it spoils many fine points of algebra and set theory. Many more subtle distinctions and questions about constructibility are obscured by Dedekind’s axiom. Too, the strength of Dedekind’s axiom becomes apparent because of its many consequences, some of which are now explained. Theorem 8.2. Dedekind’s axiom implies the Archimedean axiom. Theorem 8.3. Dedekind’s axiom implies Cantor’s axiom. Main Theorem 12 (Dedekind’s axiom implies completeness). The elements of geometry—the points, lines and planes— which satisfy the axioms of incidence, order, congruence, and Dedekind’s axiom have no extension to any larger system, for which all these axioms still hold. To facilitate the proofs of theorem 8.2 as well as theorem 8.3, we introduce still another suggested axiom: Weierstrass’ Axiom. Let Ai for i = 2, 3, . . . be sequence of a points , and B a point, lying all on one line, such that Ai−1 ∗ Ai ∗ B for i = 2, 3, . . . (8.16) Then there exists a point K ∗ such that (i) either K ∗ = B or
Ai ∗ K ∗ ∗ B for i = 1, 2, 3, . . .
(8.17)
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(ii) Furthermore, every other point X such that Ai ∗ X ∗ B for i = 1, 2, 3, . . .
(8.18)
Ai ∗ K ∗ ∗ X ∗ B for i = 1, 2, 3, . . .
(8.19)
satisfies
Proposition 8.1. Dedekind’s axiom implies Weierstrass’ axiom. Proof that Dedekind’s axiom implies Weierstrass’s axiom. We define a Dedekind cut as follows: Let Σ1 consist of all points P such that either P ∗ A1 ∗ B or P = A1 or there exists i ≥ 1 such that A1 ∗ P ∗ Ai As follows logically from Hilbert’s four-point theorem (5.5), the complement Σ2 on the line A1 B consists of all points such that A1 ∗ Ai ∗ P for all i = 2, 3 . . . The cut point K ∗ of the Dedekind cut {Σ1 , Σ2 } needs to lie in Σ2 . It is either K ∗ = B or satisfies Ai ∗ K ∗ ∗ B for i = 1, 2, 3, . . . . This confirms item (i) of the Weierstrass axiom. Furthermore, if one assumes that any point X satisfies Ai ∗ X ∗ B for i = 1, 2, 3, . . .
(8.20)
then X ∈ Σ2 . Hence K ∗ ∗ X ∗ B because K ∗ is the vertex of the ray producing Σ2 . Now relation (8.20) and the four-point theorem imply Ai ∗ K ∗ ∗ X ∗ B for i = 1, 2, 3, . . .
(8.19)
as to be shown, in order to confirm item (ii) of Weierstrass’ axiom.
Proposition 8.2. Weierstrass’ axiom implies Cantor’s axiom. Proof that Weierstrass’s axiom implies Cantor’s axiom. Within the assumptions to setup Cantor’s axiom, Hilbert’s n-point theorem (5.7) implies inductively A1 ∗ Ai ∗ B j ∗ B1 for all i, j ≥ 2 The first item (i) of Weierstrass’ axiom implies Ai ∗ K ∗ ∗ B1 for i = 1, 2, 3, . . . Fix some index j. Since Ai ∗ B j ∗ B1 for all i = 1.2. . . . , the second item (ii) from Weierstrass’ axiom with X := B j implies Ai ∗ K ∗ ∗ B j ∗ B1 for i, j = 1, 2, 3, . . . as to be shown.
Proposition 8.3. Weierstrass’ axiom implies the Archimedean axiom. Proof. As in the Archimedean axiom, two segments CD and AB are given. We use CD as unit of measurement and make the following
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−−→ Definition 8.2. We say that a point P on the ray AB can be reached with the unit of measurement CD if and only if—as in the Archimedean axiom—, there exists a natural number n for which construction of the finite sequence of points A0 = A, A1 , A2 , . . . , An such that −−→ −−→ CD Ak−1 Ak and AAk = AB for k = 1, 2, . . . n
(8.4)
leads up to a point An such that point P either lies between An−1 and An , or P = An−1 . Let A1 , A2 , A2 , . . . be the sequence of points constructed by the Archimedean measurement (8.4). If the Archimedean axiom would not hold, then there would exist a point F which cannot be reached with the unit measurement CD. We can now apply Weierstrass’ Axiom for the sequence Ai and point F. Hence there exists a point K ∗ such that Ai ∗ K ∗ ∗ F for i = 1, 2, 3, . . .
(8.21)
In other words, the point K ∗ cannot be reached by measurement. Let K− and K+ be the points on both sides of K ∗ such that K ∗ K− K ∗ K+ CD, and K− ∗ K ∗ ∗ F. It is clear that these points K+ and K− cannot be reached by measurement, neither. Because of item (ii) of Weierstrass’ axiom, K ∗ is the point most to the left that cannot be reached by measurement: every other point X such that Ai ∗ X ∗ B for i = 1, 2, 3, . . .
(8.20)
Ai ∗ K ∗ ∗ X ∗ B for i = 1, 2, 3, . . .
(8.19)
satisfies Such a reasoning would now imply both for X := K+ and X := K− , which is impossible. This contradiction implies that the Archimedean axiom does hold. In order to defend my preference of Cantor’s axiom—and trying to take some of the deus ex machina image away from Dedekind—I prove now: Main Theorem 13. Assuming both the Archimedean axiom, as well as Cantor’s axiom, Dedekind’s axiom follows. Proof. Given is a line AB with a Dedekind cut {Σ1 , Σ2 } of it. Let OI be any measurement unit. We may assume A ∈ Σ1 and B ∈ Σ2 . As explained in the theorem of measurement, one measures the distance from A ∈ Σ1 to the cut point. Thus one can construct a sequence of approximations to the cut point: L s ∈ Σ1 and U s ∈ Σ2 such that |AL s | = n − 1 +
d1 d2 d3 ds + + + ··· + s 2 4 8 2
and |L s U s | = 2−s for s = 0, 1, 2, . . . . Cantor’s principle of boxed intervals yields a point X such that L s ∗ X ∗ U s , which is the cut point. 8.2.3. Hilbert’s axiom of completeness The following axiom of completeness is suggested in the millenium edition of Hilbert’s foundations: V.2 (Axiom of linear completeness) An extension of a set of points on a line, with its order and congruence relations existing among the original elements as well as the fundamental properties of line order and congruence that follow from Axioms I-III and from V.1, is impossible.
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Remark. The insight that it is enough to require the extension of a set of points on a line goes back to Paul Bernays. (Theorem of completeness, Hilbert’s Proposition 32). The elements of geometry—the points, lines and planes— have no extension to any larger system, under the assumptions that the axioms of incidence, order, congruence, and the Archimedean axiom still hold for the extension. The parallel axiom may be assumed or not, it does not interact with completeness at all. The axiom of completeness is not a consequence of the Archimedean axiom. But, on the contrary, the Archimedean axiom needs to be assumed for completeness to be meaningful to hold. Main Theorem 14. Together, the Archimedean axiom (V.1), and the axiom of linear completeness (V.2) imply Cantor’s axiom and Dedekind’s axiom. There exist infinitely many models for the axioms I through IV, and V.1. But only one model satisfies the axiom of completeness, too—this the Cartesian geometry. Historic context. In the very earliest edition, Hilbert proposed the conclusion of the Theorem of completeness (8.2.3) as an axiom. Only the German edition of the foundation of 1903 contains the axiom of linear completeness. Even earlier, the axiom of linear completeness had already appeared in May 1900 in the French translation of the foundation. It appeared for the very first time on October 12th 1899, in Hilbert’s lecture "Über den Zahlbegriff". Hilbert’s axiom of completeness has given rise to many positive as well as negative comments by important mathematicians. "An axiom about axioms with a complicated logical structure" (Schmidt) "An unhappy axiom" (Freudenthal) "The axioms of continuity are introduced by Hilbert, to show that they are really unnecessary." (Freudenthal) Hilbert’s completeness axiom is obviously not a geometric statement, and not a statement formalizable in the language used previously—so what does it accomplish? (M.J. Greenberg, 2010) "The foundations of geometry contain more than insight in the nature of axiomatic." (Freudenthal) "The most original creation in Hilbert’s axiomatic" (Baldus) "Hilbert has made the philosophy of mathematics take a long step in advance." (H. Poincaré)
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9. Legendre’s Theorems Recall that a Hilbert plane is a geometry, where the axioms of incidence, order, and congruence are assumed, as stated in Hilbert’s Foundations of Geometry. Neither the axioms of continuity (Archimedean axiom and the axiom of completeness), nor the parallel axiom needs to hold for an arbitrary Hilbert plane. The proposition numbering is taken from Hilbert’s Foundations of Geometry. For simplicity, I am using the letter R to denote a right angle. 9.1.
The First Legendre Theorem
Proposition 9.1 (The First Legendre Theorem, Hilbert’s Proposition 35). In a Hilbert plane, for which the Archimedean Axiom holds, every triangle has angle sum less or equal two right angles. Plan of proof. The proof relies on three ideas: (a) The exterior angle theorem Euclid I.16 given by Proposition 7.40, from which one concludes Euclid I.17 (Proposition 7.41): the sum of any two angles of a triangle is less than two right angles. (b) The construction from Euclid’s proof 7.9 of the exterior angle theorem. From a given triangle, this construction yields a second triangle, with the same angle sum as the first one, as explained in Lemma 9.1. Additionally, one of its angles is less or equal half of an angle of the original triangle. (c) The Archimedean property for angles given in Hilbert’s Proposition 8.1. Lemma 9.1. For any given 4ABC, there exists a 4A0 B0C 0 such that α0 + β0 + γ0 = α + β + γ
and
α0 ≤
α 2
(9.1)
Figure 9.1. Two triangles with same angle sum—and area.
−−→ Proof of the Lemma. Let D be the midpoint of side BC. Extend the ray AD and transfer segment AD to get point E such that AD DE. We need to consider two cases:
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(i) If ∠EAB ≤ ∠CAE, the new 4A0 B0C 0 has vertices A0 = A, B0 = B and C 0 = E. (ii) If ∠EAB > ∠CAE, the new 4A0 B0C 0 has vertices A0 = A, B0 = C and C 0 = E. We explain the details for case (i). By SAS congruence, 4ADC 4EDB, because the vertical angles at vertex D are congruent, and the adjacent sides are pairwise congruent by construction. The congruence of the two triangles yields two pairs of congruent angles γ = ∠ACD ∠DBE
and
γ0 = ∠DEB ∠DAC
From angle addition at vertices A and B, respectively, and a final addition of formulas, one concludes α = α0 + γ0 β + γ = β0 α + β + γ = α0 + β0 + γ0 A similar result is concluded in the second case (ii) via the congruence 4ADB 4EDC.
Figure 9.2. Two triangles with same angle sum, cases (i) and (ii).
Remark. Equivalent conditions all leading to case (i) are β ≤ γ ⇔ AC ≤ AB ⇔ EB ≤ AB ⇔ α0 ≤ γ0 Reason. The first equivalence is proved by Euclid I.18 and Euclid I.19 (Propositions 7.48 and 7.49) for triangle 4ABC. Moreover AC EB because of the triangle congruence 4ADC 4EDB. The third equivalence is proved by Euclid I.18 and Euclid I.19 (Propositions 7.48 and 7.49) for triangle 4ABE. Corollary 18. The two triangles 4ABC and 4A0 B0C 0 have the same area. Reason. To produce the new triangle 4A0 B0C 0 from the old triangle 4ABC, one needs to take away triangle 4ADC and add the congruent triangle 4EDB. End of the proof of the First Legendre Theorem. Let 4ABC be any triangle. We use the first Lemma repeatedly to get a sequence of triangles 4A0 B0C 0 = 4A1 B1C1 , 4A2 B2C2 , 4A3 B3C3 , . . . 4An BnCn , . . .
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such that α + β + γ = αn + βn + γn and αn ≤ theorem implies βn + γn < 2R and hence
α 2n
for all natural numbers n ≥ 0. The exterior angle
α + 2R (9.2) 2n for all n ≥ 0. Now a limiting process with n → ∞ implies the result. An accurate version of this part of the argument uses the Archimedean property for angles. We argue by contradiction, assuming that the angle sum would be α + β + γ > 2R. Let α + β + γ = αn + βn + γn <
ε := α + β + γ − 2R
(9.3)
which, because of the assumption α + β + γ > 2R, would be an angle ε > 0. By the Archimedean property for angles Proposition 8.1, there would exist a natural number r such that α <ε (9.4) 2r Now (9.2), (9.4) and (9.3) together would imply α α + β + γ = αr + βr + γr < r + 2R ≤ ε + 2R = α + β + γ 2 which is impossible. Because any two angles are comparable, α + β + γ ≤ 2R is the only possibility left, as was to be shown. 9.2.
The Second Legendre Theorem
Proposition 9.2 (The Second Legendre Theorem, Hilbert’s Proposition 39). Given is any Hilbert plane. If one triangle has angle sum 2R, then every triangle has angle sum 2R. It turns out to be easier, to prove at first a Proposition about quadrilaterals. We define some special quadrilaterals. Definition 9.1. A Saccheri quadrilateral ABCD has right angles at two adjacent vertices A and B, and the opposite sides AD BC are congruent. A Lambert quadrilateral has three right angles. A rectangle has four right angles. Proposition 9.3 (Saccheri and Lambert quadrilaterals, Hilbert’s Proposition 36). Assume that ABCD is a Saccheri quadrilateral with right angles at vertices A and B—and congruent opposite sides AD BC. Let M be the midpoint of segment AB. The perpendicular bisector m of segment AB intersects the opposite side CD at right angles, say at point N. One gets two congruent Lambert quadrilaterals AMND and BMNC. Proof. To show that m intersects the opposite side CD, one can use plane separation with line m. Indeed, points A and B lie on different sides of m by construction. By Euclid I.27, the three lines AD and m and BC are parallel. Hence points A and D lie on the same side of m. By the same reasoning, B and C lie on the same side of line m. Hence C and D lie on different sides of m. Thus segment DC intersects m. By SAS congruence, 4MAD 4MBC, since the right angles at vertices A and B and the adjacent sides match pairwise. Next we see that the triangles 4MDN 4MCN, again by SAS congruence, since the angles at the common vertex M and the adjacent sides match. Hence we conclude that ∠ADM ∠BCM and ∠MDN ∠MCN. Now angle addition yields ∠ADC ∠BCD Hence the quadrilateral ABCD has two congruent angles at vertices C and D. At vertex N, the angles ∠MND ∠MNC are congruent supplementary angles. Hence they are both right angles. This finishes the proof that AMND and BMNC are congruent Lambert quadrilaterals.
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Figure 9.3. A Saccheri quadrilateral is bisected into two Lambert quadrilaterals.
Figure 9.4. The steps to get symmetry of a Saccheri quadrilateral.
Definition 9.2. The reflection by line l is defined as follows: From a given point P the perpendicular is dropped onto l, and extended by a congruent segment FP0 PF beyond the foot point F. Then P0 is called the reflected point of P. Corollary 19. By the perpendicular bisector of its base line, a Saccheri quadrilateral is bisected into two Lambert quadrilaterals, which are reflection symmetric to each other. Proposition 9.4 (Hilbert’s Proposition 37). Assume ABCD is a rectangle. Drop the perpen←→ dicular from a point E of line CD onto the opposite side AB. Let F be the foot point. Then the quadrilaterals ADEF and BCEF are both rectangles. ← → ← → Proof. One reflects the segment EF, both by line AD and BC. Let Ei Fi for i = 1, 2 be the mirror images. By Proposition 9.3, both segments are congruent to EF. Because ABCD was assumed to be a rectangle, both points Ei lie on line CD and both Fi lie on line AB. The assumptions of Proposition 9.3 hold for all three quadrilaterals EFF1 E1 , EFF2 E2 and E1 F1 F2 E2 . Hence all three are Saccheri quadrilaterals. We get four congruent angles with vertices E1 , E2 and E—denoted in figure 9.2 by ε, θ, ϕ and κ. One of these three points E1 , E2 and E lies between the other two. At that vertex, one gets a pair of supplementary angles, which are congruent. Hence they are right angles, and the other angles just mentioned are right angles, too. Proposition 9.5 (Hilbert’s Proposition 38). If there exists a rectangle ABCD, then every Lambert quadrilateral is a rectangle.
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Figure 9.5. Getting rectangles of arbitrary width— Is ζ a right angle?
Figure 9.6. ε is congruent to both θ and ϕ, hence there are two supplementary right angles θ and ϕ!
Figure 9.7. If one rectangle exists, why are all Lambert quadrilaterals rectangles?
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Proof. Assume that the Lambert quadrilateral A1 B1C1 D1 has its three right angles at vertices A1 , B1 and D1 . One may assume that A = A1 , and the three points A, B, B1 as well as A, D, D1 lie on a line. We have drawn the case that A ∗ B ∗ B1 and A ∗ D1 ∗ D. The easy modification of the proof to the other possible orders of points A, B, B1 and A, D1 , D is left to the reader. As in Proposition 36, we show that segment D1C1 and line BC do intersect, say at point F. Now Proposition 37 implies that these two lines are perpendicular to each other. Hence ABFD1 is a rectangle. Applying Proposition 37 a second time, we conclude that the quadrilateral AB1C1 D1 is a rectangle, too. Corollary 20. If there exists a rectangle ABCD, then every Saccheri quadrilateral is a rectangle. There exists rectangles of arbitrarily prescribed height and width. Reason. Because of Proposition 9.3, a Saccheri quadrilateral is subdivided into two Lambert quadrilaterals by its midline. By Proposition 9.5, those are rectangles. Hence the Saccheri quadrilateral is a rectangle, too. The height and width of a Saccheri quadrilateral can be prescribed arbitrarily, and they are all rectangles. Hence there exists rectangles of arbitrarily prescribed height and width. Note of caution. In hyperbolic geometry, the height and width of a Lambert quadrilateral cannot be prescribed arbitrarily. Indeed if one chooses the width for a Lambert quadrilateral, the height has an upper bound. Proposition 9.6 (Hilbert’s Proposition 39). To every 4ABC with angle sum 2ω, there corresponds a Saccheri quadrilateral with two top angles ω.
Figure 9.8. To every triangle corresponds a Saccheri quadrilateral.
Proof. The Saccheri quadrilateral is ABGF, with right angles at F and G and congruent opposite sides AF BG, and congruent angles ω at A and B. The drawing indicates, how the Saccheri quadrilateral is obtained. One connects the midpoint D of segment AC and midpoint E of segment BC, by line l. Then one drops the perpendiculars from all three vertices A, B, C onto line l. Let F, G and H be the foot points of the perpendiculars, respectively.
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Figure 9.9. How to get two pairs of congruent triangles, and how to get the top angles.
By the SAA congruence theorem, 4AFD 4CHD, because of the right angles at vertices F and H, the congruent vertical angles at vertex D, and because segments AD DC are congruent by construction. By similar reasoning we get 4BGE 4CHE. From these triangle congruences we conclude that HC FA, and HC GB. Hence FA GB, which implies that ABGF is a Saccheri quadrilateral. Its top angles at vertices A and B were shown to be congruent in Proposition 36. We denote them by ω. From the triangle congruences, too, we get γ1 = ∠DCH ∠DAF and γ2 = ∠ECH ∠EBG . The sum of the angles of 4ABC is α + β + γ = α + γ1 + β + γ2 = ∠FAB + ∠GBA = 2ω as to be shown.
Corollary 21. A triangle has angle sum 2R if and only if the corresponding Saccheri quadrilateral is a rectangle. Proof. This is an immediate special case of Proposition 9.6.
End of the proof of the Second Legendre Theorem. Assume triangle 4ABC has angle sum 2R. Question. What can you say about the corresponding Saccheri quadrilateral? Answer. The Saccheri quadrilateral corresponding to 4ABC is a rectangle by the Corollary. Hence, by Proposition 38, every Lambert quadrilateral is a rectangle. Now let 4XYZ be any triangle. Question. What can you say about the corresponding Saccheri quadrilateral? Answer. As explained in Proposition 9.3, the Saccheri quadrilateral corresponding to 4XYZ is bisected into two congruent Lambert quadrilaterals. By Proposition 9.5, these are both rectangles, since a rectangle already exists. Hence the Saccheri quadrilateral corresponding to 4XYZ is a rectangle. We have just shown that the Saccheri quadrilateral corresponding to the arbitrarily chosen 4XYZ is a rectangle, too. Hence by Proposition 9.6, the angle sum of 4XYZ is 2R.
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9.3. The alternative of two geometries Finally, one wants to link the angle sum of triangles to uniqueness of parallels. Proposition 9.7. If a triangle 4ABC has angle sum α + β + γ < 2R, then there exist two parallels to line AB through point C. Conversely, if there is a unique parallel to line AB through point C, then the triangle 4ABC has angle sum 2R. Corollary 22. In a Pythagorean plane, each triangle has angle sum 2R. In other words, each Pythagorean plane is semi-Euclidean. Proof. We form two congruent z-angles α = ∠BAC by transferring that angle with one side ray −−→ −−→ CA as transversal. The second side m1 of the new angle α has to lie the opposite side of CA as −−→ point B. Similarly one transfers angle β = ∠ABC with one side ray CB, and gets as second side the ray m2 . The angle formed by the rays m1 and m2 is α + β + γ < 2R. Hence m1 and m2 do not lie on the same line. On the other hand, the rays m1 and m2 are both (parts of) parallels to line l through point P. This follows from Euclid’s I.27 (Alternate interior angles imply parallels). We have thus constructed two different parallels to line l = AB through point C. Proposition 9.8 (The Third Legendre Theorem). Given is a Hilbert plane, for which the Archimedean Axiom is assumed to hold. If the angle sum of every triangle is 2R, then the Euclidean Parallel Postulate holds. Idea of the proof. Given is line l and a point P not on l. We need to check the uniqueness of the parallel to line l through point P. Inductively, there is constructed a sequence of isosceles triangles 4PFn−1 Fn . To start, let F0 = F, and let F1 be any of the two points on line l such that F0 F1 PF. Next let F2 be −−−→ the point on ray FF1 such that F1 F2 PF1 . Inductively, assume that F1 , F2 , . . . , Fn−1 have been −−−→ constructed, and let Fn is the point on ray FF1 such that Fn−1 Fn PFn− 1 and F ∗ Fn−1 ∗ Fn .
Figure 9.10. The angle sum 2R implies uniqueness of the parallel.
The angles φn = ∠Fn−1 PFn can all be calculated by means of the following
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Figure 9.11. Halfing an angle with an isosceles triangle.
Proposition 9.9. Assume that the angle sum for every triangle is 2R. Then the base angle of an isosceles triangle is half of the exterior angle at the top. Reason for proposition 9.9. Let δ be the exterior angle at the top vertex A of triangle 4ABC. Thus δ is the supplement of the interior angle α at that vertex. It was assumed that the angle sum of any triangle is 2R. Hence δ = 2R − α = α + β + γ − α = β + γ which is the sum of the two nonadjacent interior angles. For an isosceles triangle with top F = A, the two base angles β and γ are congruent by Euclid I.5. Hence the exterior angle at the top is δ = 2β double the base angle. Hence the base angle β = 2δ is half of the exterior angle δ. End of the proof of the third Legendre Theorem. 4F0 PF1 has a right angle on top. (It is a right isosceles triangle, for which all three angles can be calculated in Euclidean geometry.) Indeed the Proposition yields R φ0 = ∠F0 PF1 = (1.1) 2 The other base angle of 4F0 PF1 is the exterior angle on the top of 4F1 PF2 . Since this triangle is isosceles, too, the Lemma implies that its base angle is φ1 = ∠F1 PF2 =
φ0 R = 2 4
(1.2)
Inductively, we get that φn = ∠Fn−1 PFn =
R 2n
(1.n)
for all n = 0, 1, 2, . . . . Here is the induction step: Assume that (1.n-1) has been shown. The exterior angle on the top of 4Fn−1 PFn is also a R base angle of 4Fn−2 PFn−1 , which is φn−1 = 2n−1 by the induction assumption. Since 4Fn−1 PFn is isosceles, too, the Proposition implies that its base angle is half of that angle. Hence φn−1 R R = n−1 = 2 2 · 2 2n which confirms (1.n). By angle addition, formulas (1.n) imply φn =
∠FPFn = φ1 + φ2 + · · · + φn = R
! ! 1 1 1 1 + + ··· + n = R 1 − n 2 4 2 2
(2)
One parallel m to l through point P is conveniently constructed as "double perpendicular", as explained above. We now give an argument to show that m is the unique parallel to l through
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− → P. We assume towards a contradiction that m0 , m is a different parallel to l through P. Let m0 be −−→ one of the two rays on m0 starting at point P that forms an acute angle with the perpendicular PF. − → − Let → m be the ray on m starting at point P that lies on the same side of PF as m0 . − Legendre’s triangle construction needs to be done on that side of PF where the two rays → m and − →0 − → → − 0 m lie. Let ε > 0 be the angle between m and m . By the Archimedean axiom for angles, there exists a natural number r such that R <ε 2r − → −−→ Hence the (complementary) angle between the rays PF and m0 satisfies R−ε
R = ∠FPFr 2r
− → This shows that the ray m0 lies in the interior of ∠FPFr . Hence, by the Crossbar Theorem 5.9, the − →0 ray m intersects the segment FFr , say at point Q. Thus the line m0 is not a parallel to line l, which is a contradiction. Hence there exists only one parallel to line l through point P. For completeness, we restate Proposition 9.10 (The Crossbar Theorem). A segment with endpoints on the two sides of an angle and a ray emanating from its vertex into the interior of the angle intersect. Taking the three Legendre Theorems together one gets the Main Theorem 15 (Alternative of Two Geometries). In a Hilbert plane, for which the Archimedean Axiom is assumed, there occurs one of the two possibilities: (a) All triangles have angle sum two right angles. All Lambert and all Saccheri quadrilaterals are rectangles. The Euclidean parallel postulate holds. One has arrived at the Euclidean geometry. (b) All triangles have angle sum less than two right angles. All Lambert quadrilaterals have an acute angle. All Saccheri quadrilaterals have two congruent acute top angles. Rectangles do not exist. For every line l and point P not on l, there exist two or more parallels to line l through point P. In this case, one gets the hyperbolic geometry.
Figure 9.12. For angle sum less 2R, there exist two different parallels.
Reason. Pure logic tells that either (1) or (2) holds: (1) All triangles have angle sum two right angles.
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(2) There exists a 4ABC with angle sum not equal than two right angles. By the third Legendre Theorem, alternative (1) implies the Euclidean parallel postulate. By Proposition 39, we conclude that all Lambert and Saccheri quadrilaterals are rectangles. Thus case (a), the usual Euclidean geometry occurs. Now we assume alternative (2), and derive all the conclusions stated in (b). By the first Legendre Theorem, the 4ABC has angle sum less than two right angles. By the contrapositive of the second Legendre Theorem, we conclude that no triangle can have angle sum two right. Hence, again by the first Legendre Theorem, all triangles have angle sum less than two right angles. The statement about quadrilaterals follows from Proposition 9.6. The Proposition below gives the well known construction, leading two different parallels. Thus all assertions of case (b) hold—we have arrived at the hyperbolic geometry. 9.4.
What is the natural geometry? A legitimate question remains open at this point: Is there a clear cut, suggestive or self-evident postulate that would better replace the Euclidean postulate?
Legendre’s contribution to this discussion is his investigation of the following postulate: Definition 9.3 (Legendre’s Postulate). The exists an angle such that every point in its interior lies on a segment going from one side of the angle to the other one. Proposition 9.11 (The Fourth Legendre Theorem). Given is a Hilbert plane, for which the Archimedean Axiom is assumed to hold. If Legendre’s postulate holds, there exists a triangle with angle sum two right. By combining the result with Legendre’s second Theorem, we conclude that every triangle has angle sum two right. Corollary 23. A Hilbert plane were the Archimedean Axiom and Legendre’s postulate hold is semi-Euclidean. Proof. The proof relies on three ideas: (a) The additivity of the defect of triangles. (b) A construction doubling the defect of a triangle. (c) The Archimedean property for angles given in Hilbert’s Proposition 34.
Definition 9.4 (Defect of a triangle). The defect of a triangle is the deviation of its angle sum from two right angles. We write δ(ABC) = 2R − α − β − γ. Lemma 9.2 (Additivity of the defect). Let 4ABC be any triangle and D be a point on segment AB. Then δ(ABC) = δ(ADC) + δ(DBC)
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Figure 9.13. The defect and the area of triangles are both additive.
Figure 9.14. Additivity of the defect.
Proof. With the notation of the figure on page 251, and the fact that δ1 and δ2 are supplementary angles. δ(ABC) = 2R − α − β − γ = 2R − α − β − γ1 − γ2 = (2R − α − γ1 − δ1 ) + (2R − β − γ2 − δ2 ) = δ(ADC) + δ(DBC) Suppose that a triangle is partitioned into four smaller triangles by three points lying on its sides. Then the defect of larger triangle is the sum of the defects of the four smaller triangles. Lemma 9.3 (Doubling the defect). Given is a triangle 4BAC where Legendre’s postulate holds for the angle ∠BAC. Then there exists a triangle 4B0 AC 0 , with the same vertex A and angle at A such that δ(B0 AC 0 ) ≥ 2 δ(BAC) (9.5)
Proof. One starts similarly to the procedure for the first Legendre Theorem. Let D be the midpoint −−→ of side BC, and extend the ray AD and transfer segment AD to get point E such that AD DE. The congruences 4ADC 4EDB and 4ADB 4EDC (9.6)
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Figure 9.15. Doubling the defect of a triangle.
are shown by SAS congruence. Next, the congruence 4ABC 4ECB
(9.7)
−−→ is shown by ASA congruence. By Legendre’s postulate, there exists a point B0 on ray AB and a − − → point C 0 on ray AC such that point E lies on the segment B0C 0 . We next claim point B lies between A and B0 . This follows from the exterior angle theorem: Indeed, the triangle 4AEC 0 has the interior angle ∠EAC 0 , which is less than a non adjacent exterior angle, thus ∠EAC 0 < ∠AEB0 On the other hand, the triangle congruence (9.6) implies ∠EAC 0 = ∠DAC ∠BED = ∠AEB Hence ∠AEB < ∠AEB0 . Since B and B0 lie on the same side of line AE, this implies that B lies between A and B0 . Similarly, one shows that C lies between A and C 0 . The larger triangle 4B0 AC 0 is partitioned into four smaller triangles by the segments between the three points B, E and C lying on its sides. The additivity and positivity of the defect, and finally the congruence (9.11) yield δ(AB0C 0 ) = δ(ABC) + δ(ECB) + δ(EBB0 ) + δ(ECC 0 ) ≥ δ(ABC) + δ(ECB) = 2 δ(ABC)
(9.8)
as to be shown.
End of the proof of the Fourth Legendre Theorem. Let 4ABC be such that Legendre’s postulate holds for the angle ∠BAC. We use Lemma 9.3 repeatedly to inductively get a sequence of triangles 4AB0C 0 = 4AB1C1 , 4AB2C2 , 4AB3C3 , . . . 4ABnCn , . . . all with the same vertex A and angle ∠BAC. Since the defect has obviously the upper bound 2R, one concludes 2R ≥ δ(ABnCn ) ≥ 2n δ(ABC) and after dividing one gets 2R 2n for all natural numbers n. Now the Archimedean property for angles implies δ(ABC) = 0 and hence α + β + γ = 0, as to be shown. δ(ABC) ≤
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Proof of the Corollary. If Legendre’s postulate holds for one angle ε, one easily checks that it holds for any smaller angle, For the doubled angle 2ε, the postulate holds, too. Given any other angle α, we use the Archimedean property. There exists a natural number r such that α <ε (9.9) 2r Now we successively conclude that the Legendre postulate holds for the angles α α α , 2 r ,..., , α r 2 2 2 . Hence Legendre’s postulate holds for every angle. By the proof above, we see that every triangle has angle sum two right.
Figure 9.16. Hyperbolic geometry produces an angle at vertex E.
Problem 9.1. Start a similar construction in hyperbolic geometry. Instead of relying on Legendre’s postulate, it is natural to choose the points B0 and C 0 on the two sides of angle ∠BAC such that BB0 CE and CC 0 BE. Which congruences hold among the four triangles appearing in formula (9.8)? Show that the quadrilateral AB0 EC 0 is convex and that its total defect is 2(2R − ∠B0 EC 0 ). Solution. As above, 4ABC 4ECB
(9.10)
4BB0 E 4CEC 0
(9.11)
is shown by ASA congruence and is shown by SAS congruence. The total defect is δ(AB0 EC 0 ) = δ(ABC) + δ(EBC) + δ(EBB0 ) + δ(ECC 0 ) = 2δ(ABC) + 2δ(EBB0 )
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Angle addition at vertex B yields 2R = α0 − β − γ and hence δ2 :=
δ(AB0 EC 0 ) = (2R − α − β − γ) + (2R − α0 − β0 − γ0 ) 2 = (2R − α0 − β − γ) + (2R − α − β0 − γ) = 2R − α − β0 − γ0 = 2R − ∠B0 EC 0
Half of the defect of quadrilateral AB0 EC 0 equals its exterior angle at vertex E.
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10. Neutral Geometry of Circles and Continuity Definition 10.1 (Circle). A circle with center O through point R in a plane A is the set of all points P in the plane A such that OR OP. A point P from the plane A lies inside the circle C if OP < OR. A point Q from the plane A lies outside the circle C if OQ > OR. The set of points inside a circle is called the interior or disk of the circle. The interior of circle C is denoted by Int(C). 10.1. Immediate consequences of neutral geometry We begin by stating some properties of circles that follow simply from Hilbert’s axioms. We make repeated use of the exterior angle theorem, its consequences, and the triangle inequality.
Proposition 10.1.
(i) A circle intersects a line in at most two points.
(ii) If a line contains points inside a circle, the foot point of the perpendicular dropped from the center onto the line lies inside the circle, or the line goes through the center. Question. How does the assumption that three points A, B and C lie both on a circle and a line lead to a contradiction. Give a detailed reasoning, based on Proposition 7.46(iv). Answer. Of any three points on a line, one lies between the two others (Theorem 4 in Hilbert’s foundations). We can suppose that A ∗ B ∗ C. Let the isosceles 4OAB and 4OBC have base angles α and γ, respectively. Because these are the base angles of the isosceles 4OAC, too, they are congruent and acute. Congruent base angles α γ occur as supplements at the middle vertex B. This would imply the base angles are right angles, which is impossible. Proof of item (ii). One may assume that the center O does not lie on the line l, for otherwise the assertion is clear. We drop the perpendicular from O onto l. As assumed, let P be a point on the line l inside the circle. To show that the foot point F lies inside C, consider the right triangle 4OPF. The right angle is the largest angle of any right triangle, as follows from the exterior angle theorem. Hence by Euclid I.19, the hypothenuse OP is the longest side. Hence OF < OP < OR with R any point on circle C. This shows that the foot point F lies inside circle C. Proposition 10.2. All points between two distinct points inside or on a circle lie inside the circle. Proof. Let A and B be two points inside or on the circle. Let P be a point inside the segment AB. One of the two triangles 4OAP or 4OBP is obtuse or right, because of the supplementary angles at vertex P. Because a triangle can have only one obtuse angle, the side opposite to the obtuse angle is the longest side of a triangle. We conclude that either OB > OP or OA > OP, which both mean that point P lies inside the circle. The proof for the special case that segment AB goes through the center uses Hilbert’s Proposition 5.7. Any finite set of points on a line can be put into an ordered list such that their order relations hold according to the ordering of the list. Possible cases for such a list are O ∗ A ∗ P ∗ B, A ∗ O ∗ P ∗ B, A ∗ P ∗ O ∗ B, A ∗ P ∗ B ∗ O. In each of these cases, it is easy to see that point P lies inside the circle. Definition 10.2 (Chord and diameter). The segment between two points of a circle is called a chord or secant. A chord through the center of the circle is called a diameter. Corollary 24. All points of a chord besides the endpoints lie inside the circle.
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Proposition 10.3. Let Q1 and Q3 be two points outside or on circle C, and let F be the foot point of the perpendicular dropped from center O onto the line Q1 Q3 . (i) If the foot point lies on or outside the circle, then all other points of line Q1 Q3 lie outside the circle. (ii) If the foot point F does not lie between the points Q1 and Q3 , then all points inside the segment Q1 Q3 lie outside the circle. Proof. To check item (i), consider the right triangle 4OFQ. The hypothenuse OQ is the longest side. Hence OQ > OF. To check item (ii), let Q be a point in segment Q1 Q3 . The order of the four points is F ∗ Q1 ∗ Q ∗ Q3 or F ∗ Q3 ∗ Q ∗ Q1 . Take the first case. Angle ∠OQ1 Q is obtuse, because it is supplementary to an acute angle of the right triangle 4OFQ1 . Because a triangle can have at most one obtuse angle, the side OQ opposite to the obtuse angle is the longest side. Hence OQ > OQ1 , and point Q lies outside the circle C. Proposition 10.4. Suppose a circle and a line intersect in two points A and B. Recall that the segment between the two intersection points is called a chord. (iii) The foot point of the perpendicular dropped from the center to the line bisects the chord. (iv) The chord has at most the double length as the radius. (v) The points on the line AB outside the chord AB lie outside the circle, and the points inside the chord lie inside the circle.
Figure 10.1. Some statements about a line intersecting a circle in two points.
Proof. Item (iii) uses once more the figure on page 256. The two triangles 4OFA and 4OFB are congruent by the hypothenuse leg theorem. Hence AF BF, confirming that the foot point F is the midpoint of segment AB. To confirm item (iv), note that the hypothenuse is the longest side of a right triangle. Hence OA > AF BF, and 2OA > AF + FB AB. The equality occurs in the exceptional case that AB is a diameter. Item (v) is left to the reader.
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Definition 10.3 (Short and long arc). Given is a circle with center O and a chord AB that is not a diameter. The long arc is the part of the circle lying in the same half plane of line AB as the center of the circle. The short arc is the part of the circle lying in the opposite half plane of line AB as the center of the circle.
Figure 10.2. The short/long arc lies in the interior/exterior of the central angle.
Proposition 10.5. Given is a circle with center O and a chord AB that is not a diameter. A point S on the circle which lies in the interior of the central angle ∠AOB lies on the short arc, and a point L on the circle which lies in the exterior of the central angle lies on the long arc of chord AB. Proof. Take a point S lying at the intersection of the circle with the interior of angle ∠AOB. By −−→ the Crossbar Theorem, the ray OS intersects the segment AB. Let P be the intersection point. By the previous Proposition 10.4 item (v), all points inside the segment AB lie inside the circle. Hence OP < OA OS and O ∗ P ∗ S . In other words, point S lie in the half-plane of line AB opposite to the center O, and hence on the short arc. Take a point L lying at the intersection of the circle with the exterior of angle ∠AOB. Either the −−→ ray OL does not intersect the line AB, or there is an intersection point Q. In the first case, point L lie in the same half-plane of line AB as the center O, and hence on the long arc. In the second case, we use the previous Proposition 10.4 item (v). We know that all points on the line AB outside the segment AB lie outside the circle. Hence OP > OA OL and O ∗ L ∗ Q. Once more, point L lie in the same half-plane of line AB as the center O, and hence on the long arc. Problem 10.1 (A chord of given length). Given is a circle C with center O, a point P inside the circle and a chord A0 B0 . (i) Construct a chord congruent to A0 B0 through point P. (ii) Describe your construction. (iii) Construct all solutions—none, one or two—and explain how you decide which case occurs. Draw examples for all three cases.
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Answer. (i) The construction of a chord AB A0 B0 through the point P is done in the figure on page 258. (ii) Through the given point P, we draw a second circle D around O. Assume this circle intersects the chord A0 B0 , and let P0 be an intersection point. The axial reflection which maps P to P0 has the perpendicular bisector of PP0 as axis of symmetry. For the construction of this axis, we draw two further circles around P and P0 , both through point O. They intersect at a second point Q, and the axis is the line r = OQ. The chord AB to be constructed is the mirror image of the given chord A0 B0 . To obtain AB, let Q be the intersection point of line A0 B0 with the axis of symmetry r. The line QP intersects the given circle at points A and B. (iii) The circle D can either intersect the chord A0 B0 at two points, touch it at one point, or not intersect it at all. Depending on these cases, we get two, one or no solution of the construction problem.
Figure 10.3. Construction of a chord through the given point P of given length A0 B0 .
Proposition 10.6. The perpendicular t to a radius OT , erected at point T , has only one point with a circle in common. The circle lies on one side of this line. Proof. By Proposition 7.51, the foot point T of the perpendicular from O to the line t is the unique point on the line t, with shortest distance from O. Hence all points P , T on the tangent line t have distance OP > OT from the center of the circle. Hence any radial segment OR on a line OP is shorter than segment OP. Because of OR < OP, these radial segments do not intersect the tangent t. By the plane separation theorem 5.4, this fact implies that the center O and all points R , T of the circle lie on the same side of tangent t. Too, we see that the tangent t has only the foot point T in common with the tangent line t.
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Figure 10.4. Can you see why the construction works?
Definition 10.4 (Tangent to a circle, an ad hoc definition). The perpendicular erected at a point of the circle to the radius is called the tangent of the circle at this point. Proposition 10.7. A line which has a point in common with a circle, and is different from the tangent, intersects the circle in exactly two points. Proof. Take a circle C with center O, a point T on the circle, the perpendicular t erected at point T onto the radius OT , and a second line l , t. We drop the perpendicular from the center O onto line l, and let F be the foot point. Clearly F , T , since l , t. We transfer the segment T F to get a congruent segment T F FA with T ∗ F ∗ A. The two triangles 4OFT 4OFA are congruent by SAS congruence. Hence OA OT , and point A , T lies on the circle. Existence of a third intersection point has been ruled out above by item (v) from proposition 7.46. Proposition 10.8. A line has only one point in common with a circle if and only if it is a tangent. Construction 10.1 (A neutral construction of the tangent to a circle). Let point A be the intersection of the segment OP with the given circle C. At point A, we erect the perpendicular p onto segment OP. We need a second circle around O through point P. Let B be an intersection point of this second circle with the perpendicular p. Finally, we need an intersection point T of circle C with segment OB. This intersection point is the touching point of a tangent from point P to the given circle C. Validity of the Construction. The triangles 4OAB 4OT P are congruent. This can be seen by SAS congruence, because of the common angle ∠AOB = ∠POT = ∠T OP and two pairs of congruent adjacent sides OA OT and OB OP.
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Figure 10.5. How to get the second intersection point of a circle with a line.
Figure 10.6. A neutral construction of the tangents to a circle
Hence we get congruent right angles ∠OT P ∠OAB = R. Since point T lies on the circle C, too, the segment OT is a radius of that circle. Since T P is perpendicular to the radius OT , it is a tangent of circle C. Proposition 10.9. For any given Hilbert plane, we assume that an ad hoc construction of the
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Figure 10.7. If any ad hoc construction of the tangents to a circle is available, the circle-line intersection property holds.
tangent to a circle is possible and legitimate. Then the circle-line intersection property 10.5 holds. Proof. We proceed as shown in the figure on page 261. Given are the circle circ and the line l that contains a point inside the circle. We drop the perpendicular p from center O onto the line l and get the foot point A. −−→ Let P be the intersection point of the circle with the perpendicular ray OA. Because of axiom (III.1), the point P can be obtained by transfer of any radial segment which has the radius of the circle as length. We draw a second circle c through the foot point A around the same center O. By Proposition 10.1 item(ii), the foot point A lies inside the circle circ. Hence the order of points O ∗ A ∗ P occurs, and the point P lies outside the (new smaller) circle c. By assumption, an ad hoc construction of the tangent from point P to the circle c is possible and legitimate. We drop the perpendicular from center O onto the tangent t and obtain the foot point T . −−→ Let point B be the intersection of the radial ray OT with the given line l. The triangle congruence 4OT P 4OAB holds by ASA congruence. Indeed, the existence of the intersection point B is confirmed by the extended ASA-congruence, see Proposition 7.13. Hence OP OB, thus point B lies on the given circle circ and is an intersection point of circle circ with line l. 10.2. The tangent is the limiting position of a secant Imagine that a point P on the circle "moves" towards a fixed point T on the circle. We claim that the secant T P moves towards the tangent t. A shorter, but more precise claim is: with the tangent at its endpoint, a secant which is short enough forms an angle smaller than any prescribed angle. We need the following theorem to justify the above claim.
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Theorem 10.1 (The tangent to a circle is the limiting position of a short secant). Take any small −−→ angle ε := ∠(T A, t) between the secant T A and the tangent t. If for another point P on the circle, −−→ the secant T P < T A is shorter, then the point P lies inside the angle ∠(T A, t): −−→ −−→ T P < T A implies ∠(T P, t) < ∠(T A, t) We see now that the claim from above holds. Given is any (small) angle ε by which we want −−→ to rotate the tangent. We draw the secant T A which forms this angle ε := ∠(T A, t) with the tangent t at touching point T . We obtain the second intersection point A, and put δ equal to the length of the chord T A. If any other secant T P is shorter than T A, then the point P lies inside the angle −−→ ∠(T A, t). Indeed, the tangent to a circle, as given by the perpendicular to the radius T O is the limiting position of a short secant. Theorem 10.2. Given is a circle around O, a point P on the circle, and a line l through point P. The following three statements are equivalent. All three can be used as a definition of the tangent to a circle. The line l is perpendicular to the radius OP. The point P is the only common point of the line and the circle. The line is the limiting position of a secant PQ in the limit Q → P.
Figure 10.8. The tangent is the limiting position of a short secant. Hence the shorter secant T P forms with the tangent the smaller angle ∠RT P.
Plan of proof for Theorem 10.1. Given is a circle C with center O, a point T on the circle, and a −−→ ray t = T R perpendicular to the radius OT . Take any small angle ε. Let e be a second ray with vertex T , lying on the same side on tangent t as the center O, and such that ∠(t, e) = ε.
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Question. Why does the ray e intersect the circle in a second point A. Answer. The rays e and t lie on different lines, since they are assumed to form an angle ε = ∠(e, t). By Proposition 10.7, the line on e which has a point in common with a circle, and is different from the tangent t, does intersect the circle in exactly two points. We claim that any secant T P < T A forms with the tangent at T an angle smaller than ε. It is enough to consider the case that points P and A lie on the same side of radius OT . In that case, we show that the secant T P lies inside the angle ∠(e, t), and ∠PT R < ∠AT R. The argument is done in several steps: (i) At first, we show that T P < T A is equivalent to ∠T OP < ∠T OA. In other words, the length of a chord depends monotone on the central angle. (ii) Secondly, we show that ∠T OP < ∠T OA implies ∠OT A < ∠OT P. −−→ (iii) Finally, ∠OT A < ∠OT P implies the ray T P lies inside the angle ∠RT A, and hence ∠RT P < ∠RT A. Detailed proof of Theorem 10.1. Question. By which proposition do we know that ∠T OP < ∠T OA is equivalent to T P < T A ∠T OP > ∠T OA is equivalent to T P > T A ∠T OP ∠T OA is equivalent to T P T A
Figure 10.9. We use the Hinge theorem.
(10.1)
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Answer. The Hinge Theorem 7.55 tells that increasing the angle between two constant sides increases the opposite side of a triangle [Euclid I.24]. Too, Corollary 14 states the equivalence: if and only if the angle between the two constant sides increases, is constant or decreases, then the opposite side does the same. For completeness, we give the exact detailed reasoning in the present case. Proof of the Hinge Theorem in the present case. Assume that the triangles 4T OP and 4T OA satisfy ∠T OP < ∠T OA. Question. Show that the segments T A and OP intersect each other. −−→ Answer. Since points P and A are assumed to lie on the same side of line T O, the ray OP lies inside the angle ∠T OA. Hence the Crossbar Theorem implies that this ray intersects the segment T A. Let Q be the intersection point. By Corollary 24, all points of a chord besides the endpoints lie inside the circle. Hence, we have checked that O∗Q∗P
and T ∗ Q ∗ A
(10.2)
Question. Use Euclid I.19 to give a reasoning why T P < T A. Answer. To compare segments T P and T A, we now use Euclid I.19 for the triangle 4T AP. We need to compare the angles σ = ∠T AP and σ0 = ∠T PA opposite to these two segments. Let φ be the base angle of the isosceles triangle 4AOP. O∗Q∗P T ∗Q∗A
implies σ < φ implies φ < σ0
Hence σ < σ0 follows by transitivity, and T P < T A by Euclid I.19. Question. Why does now, conversely, T P < T A implies ∠T OP < ∠T OA. Answer. The argument above shows that ∠T OP < ∠T OA implies T P < T A. Similarly, one gets that ∠T OP > ∠T OA implies T P > T A. By SAS congruence, ∠T OP ∠T OA implies T P T A. All three statements together allow to get the corresponding converse statements—since all angles are comparable, and all segments, too. Thus the Hinge Theorem yields item (i). Next, we check item (ii) that ∠T OP < ∠T OA implies ∠OT P < ∠OT A. This is now easy, since the reasoning above shows that both ∠T OP < ∠T OA and T P < T A. Hence point Q is the intersection point of segments segments T A and OP, as stated in equation (10.2). Obviously, O ∗ Q ∗ P implies ∠OT A = ∠OT Q < ∠OT P. Finally, we check item (iii). The angle ∠OT P < ∠OT R = R, since ∠OT P is the base angle of the isosceles triangle 4OT P and hence acute, but angle ∠OT R is right by assumption. Since −−→ by assumption, points P and R lie on the same side of radius OT , the ray T P lies inside the angle ∠RT A. We see that ∠RT P < ∠RT A, as claimed. 10.3. Mutual placement of two circles We conclude this small survey with some simple facts about the mutual relations of two circles. The set of points inside a circle is called the interior or disk of the circle, The interior of circle C is denoted by Int(C). Proposition 10.10. Let C be a circle with center C and radius c, and D be a circle with center D and radius d. (i) The interiors of the two circles C and D are disjoint if and only if |CD| ≥ c + d.
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(ii) The interior of circle C is a subset of the interior of circle D if and only if |CD| ≤ d − c. (iii) The interior of circle D is a subset of the interior of circle C if and only if |CD| ≤ c − d. (iv) The interiors of the two circles intersect, but neither one is a subset of the other if and only if max{d − c, c − d} < |CD| < c + d. In that case, each circle intersects both the interior and exterior of the other one. The interior of each circle intersects the exterior of the other circle. With inequalities and equalities interpreted as segment comparison and congruence. This Proposition holds in any Hilbert plane, without assuming neither the Archimedean axiom nor the circlecircle intersection property, nor the circle-line intersection property. Proof. It is left to the reader to check these statements for the case C = D of concentric circles. We now assume C , D, and let C1 and C3 be the two points on the line CD lying on the circle C. Similarly, let D1 and D3 be the two points on the line CD lying on the circle D. These points exist, because they can be constructed by transferring any radial segment. We can assume the order relations C1 ∗ C ∗ D ∗ D3 , C1 ∗ C ∗ C3 and D1 ∗ D ∗ D3 . (i) Assume the interiors of the two circles C and D are disjoint. Then no point can lie both inside segment C1C3 , and inside segment D1 D3 . Hence either C1 ∗C ∗C3 ∗ D1 ∗ D ∗ D3 , or C3 = D1 . In the former case, |CD| > c + d. In the latter case |CD| = c + d. Conversely, assume that the interiors of the two circles C and D intersect. Let P be a point inside both circle. The weak triangle inequality implies |CD| ≤ |CP| + |PD| < c + d. (ii) Assume that the interior of circle C is a subset of the interior of circle D. In that case, every point lying inside segment C1C3 lies inside segment D1 D3 , too. Hence either C1 = D1 and D1 ∗ D ∗ D3 ∗ C3 ,
or
C1 ∗ D1 ∗ D ∗ D3 and C3 = D3 ,
or
C 1 ∗ D1 ∗ D ∗ D3 ∗ C 3 . The first two cases imply |CD| = d − c, and the last case implies |CD| < d − c. Conversely, if |CD| < d − c, then C1 ∗ D1 ∗ D ∗ D3 ∗ C3 . Similarly |CD| = d − c implies either C1 = D1 and D1 ∗ D3 ∗ C3 , or C3 = D3 and C1 ∗ D1 ∗ D3 . Now let P be any point in the interior of circle C. The weak triangle inequality implies |PD| ≤ |PC| + |CD| < c + d − c = d, and hence point P lies inside the circle D, too, and hence Int(C) ⊆ Int(D). (iii) (iv) These are just logical consequences. Proposition 10.11. Let C be a circle with center C and radius c, and D be a circle with center D and radius d. (a) If the two circles intersect in a point not on the line between the two centers, then max{d − c, c − d} < |CD| < c + d.
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(b) The two circles intersect in exactly one point if and only if they are not concentric—C , D—, and equality holds in either (i) |CD| = c + d. (ii) |CD| = d − c. (iii) |CD| = c − d. from the previous proposition. In that case, the intersection point and the two centers lie on a line. Proof.
(a) Assume the two circles intersect in point P. Item (a) follows from the strict triangle inequality for triangle 4CDP. Indeed |CD| < |CP| + |PD| = c + d c − d = |CP| − |DP| < |CD| + |DP| − |DP| = |CD| d − c = |DP| − |CP| < |DC| + |CP| − |CP| = |CD|
(b) First we assume that the two circles intersect in exactly one point. This point needs to lie on the line CD, because otherwise, one would get a second intersection point P0 symmetric to line CD. Hence the intersect point is either C3 or D1 , and equality holds in either (i), (ii) or (iii). Conversely, if equality holds in either (i), (ii) or (iii), and C , D, then the two circles intersect in either point C3 or D1 , and no further point on the line CD. Because of item (a), they cannot intersect in any point P which is not on the line CD, neither. Thus, we have shown that in the case (iv) max{d − c, c − d} < |CD| < c + d. the interiors of the two circles intersect, and the interior of each circle intersects the exterior of the other circle. Of course, this immediately suggests that the two circles intersect? But such a statement does not follow just from Hilbert’s axioms! 10.4. Continuity principles for circles Much more cannot be deducted about intersection of circles and lines, simply from Hilbert’s axioms. Indeed, in theorem 10.3 below, we give a precise statement of this sharply negative result. One needs to introduce and use some continuity principle to obtain sharper, and more intuitive results about circles and lines. We discuss the line-circle and the circle-circle intersections. Definition 10.5 (Line-circle intersection property). A line that contains a point inside a circle does intersect the circle. Proposition 10.12. Assume that the line circle intersection property holds. (i) A line that contains a point inside a circle intersect it in exactly two points. (ii) Let P be a point inside circle C and Q a point outside the circle. Then the segment PQ intersects the circle in exactly one point, and the line PQ intersects the circle in two points.
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Proof. In the special case that the center O of the circle lies on the line l, the result follows easily—even just from Hilbert’s axioms. Indeed, the two intersection points A, B can be obtained by transferring the radial segment OR onto the two opposite rays on line l with vertex O. Following Hilbert’s Proposition 5.7, any finite set of points on a line can be put into an ordered list such that their order relations hold according to the ordering of the list. Possibly by exchanging the names of A and B, such a list is either A ∗ O ∗ P ∗ B ∗ Q or A ∗ P ∗ O ∗ B ∗ Q. Let A be one intersection point of the given circle and line l. Reflecting point A by the line OF ⊥ l yields a second intersection point B. In proposition 10.1, it was shown that a line and a circle cannot intersect in three points. Hence the given line l intersects the circle in two points A and B. We now check item (ii). Recall the order of the points A, P, B, Q on line l. Proposition 10.1 implies that point P lies inside the segment AB, and point Q lies outside that segment. By Hilbert’s four-point theorem, we conclude that either the list A ∗ P ∗ B ∗ Q or Q ∗ A ∗ P ∗ B yields the order of these points. Hence the segment PQ contains exactly one of the intersection points A and B of the given circle C and the line PQ. Proposition 10.13. Assume that the line circle intersection property holds. If a circle has points on both sides of a line, then it intersects the line in two points.
Figure 10.10. A circle with points on both sides of a line.
Proof. Let A and B be points on different sides of line l, both lying on a circle around center O. There are some special cases: In the special case that the center O lies on the line, the result follows immediately, by transferring radial segments. In the special case the three points A, O and B lie on a line, two of them, say A and O, lie on different sides. By the plane separation theorem, the line intersects the segment OA. The intersection point P of line l and segment OA lies inside the circle. Hence the line-circle intersection property implies that the line and the circle intersect. Now we deal with the generic case: assume that center O does not lie on the line, and the three points A, O and B do not lie on a line. By the plane separation theorem, the segment AB intersects the line l. Hence, by Pasch’s axiom, this line intersects a second side of the triangle OAB, say segment OA. Again, the intersection point P of line l and segment OA lies inside the circle. Hence the line-circle intersection property implies that the line and the circle intersect.
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Definition 10.6 (Circle-circle intersection property). Given are two circles C and D. Let P, Q be two points on circle D. Assume that point P lies inside circle C and Q lies outside circle C. Then the two circles C and D intersect each other. Proposition 10.14. The two circles considered in the circle-circle intersection property have exactly two intersection points. Proof. Existence of the second intersection point follows using reflection across the line OO0 connecting the two centers. The existence of a third intersection point was excluded in Lemma 7.2 for the SSS congruence theorem. Corollary 25. Let C be a circle with center C and radius c, and D be a circle with center D and radius d. The two circles intersect in exactly two points if and only if max{d − c, c − d} < |CD| < c + d, where the inequalities are to be interpreted as segment comparison. Proposition 10.15. The circle-circle intersection property implies the line-circle intersection property.
Figure 10.11.
A segment from inside to outside a circle intersects the circle—proved from the circle-circle intersection.
Proof. Assume that the circle-circle intersection property holds. Let l = AB be the line between −−→ the two given points A inside, and B outside the circle C. The ray OB intersects the circle in a point Q inside the segment OB. A special situation occurs in case that the center O of the circle −−→ lies on the line l. In this case, the intersection point Q = X of the ray OB with the circle is already
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the required intersection of line and circle. We now exclude that special case and assume that the −−→ center O does not lie on line l. The ray OA from the center intersects the circle in a point P outside the segment OA (since O , A). Let O0 and P0 , Q0 be the reflection images of points O and P, Q across the axis l = AB. We construct the circle C0 with center O0 through point P0 . Points P0 and Q0 both lie on the circle C0 —which is the reflection image of circle C. In order to apply the circle-circle intersection property, we check that point P0 lies inside and point Q0 lies outside the original circle C. Indeed |OP0 | = |O0 P| ≤ |O0 A| + |AP| = |OA| + |AP| = |OP| We have used that the lengths of symmetric segments OP0 O0 P are equal (Proposition 7.61), the weak triangle inequality for 4O0 AP (which may degenerate), the fact that the symmetric points O and O0 have the same distance from point A on the axis, and the assumption that point A lies inside circle C. We need to exclude the equality |OP0 | = |OP|. But that assumption would imply |O0 P| = |OP|. Hence point P = P0 and hence the center O would lie on the axis AB, too, which we did exclude. This contradiction confirms |OP0 | < |OP|, and P0 lies inside the circle C. With a bid modified argument we check that Q0 lies outside circle C. |OQ| + |QB| = |OB| = |O0 B| < |O0 Q| + |QB| and hence |OQ| < |O0 Q| = |OQ0 | which implies that Q0 lies outside circle C. In this case, the triangle inequality holds strictly, since points O0 , Q and B do not lie on a line. (We would get a contradiction: all five points O, O0 , Q, Q0 , B would lie on this line, which would be line l. Hence we would have the case excluded above.) Because point P0 lies inside, but point Q0 lies outside the circle C, and both points lie on the reflected circle C0 , the circle-circle intersection property implies that the two circles intersect in some point X. Since it has the same distance from both centers O and O0 , by proposition 11.1 point X lies on the axis l. Thus we have obtained an intersection point of the given circle C and line AB. A second intersection point Y is obtained by reflecting X across the perpendicular p to line AB through point O—which is in the generic case just line OO0 . One of the two points X and Y lies in the same half plane of p as either point A or point B. It is left to the reader to check that this intersection lies between A and B. Proposition 10.16. The line-circle intersection property together with the parallel axiom imply the circle-circle intersection property. A proof using analytic geometry is given in the section about coordinate planes. Problem 10.2 (Construction with classical tools in neutral geometry). Give a construction of an equilateral triangle, and an angle of 60◦ with compass and straightedge in neutral geometry. (a) Do the construction and give a description. (b) What can one say about the angles at the vertices of the triangle, in neutral geometry? Why are extra steps needed for the construction of the 60◦ angle? (c) At which vertex can you get the angle of 60◦ even in neutral geometry, nevertheless? (d) Justify your answer.
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Figure 10.12. The construction of a 60◦ angle at the center O is possible with straightedge compass in neutral geometry.
(e) Convince yourself once more that all reasoning has been done in neutral geometry. Additional to Hilbert’s axioms, which intersection property do you need. Answer. (a) Description of the construction. Draw any segment AB. The circles about A through B, and about B through A intersect in two points C and D. We get an equilateral triangle 4ABC. Now we draw a third circle about C through point A, which passes through point B, too. With the two circles drawn earlier, the third circle has additional intersection points H and F. Finally, we draw the segments AF, BH and CD. These are the perpendicular bisectors of the sides of triangle 4ABC. All three intersect in one point O inside triangle 4ABC. At vertex O, one gets six congruent angles which add up to 360◦ , hence they are all 60◦ . (b) The angles at the vertices may not be 60◦ . Extra steps are needed, because, in neutral geometry, the angle sum of a triangle may be less than two right angles. All we can conclude about the angles at the vertices A, B, C, is their congruence. This follows because the angles opposite to congruent sides are congruent. Still they may all three be less than 60◦ . (c) The 60◦ angles appear at the center. At vertex O, one gets six congruent angles which add to 360◦ , hence they are all 60◦ . (d) Justification. The construction of the perpendicular side bisectors for triangle 4ABC still works in neutral geometry because of proposition 11.1. By plane separation, they intersect the corresponding sides of the triangle. Too, they lie inside the angles of the triangle. Hence
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the Crossbar Theorem implies that they intersect each other, as explained in theorem 11.3 (about the in-circle). The intersection point O is the center of the in-circle of triangle 4ABC, because it has equal distances from the three sides. The same point is the center of the circum-circle, because it has equal distances from the vertices. −−→ The three side bisectors form six congruent angles at vertex O. Indeed, the ray OD bisects the angle ∠AOB. since SSS-congruence yields the triangle congruence 4AOD 4BOD. −−→ The opposite ray OC bisects the angle ∠HOF. This can for example be obtained from the triangle congruence 4HOC 4HOA 4BOF 4OCF (e) The circle-circle intersection property is needed. All these justifications above have been done in neutral geometry. Besides Hilbert’s axioms, we only need the circle-circle intersection property 10.6. Corollary 26. Assume Hilbert’s axioms (I.1)–(I.3), (II.1)–(II.4) and (III.1)–(III.5), together with the circle-circle intersection property holds. Then for each given segment AB and half plane of line AB, there exists a unique equilateral triangle ABC. An angle of 60◦ is constructible. Problem 10.3 (Construction with Hilbert tools in Euclidean geometry). Give a construction of an equilateral triangle, and an angle of 60◦ with Hilbert tools in Euclidean geometry. Use the Theorem of Pythagoras.
Figure 10.13. This construction of an equilateral triangle with Hilbert tools works only in Euclidean geometry
Answer. We construct an equilateral triangle with side length |AB| √ = |BC| = |CA| = 2. The Theorem of Pythagoras implies that the altitude of this triangle has length 3. We begin by constructing the perpendicular bisector of segment AB. With its midpoint M as vertex, we transfer segment MA onto the bisector and get segment MP MA.
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−−→ Next we transfer segment AP onto the ray MP and get a segment MQ AP. The Theorem of Pythagoras implies and congruence imply √ |MQ| = |AP| = 2 −−→ In a further similar step, we transfer segment BQ onto the ray MP and get a segment MC BQ. The Theorem of Pythagoras implies and congruence imply p √ |MC| = |BQ| = |BM|2 + |MQ|2 = 3 Using, the Theorem of Pythagoras a third time, we check that p |AC| = |AM|2 + |MC|2 = 2 Hence the triangle 4ABC is equilateral. Corollary 27. Assume Hilbert’s axioms (I.1)–(I.3), (II.1)–(II.4) and (III.1)–(III.5), as well as the Euclidean parallel axiom. The circle-circle intersection property needs not be assumed to hold. Then for each given segment AB and half plane of line AB, there exists a unique equilateral triangle ABC. An angle of 60◦ is constructible. Proof. Note that the Euclidean parallel axiom implies the Pythagorean theorem. Now the construction of the equilateral triangle ABC over a given segment AB can be done with Hilbert tools. The details are left to the reader. Problem 10.4. Is an angle of 60◦ constructible just with Hilbert tools is neutral geometry? Neither the parallel axiom, nor the circle-line intersection, nor the circle-circle intersection property is assumed. 10.5. Continuity principles for circles are independent of Hilbert’s axioms We now give a strikingly negative result: Theorem 10.3. Neither the circle-line intersection property, nor the circle-circle intersection property hold in every Hilbert plane. They cannot be deducted simply from the axioms of incidence, order and congruence. Proof. Hilbert’s basic counterexample is given in more details in proposition 31.2 in the section on Euclidean constructions with restricted q √ means. Indeed, because the number z = 2 − 1 is not totally real, a segment of that length cannot be constructed by Hilbert tools, just starting from a given unit segment, with |AB| = 1. Let H ⊂ R2 be the Hilbert plane consisting of all points in the Euclidean plane that can be constructed with Hilbert tools in finitely many steps, with only any two points A and B being given at the start. In the figure on page 600, points A through G are in the Hilbert plane H ⊂ R2 , since they are constructible with Hilbert tools. But point H is not constructible, and hence H < H. Problem 10.5. Does the segment-circle intersection property hold in the Hilbert plane H? Which (nonempty) segment and circle provide a counterexample? Answer. Let C be the circle around G through point B. The segment AD has one endpoint A inside, and the other endpoint D outside of circle C. This segment is nonempty, too. But it does not intersect that circle, because point H is not in the Hilbert plane H. Proposition 10.17. There exist models of hyperbolic geometry where an equilateral triangles does not exist over every segment. Proof. See Hartshorne’s book p. 373. The proof is omitted. It needs a bid of abstract algebra, and Poincaré’s disk model with a virtual line of infinity.
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10.6. Derivation of continuity principles from Dedekind’s axiom What does Dedekind’s axiom imply about intersection of circles and lines? For convenience of the reader, we prove at first the weaker statement of theorem 10.4, and then the stronger theorem 10.5. Theorem 10.4. Dedekind’s axiom implies the line-circle intersection property. Theorem 10.5. Dedekind’s axiom implies the circle-circle intersection property. Proof of Theorem 10.4: Dedekind’s axiom implies the line-circle intersection property. Given is a circle C, and a line l. We assume that point P lies on the line l and inside the circle C. In the special case that the line goes through the center of the circle, existence of the intersection points of line and circle can be deducted directly via Hilbert’s axioms. Hence we may assume that the center O of the circle does not lie on the line. Let Q be a point on the line l such that PQ 2OR. Question. Why does the point Q lie outside the circle C. Answer. Indeed, the triangle inequality 7.52 for the triangle 4OPQ implies PO + OQ > PQ, and hence OQ > PQ − OP > 2OR − OR OR. We now construct the Dedekind cut {Σ1 , Σ2 } of the line l. Let F be the foot point of the line dropped from center O onto the line l. Let the set Σ1 consist of all points of the line l that lie either −−→ −−→ on the ray opposite to FQ, or inside the circle C. Let the set Σ2 consist of all points on the ray FQ that lie outside the circle C. −−→ In the case that the ray FQ and the circle C intersect, we are ready anyway. Otherwise, we show that {Σ1 , Σ2 } is a Dedekind cut. Let P2 be a point between two points P1 , P3 ∈ Σ1 . We now check that point P2 lies in Σ1 . If −−→ both P1 and P3 lie both on the ray opposite to FQ, we are ready. So we need consider the case that P1 ∗ F ∗ P3 . Hence either P1 ∗ P2 ∗ F ∗ P3 or P1 ∗ F ∗ P2 ∗ P3 . In the first P2 lies inside the circle, −−→ and hence in Σ1 . If P2 lies on the ray opposite to FQ, and P2 ∈ Σ1 —we are ready. From item (ii) of proposition 10.2, we see that a point Q2 between two points Q1 , Q3 ∈ Σ2 lies in the set Σ2 , too. All assumptions occurring in Dedekind’s axiom have been checked. Hence we conclude from Dedekind’s axiom existence of a cut point K ∗ . By its definition, the cut point does not lie between two other points of Σ1 , and neither between two other points of Σ2 . On the other hand, a point inside a circle lies between two other points inside the circle, and any point outside the circle lies between two other points outside the circle, and on the same side of foot point F. This contradiction rules out that K ∗ could lie inside or outside the circle. Hence the cut point ∗ K lies on the circle, and is an intersection point of the given circle and line. Lemma 10.1. Let P be a point inside and Q a point outside the circle C. Then any point inside the circle lies between two other points on the line PQ and inside the circle, similarly, every point outside the circle lies between two other points on the line PQ outside the circle. Especially, the segment PQ contains both points inside and outside the circle. −−→ Proof. Let R be the intersection of the circle with the central ray OP. The point R can be constructed with Hilbert tools by transferal of any radial segment. Let P1 and P3 be the points on the line PQ, on both sides of point P, such that PR PP1 and PR PP3 . We claim that these points lie inside the circle C. Indeed, this follows from the triangle inequality since OPi < OP + PPi OP + PR = OR, for i = 1, 3.
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Figure 10.14. A segment from inside to outside a circle contains both points inside and outside the circle.
A similar argument is used to get the points outside the circle. Let S be the intersection of −−→ the circle with the central ray OQ. Let Q1 and Q3 be the points on the line PQ, on both sides of point Q, such that S Q Q1 Q and S Q Q3 Q. These points both lie outside the circle C, since the triangle inequality yields OS + S Q = OQ < OQi + Qi Q and hence OS < OQi , for i = 1, 3. Proof of Theorem 10.5: Dedekind’s axiom implies the circle-circle intersection property. Given are two circles: circle C, with center O, and circle D with center D—and two points P, Q on circle D. Assume that point P lies inside circle C and Q lies outside circle C. We need to show that the two circles do intersect. To begin with, I reduce the proof to the convenient case that the two points P and Q lie on the same side of line OD, but not both on that line. Indeed, I begin by eliminating the special case that both points P, Q lie on line CD. One −−→ transfers an arbitrary angle at vertex D and produces a third ray with vertex D, different from DP −−→ and DQ. Transferring the radius of circle D onto this ray yields a third point R on this circle, which is not on line CD. Depending on whether R lies outside or inside circle C, we now can use points P, R or Q, R for the construction to follow. Next I eliminate the case that points P, Q lie on opposite sides of line CD. Let P0 be the reflection of point P across line CD. Because P and P0 lie on opposite sides of CD, plane separation implies that P0 and Q lie on the same side of CD. Furthermore, because CP CP0 and DP DP0 , the point P0 lies on circle D and inside circle C, same as point P. The points P0 and Q are used for the construction to follow. −−→ For any point K , D, let S (K) be the unique intersection point of circle D with ray DK. Let l be the line through points P and Q and let K be a generic point on line l. Take any three points Ki
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Figure 10.15. Dedekind’s axiom implies existence of the intersection point K ∗ of two circles.
for i = 1, 2, 3 and assume P ∗ K1 ∗ K3 ∗ K2 ∗ Q. Since those points lie all on the same side of line CD, we get ∠CDP < ∠CDK1 < ∠CDK2 < ∠CDK3 < ∠CDQ. Now we apply the Hinge theorem to the triangles 4CDS (P), 4CDS (Ki ), with i = 1, 2, 3, and 4CDS (Q). All these triangles have the common side CD, and congruent sides with vertex D. Hence the Hinge Theorem implies CS (P) < CS (K1 ) < CS (K2 ) < CS (K3 ) < CS (Q) In the special case that point P or Q lie on the line CD, the first or last inequality follows from the triangle inequality. A Dedekind cut Σ1 , Σ2 of line l is now defined in the following way: K ∈ Σ1 if K is a point of the segment PQ and S (K) lies inside or on the circle C, or if K ∗ P ∗ Q. K ∈ Σ2 if K is a point of the segment PQ and S (K) lies outside of circle C, or if P ∗ Q ∗ K. First we check that this definition does produce a Dedekind cut. It is enough to explain one case. Let r(C) be the radius of circle C. Assume that K1 , K3 ∈ Σ1 , and K1 , K3 lie both in the segment PQ. By definition that means that CS (K1 ) ≤ r(C) and CS (K3 ) ≤ r(C). Hence formula (*) implies CS (K2 ) ≤ r(C) and hence K2 ∈ Σ1 , as to be shown. The remaining cases are even more easy to dispose. For example, assume that K1 ∗ P ∗ Q and only K3 ∈ Σ1 lies in the segment PQ. In that case, either K2 ∗ P ∗ Q or K2 = P and hence K2 ∈ Σ1 and we are finished. Or, P ∗ K2 ∗ K3 ∗ Q, and we proceed as above, with K1 := P.
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Because Σ1 , Σ2 is a Dedekind cut, by Dedekind’s axiom there exists a cut point K ∗ on the line l. Now Σ1 ∪{K ∗ } and Σ2 ∪{K ∗ } are the two opposite rays on the line l with vertex K ∗ . Let S ∗ := S (K ∗ ). −−−→ Recall that this is the point on circle D and the ray DK ∗ . The claim is that S ∗ lies on the circle C, too. To confirm this claim, we need the following Lemma.
Figure 10.16. A circular arc from inside to outside a circle C contains both points inside and outside that circle.
Lemma 10.2. Let S and S out be points be a point inside and outside the circle C, both lying on the arc of the second circle D between points P and Q. Every such point S out lies between two other points S 1 and S 3 on the circle D and outside circle C. Similarly, point S lies between two other points S 4 and S 2 on the circle D and inside circle C. Moreover, the corresponding projections onto the line PQ have the order K4 ∗ K ∗ K2 ∗ K1 ∗ Kout ∗ K3 . −−−−→ Proof. Let Kout be the intersection of the circle C with the central ray CS out —which can be constructed with Hilbert tools by transferal of any radial segment. The tangent line to circle D at point S out is the perpendicular to the radius DS out and can easily be constructed. Let T 1 and T 3 be the points on the tangent line, on both sides of point S out , such that S out T 1 < S out Kout and S out T 3 < S out Kout . We claim that T 1 and T 3 lie outside the circle C. Indeed, this follows from the weak triangle inequality since CT i ≥ CS out − S out T i CKout + Kout S out − S out T i > CKout
(10.3)
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for i = 1, 3. −−→ Let S 1 and S 3 be the points where the central rays DT i intersect the circle D. We claim that these points S i lie outside the circle C, too. Indeed, the triangle 4CS out S i is isosceles, and hence has acute base angles. Hence the supplementary angles ∠S i S out T i are obtuse. Because the side across an obtuse angle is the longest side in any triangle, we conclude S out S i < S out T i . Hence these points lie inside the circle drawn around S out through point Kout . By the same reasoning as used in equation (10.3), they lie outside the circle C. −−→ Finally, the intersection points Ki of the central rays DT i with the line PQ lie on different sides of point K. A similar argument is used to get the points S 4 and S 2 on the circle D and inside circle C, lying on different sides of line DS . Finally, we confirm that point S ∗ lies on the circle C, too. The cut point satisfies P ∗ K ∗ ∗ Q. Now, assume towards a contradiction that point S ∗ would lie inside circle C. Then Lemma (10.2) would yield existence of points P, K2 ∈ Σ1 such that P ∗ K ∗ ∗ K2 ∗ Q. Then P and K2 would be two points in Σ1 lying on opposite sides of the cut point K ∗ . This is impossible. Similarly, using Lemma (10.2), we confirm that it is impossible that S ∗ lies outside circle C. The only remaining possibility is that S ∗ lies on circle C, and hence it is a intersection point of the two circles C and D. A second intersection point of the two circles is constructed by reflection across the line CD.
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11. Neutral Triangle Geometry 11.1.
Introduction from neutral geometry
Before we start. These exercises deal with some interesting questions from triangle geometry. I could not resist the temptation to discuss neutral and Euclidean triangle geometry together. Recall that in neutral geometry, only the axioms of incidence, order, congruence are assumed. The attempts of Farkas Bolyai (the father) as well as Legendre to prove the parallel axiom—starting just with neutral geometry—were in vain. But reviewing them from the standpoint of today, they brilliantly show in which tricky and surprising way the parallel axiom is linked to triangle geometry. Some of the theorems about a triangle remain valid in hyperbolic geometry, others need to be weakened and modified. Therefore I have tried to prove as much as possible in neutral geometry. Then I specialize, at first to Euclidean geometry. Finally, hyperbolic geometry is treated within Klein’s and Poincaré’s models. From the section on congruence, recall definition 7.3 of a triangle, and Euler’s conventional notation: in triangle 4ABC, let a = BC, b = AC, and c = AB be the sides and α := ∠BAC, β := ∠ABC, and γ := ∠ACB be the angles. Too, we need to recall the characterization of a perpendicular bisector.
Figure 11.1. A point lies on the perpendicular bisector if and only if its distances from both endpoints are congruent.
Proposition 11.1 (Characterization of the perpendicular bisector). A point P lies on the perpendicular bisector p of a segment AB if and only if it has congruent distances to both endpoints of the segment. Point P lies on the same side of the perpendicular bisector p of a segment AB as point B if and only if |PB| < |PA|. Proof. Assume point P lies on the perpendicular bisector. Let M be the midpoint of segment AB. The congruence 4AMP 4BMP follows by SAS congruence. Indeed, PM PM since a segment is congruent to itself, and AM BM as well as ∠AMP ∠BMP 90◦ since PM is the perpendicular bisector. Hence AP BP. The converse is a consequence of the second statement. Indeed a point P with congruent distances AP BP from points A and B can neither lie on the same side of bisector p as A nor on the same side as B. Hence the point P lies on the bisector p. To check the second statement, we assume that point P lies on the same side of the bisector p as point B. Under this assumption, segment AP intersects the bisector—as follows from Pasch’s
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axiom, applied to triangle 4ABP and the bisector p. Let Q be the intersection point of p and segment AP. Because Q lies on the bisector, the first part of the proof implies that the triangle 4ABQ is isosceles. We apply the triangle inequality for 4BQP and conclude |PB| < |PQ| + |QB| = |PQ| + |QA| = |PA| as to be shown.
11.2. The circum-circle The construction of the circum-circle of a triangle depends on the perpendicular bisectors of its sides. Proposition 11.2 (Conditional circum-circle—neutral version). For any triangle in a Hilbert plane, the following three statements are equivalent: (a) The perpendicular bisectors of two sides of a triangle intersect. (b) The triangle has a circum-circle. (c) The bisectors of all three sides intersect in one point. Proof. We show that (c) implies (a), (a) implies (b), and finally (b) implies (c). The first claim is obvious. Reason for "(a) implies (b)". Let O be the intersection point of the perpendicular bisector pb of segment AC, and the bisector of segment BC, which is called pa . By Proposition 11.1 one concludes OA OC and OB OC Hence transitivity implies OA OB, and point O is the center of the circum-circle.
Reason for "(b) implies (c)". Let point O be the center of the circum-circle. Thus O has congruent distances to all three vertices A, B and C. By Proposition 11.1, congruent distances from A and B imply that center O lies on the bisector pc . Similarly, congruent distances from B and C imply that center O lies on the bisector pa , and congruent distances from A and C imply that center O lies on the bisector pb . Hence center O is the intersection point of all three bisectors. Obviously (c) implies (a). Hence we conclude that all three assumptions are equivalent. Note that this means that either all three are true, or all three are false. Let a, b, c be any three different mutually parallel lines in an ordered incidence plane. For their mutual placement, the following two cases may occur: (a) there exists one line among them such that the two other lines are in opposite half planes of the first line; (b) as seen from any one of the three lines, the two others lie in the same half plane. We shall see that actually both cases are possible. In some situations it is possible to decide which one of cases (a) or (b) occurs. For easier communication, I have invented the following terms. Definition 11.1 (Three parallel lines forming stripes). I say that three different mutually parallel lines in an ordered incidence plane are forming stripes iff case (a) occurs,— there exists one line among them such that the two other lines are in opposite half planes of the first line.
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Definition 11.2 (Three parallel lines in triangular shape). I say that three different mutually parallel lines in an ordered incidence plane are in triangular shape iff case (b) occurs,—as seen from any one of the three lines, the two others lie in the same half plane. Lemma 11.1. In an ordered affine plane, any three different mutually parallel lines are forming stripes. In other words, there do not exist three parallel lines in triangular shape. Proof. Let a, b, c be any three different mutually parallel lines in an ordered affine plane. Take any point A on line a, and point B on line b. The line AB intersects line a at the unique point A. By Proclus’ lemma3.3, the line AB intersects the parallel c k a, too. Let C be the intersection point. By the three-point Theorem 5.2 (Hilbert’s Proposition 4), we know that among any three different points lying on a line, there exists exactly one lying between the two other points. We may assume the case A ∗ B ∗C to occur, without loss of generality. Then points A and C, and by plane separation the entire lines a and c lie in opposite half -planes of the line b. Remark. It is very easy to make the mistake to conjecture that lemma 11.1 would hold more generally in any ordered incidence plane. But this is not true. Counterexamples can be constructed in Hilbert’s hyperbolic geometry as well as in the Poincaré disk model of hyperbolic geometry. • Any improper triangle with three improper vertices, has as sides three parallel lines which are in triangular shape according to definition 11.2. • In Hilbert’s hyperbolic geometry, three line in triangular shape may be obtained as follows. One draws any three different rays a, b, c from a common vertex P. By proposition 45.9, there exists an inclosing line for each one of the angles ∠(a, b), ∠(b, c) and ∠(c, a). In case that two among the rays are opposite, the inclosing line is replaced by the line on these opposite rays. These three enclosing lines are parallels, and they form a triangular shape. • Another easy example is constructed from a hexagon with six right angles. It is known that such hexagons exist in the hyperbolic plane. Any three nonadjacent sides of such a hexagon lie on three parallel lines which are in triangular shape. The remaining three sides of the hexagon are their common perpendiculars. • The figure on page 281 gives a most astonishing example. In the hyperbolic plane exist triangles, and perpendiculars erected on their three sides such that all three perpendiculars turn out to be parallel to each other,— and even form a triangular shape according to my definition 11.2. Proposition 11.3 (About triangles without circum-circle). In any Hilbert plane is given the triangle 4ABC. Let Ma , Mb , Mc be the midpoints of sides a, b, c, and let ma , mb , mc be their perpendicular bisectors. After a permutation of the vertices, we may assume c ≤ a ≤ b. Under the assumption that the bisectors do not intersect neither in the interior of the triangle nor at point Mb , one concludes that a < b, and gets the following situation illustrated in the figure on page 281: ma = Ma Qb , mb = Mb Pa , mc = Mc Pb B ∗ Pa ∗ Ma ∗ C if c < a. The triangle is not isosceles. B = Pa ∗ Ma ∗ C if c = a. The triangle is isosceles. A ∗ Pb ∗ Mb ∗ Qb
(11.1) (11.2) (11.3) (11.4)
Proof. By Pasch’s axiom, the bisector mb of the longest side AC either intersects a further side of triangle 4ABC, say at point Pa ;—or the opposite vertex B lies on the bisector mb and Pa = B. By corollary 9, a point lies on the perpendicular bisector if and only if its distances to the endpoints of
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Figure 11.2. Strange perpendiculars to the three sides of a triangle.
Figure 11.3.
In hyperbolic geometry, the perpendicular bisectors can be parallel as shown—but this figure is impossible in Euclidean geometry.
the segment are congruent. Hence Pa = B if and only if c = a and the triangle is isosceles. I leave it to the reader to complete the reasonings in this special case. We now assume the case c < a to occur. From proposition 11.1 follows that vertices A and B of the shortest side c lie on the same side of bisector mb , but vertices B and C lie on different sides of mb . By plane separation, the bisector mb intersects the side BC at some point Pa . In the right
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triangle 4Pa MbC, the hypothenuse is the longest side. Hence CMa = a/2 ≤ b/2 = CMb < CPa and hence C ∗ Ma ∗ Pa ∗ B We use Pasch’s axiom for triangle 4Pa MbC to conclude that the bisector ma intersects either side Pa Mb or side MbC. In the first case, the bisectors mb and ma intersect either in the interior of the triangle or at point Mb ,—which cases are ruled out by assumption. Hence bisector ma intersects side MbC at some interior point Qb . We have obtained the order relations A ∗ Mb ∗ Q b ∗ C Next we rule out the case b = AC = BC = a. In that case the bisector mc of side AB would go through vertex C. The bisectors ma = Ma Qb and mb = Mb Pa would have points in both half-planes of mc . Hence the three bisectors intersect in the interior of the triangle 4ABC,—which case we ruled out by assumption. Since b = AC > BC = a, from proposition 11.1 follows that points A and C lie on different sides of the bisector mc . Hence mc intersects the sides AC in some point Pb . From the assumption that the bisectors mc and mb do not intersect, we obtain APb < AMb , and hence A ∗ Pb ∗ Mb ∗ Qb ∗ C The segments Qb Ma ⊂ ma and Pb Mc ⊂ mc lie in different half planes of bisector mb . The right triangle 4Qb MaC has an acute angle at vertex Qb . The right triangle 4Pb Mc A has an acute angle at vertex Pb . Hence the bisectors ma and mc may • either intersect in the the half-plane Hb of the longest side b opposite to vertex B; • or all three bisectors are parallel forming stripes. Corollary 28. In neutral geometry, the following five cases are all possible. (a) The three bisectors intersect in the interior of the triangle. (b) The three bisectors intersect at the midpoint Mb of the longest side b. (c) The three bisectors intersect at a point lying in the half-plane Hb of the longest side b opposite to vertex B. (d) The three bisectors have a common perpendicular, which lies in the half-plane Hb . (e) Any two of the bisectors are asymptotically parallel. In that case, the rays of all three bisectors converge in the half-plane Hb . In the cases (d) and (e), the bisectors all mutually parallel and are forming stripes, according to definition 11.1. The triangle has no circum-circle. Corollary 29. For any triangle in a Pythagorean plane, only cases (a), (b) or (c) from corollary 28 can occur. Hence every triangle has a circum-cirle. Proof. In a Pythagorean plane, we may use proposition 7.44, in short "parallels imply congruent z-angles". As shown above, the three angular bisectors do not have congruent z-angle with the line of the longest side. In a Pythagorean plane, this implies that all three side bisectors pairwise intersect. Thus cases (d) and (e) are excluded.
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Figure 11.4. The perpendicular bisectors of the sides of triangle 4ABC are all parallel. The triangle has no circumcircle—true, but hard to believe!
Example 11.1 (Not every triangle needs to have a circum-circle). Indeed, for the construction of a counterexample, we use Saccheri quadrilaterals with a baseline far away. From hyperbolic geometry, one needs only the feature that underbarno rectangle exists. The construction starts with baseline l, and three points X, Y and Z lying on l. Next one sets up two adjacent Saccheri quadrilateral Y XAB and YZCB with common side Y B. Hence we get three congruent segments Y B XA ZC, and right angles at vertices X, Y and Z. Question. Why do the three points A, B and C not lie on a line? Answer. Note that Ma , Mb , otherwise A = B = C and X = Y = Z. Now assume towards a contradiction that points A, B and C lie on a line. The three midpoints would lie on that line, too. Hence the quadrilateral EDMa Mb would be a rectangle, contrary to the assumption that no rectangle exists. The line l = XYZ is the common perpendicular of all three perpendicular bisectors of triangle 4ABC. Hence all three perpendicular bisectors are parallel. By Proposition 11.2, the triangle has no circum-circle. But indeed, one realizes a different new feature: Definition 11.3 (Equidistance line). Given is a baseline l and a distance AX. The set of all points with distance from a baseline l congruent to AX, and lying on one side of this line, are called an equidistance line. In the example just constructed, all three distances Y B XA ZC are congruent, and the three vertices A, B and C lie on one side of line l = XYZ. Hence the three vertices of the triangle lie on an equidistance line. One is thus lead to an analog to Proposition 11.2: Proposition 11.4 (Conditional equidistance line). For any triangle, the following three statements are equivalent: (a) The perpendicular bisectors of two sides have a common perpendicular. (b) The three vertices lie on an equidistance line. (c) The bisectors of all three sides have a common perpendicular. Proof of Proposition 11.4. This proof is very similar to the well known proof of Proposition 11.2. The missing link is Proposition 11.5, which is stated and used now. We postpone its proof to the next paragraph.
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Proposition 11.5. A line is perpendicular to the perpendicular bisector of a segment if and only if the two endpoints of the segment have the same distance to the line, and lie on the same side of it. Reason for "(a) implies (b)" from Proposition 11.4. Assume that the perpendicular bisectors of sides AB and BC have the common perpendicular l. Because l and pc intersect perpendicularly, Proposition 11.5 yields that vertices A and B have the same distance from l, and lie on the same side of l. By the same reasoning, Proposition 11.5 yields that the vertices B and C have the same distance from l, and lie on the same side of l, because l and pa intersect perpendicularly. Hence all three vertices A, B and C lie on an equidistance line. Reason for "(b) implies (c)". We assume that all three vertices A, B and C have congruent distances AX BY CZ to baseline l, and lie on the same side of l. Here X, Y and Z are the foot points of the perpendiculars dropped from the vertices A, B and C onto line l. By Proposition 11.5, equal distances AX BY implies that the baseline l = XY is perpendicular to the bisector pc . Similarly, equal distances BY CZ imply l = YZ is perpendicular to the bisector pa , and CZ AX implies that l = XZ is perpendicular to pb . Hence their common baseline l = XYZ is perpendicular to all three bisectors. Obviously (c) implies (a). Hence we conclude that all three assumptions are equivalent. Once again, this means that either all three are true, or all three are false. Remark. We have just seen that l is a common perpendicular to pa , pb and pc . Now one can look at the figure in a different way: Each one of the three sides of the triangle 4ABC has as a common perpendicular with l— pc is the common perpendicular of AB and l, pa is the common perpendicular of BC and l, finally pb is the common perpendicular of AC and l. Hence not only all three sides of the triangle, but even their extensions lie on one side of line l.
Figure 11.5. The generic situation for Proposition 11.5, for which the following statement is proved: segments AX and BY are congruent if and only if lines l and pc are perpendicular.
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11.3.
Interlude about points of congruent distance from a line
Proof of Proposition 11.5. Let M be the midpoint of segment AB and pc be its perpendicular bisector. From both points A and B we drop the perpendiculars onto line l. Let X and Y be the foot points.
Figure 11.6. Constructing the Saccheri quadrilateral from its top: if the lines pc and l are perpendicular to each other, then the segments AX and BY are congruent.
At first, assume that the lines l and pc intersect perpendicularly, and let Q be their intersection point. Reversing the steps explained in the proof of Hilbert’s Proposition 36, now the first step is to obtain the triangle congruence 4AMQ 4BMQ (flier) This is done via SAS congruence. As a second step, we get the congruence 4AQX 4BQY
(wedge)
by SAA congruence. Hence AX BY, and XY BA is a Saccheri quadrilateral. To prove the converse, we assume that the two endpoints A and B have the same distance to line l and lie on the same side of l. Because of AX BY, we get the Saccheri quadrilateral XY BA. As in the proof of Hilbert’s Proposition 36, it is shown that the perpendicular bisector q of segment XY intersects segment AB, too—and indeed in its midpoint M. Question. How does one get these properties of bisector q? To repeat the details: one begins by checking the congruence (wedge), which is obtained via SAS-congruence. The second step is to get congruence (flier), again by SAS-congruence. Hence line q = pc is the perpendicular bisector of both segments AB and XY. Clearly this means that the lines pc and l = XY are perpendicular. 11.4.
The midpoint triangle and its altitudes
Definition 11.4. Let Ma , Mb and Mc denote the midpoints of the three sides of the triangle. The three segments connecting the vertices to the midpoint of the opposite side are called the medians of the triangle. The triangle 4Ma Mb Mc is called the midpoint-triangle. Theorem 11.1. The line through the midpoints of two sides and the perpendicular bisector of the third side of a triangle are perpendicular. Remark. This is a theorem of neutral geometry!
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Figure 11.7. To triangle 4ABC corresponds the Saccheri quadrilateral AFGB.
Reason, given in neutral geometry. The theorem follows from Hilbert’s Propositions 36 and 39 in the section about Legendre’s theorems. We construct the Saccheri quadrilateral AFGB corresponding to triangle 4ABC, To this end, one drops the perpendiculars from the three vertices A, B and C onto line l = Ma Mb , and proves the congruence AF CH BG. By Hilbert’s Proposition 36, the perpendicular bisectors pc partitions the Saccheri quadrilateral into two Lambert quadrilaterals. Hence pc is the symmetry line p of this Saccheri quadrilateral. It is perpendicular to both lines Ma Mb and line AB.
Figure 11.8. The altitudes of the midpoint triangle are the side bisectors and form six right angles.
Corollary 30 ("The altitudes of the midpoint-triangle are the side bisectors, and thus form six right angles."). The altitudes of the midpoint-triangle are the perpendicular bisectors of the
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sides of the original triangle. Each of the three bisectors is perpendicular to both one side of the original triangle, and one side of the midpoint triangle, thus forming six right angles.
Figure 11.9. For an obtuse triangle, too, the altitudes of the midpoint triangle are the side bisectors, and form six right angles.
Proof. The bisector pc of side AB, is perpendicular to line Ma Mb , and hence is an altitude of the midpoint triangle Mc Ma Mb , too. Similar statements hold for the other two bisectors. Proposition 11.6. Any two altitudes of an acute triangle insect inside the triangle. Two altitudes of an obtuse triangle may or may not intersect. Any possible intersection point lies outside the triangle, inside the vertical angle to the obtuse angle of the triangle. Proof. For an acute triangle, the foot points of the altitude lie on the sides of the triangle, not the extensions—as follows from the exterior angle theorem. Hence the Crossbar Theorem implies that any two altitude do intersect. The remaining details are left as an exercise. Here is stated what we have gathered up to this point. It is little, but more than nothing! Proposition 11.7. For any triangle either one of the following three cases occur: "As seen in Euclidean geometry" The triangle has a circum-circle. The bisectors of all three sides intersect in its center, called circum-center. The circum-center of the larger original triangle 4ABC: is the orthocenter of the midpoint triangle 4Ma Mb Mc . H2 = O (H2O) This case always occurs for an acute or right midpoint-triangle. It can but does not need to happen in case of an obtuse midpoint-triangle. "The genuine hyperbolic case" The three vertices of the triangle lie on an equidistance line. The bisectors of all three sides have a common perpendicular l. All three extended sides of the triangle are parallel to the baseline l.
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"The borderline case" The three vertices lie neither on a circle nor an equidistance line. Neither two of the three bisectors intersect, nor do any two of them have a common perpendicular. Corollary 31. In neutral geometry, a triangle 4ABC has a circum-circle if and only if two altitudes of the midpoint triangle do intersect. In that case all three altitudes of the midpoint triangle intersect in one point. Reason. Suppose now that two altitudes of 4Ma Mb Mc do intersect. By Proposition 11.6, this happens always for an acute or right midpoint-triangle. It can happen, but need not happen in case of an obtuse midpoint-triangle. Since the two intersecting altitudes are side bisectors for the original triangle, Proposition 11.2 implies that all three side bisectors intersect in one point. This point is the circum-center of the original triangle and the orthocenter of the midpoint-triangle as stated in formula (H2O). On the other hand, suppose that all three altitudes of the midpoint-triangle are parallel, but any two of them have a common perpendicular. In that case, we proceed analogously to above: Since the two divergently parallel altitudes are side bisectors for the original triangle, Proposition 11.4 implies the bisectors of all three sides have a common perpendicular. Furthermore, all three vertices lie on an equidistance line. Thus we arrive at "The genuine hyperbolic case". The third logical possibility is that all three bisectors are asymptotically parallel—leading to the borderline case. 11.4.1. Immediate consequences for the altitudes At this point, it is tempting to look for a more simple result which deal only with the altitudes of the original triangle, without the need to argue via the midpoint triangle. That seems to be astonishingly difficult! Before leaving everything open, I better state my rather awkward partial result: Proposition 11.8 (Preliminary and conditional orthocenter—neutral version). Suppose that the given triangle 4ABC is the midpoint-triangle of a larger triangle 4A0 B0C0 , and the altitudes of two sides of the triangle 4ABC intersect. Then the altitudes of all three sides intersect in one point. If the given triangle 4ABC is acute, and it is the midpoint-triangle of a larger triangle 4A0 B0C0 , then the three altitudes intersect in one point. Remark. Actually the additional assumption that the given triangle is the midpoint triangle of another (larger) triangle does not need to hold in hyperbolic geometry. Nevertheless the statements about the altitudes of the original triangle remain true! 1 Proof of Proposition 11.8. We use the Corollary about the six right angles for triangle 4A0 B0C0 . The perpendicular bisectors of triangle 4A0 B0C0 are the altitudes of the midpoint-triangle 4ABC. Now Proposition 11.2 —about the conditional circum-circle— is applied to larger triangle 4A0 B0C0 . We conclude that all three perpendicular bisectors of that triangle intersect. But these are just the altitudes of the original given triangle 4ABC. 11.5. The Hjelmslev Line Given are two different lines l with the different points A, B, C, D, . . . and l0 with points A0 , B0 , C 0 , D0 , . . . with congruent distances and same order along these two lines. Let K, L, M, . . . be the midpoints of the segments AA0 , BB0 , CC 0 , . . . . Theorem 11.2. Either one of the following two cases holds: 1
The section on hyperbolic geometry contains a proof using Klein’s model. A proof in neutral geometry is clearly valid a real bottle of wine.
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(a) the exceptional case: All midpoints are equal. Furthermore, this point is the midpoint of the common perpendicular of the two lines. They are parallel, and the points A, B, C, D, . . . and A0 , B0 , C 0 , D0 , . . . are ordered in opposite directions along the two lines. (b) the generic case: All midpoints are different. They lie on a third line, which is called the Hjelmslev line.
Figure 11.10. The Hjelmsjev line in the generic case.
Proof. We distinguish the following two cases, from which all others are obtained by renaming: (a) The mid points K of segment AA0 and L of segment BB0 are equal. (b) The mid points of segments AA0 , BB0 , CC 0 are all different. Problem 11.1. Show that in the first case (a) all assertions of the theorem hold. We now consider case (b). We define point B00 such that K is the midpoint of segment BB00 . The assumption K , L implies B0 , B00 . Since AB A0 B0 A0 B00 , the perpendicular bisector p of segment BB00 goes through point A0 = l0 ∩ l00 and bisects an angle between the lines l0 = A0 B0 and l00 := A0 B00 . By theorem 11.1, the line through the midpoints of two sides and the perpendicular bisector of the third side of a triangle are perpendicular. For the triangle 4BB0 B00 , we conclude that the lines KL and p are perpendicular. In the special case that points B, B0 and B00 lie on a line, we get the same conclusion. Now we can argue similarly,after replacing B by C and B0 by C 0 . Since A0C 00 A0C 0 and the orders A0 ∗B0 ∗C 0 and A0 ∗B00 ∗C 00 , the perpendicular bisector q of segment CC 00 is the angle bisector of the same angle ∠B0 A0 B00 = ∠C 0 A0C 00 and hence p = q. The line K M is again perpendicular to p. Hence the points K, L and M lie on one line, as to be shown. 11.6. The in-circle and the ex-circles The construction of the in-circle of a triangle depends on the angular bisectors of its angles. As already stated by definition 7.12, the interior angular bisector of an angle is the ray inside the angle which forms congruent angles with both sides. Definition 11.5. The interior and exterior bisecting lines of an angle are the two lines which forms congruent angles with either sides of the angle, or the opposite rays.
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Figure 11.11. The ray opposite to the bisecting ray b bisects the vertical angle, because all four red angles are congruent.
Question. Show that the bisectors of vertical angles are opposite rays. Solution. Take the vertical angles ∠(h, k) and ∠(h0 , k0 ). Let ray b be the interior bisector of angle ∠(h, k) and b0 be the ray opposite to it. Congruence of vertical angles implies ∠(h, b) ∠(h0 , b0 ) and ∠(b, k) ∠(b0 , k0 ) The definition of the angular bisector yields ∠(h, b) ∠(b, k) From these three formulas, using transitivity and symmetry of angle congruence, one gets ∠(h0 , b0 ) ∠(h, b) ∠(b, k) ∠(b0 , k0 ) Hence ray b0 ,which was chosen to be opposite to the bisecting ray b is the angular bisector of the vertical angle ∠(h0 , b0 ). Question. Show that the bisectors of supplementary angles are perpendicular to each other.
Figure 11.12. Angle addition of two pairs of congruent angles yields congruent supplementary angles—proving that the bisectors of supplementary angles are perpendicular.
Solution. We use the same notation as in the last question, and let ray c be the bisector of the angle ∠(k, h0 ), which is supplementary to angle ∠(h, k). From the last question we get ∠(b, k) ∠(b0 , k0 ) ∠(h0 , b0 ) and, by the definition of bisector c, we get ∠(k, c) ∠(c, h0 )
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Now we use that sums of congruent angles are congruent. Hence the two formulas yield ∠(b, c) = ∠(b, k) + ∠(k, c) ∠(h0 , b0 ) + ∠(c, h0 ) ∠(c, b0 ) Thus we see that ∠(b, c) and ∠(c, b0 ) are supplementary congruent angles, and hence they are right angles. Here is the characterization of the angular bisectors. Proposition 11.9. A point in the interior of an angle has congruent distances from both sides of the angle if and only if it lies on the interior angular bisector.
Figure 11.13. A point P on the interior angular bisector has congruent distances from both sides the angle ∠(h, k).
Proposition 11.10. A point has congruent distances from both lines extending the sides of an angle, if and only if it lies on the interior or the exterior bisecting line. Remark. Indeed the characterization of the angular bisectors from proposition 11.9 and 11.10 fails for a degenerate angle. The assertion of these propositions are wrong for an angle degenerating to zero or two right! Proof. Given is the angle ∠(h, k) with vertex A, and a point P. The two lines of the sides h and k of the angle ∠(h, k) are assumed to be different, by definition of an angle. Let X and Y be the foot points of the perpendiculars dropped from point P onto the two different lines extending h and k. In the figure on page 291, we give a case were point P neither lies on an angular bisector nor has congruent distances to the sides of the angle—and, in a separate drawing, a case were point P both lies on an angular bisector and has congruent distances to the sides of the angle. Claim 0. If point P lies on either one of the two lines extending h and k, the assertion of proposition 11.10 is true. Reason. For P , A, neither are the distances from point P to the two lines equal—only one is zero—, nor does point P lie on any one of the bisectors, which lie in the interiors of the four angles with vertex A we have obtained. If P = A, all is clear, too. We shall now assume that point P does not lie on either one of the lines extending rays h and k. We begin with the case were a triangle congruence characterizes the points P which either lie on the angular bisector or have congruent distances to the sides of the angle.
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Claim 1. Assume there exist the triangles 4PAX and 4PAY. In that case, PX PY
implies
∠PAX ∠PAY
Reason. The two triangles 4PAX 4PAY
(11.5)
are congruent by the hypothenuse-leg theorem 7.32. Hence ∠PAX ∠PAY.
Claim 2. Assume there exist the triangles 4PAX and 4PAY. In that case, ∠PAX ∠PAY
implies
PX PY
Reason. The two triangles (11.5) are congruent by SAA-congruence, given in proposition7.54. Hence PX PY. Claim 3. If the triangles 4PAX and 4PAY exist, the assertion of proposition 11.10 is true. Claim 4. If A , X, and point P does not lie on the line of ray h, then the triangle 4PAX exists. Hence, if A , X and A , Y, the assertion of proposition 11.10 is true. Claim 5. Either A , X or A , Y. Reason. If A = X = Y, then both rays h and k would be perpendicular to the segment PA. Hence they would be identical or opposite rays. Hence A = X = Y imply the rays h and k lie on the same line. But this contradicts the definition of an angle, and hence cannot occur. What happens in the awkward case A = X and A , Y? In this case the triangle 4PAX is nonexistent (degenerate). But it is easy to deal with the case separately, since the perpendiculars to the sides of an angle cannot be any angular bisector. Let K be any point on the ray k, and H be any point on the ray h. Claim 6. If A = X and A , Y, and the rays h and k lie on two different lines, then R = ∠PAH > ∠PAY and PX = PA > PY. Similarly, if A = Y and A , X, then ∠PAX < R = ∠PAK and PX < PA = PY. In these cases, point P neither lies on an angular bisector nor has congruent distances to the sides of the angle. Hence, if A = X or A = Y, the assertion of proposition 11.10 is true. Reason. Because of claim 5, the assumption A = Y implies A , X. Hence by claim 4, the right triangle 4PAX exists. The right angle across to the hypothenuse is larger than any of the other two angles, with vertices A or X. Hence ∠PAX < R = ∠PAK. Its longest side is the hypothenuse PA. Claim 7. If the segments PX PY to the foot points are congruent, then point P lies in the interior of the angle ∠XAY.
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Figure 11.14. Any two angular bisectors intersect at the center of the in-circle.
Theorem 11.3 (The in-circle). All three interior angular bisectors of a triangle intersect in one point. This point has congruent distances from all three sides of the triangle, and hence it is the center of the in-circle. Problem 11.2 (The in-circle). We want to construct the in-circle of a triangle. Explain the steps of the reasoning and provide drawings. Use the proposition 11.9 given above, and still other basic facts. (a) Explain why any two interior angle bisectors of a triangle intersect. (b) Explain why this point has congruent distances from all three sides of the triangle, and why it is the center of the in-circle. (c) Explain why all three interior angular bisectors of a triangle intersect in one point. (d) Explain how one can draw the in-circle of a triangle. Answer. Here is some rather elaborate recollection: (a) Why any two interior angle bisectors of a triangle intersect: By the Crossbar Theorem, an interior angular bisector intersects the opposite side. Let Wa be the intersection point of the angular bisector of angle ∠BAC with side BC. Once more, by the Crossbar Theorem, the angular bisector wb of angle ∠ABC intersects segment AWa , since points A and Wa lie on the two sides of this angle. The bisector wb intersects segment AWa of bisector wa , say in point I. (b) Point I has congruent distances from all three sides: Since point I lies one the bisector of angle ∠BAC, it has congruent distances from the sides AC and AB. Since point I lies one the bisector of angle ∠ABC, it has congruent distances from the sides BC and BA. By transitivity, it has congruent distances from all three sides. (b) Point I is the center of the in-circle: Let X be the footpoint of the perpendicular dropped from point I onto the side BC. This side is tangent to the circle around I through X, since the tangent of a circle is perpendicular to the radius IX at the touching point. Indeed, we
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have already obtained a circle to which all three sides of the triangle are tangent. Let Y, Z be the footpoints of the of the perpendiculars dropped from point I onto the other two sides CA and AB. Since IX IY IZ, the same circle goes through Y and Z. Again, the sides CA and AB are tangent to this circle since they are perpendicular to IY and IZ, respectively. (c) All three interior angular bisectors of a triangle intersect in one point: Since point I has congruent distances from the sides AC and BC, the point lies on either one of the bisectors of the angles at C—which we did not even use in the construction. Since point I lies in the interior of the angle ∠BCA, it lies on the interior angular bisector of this angle. (d) How to draw the in-circle of a triangle: As seen above, one needs to construct the angular bisectors of any two of the three angles of the triangle. One drops the perpendicular from their intersection point I to any side of the triangle. Finally, one draws the circle around I through the foot point. The exterior bisectors are bisecting the exterior angles of the triangle. They are the perpendiculars to the interior angular bisectors, erected at the vertices. They are used in the construction of the ex-circles.
Figure 11.15. The in-circle and one ex-circle.
Proposition 11.11 (Conditional ex-circle—neutral version). If any two of the exterior bisectors at vertices A and B, and the bisector of the third angle ∠BCA intersect, then all these three bisectors intersect in one point. In that case, the triangle has an ex-circle touching side AB from outside, and the extensions of the two other sides. Reason. This is an easy consequence of Proposition 11.10. The details are left to the reader. 11.7.
The Hjelmslev quadrilateral in neutral geometry
Definition 11.6. A quadrilateral with two right angles at opposite vertices is called a Hjelmslev quadrilateral. In a Hjelmslev quadrilateral quadrilateral, we draw the diagonal between the two right angles and drop the perpendiculars from the two other vertices. Their occur some remarkable congruences.
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Figure 11.16. The Theorem of Hjelmslev.
Proposition 11.12 (Hjelmslev’s Theorem). In a Hjelmslev quadrilateral, (a) at each of the two vertices with arbitrary angle, there is a pair of congruent angles between the adjacent sides, and the second diagonal and the perpendicular dropped onto the first diagonal, respectively; (b) there is a pair of congruent segments on the diagonal between the right vertices, measured between these vertices and the foot points of the perpendiculars. For the proof in neutral geometry, we need some basic lemmas about reflections. These are facts of neutral geometry. Definition 11.7 (Rotation). The composition of two reflections across intersecting lines is a rotation. The intersection point O of the reflection lines is the center of the rotation. If point P is mapped to point P0 , the angle ∠POP0 is independent of the point P. It is called the angle of rotation. Lemma 11.2. Let four lines p, q, r, s intersect in one point. The compositions of reflections across these lines satisfy p ◦ q = r ◦ s if and only if ∠(q, p) = ∠(s, r) for the directed angles. The angle of rotation is twice the angle between the two reflection lines. Definition 11.8 (Point reflection). The reflection across the center of reflection O maps any point P into a point P0 , such that the center of reflection O is the midpoint of segment PP0 . Lemma 11.3. Two reflections commute if and only if the reflection lines are either identical or perpendicular to each other. The composition of two reflections across perpendicular lines depends only on the intersection point of the two lines. It is the reflection across their intersection point.
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Definition 11.9 (Translation). The composition of two reflections across lines with a common perpendicular t is a translation along t. Lemma 11.4. Let four lines p, q, r, s have the common perpendicular t, and let P, Q, R, S be the intersection points. The compositions of reflections across these lines satisfy p ◦ q = r ◦ s if and only if QP = S R for the directed segments. Problem 11.3. Prove the segment congruence of Hjelmslev’s Theorem in neutral geometry. Assume that the angle congruence has already been shown. Use compositions of reflections across the lines marked in the figure on page 296 and check that c◦g=h◦a
Figure 11.17. The segment congruence in neutral geometry.
Proof of the segment congruence in neutral geometry. e ◦ c ◦ g = c1 ◦ b ◦ g = c1 ◦ f ◦ a1 = h ◦ d ◦ a1 =h◦e◦a =e◦h◦a
by Lemma 11.3 at point C by Lemma 11.2 at point B by Lemma 11.2 at point D by Lemma 11.3 at point A by Lemma 11.3 at point H
Hence c ◦ g = h ◦ a, and finally Lemma 11.4 yields CG = HA, as to be shown.
Proposition 11.13 (Translation and Saccheri quadrilaterals). Let point A be mapped to point A0 by a translation along the line t, and let F and F 0 be the foot points of the perpendiculars dropped from the points A and A0 onto the line t of translation. The quadrilateral FAA0 F 0 is a Saccheri quadrilateral. Hence the segments AA0 and FF 0 are parallel. The base segment FF 0 is double the segment QP between the intersection points of the lines of reflection with their common perpendicular, independently of the point A.
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Remark. In Euclidean geometry, the segment AA0 from any point A to its image A0 is double the segment QP between the intersection points of the lines of reflection with their common perpendicular, independently of the point A. Note that this is not true in neutral geometry! Indeed, in hyperbolic geometry, the top segment AA0 is longer than the base segment FF 0 , for any point A not on the line of translation. To prove the entire Theorem of Hjelmslev in neutral geometry, one needs the following more elaborate result. Definition 11.10 (Translation-reflection). The composition of two reflections across a line p and a point Q is called a translation-reflection. The perpendicular to the line to the point is baseline along which the translation-reflection acts. Theorem 11.4 (A translation-reflection specifies both the translation line and distance). Let two lines p, r and two points Q, S be given. The compositions of reflections across these lines and points satisfy p◦Q=r◦S if and only if the following three conditions hold: there exists a common perpendicular to lines p and r through the points Q and S ; the respective distances from point Q to the foot point P on p, and from point S to the foot point R on r are congruent; The two segments QP = S R have the same orientation on the common perpendicular. Similarly, p◦Q=S ◦r holds if and only if the two segments QP = RS are congruent and have opposite orientations on a common perpendicular of the lines p and r. Question. What does it mean in terms of order relations or rays that two segments PQ and RS on one line have the same orientation? Answer. Two segments QP and S R on one line have the same orientation if and only if of the two −−→ −−→ rays QP and S R one is a subset of the other. Problem 11.4. Prove Hjelmslev’s Theorem in neutral geometry. In the Hjelmslev configuration, you need at first to define lines g0 and h0 by angle congruences. Proof of the Hjelmslev Theorem in neutral geometry. Begin defining the lines g0 and h0 by the angle congruences ∠(b, g0 ) ∠( f, a1 ) ∠(h0 , d) ∠(c1 , f ) for the directed angles. Quite similar above, we check C ◦ g0 = c1 ◦ b ◦ g0 = c1 ◦ f ◦ a1 = h0 ◦ d ◦ a1 = h0 ◦ A
by Lemma 11.3 at point C by assumption by assumption by Lemma 11.3 at point A
By the translation-reflection theorem 11.4, he translation-reflections C ◦ g0 and h0 ◦ A have the same base line and distance. In other words, line AC is the common perpendicular of lines g0 and h0 . Hence these lines are the same as the perpendiculars: g = g0 and h = h0 . Furthermore, CG0 H 0 A as directed segments.
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Figure 11.18. Begin defining the lines g0 and h0 by the angle congruences.
11.8. Limitations of neutral triangle geometry You have for sure realized that the propositions above do not give the complete and beautiful results from Euclidean geometry—instead they contain awkward conditions and unexpected cases. This state of affairs has good reasons. Indeed many of the more complete theorems of triangle geometry are equivalent in some cases to the Euclidean parallel axiom, or in other cases to the angle sum of a triangle being two right angles. The following theorem is confirming this claim.
Figure 11.19. Triangle 4ABC has a circum-circle, hence line n intersects line l in its center O—parallels are unique.
Theorem 11.5 (Farkas Bolyai). Assume only the axioms of incidence, order, congruence—leading to neutral geometry. If every triangle has a circum-circle, then the Euclidean parallel axiom holds. Proof. As explained in Proposition 7.43 in the section on Legendre’s theorems, existence of a parallel can be proved in neutral geometry. Indeed, one parallel is conveniently constructed as "double perpendicular".
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Figure 11.20. The perpendicular bisectors of the sides of triangle 4ABC are all parallel. The triangle has no circum-circle—true, but hard to believe! There exists two parallels m and n to line l through point P.
We start with this construction from Proposition 7.43. Given is any line l and any point P not on l. One drops the perpendicular from point P onto line l and denotes the foot point by F. Next, one erects at point P the perpendicular to line PF. Thus, one gets the "double perpendicular" m. Let A be any point on the segment PF, and B be its image of reflection by line l. By definition of reflection, the foot point F is the midpoint of segment AB. The four points P, F, A and B lie on the common perpendicular p of the two lines l and m. We want to show the line m is the unique parallel to line l through point P. Assume that line n is a parallel to line l through point P, too. Now let C be the image of reflecting point A by line n. By definition of reflection, this means that segment AC and line n intersect perpendicularly at the midpoint R. If point C lies on line AB = p, then R = P and n = m, because both are the perpendicular to p erected at P. But, if one assumes n , m is really a second different parallel to line l through point P, one can show that lines n and l intersect, contradictory to being parallel. Indeed, by the argument above we see that point C does not lie on line AB. Hence the three point A, B and C form a triangle. It is assumed that every triangle has a circum-circle. Let O be the center of the circum-circle of triangle 4ABC. Because O has same distance from points A and B, it lies on the perpendicular bisector of segment AB, which is line l. Similarly, because O has same distance from points A and C, it lies on the perpendicular bisector of segment AC, which is line n. Hence the two lines l and n intersect at point O, and are not parallel. Thus a second parallel to line l through point P does not exist. Question. Why are the two lines l and n different? Answer. Line n goes through point P, but line l does not go through point P. Corollary 32. In any Hilbert plane are equivalent
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(i) for every line l and for every point P lying not on l, there exists a unique parallel m to l through point P; (ii) every triangle has a circum-circle. Reason for "(i) implies (ii)". Assumption (i) states the Euclidean parallel postulate, resulting in a Pythagorean plane. In corollary 29 is shown that in a Pythagorean plane, every triangle has a circum-circle. Remark. A simpler though indirect argument leading to existence of the circum-circle is given in the proof for theorem 25.1 below. Reason for "(ii) implies (i)" . This is the theorem 11.5 by Farkas Bolyai.
Here is another simple proposition that implies Euclidean geometry.
Figure 11.21. If the three midpoints lie on a line, a rectangle exists.
Figure 11.22. How the hyperbolic case appears in Klein’s model.
Proposition 11.14 (Three midpoints). Given are three points A, B, C on a line l and a point P not on line l. If the three midpoints U, V, W of segments AP, BP and CP lie on a line, then a rectangle exists. Conversely, if a rectangle exists, then the three midpoints lie on one line.
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Question. Provide a drawing. Explain how one gets a rectangle, in neutral geometry. Use Theorem 11.1 and once more, the material of my package on Legendre’s geometry, especially Proposition 39, to explain your reasoning. Proof. The perpendicular bisector pc of segment AB intersects both lines l and UV perpendicularly. This follows from Theorem 11.1, applied to 4ABP. By the same reasoning, the perpendicular bisector pa of segment BC intersects both lines l and VW perpendicularly. This time, one applies Theorem 11.1 to 4BCP. If the three midpoints U, V, W lie on one line, then this line—together with lines l, and the two bisectors pc and pa form a rectangle. Conversely, use that a rectangle exists. By the second Legendre Theorem, the angle sum in every triangle is 2R, in every quadrilateral 4R, in every pentagon 6R. If the three midpoints U, V, W would not lie on one line, there would exist a pentagon with angle sum different from 6R, which is impossible. Hence the three midpoints U, V, W lie on one line.
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12. Towards a Natural Axiomatization of Geometry The title begs the question: what is meant in mathematics by the attribute "natural"? By itself, this clearly is a philosophical question beyond mathematics. First of all, mathematical postulates are and will in the future surely be chosen,—even more often,—based on criteria from history, or rooted in the convictions held by the scientific community. After that being understood, which place can be left to the requirement of "natural"? Nevertheless, the author of these notes wants to express his conviction, that "natural" for a mathematical postulate is important, and means the following: • A postulate which uses notions connected to the everyday life experience, or are well motivated by the applications, may be called natural. • A postulate which connects notions used by the proofs, or which facilitates the mathematical developments, may be called natural. • A postulate which allows surprising connections between different mathematical theories, or shows an underlying unity of mathematics, may be called natural. The material of this section is selected with such considerations in mind. 12.1.
The Uniformity Theorem
Main Theorem 16 (Uniformity Theorem). Any Hilbert plane is of either one of the following three types: semi-Euclidean The angle sum of every triangle is two right angle, and every Lambert or Saccheri quadrilateral is a rectangle. semi-hyperbolic The angle sum of every triangle is less than two right angle, and every Lambert or Saccheri quadrilateral has one respectively two acute angles. semi-elliptic The angle sum of every triangle is larger than two right angle, and every Lambert or Saccheri quadrilateral has one respectively two obtuse angles. The Uniformity Theorem motivates the following definition: Definition 12.1 (Three basic types of Hilbert planes). According to the three cases occurring in the Uniformity Theorem— A semi-Euclidean plane is a Hilbert plane for which the angle sum of every triangle is two right angles. A semi-hyperbolic plane is a Hilbert plane for which the angle sum of every triangle is less than two right angles. A semi-elliptic plane is a Hilbert plane for which the angle sum of every triangle is larger than two right angles.
Lemma 12.1. Given is any quadrilateral ABCD with right angles at vertices A and B, vertices C and D lying on the same side of line AB. The angles γ at vertex C and δ at vertex D satisfy γ < δ if and only if BC > AD γ δ if and only if BC AD γ > δ if and only if BC < AD
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Figure 12.1. Across the longer side BC is the larger angle δ.
Proof. As shown in the figure, we assume BC > AD. We have to check whether γ < δ. By transferring segment AD, we produce the congruent segment BE AD such that ABED is a Saccheri quadrilateral. Let its top angles be congruent to ε. In the triangle 4DEC, the exterior angle theorem yields γ < ε. Since points A and C lie on different sides of line DE, angle comparison at vertex D implies ε < δ. Hence γ < ε < δ, and transitivity of angle comparison yields γ < δ, as to be shown. By a similar argument, we prove that BC < AD implies γ > δ. Finally BC AD implies γ δ, since the top angles of a Saccheri quadrilateral are congruent.
Figure 12.2. For an acute top angle, the perpendicular dropped from a point P inside the top segment CD is shorter than the opposite sides.
Lemma 12.2. Given is a Saccheri quadrilateral ABCD, with right angles at vertices A and B. From any point P inside the top segment CD, the perpendicular is dropped onto the segment AB, with foot point Q. The following three cases can occur: γ < R and PQ < AD γ = R and PQ AD γ > R and PQ > AD
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Proof. At vertex P there occur the supplementary angles α = ∠DPQ and β = ∠CPQ. We begin by assuming PQ < AD, and look for a result about the top angle γ. Using the previous Lemma 12.1 for the quadrilateral AQPD, we conclude δ < α. Using the Lemma 12.1 once more for the quadrilateral QBCP, we conclude γ < β. Hence angle addition yields 2γ = γ + δ < β + α = 2R and hence γ < R. By a similar argument, the assumption PQ > AD implies γ > R, and finally, indeed, the assumption PQ AD implies γ = R.
Figure 12.3. For an acute top angle γ, the perpendicular PQ dropped from a point outside the top segment is longer than the opposite sides BC.
Lemma 12.3. Given is a Saccheri quadrilateral ABCD, with right angles at vertices A and B. −−→ From any point P on the ray DC outside of the top segment CD, the perpendicular is dropped onto the segment AB, with foot point Q. The following three cases can occur for the top angle: γ < R and PQ > AD γ = R and PQ AD γ > R and PQ < AD Proof. By transferring segment AD, we produce the congruent segment QE AD such that AQED is a Saccheri quadrilateral. Let its top angles be congruent to ε. We get the third Saccheri quadrilaterals BQEC, denote its top angles by β. The line BC intersects segment DE (Why?). Let F be the intersection point. At first, we assume QP > AD, and check that γ < R. The three points A, B and Q on the base line lie on the same side of top line DE, since baseline and top line of the Saccheri quadrilateral AQED have the middle line as their common perpendicular, and hence are parallel. Since QP > AD QE, points Q and P lie on different sides of line DE. Hence points P and C lie on the (upper) side of DE, whereas A, B and Q lie on the lower side. The intersection point F of line DE with line BC lies between B and C. Angle addition at vertex C yields γ + β + χ = 2R. In the triangle 4DEC, the exterior angle χ > δ − ε = γ − ε. Finally, comparison of angles at vertex E shows that β > ε. Put together, we get 2R = γ + β + χ > γ + β + γ − ε > 2γ
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Figure 12.4. For an obtuse top angle γ, the perpendicular PQ dropped from a point outside the top segment is shorter than the opposite sides BC and AD.
and hence γ < R, as to be shown. Under the assumption PQ < AD, several modifications occur, as can be seen in the figure on page 305. Indeed, because of QP < AD QE, points Q and P lie on the same side of line CE. Hence all five points A, B, Q, E and D lie on the same (lower) side of top line DE. The intersection point F of line DE with line BC lies outside segment BC. Angle addition at vertex C yields γ + β − χ = 2R. In the triangle 4DEC, the exterior angle χ > −δ + ε = −γ + ε. Finally, comparison of angles at vertex E shows that β < ε. Put together, we get 2R = γ + β − χ < γ + β + γ − ε < 2γ and hence γ > R, as to be shown.
Figure 12.5. Two Saccheri quadrilaterals with a common middle line MN have either both acute, both obtuse, or both right top angles.
Lemma 12.4. Given are two Saccheri quadrilaterals ABCD and A0 B0C 0 D0 with a common middle segment MN. Then there top angles γ and γ0 are either both acute, both right, or both obtuse.
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Proof. Without loss of generality we may assume A ∗ A0 ∗ M ∗ B0 ∗ B as order of the vertices on the base line. Using the previous Lemma 12.2 for the Saccheri quadrilateral ABCD, we get the three equivalences γ < R if and only if A0 D0 < BC γ R if and only if A0 D0 BC γ > R if and only if A0 D0 > BC Using Lemma 12.3 for the Saccheri quadrilateral A0 B0C 0 D0 , we get the three equivalences γ0 < R if and only if A0 D0 < BC γ0 R if and only if A0 D0 BC γ0 > R if and only if A0 D0 > BC Put together, we see that angles γ and γ0 are either both acute, both right, or both obtuse, as to be shown.
Figure 12.6. Any two Saccheri quadrilaterals can be put into a position such that the middle line of the first one is the base line of the second one, and vice versa.
End of the proof of the Uniformity Theorem. Given a Saccheri quadrilateral ABCD with middle line MN, and any second Saccheri quadrilateral. We transfer the middle line of the second −−→ Saccheri quadrilateral onto the ray MB. By means of congruent triangles, it is straightforward to verify that we can produce a Saccheri quadrilateral QPRS which is congruent to the second given one, and has the bottom line S Q = MN and the middle line MK = AB. The two lines CD and PR do intersect (Why?). With the intersection point L, we get a Lambert quadrilateral MKLN. Let λ = ∠KLN be its top angle.
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By Lemma 12.4, we conclude that the top angles γ and λ of the quadrilaterals ABCD and MKLN are either both acute, right, or obtuse. Similarly, we conclude that the top angles ϕ and λ of the quadrilaterals QPRS and MKLN are either both acute, right, or obtuse. Hence the top angles γ and ϕ of the two given Saccheri quadrilaterals are either both acute, right, or obtuse.
Figure 12.7. Does the figure close to a rectangle?
Figure 12.8. Does the figure close to a rectangle?
12.2.
Some strange polygons
Problem 12.1. In the figure on page 307 is shown how to start a construction of a rectangle, in neutral geometry. In Euclidean geometry, the two rays r and s have a common segment, and the rectangle closes. But in neutral geometry, there are other possibilities. Indeed, it is both possible that these two rays intersect, or they lie on parallel lines. I claim: if the two rays r and s intersect but do not lie on the same line, their intersection point lies on the perpendicular bisector of segment AB. Give a convincing argument to confirm this statement. Answer. Assume that these two rays do not lie on the same line and intersect at a point P = r ∩ s. We reflect both rays across the perpendicular bisector p of segment AB. Because of the symmetry of the figure, one obtains the rays r0 = s and s0 = r,— the two rays have simply been exchanged by the reflection. The intersection of the reflected rays is P0 = r0 ∩ s0 . The points P and P0 are mirror images of each other. Since the two rays lie on different lines, by axiom (I.2), their intersection point is unique. Hence P = r ∩ s = r0 ∩ s0 = P0 and thus the two rays intersect on the perpendicular bisector.
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Problem 12.2. In the figure on page 307 is shown how to start a construction of a rectangle, in neutral geometry. Indicate in several figures all different possible cases. Answer. We draw the segment CD and obtain the Saccheri quadrilateral ABCD. By the Uniformity Theorem, this Saccheri quadrilateral has either right, acute, or obtuse top angles. In the case of a right top angle, one obtains a rectangle ABCD. In this case, both rays r and s lie on the line CD. In the case of acute top angles, the ray r lies outside the angle ∠BCD. Ray r may or may not intersect the perpendicular bisector. In the case of obtuse top angles, the ray r lies inside the angle ∠BCD, and ray s inside the angle ∠ADC. Together, one obtains either one of the cases indicated in the figure on page 308.
Figure 12.9. Still not always a rectangle.
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Figure 12.10. Following the ray r.
Problem 12.3. While starting the construction of a rectangle for the case where Saccheri quadrilaterals have obtuse top angles, the "accident" shown in the figure on page 309 is happening. Complete the paragraph below. Indicate in the figure the three points X, Y, Z mentioned in the paragraph below.
Figure 12.11. The points X, Y, Z on the ray r.
Answer. We draw the segment CD and obtain the Saccheri quadrilateral ABCD. In the case of obtuse top angles, the ray r lies inside the obtuse angle ∠BCD, and ray s inside the angle ∠ADC. Indeed the rays r and s intersect the perpendicular bisector p of segment AB, as one may confirm as follows. By the Crossbar theorem , the ray r intersects side BD of triangle 4BCD, say at point Y. Hence by Pasch’s axiom , ray r intersects two sides of triangle 4ABD. But ray r and segment AB cannot intersect, otherwise one would obtain a triangle with two right angles at vertices B and C, which is impossible. Hence ray r intersects segment AD, say at point Z. Since the points C and Z lie on different sides of the perpendicular bisector p, the ray r intersects the bisector p, say at point X. Since ray s is the mirror image of ray r, and the intersection point r ∩ s is unique, point X = p ∩ r = p ∩ s = r ∩ s is the intersection of the rays r and s, too. Problem 12.4. Suppose a quadrilateral ABCD has right angles at vertices A and B, and congruent sides AB CD. Moreover, we may assume d = AD ≤ BC = c. (i) We drop the perpendicular from point C onto side AD, and let F be the foot-point. Give a reason why either (a) F = D or (b) CF < AB. (ii) Use Lemma 12.1 to conclude that in case (a) the quadrilateral is a rectangle.
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Figure 12.12. What kind of quadrilateral does one get?
(iii) Use Lemma 12.1 to conclude that in case (b) the angle ∠FCB is obtuse. (iv) Use the Uniformity Theorem to conclude that in case (a) the Hilbert plane is semiEuclidean, and in case (b) the Hilbert plane is semielliptic. Exclude the semihyperbolic case. (v) We may assume d = AD ≤ BC = c. Use Lemma 12.1 to conclude that in case (b) A ∗ D ∗ F. (vi) Use Lemma 12.1 to conclude that in case (b) c = AD ≤ BC = d < AF. Answer. (i) If F = D we are ready, so we assume F , D. Hence the right triangle 4FCD exists. We know that the hypothenuse is the longest side. Hence CF < DC AB which is case (b). (ii) In case (a), we get δ = 90◦ and d = c. Hence Lemma 12.1 yields γ = δ, and the quadrilateral ABCD is a rectangle. (iii) In case (b), Lemma 12.1 yields ∠FCB > ∠CFA. Since ∠CFA = 90◦ , we conclude that ∠FCB is obtuse. (iv) The Uniformity Theorem yields that in case (a) the Hilbert plane is semi-Euclidean, and in case (b) the Hilbert plane is semi-elliptic. Since all possibilities have been exhausted, we have excluded the semi-hyperbolic case. (v) We may assume d = AD ≤ BC = c. The angle ∠CDF is an interior angle of a right triangle and hence acute. But the angle δ is obtuse in case (b). Hence the angles ∠CDF and δ are distinct. This excludes the order A ∗ F ∗ D. Since points F and D are on the same side of line AB, only the order A ∗ D ∗ F is possible. (vi) We know by item (iii) that ∠FCB is obtuse. Hence Lemma 12.1 implies BC < AF, and indeed c = AD ≤ BC = d < AF. 12.3.
Defect and AAA congruence
Proposition 12.1 (Semi-hyperbolic AAA-Congruence Theorem). In a semi-hyperbolic plane, two triangles having three pairs of congruent angles are congruent.
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Figure 12.13. About the semi-hyperbolic AAA congruence.
Proof. The Uniformity Theorem 16 implies for a semi-hyperbolic geometry, any triangle 4ABC has a positive defect δ(ABC) = 180◦ − α − β − γ Moreover we use Lemma 9.2 about the additivity of the defect for non-overlapping triangles. By transfer of one of the triangles, we reduce the theorem to the case where the two triangles have a common vertex and the adjacent sides lie on the same two rays. Hence we get two triangles 4ABC and 4AB0C 0 such that,—possibly after some renaming,—either one of the cases (0), (i), (ii) or (iii) occurs: (0) B = B0 and C = C 0 ; (i) A ∗ B ∗ B0 and A ∗ C ∗ C 0 ; (ii) A ∗ B ∗ B0 and C = C 0 ; (iii) A ∗ B ∗ B0 and A ∗ C 0 ∗ C. •
Case (i) is the most interesting one to be excluded. We use Lemma 9.2 about the additivity and positivity of the defect to get δ(AB0C 0 ) = δ(ABC) + δ(BCC 0 ) + δ(BB0C 0 ) > δ(ABC)
and hence β0 + γ0 > β + γ, contradicting the assumption that the two triangles are equiangular . • The case (ii) is straightforward to exclude and is left out. • In case (iii), as explained in Problem 5.2, the segments BC and B0C 0 intersect, say at point D. For the triangle 4BDB0 , the exterior angle theorem tells β > β0 . This contradicts the assumption β = β0 and excludes this case. Hence only the case (0) is possible, showing that the triangles 4ABC and 4AB0C 0 are indeed congruent . Remark. Alternatively, the cases (ii) and (iii) are excluded since the z-angle theorem implies that the sides BC k B0C 0 are parallel. As already seen by the addendum to the extended ASA-Congruence, Proposition 7.47 only cases (0) and (i) are possible. Remark. In any semi-elliptic plane, too, AAA-congruence holds. In this case, the Uniformity Theorem 16 implies that all triangles have a positive excess of the angle sum ε(ABC) := α + β + γ − 180◦ Again, the additivity lemma holds. Since the excess is strictly positive and additive, the AAAcongruence holds in the semi-elliptic geometry, too. We see that AAA-congruence holds in the semi-elliptic as well as the semi-hyperbolic geometry, but not in the semi-Euclidean geometry.
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12.4.
A hierarchy of planes As we have learned in definition 2.1:
Definition (2.1). A Hilbert plane is any model for two-dimensional geometry where Hilbert’s axioms of incidence (I.1)(I.2)(I.3a)(I.3b), order (II.1) through (II.4), and congruence (III.1) through (III.5) hold. Neither the axioms of continuity—Archimedean axiom and the axiom of completeness—nor the parallel axiom need to hold for an arbitrary Hilbert plane. The Uniformity Theorem brings confidence that these few axioms are a correct start towards a natural axiomatization of full-blown Euclidean geometry. Which postulates do we want to be added to the Hilbert plane axioms? As the next step towards a natural axiomatization of Euclidean geometry, we investigate the Euclidean parallel property. This section gives an account of some recent achievements, mainly of M.J. Greenberg in the axiomatization of geometry. We begin with the axioms of a Hilbert plane, and successively postulate few additional axioms with the effort to arrive at a system as close as possible to Euclid’s system. By further axioms—mainly of continuity—the system is narrowed to a system that has the real Cartesian plane as its unique model. 12.5. Wallis’ axiom Already in the section on Legendre’s theorem, we have addressed the question to find a clear cut, suggestive or self-evident postulate that would better replace the Euclidean postulate. John Wallis (1616-1703), in a treatise on Euclid published in 1693, was astute enough to propose a new postulate that he believed to be more plausible than Euclid’s parallel postulate. He phrased it as follows: Finally (supposing the nature of ratio and of the science of similar figures already known), I take the following as a common notion: to every figure there exists a similar figure of arbitrary magnitude. We want to make Wallis’ idea precise in the context of axiomatic geometry, beginning with Hilbert’s axiom. Euclid’s theorems about similar triangles depends on the Archimedean axiom— via his definition of equality of ratios. Hence it is better to restrict attention to equiangular triangles instead of similar figures. Recall that equiangular triangles have three pairs of congruent angles. Both Greenberg and Hartshorne suggest a modification of Wallis’ axiom along these lines. In a more common language, the first postulate says that equiangular triangles exist in any size, the second one that equiangular triangles exist in any arbitrarily large size. (Wallis-Greenberg Postulate). Given any triangle 4ABC and segment DE there exists a triangle with DE as one side such that the triangles 4ABC and 4DEF are equiangular. (Wallis-Hartshorne Postulate). Given any triangle 4ABC and segment DE there exists a triangle 4A0 B0C 0 have side A0 B0 ≥ DE such that the triangles 4ABC and 4A0 B0C 0 are equiangular. In this paragraph, we show the following theorem: Theorem 12.1. The following three postulates are equivalent in any Hilbert plane: (a) the Euclidean parallel postulate (b) the Wallis-Greenberg postulate (c) the Wallis-Hartshorne postulate
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Proposition 12.2 (Equiangular triangles exist in all given sizes). The Euclidean parallel postulate implies that equiangular triangles exist in any given size, and hence the Wallis-Greenberg postulate. In particular, for any triangle 4ABC and segment A0 B0 there exists a triangle with A0 B0 as one side, and lying in a given half plane of this side, such that the triangles 4ABC and 4A0 B0C 0 are equiangular.
Figure 12.14. Uniqueness of parallels implies Wallis’ postulate.
Proof. Given is triangle 4ABC and segment A0 B0 , as shown in the figure on page 313. We transfer −−−→ segment AB onto the ray A0 B0 and get the congruent segments AB A0 E 0 . As in the extended ASA-theorem, we construct the congruent triangles 4ABC 4A0 E 0 F 0 . Indeed, as already explained in the proof of the extended ASA-theorem, the ray rA0 forming the −−−→ −−−→ angle ∠BAC with ray A0 B0 and the ray rE0 forming the angle ∠ABC with ray E 0 A0 , both constructed in the same half-plane of line A0 B0 , intersect at point F 0 . −−−→ Additionally, we construct the ray r0B forming the same angle ∠ABC with ray B0 A0 , again in the same half-plane of line B0 A0 . The rays r0B and rE0 are parallel by [Euclid I.27], see 7.42. The ray rA0 intersects one of these parallel rays. Hence by Proclus’ Lemma 3.3, it intersects the line of the second ray r0B , too. Indeed, otherwise, the line of r0B would have the two different parallels rA0 and rE0 both through point F 0 , which is impossible. Let now C 0 be the intersection point of rays rA0 and r0B . The triangles 4ABC and 4A0 B0C 0 have two pairs of congruent angles at vertices A and A0 , as well as vertices B and B0 , by construction. We have seen in Proposition 9.7 above that uniqueness of parallels implies that every triangle has angle sum 180◦ —every Pythagorean plane is semi-Euclidean. Hence the triangles 4ABC and 4A0 B0C 0 have a third pair of congruent angles at vertices C and C 0 and are equiangular, as to be shown. Proposition 12.3. The Wallis-Hartshorne postulate implies the Euclidean parallel postulate.
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Figure 12.15. Wallis’ postulate implies uniqueness of parallels.
Proof. Given is line l and a point P not on l. One parallel can be obtained as the "double perpendicular" as already explained in Proposition 7.43. One drops the perpendicular p from point P onto line l and denotes the foot point by F. Next, one erects at point P the perpendicular to line PF. Thus, one gets a line m parallel to the given line l. We now suppose towards a contradiction that there exists a second line t through point P which → − is parallel to line l, too. Let t be the ray on line t with vertex P and lying between the two parallels l and m. We choose any point Q , P on this ray, and drop the perpendicular onto the line PF. The foot point R lies in the segment PF, since lines t and l do not intersect. We now apply the Wallis-Hartshorn axiom to the triangle 4PQR and the segment PF. Hence there exists an equiangular triangle 4P0 Q0 R0 with side P0 R0 ≥ PF. −−→ We transfer this segment and then the triangle onto the ray PF and get a triangle 4PQ00 R00 , which is congruent to 4PQ0 R0 and hence equiangular to triangle 4PQR. We can put points Q and Q00 on the same side of line PF. Because of congruence of the angles α = ∠R00 PQ00 ∠RPQ, the −−→ −−−→ uniqueness of angle transfer imply that these rays are equal: PQ = PQ00 . Hence point Q00 lies on the line t = PQ. Since PR00 ≥ PF, we conclude that either R00 = F or R00 and P lie on different sides of F. We consider the second case. The foot points R and R00 lies on different sides of F and hence the line l. The lines RQ, l and R00 Q00 are parallel, being all three perpendicular to PF. Hence points R and Q lie on one side of line l and points R00 and Q00 lie on the other side. Hence the segment QQ00 intersects the line l, say in point T . In the first case, we conclude that point Q00 =: T lies on line l. Hence, in both cases, the lines l and t do intersect. We have confirmed no second parallel t to line l through point P can exist. 12.6.
Proclus’ Theorem We shall repeatedly use
Lemma 12.5 (Proclus lemma). If one of two parallel lines is intersected by a third line, the other one is intersected, too. What is the missing link leading from the semi-Euclidean plane to a plane with uniqueness of parallels? Based on Proclus commentaries to Euclid, Greenberg has suggested the following angle unboundedness axiom. This axiom goes back to Aristole’s book I of the treatise De Caelo ("On the heavens").
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Definition 12.2 (Aristole’s Angle Unboundedness Axiom). For any acute angle θ and any segment PQ, there exists a point X on one side of the angle such that the perpendicular XY dropped onto the other side of the angle is longer than the given segment: XY > PQ. Main Theorem 17 (Proclus’ Theorem). In any Hilbert plane, the Euclidean parallel postulate is equivalent to the following two requirements: • Every triangle has angle sum two right angles. • For every acute angle and every given segment, there exists a right triangle with the given angle and the leg across from it longer than the given segment. Corollary 33. A Hilbert plane is Pythagorean if and only if it is semi-Euclidean and Aristole’s axiom holds. We have seen in Proposition 9.7 above that uniqueness of parallels implies that every triangle has angle sum 2R. Hence every Pythagorean plane is semi-Euclidean. Indeed, uniqueness of parallels implies Aristole’s axiom, too. Proposition 12.4. Hilbert’s parallel postulate implies Aristole’s axiom.
Figure 12.16. Uniqueness of parallels implies unbounded opening of an angle.
Proof. Let the acute angle θ = ∠(m, n) and a segment be given. We transfer the segment onto the perpendicular erected at the vertex of the angle onto the side m and get the segment PQ. Next we erect the perpendicular l onto line PQ at Q. The lines l and m are parallel since they are both perpendicular to line PQ. Because uniqueness of parallels has been assumed, the lines n and l are not parallel. Let S be their intersection point. Take any point X on the line n such that S is between P and X. We drop the perpendicular from X onto line m and let Y be the foot point. Since X and Y are on different sides of line l, the segment XY intersects the line l. Call T the intersection point. The Lambert quadrilateral QT Y P is indeed a rectangle, as shown by the Uniformity Theorem. Its opposite sides are congruent by Lemma 12.1. Hence we get PQ YT < Y X as claimed by Aristole’s angle unboundedness axiom.
By Proposition 12.4 and Proposition 9.7, we know that Aristole’s Axiom and the angle sum 2R for triangles are both necessary for uniqueness of parallels to occur. To complete the proof of Proclus’ Theorem, we now show that these two assumptions are sufficient.
316
Figure 12.17. The second line n , m, different to the double perpendicular m, drawn through point P is along the side of a triangle 4PXR, and hence by Pasch’s axiom intersects line l.
Proposition 12.5. A semi-Euclidean Hilbert plane, for which the Aristole’s Axiom holds is Pythagorean.
Proof. Given is a line l and a point P not on this line. As explained in Proposition 7.43, we use the standard "double perpendicular" m to get a parallel to line l through point P. Let Q be the foot point of the perpendicular dropped from point P onto line l, and let m be the perpendicular erected at point P onto the first perpendicular PQ. We need to show that m is the unique parallel through point P to line l. We take any other distinct line n , m through point P and check whether it is parallel to line l. For the case that lines n and m are perpendicular, we get n = PQ, and hence Q = n ∩ l. So lines n and l are not parallel. We exclude this case and assume that lines n and m are not perpendicular. There exists rays nm and rn on lines m and n such that the angle θ = ∠(rm , rn ) is acute, has vertex P, and secondly, the interior of the angle θ lies in the same half plane of line m as line l. We now explicitly prove existence of an intersection point of ray rn and line l. Let XY > PQ be the segment between the sides of angle θ = ∠XAY, as postulated by Aristole’s Axiom. We can choose point X to lie on the ray rn , and drop the perpendicular from point X onto the line PQ, obtaining the foot point R. We have obtained a Lambert quadrilateral RXY P. In a semi-Euclidean plane, this is a rectangle. Its opposite sides are congruent by Lemma 12.1. Hence we conclude PR XY > PQ and points P and R lie on different sides of point Q. We now use Pasch’s axiom for line l and triangle 4PXR. Line l intersects the side PR in point Q and is parallel to side RX, since the lines l and RX are both perpendicular to PQ. Hence by Pasch’s axiom, the line l intersects the third side PX of the triangle, say in point S . Thus we have explicitly checked existence of an intersection point lines n = PX with line l, as to be shown. 12.7.
More about Aristole’s axiom
Proposition 12.6. In a semi-elliptic Hilbert plane, Aristole’s Axiom does not hold. Especially, it does not hold for the angle between any two parallels to the same line, nor for any angle given by the excess of the angle sum of a triangle over two right angles.
317
Figure 12.18. For the angle between two parallels m and n to line l, intersecting at P, the distance from a point on one side to the other side of angle ∠(m, n) is always smaller than the perpendicular PQ.
Proof. Given is a line l and a point P not on this line. Again, let Q be the foot point of the perpendicular dropped from point P onto line l, and let m be the perpendicular erected at point P onto the first perpendicular PQ. As explained in Proposition 7.43, "double perpendicular" m is a parallel to line l through point P. Let n be a second parallel to line l through point P. Take any point X on the line n such that X and Q lie on the same side of line m. We drop the perpendicular from point X onto the line PQ and let Z be its foot point. All three line m, ZX and l are perpendicular to PQ, and hence parallel. Hence points X and Z both lie between the parallels l and m. Hence point Z lies between P and Q, and hence. PQ > PZ The perpendicular from point X onto the line m has foot point Y. Thus we have obtained the Lambert quadrilateral ZXY P. In a semi-elliptic plane, its fourth angle χ = ∠ZXY is obtuse. From Lemma 12.1 we conclude PZ > XY Thus we have obtained that PQ > PZ > XY holds for any point X on one of the rays from P on line n. Hence the distance from any point of line n to line m is strictly smaller than PQ, and Aristole’s angle unboundedness axiom does not hold. Proposition 12.7. In a semi-Euclidean as well as in a semi-hyperbolic Hilbert plane, Archimedes’ Axiom implies Aristole’s Axiom. −−→ Proof. Let the acute angle θ = ∠(m, n) be given. We choose any point C on side m = AC, and drop the perpendicular onto n to get the foot point D. In order to double the segment CD, we construct point E such that C is the midpoint of segment DE and point F on ray m such that C is the midpoint of segment AF. The triangles 4ACD 4FCE are congruent by SAS-congruence, because of the vertical angles at point C and two pairs of congruent sides AC CF and DC CE. Because of the triangle congruence, ∠CDA ∠CEF = ∠DEF are both right angles. We drop the perpendicular from point F onto the other side n and get foot point G.
318
Figure 12.19. Doubling the distance from a point on one side to the other side of angle θ.
Now DEFG is a Lambert quadrilateral, with an acute or right angle ∠EFG. By Lemma 12.1 either side adjacent to the acute angle is longer than the respectively opposite side. Hence we conclude FG ≥ ED = 2 · CD We need to assure that doubling the original segment CD is sufficient to get another segment XY longer than any arbitrarily segment PQ. To this end, we use the Archimedean Axiom. Let FG := C1 D1 . By induction, we can construct segments Cn Dn ≥ 2n · CD > n · CD for all natural numbers n. Given any segment PQ, the Archimedean axiom tells there exists a natural number n such that n · CD > PQ Define XY := Cn Dn for such a number n. Both inequalities together imply XY > n · CD > PQ as required for Aristole’s angle unboundedness axiom to hold.
From Proposition 12.6, the first Legendre Theorem and the Uniformity Theorem we can recapitulate: Corollary 34. In a semi-elliptic Hilbert plane, neither Aristole’s Axiom nor Archimedes Axiom does hold. Together with Proposition 12.7, we get the remarkable result—for which it would be nice to have a direct proof: Corollary 35. In every Hilbert plane, the Archimedean Axiom implies Aristole’s Axiom. Too, from Proposition 12.7 and Proposition 12.5, we get once more the third Legendre Theorem—the direct proof of which we have given earlier. Corollary 36. A semi-Euclidean plane for which the Archimedean axiom holds is Pythagorean.
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13. Area in neutral geometry 13.1.
Equidecomposable and equicomplementable figures
Definition 13.1. A rectilinear figure, or simply figure, is a finite union of non-overlapping triangles. To express that the figure P is the union of non-overlapping triangles T i for i = 1 . . . n, we write P=
n ]
Ti
i=1
U We call the union T i a dissection of the figure P. Two triangles (or figures) are non-overlapping if their intersection is at most a finite union of points and segments. Definition 13.2. Two rectilinear figures P and Q are equidecomposable means they are finite unions of non-overlapping, pairwise congruent triangles. Too, we say that the figures P and Q can be dissected into congruent triangles, or simply, have congruent dissections. In this case, we write P=
n ]
Ti
and
i=1
Q=
n ]
Si
with T i S i
for i = 1 . . . n
i=1
Definition 13.3. Two rectilinear figures P and Q are equicomplementable means there exist figures R and S such that (1) R and S are equidecomposable (2) P and R are non-overlapping (3) Q and S are non-overlapping U U (4) P R and Q S are equidecomposable Remark. As a short hand, I write
P ∼ Q for figures P and Q being equidecomposable . P , Q to indicate that the figures P and Q are equicomplementable . Remark. More in everyday language, one may explain as follows: Two figures are equicomplementable means that one can obtain equidecomposable figures from them, by adding to each one an appropriately chosen extra figure. These two extra figures need to be equidecomposable and may not overlap with the two originally given figures. Remark. In German, Hilbert uses the words "zerlegungsgleich" for equidecomposable , and "ergänzungsgleich" for equicomplementable . Lemma 13.1. Any two dissections of the same figure have a common refinement. Proof. Question. What kinds of polygons occur by intersecting two triangles. How can they be dissected into triangles.
320
Answer. The intersection of two overlapping triangles is a convex polygon with at most six vertices. In the cases of four, five and six vertices, it can be dissected into triangles by one, two, or three diagonals, respectively. Question. Explain how to construct a common refinement of two arbitrary dissections of a figure P. U U Answer. Given are the two dissections P = ni=1 T i and P = mj=1 S j of the same figure. A common refinement is easily obtained by dissecting all overlapping intersection S j ∩ T i into triangles, since these are convex polygons with at most six vertices. Theorem 13.1. Any two congruent figures are equidecomposable . Any two equidecomposable figures are equicomplementable . All four notions—equality, congruence, being equidecomposable as well as being equicomplementable —are equivalence relations among figures. Indeed these are four different equivalence relations. Problem 13.1. Prove that two figures which are both equidecomposable to a third figure are equidecomposable . You need to use that any two dissections of the same figure have a common refinement, as shown in Lemma 13.1 above. Problem 13.2. Prove that two figures which are both equicomplementable to a third figure are equicomplementable . You can use that the relation of being "equidecomposable " is a equivalence relation. Remark. Euclid just uses in all four cases the same word "equal", and tells in his common notions that two things equal to a third are equal. That is a good postulate, once there is a clear cut notion of equality. But to prove the Theorem 13.1 above, we had to deal with a well-defined different equivalence relation, not with equality! Given is a triangle 4ABC, a side AB of which has been chosen as base. I call the line through the midpoints of the two remaining sides a midline. Proposition 13.1 (A neutral version of Euclid I.37). Two triangles with the same base and the same midline are equicomplementable . Two triangles with the same base, and the midpoints of the four remaining sides lying on one line, are equicomplementable . Remark. Astonishing enough, proposition 13.1 is even valid in neutral geometry. On the other hand, Euclid’s original proposition I.37 is not valid in hyperbolic geometry. Proof. The proof exploits the correspondence between a triangle and a Saccheri quadrilateral given in Hilbert’s Proposition 39. Complete details are given in Proposition 9.6 from the section about Legendre’s geometry. For a given triangle 4ABC and base AB, we obtain the corresponding Saccheri quadrilateral FGBA by the construction given in the figure on page 321. We have shown by SAA congruence that 4AFD 4CHD and 4BGE 4CHE Problem 13.3. Prove that the segment DE which the triangle cuts out of the middle line has length half the base FG of the corresponding Saccheri quadrilateral.
321
Figure 13.1. To every triangle corresponds a Saccheri quadrilateral.
Question. Take the case of an acute triangle and convince yourself that the triangle 4ABC and the Saccheri quadrilateral FGBA are equidecomposable . We can leave a short answer to this question to the reader. The Saccheri quadrilateral depends only on the base AB and the midline DE of the triangle. Hence we obtain the same Saccheri quadrilateral from another triangle with the same base and same midline. Using the positive answer to the question above, by transitivity (see Theorem 13.1) we conclude that two acute triangles with the same base and midline are equidecomposable . In the case of an obtuse triangle, it can happen that the triangle 4ABC and the Saccheri quadrilateral FGBA are only equicomplementable . The details are explained below. By transitivity, now used for the relation of equicomplementarity, we conclude that any two triangles with the same base and midline are equicomplementable .
Figure 13.2. Even an obtuse triangle is equicomplementable to its corresponding Saccheri quadrilateral.
Question. Take the case of an obtuse triangle, as given in the figure on page 321. Explain why the triangle 4ABC and the Saccheri quadrilateral FGBA are equicomplementable . Let the numbers denote non-overlapping regions.
322
Answer. Triangle congruences yield ] ] 1 54 6 The Saccheri quadrilateral and the triangle ] ABGF ∼ 1 2
and
and
3
]
56
4ABC ∼ 2
] ] 3 4
are complemented by the triangular region 5 as additional figure. We conclude ] ] ] ] ] ] ] ABGF 5∼ 1 2 5∼2 1 4 5 ∼2 6 ] ] ] ] ] ] ] ] 4 4 3 5 ∼ 2 4 3 ∼ 2 6∼ 2 5 ] ∼ 4ABC 5 Hence the Saccheri quadrilateral and the triangle are equicomplementable . 13.2. The winding number In every ordered incidence plane, it is possible to define an orientation. The orientation is fixed, as soon as one has agreed which are the left and right half plane for one fixed ray. We denote the congruence class of right angles by R. Definition 13.4 (Winding number function). Given a triangle 4ABC with positive (counterclockwise) orientation, we define the winding number function hABCi (X) around any point X to be 4R if X lies inside 4ABC. 2R if X , A, B, C lies on a side of 4ABC. α if X = A. hABCi (X) = β if X = B. γ if X = C. 0 if X lies outside the 4ABC. which is greater or equal zero. The winding number function for a triangle 4ABC with negative (clockwise) orientation is the negative of the above. If the three points A, B, C lie on a line, the winding number function is identical zero. Lemma 13.2. The winding number function hABCi is identically zero if and only if the three points A, B, C lie on a line. The winding number functions satisfy hOABi + hOBCi = hOACi if and only if point B lies between the points A and C. Proposition 13.2. For any point X , O, A, B, C the winding number functions satisfy the permutation property (13.1) hABCi = hBCAi = hCABi = − hBACi = − hACBi = − hCBAi and the cyclic additivity properties: hABCi = hOABi + hOBCi + hOCAi Problem 13.4. Check that hABCi = hPABi + hPBCi + hPCAi still holds for an arbitrary vertex P , O.
(13.2)
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13.3. Area of rectilinear figures As well known from everyday life, it is more practical to measure the area of figures, instead of only comparing different figures. As a first step in the discussion of area, we state the properties which need to hold for a useful notion of area. Secondly, we give a construction of the area function and thus prove its existence. Definition 13.5. The area or measure is a function assigning to any rectilinear figure P a number area(P) such that the following properties hold: (1) area(T ) > 0 for all triangles T (2) T S implies area(T ) = area(S ) for any two triangles U (3) area(P Q) = area(P) + area(Q) for any non-overlapping figures P and Q. More generally, the value area(P) can be an element of any ordered Abelian group. Step 1. As the first step, we define the area for any triangle 4ABC. The is done quite differently for Euclidean geometry on the one side and for semi-hyperbolic or semi-elliptic geometry on the other side. In any Pythagorean plane the area of a triangle is defined to be half of base times height. area(4ABC) =
1 1 1 aha = bhb = chc 2 2 2
In any semi-hyperbolic plane the area of a triangle is defined to be the deviation of its angle sum from two right angles δ(ABC) = 2R − α − β − γ Recall that δ(ABC) is the defect of the triangle, according to definition 9.4. In any semi-elliptic plane the area of a triangle is defined to be the excess of its angle sum over two right angles α + β + γ − 2R
Remark. We see from this definition that the values of area are in different Abelian groups depending on the geometry and are justified in different ways. • The former definition satisfies the positivity requirement (1) because, as guaranteed by Theorem 18.1, the segment lengths in any Pythagorean plane are an ordered field. Moreover we have to use the similarly of triangles with opposite orientation. • The latter two definitions satisfy the positivity requirement (1) because of the Uniformity Theorem 16. Step 2. We define the left- and right half-planes for any ray, and the clockwise or counterclockwise orientation for any triangle. Note that the orientation depends on the order of the vertices of the triangle. We define the signed area [ABC] of 4ABC as if 4ABC has counterclockwise orientation +area(4ABC) [ABC] = −area(4ABC) if 4ABC has clockwise orientation Lemma 13.3. For any base point O, the signed triangle areas satisfy [ABC] = [OAB] + [OBC] + [OCA]
(13.3)
324
Lemma 13.4. [PAB] + [PBC] + [PCA] = 0 if and only the three points A, B and C lie on a line. Lemma 13.5. If the points X and Y lie on the same side of line EF, the two signed areas [XEF] and [Y EF] have the same signs. If the points X and Y lie on opposite sides of line EF, the two signed areas [XEF] and [Y EF] have the opposite signs.
Step 3. We define the area of a figure as the sum of the areas of the triangles in any dissection. We show that such a definition cannot lead to a contradiction. Lemma 13.6 (Main Additivity Lemma). If any triangle is dissected into finitely many nonoverlapping triangles, its area is the sum of the triangles of the dissection. T=
n ]
Ti
implies area(T ) =
i=1
n X
area(T i )
i=1
Proof. We give the triangle T = 4ABC and all triangles of the dissection the positive orientation. P Any reference point O is chosen. We use the formula (13.3) for each term of the sum ni=1 area(T i ). There occur contributions [OEF] of two different types: • from a side of some triangle T i in the dissection, which does not entirely lie on one of the sides AB, BC or CA of the big triangle T • from a side of some triangle T i lying entirely on a side of the triangle T . Assume that for side EF from the dissection, the first case applies. There exist two triangles in the dissection having side EF, say T 1 = 4EFX and T 2 = 4FEY, lying in the two half-planes of EF. Hence they do not overlap. They have the same orientation, which we have assumed to be positive. The total area of the quadrilateral formed by the two triangles is area(EY FX) = area(4EFX) + area(4FEY) = [EFX] + [FEY] = [OEF] + [OFX] + [OXE] + [OFE] + [OEY] + [OY F] = [OFX] + [OXE] + [OEY] + [OY F] We see that the two contributions from the common side EF cancel. The same cancelation occurs for all interior sides of the dissection. The contributions from sides of some triangle T i lying entirely on one of the sides of the big triangle T add up to [OAB] + [OBC] + [OCA] = area(4ABC) Theorem 13.2. Assume that the area for triangles has the properties (1) area(T ) > 0 for all triangles T (2) T S implies area(T ) = area(S ) for any two triangles (3t) If two non-overlapping triangles P and Q are joined to a larger triangle, this has the U area(P Q) = area(P) + area(Q).
325
Then the area of any rectilinear figure is well defined. Indeed, one gets the same sum n X
area(T i ) =: area(P)
i=1
U independent of the dissection P = T i one has used. U Furthermore, area(P Q) = area(P) + area(Q) holds for any non-overlapping figures P and Q. U U Proof of Theorem 13.2. Given are the two dissections P = ni=1 T i and P = mj=1 S j of the same figure. Let Rk with k = 1 . . . K ≤ 3nm be a common refinement. By applying the additivity lemma to each T i , and each S j , we obtain n X
area(T i ) =
i=1
K X
area(Rk ) =
m X
area(S j )
j=1
k=1
Hence any two dissections yield the same value of sum of area for the given figure, which is thus well defined. The proof of the second item can be left to the reader. Corollary 37. The interior domain R of a simple closed polygon P1 , P2 , . . . Pn , Pn+1 = P1 has the area n X area(R) = [OPi Pi+1 ] i=1
We have put Pn+1 = P1 , and assumed positive orientation. We can dissect the interior region R into triangles, and let all triangles have positive orientation. Then the circumference curves of these triangles add up to the positively oriented polygon surrounding them all. Theorem 13.3. (a) Any two equidecomposable figures have the same area. (b) Any two equicomplementable figures have the same area. Proof. Let P and Q be two equidecomposable figures. By definition, they have dissections P=
n ]
Ti
and
Q=
i=1
n ]
Si
such that
i=1
Ti S i
for i = 1 . . . n
From the definition of area, we get area(P) =
n X
area(T i ) =
i=1
n X
area(S i ) = area(Q)
i=1
as to be shown. Now let P and Q be two equicomplementable figures. By definition, there exist figures R and S such that (1) R and S are equidecomposable (2) the non-overlapping unions P
U
R and Q
U
S are equidecomposable
326
Hence part (a) and the additivity from theorem 13.2 imply area(R) = area(S ) ] ] area(P R) = area(Q S) area(P) + area(R) = area(Q) + area(S ) Since substraction is defined in the Abelian group from which the value of area are taken, we get area(P) = area(Q), as to be shown. Proposition 13.3. If the figures Pi are equidecomposable to the figures Qi for i = 1 . . . n, and U neither the figures Pi nor the figures Qi overlap, then the union ni=1 Pi is equidecomposable to the Un union i=1 Qi . The corresponding statement holds in the case of equicomplementable figures.
Figure 13.3. deZolts postulate tells that P and Q are not equicomplementable .
Proposition 13.4 (The postulate of de Zolt and Stolz). Let Q ⊂ P be two figures such that there exists a triangle T inside P which does not overlap with the smaller figure Q. Then the figures P and Q are not equicomplementable . U Proof. From dissections of the figures P and Q and Q T , we construct a common refinement Rk such that K L M ] ] ] Q= Rk , T = Rk , P = Rk k=1
k=K+1
k=1
with K < L ≤ M. By the additivity of area (Theorem 13.2), we get area(Q) =
K X k=1
area(Rk ) <
M X
area(Rk ) = area(P)
k=1
Hence area(Q) , area(P). By part (b) of Theorem 13.3, the figures P and Q are not equicomplementable .
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13.4. A standard equicomplementable form for a figure There remains the question—are any two figures with the same area equicomplementable ? To prove this converse of Theorem 13.3, one needs a standard equicomplementable form for any given figure. In Euclidean geometry, a rectangle with one side unit length, the other side the given area, is a convenient standard form. In hyperbolic geometry, no rectangles exist. Moreover, the area of Saccheri quadrilaterals with a unit base turns out to be bounded above. A way out is to use for the purpose of comparison a finite set of non-overlapping congruent triangles. We present this approach in a way valid for any Hilbert plane for which the circle-line intersection property holds. Lemma 13.7. To any given triangle and a segment longer than its shortest side, there exists an equicomplementable triangle which has the given segment as one of its sides. Proof. Let the given triangle 4ABC have side AB < c0 where c0 is the given segment. We bisect sides AB and AC and draw the middle line DE connecting the midpoints D and E, respectively. Let D0 be an intersection point of the middle line with the circle around B of radius c0 /2. We extend −−−→ the ray BD0 to get the point A0 such that D0 is the midpoint of segment BA0 . The triangle 4A0 BC and the given triangle have the common base BC and the common middle line DE. The triangles 4ABC , 4A0 BC are equicomplementable by Proposition 13.1, and obviously the triangle 4A0 BC has the side A0 B of length c0 as required. Lemma 13.8. Any triangle is equicomplementable to an isosceles triangle with the same base and the same midline. Proof. For a given triangle 4ABC and side AB, we obtain the corresponding Saccheri quadrilateral FGBA by the construction given in the figure on page 321. Let DE be the middle line and p be the perpendicular bisector of side AB. The top of the isosceles triangle 4ABC 0 is the point on p which has the same distance from line DE as the points A and B, but lies on the other side. The two triangles 4ABC , 4ABC 0 have the common side AB and the same midline DE. By Proposition 13.1 they are equicomplementable . Lemma 13.9. For any two given triangles T 1 and T 2 there exists a pair of non-overlapping congruent triangles T 0 T 00 such that the unions ] ] T1 T2 , T 0 T 00 are equicomplementable . Proof. If T i are two congruent equilateral triangles, we are ready. Suppose this is not the case and that triangle T 2 has a side a0 longer than one side of T 1 . By Lemma 13.7 there exists a triangle T 10 equicomplementable to T 1 and having the side a0 . We have obtained two equicomplementable triangles T 10 = 4A0 BC and T 2 = 4A2 BC which we can put into the two opposite half planes of their common side a0 = BC. By Lemma 13.8 there exist isosceles triangles T ∗ = 4A∗ BC and T ∗∗ = 4A∗∗ BC on opposite sides of their common base BC such that T ∗ , T 10 and T ∗∗ , T 2 . The kite A∗ BCA∗∗ has be constructed as the non-overlapping union of T ∗ with T ∗∗ but clearly ] ] A∗ BCA∗∗ = T ∗ T ∗∗ = 4BA∗ A∗∗ 4CA∗ A∗∗ and the latter are two congruent triangles. We put T 0 := 4BA∗ A∗∗ and T 00 := 4CA∗ A∗∗ , which are congruent triangles and are ready.
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Lemma 13.10. Any given figure is equicomplementable to a disjoint union 2n T 0 of 2n congruent non-overlapping triangles T 0 . Proof. By further subdivisions, any figure P is equicomplementable to a union of 2n nonoverlapping triangles T 1 , T 2 , . . . T 2n . Applying the Lemma 2n−1 times we construct triangles T i0 T i00 for i = 1 . . . 2n−1 such that 2n ] i=1
2n−1 ] ] T i0 T i00 Ti , i=1
In an inductive process for s = 1 . . . n, we get a union of 2n non-overlapping triangles which • together are still equicomplementable to the originally given figure P; • consists of 2n−s non-overlapping subsets, • each of which is the non-overlapping union of 2 s congruent triangles. In the end, we see that the given figure P is equicomplementable to a disjoint union of 2n congruent non-overlapping triangles. Lemma 13.11. For any two given triangles1 S and T there exists isosceles triangles P and Q with a common base such that S , P and T , Q. Proof. If S and T are two congruent equilateral triangles, we are ready. Suppose this is not the case and that triangle T has a side a0 longer than one side of S . By Lemma 13.7 there exists an equicomplementable triangle S 0 , S having the side a0 . We have obtained two triangles S 0 and T with the common side a0 = BC. By Lemma 13.8 there exist isosceles triangles P , S 0 and Q , T , all on the common side a0 . Proposition 13.5 (Figures with same area are equicomplementable). Assume in the given Hilbert plane that • the circle-line intersection property holds; • an area has been defined in accordance to Definition 13.5. The area takes values in an ordered and hence free Abelean group such that (1) area(T ) > 0 for all triangles T (2) T S implies area(T ) = area(S ) for any two triangles U (3) area(P Q) = area(P) + area(Q) for any non-overlapping figures P and Q. Then any two figures P and Q for which area(P) = area(Q) are equicomplementable . Remark. An additive group is called free if ng = 0 implies g = 0 for any integer n , 0 and any group element g. An ordered Abelean group is always free. Proof. By Lemma 13.10, the given figure F is equicomplementable to a disjoint union 2n S 0 of 2n congruent non-overlapping triangles S 0 and the given figure G is equicomplementable to a disjoint union 2n T 0 of 2n congruent non-overlapping triangles T 0 . The integer n can be chosen to be equal for both figures, by means of further subdivisions. By assumption, both figures have the same area. Hence 2n area(T ) = area(F) = area(G) = 2n area(S ) Since the area takes values in an ordered and hence free Abelean group, we conclude that area(T ) = area(S ). 1
they need not have the same content
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By Lemma 13.11 there exists isosceles triangles P and Q with a common base such that S , P and T , Q. Hence area(P) = area(S ) = area(T ) = area(Q) We put the isosceles triangles P = 4XBC and Q = 4Y BC on the same side of their common base BC. Let M be the midpoint of BC. We may assume that the altitude of XM ≥ Y M. Because of the assumption that the area of any triangle is positive, it is impossible that X , Y. Indeed in that case ] ] 4XBC = 4Y BC 4XY B 4XYC would imply that area(P) > area(Q). Now since X = Y, we see that actually P = 4Pt BC = 4Qt BC = Q. Wee that the originally given figures are equicomplementable since S ,P=Q,T ] ] F, 2n S , 2n T , G 13.5.
The role of the Archimedean axiom
Problem 13.5. We assume that the Archimedean axiom holds. Explain why any triangle T even equidecomposable to a corresponding Saccheri quadrilateral. Proof of Theorem ??. The assertion holds for every acute triangle. For an obtuse triangle, we reverse the process used in the first Legendre Theorem. Thus we obtain a sequence of triangles by bisection of the longest side and having all the same base. They have a common middle line. According to Problem 13.3 and the figure on page 321 , the middle line cuts these triangles in segments DE which have the length half the base FG of the Saccheri quadrilateral. By the Archimedean axiom, it takes only a finite number of steps to obtain a triangle which cuts the midline in two point between or on the base of the Saccheri quadrilateral and hence is equidecomposable to the Saccheri quadrilateral. Lemma 13.12. We assume that circle-line intersection property and the Archimedean axiom hold. To any given triangle and a segment longer than its shortest side, there exists an equidecomposable triangle which has the given segment as one of its sides. Lemma 13.13. We assume that circle-line intersection property and the Archimedean axiom hold. Then any two equicomplementable triangles are even equidecomposable . Theorem 13.4. Assume that the Archimedean axiom (V.1) holds, any two equicomplementable figures are even equidecomposable . We now turn the case that the Archimedean axiom is not satisfied. We refer to the figure on page 321 with the equicomplementable Saccheri quadrilateral and obtuse triangle AFBG , 4ABC Note that AG is the diameter of the Saccheri quadrilateral. Let M be the midpoint of its base FG. Lemma 13.14. Suppose these two figures have congruent dissections into t triangles. Then |HM| ≤
3t + 4 |AG| 2
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Proof. Let C be the sum of the circumferences of all t triangles of the dissection. |AB| + |BC| + |AB| ≤ C ≤ 3t |AG| For the circumference of triangle 4ABC, we need to get a the lower bound from the triangle inequality. 2|HM| = |HF| + |HG| ≤ 4|FA| + |AB| + |BC| ≤ 4|AG| + 3t |AG| Lemma 13.15. Suppose that the Saccheri quadrilateral and obtuse triangle from the figure on page 321 are even equidecomposable
AFBG ∼ 4ABC and that |HM| ≥ N · |AG| for some large integer N. Then the congruent dissections contain at least t≥
2N − 4 3
triangles. Proof. The assumption and Lemma 13.14 imply N · |AG| ≤ |HM| ≤
3t + 4 · |AG| 2
and solving for t yields the estimate to be shown.
Proposition 13.6. Assume the Archimedean axiom (V.1) does not hold. Then there exist a triangle and a Saccheri quadrilateral that are equicomplementable but not equidecomposable .
Proof.
The logical relations between the different concepts of content gained in this section are summed up in the following diagram: clear
−−−−−→
P and Q are equidecomposable yTh. 13.3 area(P) yArchimedes and Th. 23.4
P and Q are equicomplementable Th. 13.3 y area(Q)
Th. 23.2 y
P and Q are equidecomposable ←−−−−−−−−−−−−−− P and Q are equicomplementable Archimedean axiom
Corollary 38. Assuming the Archimedean axiom (V.1) holds, any two equicomplementable figures are even equidecomposable . Assuming the Archimedean axiom (V.1) does not hold, there exist two equicomplementable figures that are not equidecomposable .
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14. A Simplified Axiomatic system of Geometry—my own Suggestion 14.1.
A simplified axiomatization of geometry
Introduction The axiomatic system of Hilbert shows exceptional insight. On the other hand, it is unnatural in many respects. My simplification restricts attention to two dimensions, introduces circles in the natural classical way, and continuity in the way directly related to proofs. My goal is to give a logical analysis of the natural way to do geometry, and still keep up the level of rigor introduced by Hilbert. But I do not try to minimize the axioms as strictly as Hilbert. Especially the axioms dealing with the order of three points on a line, the SAS congruence, and the angle sum can be stated equivalently, but making less assumptions. 0. Undefined elements and relations Elements: • A class of undefined objects called points, denoted by A, B, C, . . . . • A class of undefined objects called lines, denoted by a, b, c, . . . . • A class of undefined objects called circles, denoted by C, D, . . . . Relations: • Incidence (being incident, lying on, containing) • Order (lying between) (for points on a line) • Congruence (for segments) • Center (for circle) I. Axioms of incidence I.1 For two points A and B there exists a line that contains both points. I.2 For two different points A and B there exists no more than one line that contains both points. I.3a Any line contains at least two points. I.3b There exist at least three points that do not lie on a line. Remark. I did separate axiom I.3 into I.3a and I.3b, in order to stress there is no direct logical connection between the two sentences intended. II. Axioms of order II.1 If a point B lies between a point A and a point C, then the points A, B, C are three distinct points of a line, and B lies between C and A. Definition 14.1 (Segment). Let A and B be two distinct points. The segment AB is the set consisting of the points A and B and all points lying between A and B. The points A and B are called the endpoints of the segment, they are assumed to be different. The points between the endpoints are called the interior points of the segment, and the remaining points on the line AB are called the exterior points of the segment. −−→ Definition 14.2 (Ray). Given two distinct points A and B, the ray AB is the set consisting of the points A and B, the points inside the segment AB, and all points P on the line AB such that point B lies between vertex A and P. The point A is called the vertex of the ray. II.2∗ Of any three different points on a line exactly one lies between the two others. −−→ II.3∗ For two points A and B, the ray AB contains a point C not lying inside the segment AB.
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Definition 14.3 (Triangle). We define a triangle to be union of the three segments AB, BC and CA. The three points A, B and C are assumed not to lie on a line. These three points are the vertices, and the segments BC, AC, and AB are the sides of the triangle. II.4* (strong Pasch’ Axiom) Given is a triangle 4ABC and a line a which does not meet any of its vertices A, B, C. If the line a passes through a point of the side AB, it also passes through a point of either one of the two other sides AC or BC, and intersects exactly two sides of the triangle. II.5 (Convention for orientation) There exist three points A, B and C such that point C lies in −−→ the left half-plane of ray AB, and the triangle 4ABC is oriented counterclockwise. III. Axioms for circles III.1 For any center O and point A , O, there exists a unique circle with center O through the point A. III.2 A circle C and a ray emanating from its center O intersect at exactly one point. Definition 14.4 (Interior and exterior of a circle). A point A lies interior of a circle if it lies −−→ between the center O and the point P where the ray OA cuts the circle. A point B lies exterior of a circle if it lies outside the segment OQ between the center O and −−→ the point Q where the ray OA cuts the circle. III.3 (Circle-circle intersection property) Let C and D be two circles. Let P, Q be two points on circle D. Assume that point P lies inside circle C and Q lies outside circle C. Then the two circles C and D intersect each other. III.4 Under the assumptions of (III.3), the two circles intersect in exactly two points, which lie on different sides of the line connecting their centers. IVa. Axioms of segment congruence IV.1 If a segment A0 B0 and a segment A00 B00 are congruent to the same segment AB, then segment A0 B0 is also congruent to segment A00 B00 . IV.1b For any two different points A and B, the segments AB and BA are congruent. IV.2 (The sums of congruent segments are congruent.) On a line, let AB and BC be two segments which except for B have no point in common. Furthermore, on the same or another line a0 , let A0 B0 and B0C 0 be two segments which except for B0 also have no point in common. In that case, if AB A0 B0 and BC B0C 0 , then AC A0C 0 IV.3a For a circle C with center O and two rays with vertex O which cut the circle at points P and Q, the segments OP and OQ are congruent.
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IVb. SAS-axiom for triangle congruence Definition 14.5 (Angle). An angle is the union of two rays with common vertex not lying on one line. We introduce a segment of unit length. Definition 14.6 (Congruences of angles). The two angles ∠POQ and ∠P0 O0 Q0 are called congruent if and only if there are four units segments OP, OQ, O0 P0 , O0 Q0 from the vertices and two congruent segments PQ and P0 Q0 between the sides of the angles. IV.4∗ Two triangles ABC and A0 B0C 0 are congruent if they have a pair of congruent angles and the two pairs of adjacent sides are pairwise congruent. In detail: The congruences AB A0 B0 , AC A0C 0 , ∠BAC ∠B0 A0C 0 imply the congruences BC B0C 0 , ∠ABC ∠A0 B0C 0 , ∠BCA ∠B0 A0C 0 V. Axioms of continuity V.1 (Axiom of Archimedes) If AB and CD are any segments, then there exists a natural number n such that n segments congruent to CD constructed contiguously from A, along a ray from A through B, will pass beyond B. V.2 (Cantor’s principle of boxed intervals) Every sequence of boxed intervals contains a common point. In detail: For a sequence of segments Ai Bi such that Ai ∗ Ai+1 ∗ Bi+1 ∗ Bi for all i = 1, 2, 3, . . . ∗
there exists a point X such that Ai ∗ X ∗ ∗ Bi for all i = 1, 2, 3, . . . VI. Axioms of parallelism VI.1 (The angle sum of a triangle is two right angles) The sum of the three interior angles of a triangle is two right angles. VI.2 (Aristotle’s Angle Unboundedness Axiom) For any acute angle θ and any segment PQ, there exists a point X on one side of the angle such that the perpendicular XY dropped onto the other side of the angle is longer than the given segment: XY > PQ. Remark. Based on Proclus commentaries to Euclid, Greenberg has suggested the angle unboundedness axiom. This axiom goes back to Aristotle’s book I of the treatise De Caelo ("On the heavens"). Remark. As shown in Corollary 35, in every Hilbert plane, the Archimedean Axiom implies Aristotle’s Axiom. Problem 14.1. Explain, how these axioms imply that each segment is congruent to itself. Problem 14.2. Explain why congruence of segments is an equivalence relation. Problem 14.3. Prove that the center of a circle is unique. Problem 14.4. Prove that every angle is congruent to itself, and to the angle with its sides switched. Explain why congruence of angles is an equivalence relation.
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14.2.
Fundamental constructions with Euclidean tools
Definition 14.7 (Traditional Euclidean tools). Constructions with traditional Euclidean tools are constructions using only straightedge and compass, done in a finite number of steps of the following types: 1. drawing a line through two different points 2. drawing a circle with a given center through a given point 3. intersecting two lines 4. intersecting a circle and a line 5. intersecting two circles 6. choosing an arbitrary point 7. choosing a point on a given line 8. choosing a point on a given circle 9. deciding whether two points are equal or different Remark. For the enumeration and count of a construction, we do not count producing the given pieces, neither pieces that were only drawn to facilitate the understanding. But producing the resulting pieces is counted. Problem 14.5. Explain how an angle is bisected with traditional Euclidean tools. Enumerate and count the steps needed, according to definition 14.7, and provide a drawing. Answer. Let ∠(h, k) with vertex A be given. 1. We choose a point B , A on the ray h. 2. Draw circle(A, B). 3. Find the intersection point 2 ∩ k. 4. Draw circle(3, A). 5. Draw circle(B, A). 6. Find the intersection point 4 ∩ 5 in the interior of the given angle ∠(h, k). −−→ 7. Draw the ray A, 6. This is the angle bisector to be constructed. We count that seven steps are needed for the construction. Problem 14.6. Given is a line l and a point lying on the line. Explain how the perpendicular is erected with traditional Euclidean tools. Answer. We draw any circle around the given point P and let A and B be the intersection points of line l with this circle. Next we draw two circles, with center A through point B, and with center B through point A. They intersect in two points, let C be one of them. The line PC is the perpendicular to line l at point P to be constructed. We get an extra check of accuracy, since this line goes through the other intersection point of the two circles, too. Problem 14.7. Given is a line l and a point P not lying on the line. Explain how to construct the reflection point P0 across the line with traditional Euclidean tools, without using line-circle intersection as a means of construction. Answer. We choose any two different points A , B on the given line l and draw the circles with center A through point B, and with center B through point A. They intersect in two points: the given point P and a point P0 , P, which is the reflection of point P across the line l to be constructed. Problem 14.8. Given is a line l and a point not lying on the line. Explain how the perpendicular is dropped with traditional Euclidean tools. Avoid line-circle intersection as a means of construction.
335
Answer. We choose any two different points A , B on the given line l and draw the circles with center A through point B, and with center B through point A. They intersect in two points: the given point P and the reflection point Q. The line PQ is the perpendicular to the line l through point P to be constructed.
Figure 14.1. Construction of the perpendicular with just two circles and their intersection.
Remark. The two points may lie on the same or different sides of the perpendicular, the construction works for any choice of the points A , B. The two circles may have different radius. In any case, they indeed intersect at two points. Problem 14.9. Explain that two equilateral triangles the common side of which is any given segment, can be constructed with traditional Euclidean tools. How can the two triangles be distinguished. Answer. Let the segment AB be given. We draw the circles with center A through point B, and with center B through point A. They intersect in two points C and D. The triangles 4ABC and 4ABD are the two equilateral triangles with sides AB to be constructed. They lie on different sides of line AB. Problem 14.10. Explain how the perpendicular bisector of a segment is constructed with traditional Euclidean tools. Answer. Let the segment AB be given. We draw the circles with center A through point B, and with center B through point A. They intersect in two points C and D. The line CD is the perpendicular bisector segment AB to be constructed. Problem 14.11. Given is a line l, a point A not lying on the line, and a point B lying on the line. Construct a circle through points A and B that is tangent to the line. Provide a drawing and explain the major steps. Answer. We erect the perpendicular at point B onto the given line l. Next we construct the perpendicular bisector of segment AB. The intersection of these two lines is the center O of the circle through A and B, tangent to line l, to be constructed.
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Figure 14.2. Construction of the circle through A and B tangent to line l.
14.3.
Euclidean tools are at least as strong as Hilbert tools
Theorem 14.1 (The traditional Euclidean tools can do at least all constructions possible with Hilbert tools). We assume only the axioms of incidence, order, circles, and congruence as stated above. Under these assumptions, it is also possible to transfer any segment as well as any angle, uniquely in the way postulated by Hilbert’s axioms (III.1) and (III.4). In other words, traditional straightedge and compass are stronger tools than Hilbert tools. Remark. The axioms of continuity and parallelism are not needed in theorem 14.1. Too, we have shown that Euclid’s collapsible compass can emulate a non-collapsible compass. Main Theorem 18. The axiom system given above implies Hilbert’s axioms for two-dimensional geometry. On the other hand, the converse is not true. Indeed, it is a more restrictive system. We now elaborate the steps needed to check Theorem 14.1. Proposition 14.1 (Transfer of a segment). Given is a segment AB and a ray a0 with vertex A0 . It is always possible to find, with traditional Euclidean tools, a point B0 on this ray such that the segment AB is congruent to the segment A0 B0 . Construction 14.1 (Transfer a segment). Construct an equilateral triangle 4AA0C, getting point C as an intersection of two circles with center A through point A0 , and center A0 through point A. −−→ A segment AB1 AB on the ray CA can be obtained from the intersection point B1 of this ray with the circle around A through point B. A second circle around C through point B1 enable one to −−→ get segment CB2 CB1 on the ray CA0 , from the intersection point B2 of ray and circle. Finally, the segment A0 B0 A0 B2 on the given ray a0 is obtained from the intersection point B0 of ray a0 with the circle around A0 through point B2 . Reason for validity of the construction. Using segment addition or subtraction, it is easy to check that AB1 A0 B2 , since by construction both CB1 CB2 and CA CA0 . Hence transitivity of congruence implies AB AB1 AB2 AB0 .
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Figure 14.3. Transfer of a segment with Euclidean tools.
Figure 14.4. Alternative construction to transfer of a segment with Euclidean tools.
Problem 14.12. An alternative construction using more circles, but less lines is shown in the figure on page 337. Describe this construction and give the reason for its validity. Proposition 14.2 (Transfer of an angle). Given is an angle ∠(h, k), a ray h0 that emanates from any point A0 , and a half plane of that ray. Then there exists a unique ray k0 such that the angle ∠(h, k) is congruent to the angle ∠(h0 , k0 ) and at the same time all interior points of the angle ∠(h0 , k0 ) lie on the given side of h0 .
338
Figure 14.5. Transfer of an angle with Euclidean tools.
Construction 14.2 (Transferring an angle). We draw a circle around the vertex A of the given angle and get congruent segments AB AD on its sides h and k. We transfer segment AB onto the ray h0 by means of the construction 14.1 above and get the segment A0 B0 AB. Finally we transfer segment BD to point B0 , and get a circle around B0 of radius B0 D2 BD. This step uses construction 14.1 once more, and hence produces a second equilateral triangle 4BB0 E. We construct the intersection points D0 and D00 of the circles around A0 of radius A0 B0 AB, and around B0 of radius B0 D2 BD. One chooses among points D0 and D00 the one lying in the half plane of h0 as required and gets the angle to be constructed: either ∠B0 A0 D0 ∠BAD, or ∠B0 A0 D00 ∠BAD. The intersections of circles and segments need not be postulated, since we can recall: Proposition 14.3. A segment between a point inside a circle and a point outside of this circle intersects the circle in exactly one point. Proof. By Proposition 10.15, the circle-circle intersection property implies the line-circle intersection property. By Proposition 10.12, the line-circle intersection property implies the segment
339
AB between the point A inside and the point B outside of the circle intersects the circle in exactly one point. We recall some details, for the convenience of the reader. Assume that the circle-circle intersection property holds. Let l = AB be the line between the two given points A inside, and −−→ B outside the circle C. The ray OB intersects the circle in a point Q inside the segment OB, as postulate by axiom. (IV.2) and given by definition 14.4 of the exterior of a circle. A special situation occurs in case that the center O of the circle lies on the line l. In this case, −−→ the intersection point Q = X of the ray OB with the circle is already the required intersection of line and circle. We now exclude that special case and assume that the center O does not lie on line l. −−→ The ray OA from the center intersects the circle in a point P outside the segment OA (since O , A). Recall that reflection images of any point across a line can be constructed with the intersection of two circles. Let O0 and P0 , Q0 be the reflection images of points O and P, Q across the axis l = AB. We construct the circle C0 with center O0 through point P0 . Points P0 and Q0 both lie on the reflected circle C0 . Thus we obtain the figure already shown on page 274. Line l is the
Figure 14.6.
A segment from inside to outside a circle intersects the circle—proved from the circle-circle intersection.
perpendicular bisector of the segment OO0 . Points O0 and P lie on the same side of line l, since points O and O0 lie on different sides of line l but points O and P lie on the same side. Because of proposition 11.1 we know that any point X lies on the same side of line l as point O if and only if |OX| < |O0 X|. Because points P and O0 lie on the same side of line l, we get |OP0 | = |O0 P| < |OP|. Hence point P0 lies inside circle C. Because points O and O0 lie on different sides of line l and points O and Q lie on the same side, the points Q and O0 lie on different sides of line l and |OQ| < |O0 Q| = |OQ0 |. Hence point Q0 lies outside circle C. Because point P0 lies inside, but point Q0 lies outside the circle C, and both points lie on the reflected circle C0 , the circle-circle intersection property implies that the two circles intersect in
340
Figure 14.7. Transfer 1 of an angle.
some point X. Since it has the same distance from both centers O and O0 , by proposition 11.1 point X lies on the axis l. Thus we have obtained an intersection point of the given circle C and line AB. Remark. In this derivation of the line-circle intersection property, we had to distinguish the cases that the given line goes through the center or not O. This is an instance were the item "deciding whether two points are equal or different" is actually needed. In the special case of a line through the center, we need to invoke the circle axiom (IV.2) directly, since the reflection across the line does not produce a different second circle. In case of a line passing closely to the center, it is still very impractical to use circle-circle intersection because of the small angle at the intersection of the two circles, and the direct application of the circle-line intersection leads to a more accurate construction. Because of that, the line-circle intersection is a practical tool useful to improve the accuracy of constructions. Too, we have shown in Proposition 10.13: Proposition 14.4. If a circle has points on both sides of a line, then it intersects the line in two points. Problem 14.13. Given is a line l and a point not lying on the line. Explain how the perpendicular is dropped using a circle-line intersection. Problem 14.14. The figure on page 14.3 shows a variant construction for the transfer of an angle. Enumerate and count the number of elementary steps according to definition 14.7 needed to perform the construction. Which additional assumption is made in this construction. Solution. 1. Draw circle(A, B). 2. Draw circle(B, A).
341
Find an intersection points X of 1 ∩ 2. Find the second intersection point Y of 1 ∩ 2. Draw the line p = XY. This is the perpendicular bisector of segment AB. Find the intersection points C = h ∩ p. Draw circle(A, C). Find the intersection D = k ∩ 7. −−→ Draw ray CB. Draw circle(C, D). Find the intersection point F = 9 ∩ 10. Draw circle(B, F). Find the intersection point G = l ∩ 12. Draw circle(B, C). Find the intersection point E = l ∩ 14. Draw circle(E, G). Find an intersection point of 14 ∩ 16. If this point does not lie in the half plane as required, find the second intersection point K of 14 ∩ 16. −−→ 19. Draw the ray BK. −−→ The angle ∠EBK is the transfer of the given angle ∠(h, k) onto the given ray l = BE to be constructed. We see that 18 or 19 steps are needed—provided that intersection point C does exist.
3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.
Remark. We see that the construction made the additional assumption that one side of the given angle intersects the perpendicular bisector of the vertices of the given angle and the given ray. In Euclidean geometry this assumption is not too restrictive, because it always holds for one side of either the given angle or its vertical angle. In hyperbolic geometry, the restriction is more severe. Problem 14.15. Consider transfer 2 of an angle shown in the figure on page 14.3. Enumerate and count the number of elementary steps according to definition 14.7 needed to perform the construction. Solution. 1. Draw circle(A, B). 2. Draw circle(B, A). 3. Find an intersection points F of 1 ∩ 2. 4. Find the second intersection point G of 1 ∩ 2. 5. Draw the line FG. 6. Draw line AB. 7. Find the intersection point M = 5 ∩ 6. 8. Find the intersection point C = h ∩ 1. 9. Find the intersection point D = k ∩ 1. −−→ 10. Draw ray CM. 11. Find the intersection point C 0 = 2 ∩ 10. −−−→ 12. Draw ray DM. 13. Find the intersection point D0 = 2 ∩ 12. 14. Draw circle(D0 , B). 15. Find the intersection point E = l ∩ 2. 16. Draw circle(E, B). 17. Find the intersection point K = 14 ∩ 16.
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Figure 14.8. Transfer 2 of an angle.
18. 19. 20. 21. 22.
Draw circle(K, C 0 ). Find an intersection point of H = 2 ∩ 18. If this point H does not lie in the half plane as required, draw circle(E, H). Find the new intersection point H := 2 ∩ 20. −−→ Draw ray BH.
−−→ The angle ∠EBH is the transfer of the given angle ∠(h, k) onto the given ray l = BE to be constructed. We see that 20 or 22 steps are needed. This construction does work always. 14.4. Birkhoff’s set of postulates In 1932, G. D. Birkhoff created a set of four postulates of Euclidean geometry sometimes referred to as Birkhoff’s axioms. These postulates are all based on basic geometry that can be confirmed experimentally with a scale and protractor. Since the postulates build upon the real numbers, the approach is similar to a model-based introduction to Euclidean geometry. Birkhoff’s axiom system was utilized in the secondary-school text Basic Geometry. Birkhoff’s axioms were also modified by the School Mathematics Study Group to provide a new standard for teaching high school geometry, known as SMSG axioms.
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(I) Postulate of Line Measure. A set of points {A, B, . . . } on any line can be put into a one to one correspondence with the real numbers {a, b, . . . } such that |b − a| = d(A, B) for all points A and B. (II) Point-Line Postulate. There is one and only one line l that contains any two given distinct points P and Q. (III) Postulate of Angle Measure. A set of rays {l, m, n, . . . } through any point O can be put into one to one correspondence with the real numbers ( mod 2π) such that for any points A , O on line l and B , O on line m, the difference am − al ( mod 2π) of the numbers associated with the lines l and m is the angle ∠AOB. Furthermore, if the point B on m varies continuously in a line r not containing the vertex O, the number am varies continuously, too. (IV) Postulate of Similarity. Given are two triangles 4ABC and 4A0 B0C 0 for which we assume d(A0 , B0 ) = k · d(A, B) , d(A0 , C 0 ) = k · d(A, C) and ∠B0 A0C 0 = ±∠BAC with some constant k > 0. Then d(B0 , C 0 ) = k · d(B, C) , ∠C 0 B0 A0 = ±∠CBA and ∠A0C 0 B0 = ±∠ACB
344
Part II
Euclidean Geometry
345
15. Some Euclidean Geometry of Circles Some interesting properties of circles and rectangles in Euclidean geometry are investigated in this section. The material is from book III of Euclid, and includes furthermore the converse of Thales’ theorem, and the basic properties of rectangles. Recall that in Euclidean geometry, the parallel axiom and its consequences are now assumed to hold.
Figure 15.1. Thales Theorem.
15.1.
Thales’ Theorem
Theorem 15.1 (Thales’ Theorem). "The angle in a semicircle is a right angle." More precisely stated: If an angle has its vertex C on a circle, and its sides cut the circle at the two endpoints A and B of a diameter, then angle γ = ∠ACB is a right angle. Thales’s lived ca. 624-546 B.C., in Miletus, a Greek island along the coast of Asia Minor. These dates are known rather precisely, because, as reported by Herodotus, he predicted a solar eclipse, which has been determined by modern methods to have occurred on May 28th of 585 B.C. Tradition names Thales of Miletus as the first Greek philosopher, mathematician and scientist. He is known for his theoretical as well as practical understanding of geometry. Most important, he is credited with introducing the concept of logical proof for abstract propositions. Traditions surrounds him with legends. Herodotos mentioned him as having predicted a solar eclipse, which put an end to fighting between the Lydians and the Medes. Aristotle tells this story about him: Once Thales somehow deduced that there would be a great harvest of olives in the coming year; so, having a little money, he gave deposits for the use of all the olive presses in Chios and Miletus, which he hired at a low price because no one bid against him. When the harvest-time came, and many were wanted all at once and of a sudden, he let them out at any rate he pleased, and made a quantity of money. Plutarch tells the following story: Solon who visited Thales asked him the reason which kept him single. Thales answered that he did not like the idea of having to worry about children. Nevertheless, several years later Thales anxious for family adopted his nephew Cybisthus.
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Thales went to Egypt and studied with the priests. While he was in Egypt, he was able to determine the height of a pyramid by measuring the length of its shadow at the moment when the length of his own shadow was equal to his height. Thales is said to have proved some simple theorems of geometry, as well as the not so obvious theorem about the right angle in a semicircle. As stressed by David Park [28], the story raises an important point, whether or not Thales really invented the proof: Babylonians and Egyptians had a number of mathematical tricks. For example, Babylonians knew this proposition, as well as the Pythagorean theorem a thousand years before Thales and Pythagoras found them. If they were known, they must have been proved, but there is no sign that anyone thought the proofs were important enough to preserve. Whoever set the process of proof at the center of the stage is the founder of all the mathematics since then, and if it is not Thales it was someone who lived not long afterwards. Proof of Thales’ Theorem in modern manners. Draw 4ABC and, as an extra for the proof, the line from the center O to the vertex C. The base angles of an isosceles triangle are equal by Euclid I.5 . In modern parlance, we say: The base angles of an isosceles triangle are congruent. We use that fact at first for 4AOC. Hence α = ∠OAC ∠OCA Secondly, we use Euclid I.5 for 4COB. Hence β = ∠OBC ∠OCB By angle addition at vertex C γ = ∠ACB = α + β
(15.1)
Next we use Euclid I.32, which tells us: The sum
of the angles in a triangle is two right angles.
Because α, β, γ are just the angles in 4ABC, we conclude that α + β + γ = 2R
(15.2)
I have used the letter R to denote a right angle. Pulling formulas (40.11) and (40.11) together yields γ + γ = α + β + γ = 2R, and hence γ = R , as to be shown. Problem 15.1 (A special triangle). Use the equilateral triangle and Thales’ Theorem to construct a triangle with angles of 30◦ , 60◦ and 90◦ . Answer. On an arbitrary segment AB, an equilateral triangle is erected. This can be done as in Euclid I.1 by finding the intersection point C of a circle around A through B with a circle around B through A. We extend the segment AB on the side opposite to B, and get the second intersection point D of the extension with the circle around A. The triangle 4DBC has the angles 30◦ , 60◦ and 90◦ at its vertices D, B and C. Problem 15.2. Why are all three triangles in the figure on page 347 equilateral. To give a valid reason, use only the following facts:
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Figure 15.2. Construction of a triangle with angles of 30◦ , 60◦ and 90◦ .
Figure 15.3. Three triangles.
• •
The base angles of a triangle are congruent if and only if the triangle is isosceles. The sum of the angles in a triangle is 180◦ .
Proof. By construction, the right-hand triangle 4AMD and the left-hand triangle 4MBC are both equilateral. Both have three congruent angles. Because of the angle sum, their angles are all 60◦ . Angle addition at the vertex M implies ∠CMD = 180◦ − 60◦ − 60◦ = 60◦ . Moreover, the middle triangle 4MCD is isosceles by construction. Hence the two base angles marked β are congruent. Finally the angle sum of the middle triangle implies 60◦ + 2β = 180◦ , and hence β = 60◦ . We see that the middle triangle has three angles of 60◦ , too. Hence all its three sides are congruent, and it is equilateral, too. Problem 15.3. Construct a right triangle with hypothenuse c = 5 and one leg a = 3. Use a construction based on Thales’ theorem and describe your construction. Answer. We draw a segment of the length |AB| = c = 5 as given. Draw a semicircle with diameter AB. Draw a circle around B with radius 3. The semicircle and the circle intersect at point C. The triangle 4ABC is a right triangle. Hypothenuse |AB| = 5, and leg |BC| = 3 have the lengths as required.
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Figure 15.4. Construction of a right triangle with hypothenuse c = 5 and leg a = 3.
Problem 15.4. Construct a right triangle with projections p = 3 and q = 4 of the legs onto the hypothenuse. Use a construction based on Thales’ theorem and describe your construction.
Figure 15.5. Construction of a right triangle with projections p = 3, q = 4.
Answer. We draw segments of the lengths as given, |AF| = q = 4 and |FB| = p = 3, adjacent to each other on one line. Erect the perpendicular on line AB at point F. Draw a semicircle with diameter AB. The semicircle and the perpendicular intersect at point C. The triangle 4ABC is a right triangle with hypothenuse AB, and the projections q = AF and FB = p have the lengths as required. Problem 15.5. Use Thales’ theorem for a Euclidean construction of the perpendicular p to a given line l through a given point P lying on l. Do and describe the construction. Answer. One chooses an arbitrary point O not on line l, and draws a circle c around it through point P. This circle intersects the given line l at point P, and a second point, which is called A. Next one
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Figure 15.6. Erect the perpendicular via Thales’ Theorem. The numbers indicate the order of the steps.
draws the line OA. It intersects the circle c at A and a second point, called B. Finally, the line BP is the perpendicular to be erected on line l at point P. Problem 15.6. Use Thales’ Theorem to drop the perpendicular from a given point P not lying on line l onto the given line l. Answer. The construction is indeed a bid awkward—and done more for fun than of practical value. Theorem 15.2 (A strengthening of Thales’ Theorem). Given is a triangle 4ABC, and a circle with its side AB as diameter. (i) If the third vertex C of the triangle lies inside the circle, the angle at vertex C is obtuse. (ii) If the third vertex C of the triangle lies outside the circle, the angle at vertex C is acute. (iii) If the third vertex C of the triangle lies on the circle, the angle at vertex C is right. Corollary 39. If the diameter of a circle is one side of a right triangle, then its third vertex lies on this circle, too. Theorem 15.3 (Converse Thales’ Theorem). The vertex of a right angle the sides of which cut a circle at the endpoints of a diameter, lies on this circle. Taking Thales’ Theorem 15.1 and the converse Theorem 15.3 together, we can state: Corollary 40 (Strong Thales’ Theorem). An angle the sides of which cut a circle at the endpoints of a diameter is a right angle if and only if its vertex lies on this circle.
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Figure 15.7. Drop the perpendicular via Thales’ Theorem. The numbers indicate the order of the steps.
Proof of Theorem 15.2. Following Euler’s convention, let α = ∠BAC, β = ∠ABC and γ = ∠ACB denote the angles at the vertices A, B and C, respectively. Let O be the midpoint of the triangle side AB. In the figures on page 351 and page 351, there are given drawings of the 4ABC and the circle for the two cases: (i) OC < OA
or
(ii) OC > OA. We use Euclid I.18: If one side of a triangle is greater than another, then the angle opposite to the greater side is greater than the angle opposite to the smaller side. First consider the case that (i) OC < OA holds. Using Euclid I.18 for 4AOC, we conclude that α < α0 = ∠OCA
(15.3)
Now OA OB, since O is the midpoint of the hypothenuse AB. Hence, because of (i), OC < OB holds, too. Using Euclid I.18 once more, this time for triangle 4BOC , we conclude that β < β0 = ∠OCB
(15.4)
Adding (15.3) and (15.4) yields α + β < α0 + β0 . Angle addition at vertex C yields α0 + β0 = γ. Now we use that the sum of the angles in a triangle is two right angles, as stated in Euclid I.32. Hence 2R = α + β + γ < α0 + β0 + γ = 2γ
(15.5)
and hence γ > R. Hence the triangle is obtuse, as to be shown. By a similar reasoning, one shows that in case (ii), the assumption OC > OA implies that the angle γ is acute.
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Figure 15.8. A strengthening of Thales’ Theorem, with vertex C inside the circle
Figure 15.9. A strengthening of Thales’ Theorem, with vertex C outside the circle
Problem 15.7. Provide a drawing for case (i). Provide two drawings for case (ii), one where the triangle 4ABC is acute, and, as a catch, a second one where the 4ABC is obtuse, nevertheless. (The triangle is obtuse because of a different obtuse angle.) Problem 15.8 (The midpoints of chords). Given is a circle C with center O, and a point P inside C. Describe the location of the midpoints of all chords through point P. Give a reason based on Thales’ Theorem or its converse. Answer. The midpoints of all chords in the given circle C through the point P lie on a circle with diameter OP. Here is the reason: The midpoints are the foot points F of the perpendiculars dropped from the center O of the circle onto the respective chords. By the converse of Thales’ Theorem, all the vertices F of the right angles with sides through the two points O and P lie on Thales’ circle with diameter OP.
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Figure 15.10. A strengthening of Thales’ Theorem, still another case with vertex C outside the circle
Figure 15.11. Where lie the midpoints of all chords through point P?
Corollary 41 (My strongest Thales’ Theorem). Given is any circle and any angle. Any two of the following statements (i) (ii) (iii) imply the third one. (i) The two sides of the angle cut the circle at the endpoints of a diameter. (ii) The vertex of the angle lies on the circle. (iii) The angle is a right angle. Problem 15.9. Give explanations how to obtain a complete proof of the three parts of the strong Thales’ Theorem 41, using all material from this section.
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Figure 15.12. The midpoints of chords lie on Thales’ circle with diameter OP.
Answer. The theorem has three parts: "(i) and (ii) ⇒ (iii)": This is simply the original Thales’ Theorem 15.1. "(i) and (iii) ⇒ (ii)": This is the Converse Thales’ Theorem 15.3. "(ii) and (iii) ⇒ (i)": This is consequence of Euclid III.20 (Theorem 15.4) and Euclid III.22 (Theorem 15.6). By Euclid III.20, the central angle ∠AOB is twice the circumference angle with vertex on the long arc. By Euclid III.22, the circumference angles on the long and the short arc add up to two right angles. Assume that A, B and C are any three points on a circle and ∠ACB is a right angle. Let D be the other endpoint of diameter CD. By Euclid III.22, the angle ∠ADB is a right angle, too. Either C or D lies on the long arc AB, or AB is a diameter. Assume that C lies on the long arc AB. By Euclid III.20, we see that the central angle ∠AOB is two right angles. If D lies on the long arc, we get the same conclusion that the central angle ∠AOB is two right angles.. Hence AB is a diameter, anyway. 15.2. Rectangles and the converse Thales Theorem As already stated in the section on Legendre’s Theorems, a rectangle is defined to be a quadrilateral with four right angles. The segments connecting the opposite vertices are called the diagonals of the rectangle. Obviously, a diagonal bisects a rectangle into two right triangles. They turn out to be congruent. Conversely, one can built a rectangle from two congruent right triangles. At first, we use this idea for an independent proof of the converse of Thales’ Theorem 39. Secondly, the same idea helps to prove the rather obvious, but important properties of a rectangle. Independent proof of the converse of Thales’ Theorem 39. Let 4ABC be the given right triangle, with the right angle at vertex C. The idea of this independent proof is constructing a rectangle from the right triangle.
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Step 1: We transfer the angle α = ∠BAC to vertex B in order to produce the congruent angle ∠ABD = α on the side of hypothenuse AB opposite to vertex C. Furthermore, we transfer the segment AC to obtain the congruent segment BD AC on the newly produced ray.
Figure 15.13. (up-down) 4ABC congruent 4BAD (left-right) 4CAD congruent 4DBC 4AMC congruent 4BMD
(scissors)
Question. Show the congruence 4ABC 4BAD
(up-down)
Which congruence theorem do you use? Answer. This follows by SAS congruence. Indeed, the two triangles have the common side AB, the two sides AC BD are congruent, and the angles ∠BAC ∠ABD are congruent, both by construction. Step 2: The quadrilateral ADBC obtained from the two triangles has remarkable symmetry. It is now bisected along its other diagonal CD to obtain a second pair of congruent triangles 4CAD 4DBC
(left-right)
Question. Explain how this congruence is shown. Answer. This follows by SAS congruence. Because of the first congruence (up-down), we have two pairs of congruent sides: AC BD by construction, and BC AD as a consequence of step 1. Furthermore, there are two pairs of congruent z-angles: the angles ∠BAC ∠ABD = α are congruent by construction. The angles ∠ABC ∠BAD = β are congruent as a consequence of (up-down). Hence angle addition yields ∠CAD ∠DBC. Finally, we get the claim (left-right) via SAS congruence. The congruence (left-right) now yields two further pairs of congruent z-angles: ∠ACD ∠BDC = α0 and ∠CDA ∠DCB = β0 Step 3: Since points C and D lie on different sides of line AB, the segment CD intersects this line at midpoint M. Question. Use the congruence 4AMC 4BMD (scissors) in order to show that the diagonals AB and CD bisect each other at their common midpoint M. By which theorem is this congruence proved?
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Answer. The congruence (scissors) is obtained with the SAA congruence theorem. To use this theorem, we need the congruent sides AC BD, the adjacent pair of congruent z-angles α0 , and the vertical angles at vertex M. Hence one gets the congruent triangles (scissors). Finally, we get the segment congruences AM BM = m0 and CM DM = m Clearly point M lies between C and D, because these two points are on different sides of line AB by construction. But it is quite hard to confirm that M lies between A and B. Indeed, the congruence AM BM implies that point M lies between A and B. 1
Figure 15.14. The congruence of 4BCA to 4CBD implies that the diagonals of a rectangle are congruent
Step 4: We now get to the part of the reasoning which is valid only in Euclidean geometry. Because the angle sum in triangle 4ABC is two right, and the angle at vertex C is assumed to be a right angle, we conclude that α + β = R. Similarly, one confirms that the quadrilateral ADBC has a right angle at vertex B. Hence it is a rectangle. The right angle at vertex A yields still another pair of congruent triangles 4ABC 4DCB
(overlap)
Question. Which theorem do you use to confirm this claim? Answer. This follows by SAS congruence. Indeed, the two triangles have the common side BC = CB, the two sides AC DB are congruent by construction, and the angles ∠BCA and ∠DBC are both right. Indeed, ∠BCA was assumed to be a right angle, whereas ∠DBC = ∠DBA + ∠ABC α + β = R because of congruent z-angles α by construction, and the angle sum. 1
See also figure on page 207 which is taken from the millenium edition of "Grundlagen der Geometrie", page 26.
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As a consequence of the congruence (overlap), we get AB CD. Thus we have shown that the diagonals of the quadrilateral ADBC are congruent. Step 5: This is the final step to get the converse Thales Theorem. Because the diagonals AB CD are congruent, and point M bisects them both: AM MB and CM MD Hence all four segments from point M to the vertices A, B, C and D are congruent. Hence vertex C lies on a semicircle with diameter AB, as to be shown. Remark. As a variant for the final step 5, we compare the distance AM BM = m0 with CM DM = m. Assume that m0 > m. Across the longer side of a triangle lies the larger angle. Using this fact for triangle 4AMC, we conclude that α0 > α. Similarly, using triangle 4BMC, we get β0 > β. Hence by addition, we get R = α0 + β0 > α + β. Hence the angle sum in the right triangle 4ABC would be less than two right angles, which is false in Euclidean geometry. Similarly, the assumption m0 < m leads to a contradiction, too. The only remaining possibility is m = m0 . Hence vertex C lies on a semicircle with diameter AB and center M, as to be shown. Problem 15.10. Which steps are still valid in hyperbolic geometry. Compare these quantities: m with m0 , α with α0 , and β with β0 . Answer. Steps 1,2 and 3 are still valid. We still get α0 + β0 = 90◦ , because the construction was started with a right triangle 4ABC with right angle at vertex C. But, because the angle sum of a triangle is less than two right angles, one gets α + β < 90◦ . Since m < m0 occurs if and only if α < α0 and β < β0 , these inequalities are all three true. The figure on page 356 indicates the differing angles and segments using different colors. Remarkably enough, one can take the proof a
Figure 15.15. What steps 1,2 and 3 yield in hyperbolic geometry
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step further without appeal to Euclidean geometry: The perpendiculars dropped from M onto the four sides of the quadrilateral ADBC lie on two lines, but these two lines are not perpendicular to each other. We shall now recapitulate the same ideas, in order to derive the basic properties of a rectangle. Corollary 42 (The basic properties of the rectangle). Given is a rectangle. This is a quadrilateral, about which it is only assumed that it has four right angles. (i) A diagonal partitions the rectangle into two congruent triangles. (ii) The opposite sides of a rectangle are congruent. (iii) The opposite sides of a rectangle are parallel. (iv) The two diagonals of a rectangle bisect each other. (v) The diagonals of a rectangle are congruent. (vi) The diagonals of a rectangle intersect at the center of its circum circle. Proof. By definition, a rectangle is just a quadrilateral with four right angles—that is all what is assumed. The diagonal AB bisects the rectangle ADBC into two right triangles 4ABC and 4ABD. Again, let α = ∠CAB and β = ∠ABC be the angles of the first of these triangles at vertices A and B, respectively. We now use the fact that the angle sum in a triangle is two right angles. 1 Hence α + β + R = 2R and α + β = R. The given rectangle has a right angle at vertex C, hence angle subtraction at vertex A now confirms that the second lower triangle 4ADB has the ∠BAD = R − α = β at vertex A. Thus we have obtained a pair of congruent z-angles β = ∠ABC ∠BAD. Similarly, we get a second pair of congruent z-angles α = ∠BAC ∠ABD. With the ASA congruence theorem, we can now confirm the two triangles 4ABC 4BAD are congruent. Finally, the construction done above can be used once more to rebuild the rectangle from these two congruent right triangles. We see that a diagonal partitions the rectangle into two congruent right triangles, and the opposite sides of the rectangle are congruent. Too, they are parallel because of the congruent z-angles. The two diagonals of a rectangle are congruent because of 4BCA 4DAC
(overlap)
Let M be the intersection point of the diagonals. The congruence (overlap) implies that the triangle 4AMC has congruent base angles α. Hence the converse isosceles theorem implies MA MC. Similarly, one can show that the triangle 4BMC has congruent base angles β and the converse isosceles theorem implies MC MB. Hence the two diagonals of a rectangle bisect each other. and their intersection point is the center of its circum circle. Problem 15.11. In Euclidean geometry, one defines a parallelogram to be a quadrilateral, the opposite sides of which are parallel. Which parts of the Corollary 42 are true for all parallelograms, in Euclidean geometry. Answer. Items (i) true (iv) are true for any parallelogram, but items (v) and (vi) are not true in general. 1
Actually, while proving the second Legendre Theorem, we have shown that the existence of rectangle already implies this fact about the angle sum.
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15.3.
Construction of tangents to a circle
Problem 15.12 (Tangents to a circle). Given is a circle C, with center O and a point P outside of C. Construct the tangents from point P to the circle C. Actually do and describe the construction!
Figure 15.16. Tangents to a circle
Construction 15.1 (Tangents to a circle). One begins by constructing a second circle T with diameter OP. (I call this circle the Thales circle over the segment OP). The Thales circle intersects the given circle in two points T and S . The lines PT and PS are the two tangents from P to circle C. Validity of the Construction. By Thales theorem, the angle ∠PT O is a right angle, because it is an angle in the semicircle over diameter OP. Since point T lies on the circle C, too, the segment OT is a radius of that circle. By Euclid III. 16, The line perpendicular to a diameter is tangent to a circle. Since T P is perpendicular to the radius OT , and hence to a diameter, it is a tangent of circle C.
15.4. A bid of philosophy If a Mathematician learns such a nice theorem, as Thales’ theorem is for sure, what does he want to do with it? 1 How does mathematics and other clever people benefit from it, after putting all issues of priority aside? Here are some general considerations: Generalize the theorem. This can mean either getting along with less assumptions, or just putting the given assumptions into a more general context. Simplify the statement of the theorem. 1
Well, in case he has discovered the theorem himself, one should publish it—with the possibility in mind that somebody else has already discovered something similar before.
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Sharpen the conclusions. Expressing the conclusions in another way and simplifying the statement can both help to sharpen them. Ask whether a converse holds. There can be more than one version for the converse, depending one how the theorem is formalized. Too, in case the converse fails to be true, one can ask for similar statements of which the converse does hold. Consider special cases. They can look surprisingly different—for example, some assumption may turn out to be too obvious to be stated explicitly. Find applications. Are there constructions or algorithms which follow from the theorem. Simplify the proof of the theorem. Build a theory. Find the natural place for the theorem in a larger context. Problem 15.13. We want to construct the tangents through a given point outside to a given circle. Which one of construction (15.1) and construction (10.1) —explained in the section on neutral geometry of circles and continuity—remains valid in hyperbolic geometry? Explain why. Answer. Construction (15.1) is no longer valid in hyperbolic geometry. On the other hand, construction (10.1) remains valid in hyperbolic geometry. Construction (15.1) is not valid in hyperbolic geometry, because the angle sum of a triangle is less than two right, and hence Thales’ theorem does not hold in hyperbolic geometry. On the other hand, construction (10.1) depends only on SAS-congruence, and the fact that tangent and radius being perpendicular. These statements hold in hyperbolic geometry, too. Problem 15.14. Once more: given is a circle and a point outside of it. We have to construct the tangent from the point to the circle. One could simply put the straightedge through this point and move it around until its line just touches the circle. What can one say on the one side in favor, and what on the other side against this "lazy-boy" construction? Gather all arguments coming to your mind. Answer. Arguments in favor. The "lazy-boy construction" is quick and easy. The accuracy with which the tangent is obtained is as good as for either one of the traditional constructions—as long as one wants to obtain only the tangent and not the touching point. After the fact—once it is known that the construction of the tangent is possible with the traditional tools—it turns out that the "lazy-boy construction" does not allow constructions for other problems than those solvable with the traditional tools. The "lazy-boy construction" is just an additional tool helping to do things simpler, in the same sense as the geometrical triangle does that. Too, the "lazy-boy construction" would still work in hyperbolic geometry. Arguments against the easy construction. The "lazy-boy construction" does not allow to obtain the touching point with any reasonable accuracy. One can of course find the touching point by an additional step—dropping the perpendicular from the center of the circle onto the tangent. But in that way, the number of steps needed is nearly the same as for the traditional construction of the tangent. Before establishing the fact that construction of the tangent is possible with the traditional tools, the only way to formally justify this procedure is by requiring a new axiom— on the same footing as it has been suggested for origamy, or the two marked ruler, or any other extended tools of construction.
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Question. Is the axiom "The tangent from a point to a circle can be obtained by appropriate placement of the straightedge (lazy-boy construction)." equivalent to the circle-line intersection property, or to the circle-circle intersection property? Or is it a weaker axiom? Question. How often does one really want new axioms? Because of such questions, the "lazy-boy construction" cannot simply replace the traditional construction, nor does it make the traditional construction and its proof superfluous. My last word. In the end, I think one can use this simple way to obtain the tangent, if it is not necessary to have the touching point. 15.5.
Common tangents of two circles
Problem 15.15 (Common tangents of two circles). How many common tangents do two circles have. Informally draw all different cases, with 0, 1, 2, 3, 4 common tangents. Describe how they arise. Answer. For any two different circles, there are five possibilities regarding their common tangents: (0) One circle lies inside the other. They have no common tangents. (1) One circle touches the other from inside. There is one common tangent, located at this touching point. (2) The two circles intersect in two points. They have two common tangents, which lie symmetrically to the axis connecting the two centers. (3) The two circles touch each other from outside. They have three common tangents. (4) The two circles lie outside of each other. They have four common tangents. These are two pairs lying symmetrically to the axis connecting the two centers. Remark. Let the circles have center O and radius a, and center Q and radius b. These cases correspond to: (0) |OQ| < |b − a|: One circle lies inside the other. (1) 0 < |OQ| = |b − a|: One circle touches the other from inside. (2) |b − a| < |OQ| < a + b: The two circles intersect in two points. (3) |b − a| < |OQ| = a + b: The two circles touch each other from outside. (4) a + b < |OQ|: The two circles lie outside of each other. (∞) 0 = |OQ| = |b − a|: The two circles are equal to each other.
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Figure 15.17. Two circles with no common tangents.
Figure 15.18. Two circles with one common tangent.
Figure 15.19. Two circles with two common tangents.
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Figure 15.20. Two circles with three common tangents.
Figure 15.21. Two circles with four common tangents.
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Problem 15.16 (Construction of the common tangents of two circles). Given are two circles, with center O and radius a, and center Q and radius b. They lie outside of each other. Explain how to construct, with Euclidean tools, the common tangents of the two circles. Actually do and describe the construction!
Figure 15.22. Construction of two of the four common tangents for two circles lying outside each other.
The drawing on page 363 constructs just two common tangents, using construction 15.2. The other two are their mirror images by line OQ connecting the centers. Construction 15.2 (Common tangents of two circles). One constructs two new circles around one of the given centers, say Q, choosing as their radii the sum a + b and difference |a − b| of the radii a and b of two given circles. Next one constructs the tangents from the second center O to these two new circles. Too, one needs the radii where these tangents touch. The common tangents are produced by parallel shifts of these tangents, to both sides, by a distance given by the radius of the original circle around O. Shifting the tangents to the circle of radius a + b yields the inner common tangents, whereas shifting the tangents to the circle of radius |a − b| results in the outer common tangents. A totally different method to construct the common tangents is described in the section on similar triangles. 15.6. Angles in a circle Given is a circle with center O and a chord AB that is not a diameter. Recall Definition 10.3 and Proposition 10.5: a point S on the circle which lies in the interior of the central angle ∠AOB lies on the short arc, and in the half plane of AB opposite to the center O. A point L on the circle which lies in the exterior of the central angle lies on the long arc of chord AB and in the same half plane of AB as the center O.
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Definition 15.1. We call the angle with vertex at the center, the central angle, and the angle with vertex on the opposite arc, the circumference angle of the given arc. Before giving detailed proofs, I state Euclid’s important results: Theorem 15.4 (Euclid III.20). The central angle is twice the circumference angle with vertex on the long arc. Theorem 15.5 (Euclid III.21). If two angles inscribed in a circle subtend the same arc, and if their vertices lie both on the same side of the chord, they are congruent. Theorem 15.6 (Equivalent to Euclid III.22). The circumference angles on the long and the short arc add up to two right angles. The supplement of the circumference angle with vertex on the short angle is congruent to the circumference angle on the long arc.
Figure 15.23. The circumference angle of a short arc is acute, of a long one is obtuse.
Corollary 43. The circumference angle of the short arc is acute, the circumference angle of the long arc is obtuse, and only the circumference angle of a semicircle is a right angle. Remark. Because of Euclid III.22, and the congruence of supplementary angles stated in Proposition 7.15 [Theorem 14 of Hilbert], we conclude that Euclid III.21 holds both for the long and the short arc. Definition 15.2 (Similar arcs). We say that the arcs in two circles are similar if and only if they are either • both short arcs and have congruent central angles; • both long arcs and have congruent central angles; • they are both semicircles. Corollary 44. The circumference angles of any congruent or similar arcs are congruent. Proof. This statement is justified by Euclid III.21 and Euclid III.22. (see theorem 15.5 and 15.6)
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Usually, we first consider the case of an arc AB less or equal half of the circle. It may be, historically Euclid wanted to stick to this situation. With the everyday presence of oscillations in our technical world, it is more natural to allow angles in the entire range from 0◦ to 360◦ , and even beyond a whole turn. As an angle in the usual sense, the central angle ∠AOB is assigned to the short arc. By allowing overobtuse angles, we assign to the long arc the central angle 360◦ − ∠AOB. The long arc is more than a half circle, and hence in this case, the central angle is greater than 180◦ . Furthermore, we want to include the special case that the arc AB is equal to a half circle, the chord AB gets a diameter. In that case, Euclid III.20 becomes Thales’ theorem. Finally, we agree to consider the circumference angle with vertex on the long arc to be the circumference angle of the short arc. Similarly, the circumference angle with vertex on the short arc is considered to be the circumference angle of the long arc. With these agreements put down, we can state Euclid III.20 in a short and more general form: Corollary 45. The central angle is twice the circumference angle of the same arc. This holds both for any arc on the circle, being either long or short. We are now ready to proceed to the proofs. Problem 15.17. Provide a drawing with appropriate notation. Mark your arc AB in some color, then choose your third point C outside that arc. Mark the circumference angle γ = ∠ACB in the same color. Too, mark the central angle ω = ∠AOB. Avoid that the center of the circle lies on any of the chords involved. Prove the claim for the situation occurring in your drawing. The simpler version of the proof uses base angles of isosceles triangles (Euclid I.5), and the exterior angle of a triangle (Euclid I.32). (There are several possibilities, with angle addition or subtraction, but it is enough to do the proof for the situation in your drawing. The other cases are all quite similar.)
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Figure 15.24. Central and circumference angle of a circular arc
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Reason for Euclid III.20. We deal with the central angle ω = ∠AOB and the circumference angle _
γ = ∠ACB of the arc AB. We use the two isosceles triangles 4BOC and 4COA. Each one of them has a pair of congruent base angles, called α and β, respectively. One extends the segment CO beyond the center O and let C 0 be any point on this extension. Now we see the exterior angle x = ∠AOC 0 of triangle 4AOC equals 2α. Indeed, we use the congruence of the base angles of the isosceles triangle 4AOC together with Euclid I.32: The exterior angle of a triangle is the sum of the two nonadjacent interior angles. Similarly, we see that exterior angle y = ∠BOC 0 of triangle 4COB is 2β. For the case drawn in the figure on page 366, the center O lies inside triangle 4ABC. Hence the two isosceles triangles do not overlap. In that case, angle addition is used at the vertices C and O. At vertex C, one gets γ = α + β. At vertex O, one gets ω = x + y. Together with the equations x = 2α
and
y = 2β
one gets ω = x + y = 2α + 2β = 2γ
(15.6)
as to be shown. In the second figure on page 368, the circumference angle is γ = α − β and the central angle is ω = x − y. Since points A and O lie on different sides of line BC, central and circumference angle are both obtained as differences. The next two problems deal with Euclid III.22. At first, I do a special case, and the other problem completes the proof of Euclid III.22. A quadrilateral the four vertices of which lie on a circle is called a circular quadrilateral. Problem 15.18. Let ABCD be a circular quadrilateral, and assume that two vertices give a diameter BD, and the other two vertices A and C lie on different sides of it. Use Thales’ theorem to show that the sum of the angles at B and D is two right angles. Provide a drawing with appropriate notation. The quadrilateral need not be a rectangle! Solution. The diameter BD partitions the quadrilateral into two triangles 4BAD and 4BCD. The sum of the angles of the quadrilateral ACBD equals the sum of the angle sums of the two triangles. Since the angle sum in a triangle is two right angles, the angle sum of a quadrilateral is 4R. By Thales’ theorem, these are both right triangles, with right angles at vertices A and C. After subtraction of the two right angles at vertices A and C, the sum of the two remaining angles of the quadrilateral at vertices B and D is 2R. Theorem 15.7 (Euclid III.22). The opposite angles of a circular quadrilateral add up to two right angles. Proof. In the circum circle, either one of the arcs ABC or ADC is less or equal the other one, and hence less or equal a semicircle. We can assume that arc ABC is less or equal a semicircle. Let B2 be the second endpoint of diameter BB2 . Now B2 lies on the arc ADC. As shown in the last problem, the angles at opposite vertices B and B2 add up to two right angles in the quadrilateral ABCB2 ∠ABC + ∠AB2C = 2R (15.7)
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Figure 15.25. Angles in a circle—still two other cases
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Figure 15.26. A special quadrilateral
Figure 15.27. Opposite angles in a quadrilateral with a circum circle
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Figure 15.28. Use isosceles triangles
By Euclid III.21, the circumference angle of arc ABC is ∠ADC ∠AB2C
(15.8)
∠ABC + ∠ADC ∠ABC + ∠AB2C = 2R
(15.9)
Together, we conclude Remark. Here is an argument to get Euclid III.22 at once. We draw the four radial segments OA, OB, OC and OD and partition the quadrilateral ABCD into four isosceles triangles. Let p, q, r and s be their respective base angles. Because of angle addition at each vertex, the sum of the two angles of the quadrilateral ABCD at the opposite vertices A and C is α + γ = (p + q) + (r + s) = (p + s) + (q + r) = β + δ
(15.10)
Since the sum of angles at all four vertices is α + β + γ + δ = 4R, we conclude β + δ = α + γ = 2R
(15.11)
Proposition 15.1 (Another equivalent formulation of Euclid III.22). An interior angle of a circular quadrilateral is congruent to the exterior angle at the opposite vertex. Proposition 15.2 (Strengthening of Euclid III.22). Assume vertices B and D lie on different sides of AC. The opposite angles β + δ of a quadrilateral ABCD add up to (i) more than two right angles if and only if vertex D lies inside the circum circle of triangle ABC,
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(ii) two right angles if and only if vertex D lies on the circum circle, (iii) less than two right angles if and only if vertex D lies outside the circum circle of triangle ABC. Problem 15.19. Draw a convex quadrilateral ABCD. The vertices A, B, C, D are named in alphabetic order as they follow each other around the quadrilateral. Draw any instance for these two examples: (a) the opposite angles have sum α + γ = 120◦ , (b) the opposite angles have sum α + γ = 270◦ . Draw a circle through A, B and C and explain how point D is located relative to the circle. Answer. The angle sum of any quadrilateral is 4R, and hence β + δ = 360◦ − α − γ. In case (a), one gets β + δ = 360◦ − 120◦ = 240◦ > 180◦ . Hence vertex D lies inside the circum circle of triangle ABC. In case (b), one gets β + δ = 360◦ − 270◦ = 90◦ < 180◦ , and hence vertex D lies outside the circum circle of triangle ABC. Given is a circle with center O. Let AB be the chord and T be a point on the perpendicular to the radius at point A. We assume that center O does not lie in the same half plane of the chord AB as point T . Proposition 15.3. The angle ∠ABT is congruent to the circumference angle ∠ACB with vertex C on the long arc.
Proof. Since we assume that center O does not lie in the same half plane of the chord AB as point T , the angle τ = ∠ABT is acute or right. We prove the statement first for the circumference angle of an appropriately chosen point D. Let BD be a diameter of the circle. By Thales’ theorem, triangle 4ABD is a right triangle. Because the angle sum of a triangle is 2R, its two angles at vertices D and B, add up to a right angle: β+δ=R On the other hand, angle addition at vertex B yields, β+τ=R From these two equation and angle subtraction, we conclude that τ = δ. Thus the angle τ = ∠T AD between the perpendicular of the radius at point B and the chord AB is congruent to the circumference angle δ = ∠ADB of that chord. By Euclid III.21: Two angles from points of the circle subtending the same arc are congruent. Hence it does not matter that we have chosen a special position for point D. The circumference angle γ = δ = τ for any point C on the long arc of AB is congruent to the angle τ between the chord and the tangent.
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Figure 15.29. The angle between tangent and chord is congruent to the circumference angle
15.7. On the nature of the tangent Euclid III.16 contains the assertion that the line erected perpendicularly onto a diameter at its end is tangent to the circle. The fact that the tangent is orthogonal to a diameter is stated repeatedly in Euclid III.16, III.18 and III.19—without further clarification how the tangent is defined. Euclid formulates our result proposition 15.3 in his proposition Euclid III.32: The angle between a tangent line and a chord is congruent to the circumference angle of the arc corresponding to the chord. I find this state of affair unsatisfactory. Hence I have defined the tangent to a circle as perpendicular to the radius (see definition 10.4). Proposition 10.8 states that a line is tangent to a circle if and only if it has exactly one point in common with the circle. This is just the second obvious property of the tangent, which does not involve limits. We now agree to use the notion of limits. We define the tangent as the limiting position of a secant line, the two endpoints of which move together at the touching point of the tangent. The circumference angle of the fixed chord AB constantly stays γ. In the limit C → B, one side of the circumference angle ∠ACB becomes the chord AB, and the other side CB becomes the tangent t to the circle at point B. Hence, in the limit we see that the angle between the chord AB and the tangent t at its endpoint B still has the same value γ. Proposition 15.4. We obtain the tangent as limiting position of a secant line, as the length of the secant approaches zero. In this setting, the angle between any chord and the tangent ray at its endpoint, pointing to the side of the short arc, is congruent to the circumference angle with vertex on the long arc.
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Figure 15.30. The angle between tangent and chord is a limiting case of the circumference angle
Corollary 46 (The tangent as limiting position of a short secant). The tangent to a circle, defined as the limiting position of a small secant, is perpendicular to the radius at the touching point. Proof. Let t be the perpendicular to the radius OB at the touching point B. Let t˜ be the tangent obtained as the limiting position of a small secant with one endpoint B. By proposition 15.3, the acute angle between chord AB and t is the circumference angle γ. On the other hand, the remark above and proposition 15.4 show that the acute angle t˜ between chord AB and tangent as limit is congruent to the circumference angle γ, too. Hence, by the uniqueness of angle transfer, we conclude t = t˜. In other words, the tangent is both the limiting position of a small secant, and the perpendicular to the radius at the touching point. Remark. Indeed, we have already obtained the same result even in neutral geometry. The most complete result is indeed Theorem 10.2. Remark. Historically, the exact nature of the tangent is not discussed at this point, or any point, by Euclid. Euclid III.16 contains the cryptic statement—I would say, the only cryptic statement in the entire work of Euclid: "The angle between the tangent line and the circle is less than any rectilinear angle." 15.8. The two circle lemma The following lemma is remarkable by itself, and shall be used again in the proof of Pappus’ and Pascal’s Theorems. Lemma 15.1 (Two Circle Lemma). If the endpoints of the common chord of two circles lie on two lines, these lines cut the two circles in two further parallel chords.
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Figure 15.31. The sides of an angle cutting two circles at the endpoints of their common chord cut them in two further parallel chords.
Problem 15.20. Explain the reason for the two circle lemma. To get a convenient picture, consider the situation drawn in the figure on page 374. The two given circles have the common chord CD. Let AB be a second chord in the left circle. Explain why the second chord PQ in the right circle is parallel to AB. Going over the proof, does it matter whether the two given lines are parallel or not? Answer. We get a circular quadrilateral ABCD inside the left circle C, and a second circular quadrilateral CDPQ inside the right circle D. As a consequence of Euclid III.22, the opposite angles of a circular quadrilateral sum up to two right angles. We check that the segments AB and PQ form congruent angles with one of the given lines. We need to compare the two angles ∠OBA and ∠OPQ. Let O be any point of line BC and O0 any point on line AD such that the points mentioned occur on the two lines in the order O ∗ B ∗ C ∗ Q, and O0 ∗ A ∗ D ∗ P. The angle ∠OBA is exterior angle in the circular quadrilateral ABCD. As a consequence of Euclid III.22, it is congruent to the opposite interior angle: ∠OBA ∠O0 DC Now the second angle is exterior angle in the second circular quadrilateral CDPQ. Hence it is congruent to the opposite interior angle: ∠O0 DC ∠OQP From these two congruences, we conclude ∠OBA ∠OQP. Hence, by Euclid I.27, the lines AB and PQ are parallel. We see that it does not matter whether the two given lines intersect or not. Question. Explain how the proof has to be modified in the case drawn in the figure on page 375. Answer. In this case, the circular quadrilateral has intersecting opposite sides AB and CD. In this case, the interior angles at opposite vertices are congruent. Hence, the exterior angles at opposite vertices are congruent, too.
375
Figure 15.32. Another example for the two-circle lemma.
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Figure 16.1. The parallel as double perpendicular.
16. Simple Euclidean Geometry These exercises deal with some interesting questions from Euclidean geometry on the level of Euclid’s classic. The parallel axiom and its consequences, and the circle intersection properties are assumed to hold. The Fundamental constructions with Euclidean tools have already been done and described in problems 14.5 through 14.10 in the section about A Simplified Axiomatic system of Geometry—my own Suggestion. 16.1. Five constructions of the parallel Given is a line l and a point P not lying on the line. We want to construct the parallel to line l through the point P. Construction 16.1 (The parallel as double perpendicular). We draw a circle with center P through point A on the line l, and obtain a second intersection point B. We draw two further circles around A and B, both through point P. They intersect in a second point Q, too, which is the the reflection image of P across the line l. We draw the line PQ ⊥ l and get the intersection points C and D with the first circle. Finally, we need the two circles with center C through point D, as with center D through point C. They intersect in two further points R and S . The line RS k l is the parallel to line l through point P. Remark. As an extra check, we have obtained indeed three points P, R and S on the parallel. Problem 16.1. Given is a line l and a point P not lying on the line. We want to construct the parallel to line l through the point P. Provide drawing with notations to illustrate the construction 16.1 given above. Remark. As an extra check, we can get the second endpoint F of the diameter BF of the first circle. The two circles with the centers F and A, both through point P. They intersect in a second point S . In this way, we get three points P, R and S on the parallel.
377
Figure 16.2. The parallel using Thales’ circle.
Construction 16.2 (The parallel using Thales’ circle). We draw a circle with center P through point A on the line l, and obtain a second intersection point B. We draw a further circle around B through point P. Let point E be the other endpoint of the diameter AE of the first circle. Finally, we draw two circles with the centers E and B, both through point P. Finally, we draw two circles with the centers E and B, both through point P. They intersect in a second point R, too. The line RP k l is the parallel to line l through point P.
Justification of the construction. The angle ∠ABE is a right angle by Thales’ theorem since it is the angle in a semicircle. All three circle drawn have congruent radius PA PB PE. Hence the intersection points P and R of the circles around points B and E lie on the perpendicular bisector of segment BE. Hence lines BE and PR are perpendicular. Thus the line given line l = AB and the line PR have a common perpendicular, and hence they are parallel. Construction 16.3 (The parallel obtained from a rhombus). We draw a circle with center P through point A on the line l. (We obtain a second intersection point B, too.) We draw the circle with center A through point P. It intersects the line l in the points G and H. We draw a third circle around G through A. It intersects the very first circle in point R. The line RP is parallel to the given line l since RP and GA are opposite sides of the rhombus GAPR.
378
Figure 16.3. The parallel obtained from a rhombus.
Remark. We can draw a forth circle around H through A. It intersects the very first circle in point S . The line RP is parallel to the given line l since The segments S P and HA are opposite sides a rhombus and hence parallel. In this way, we get three points P, R and S on the parallel, which provides an extra check. Problem 16.2. Given is a line l and a point P not lying on the line. We want to construct the parallel to line l through the point P. Provide drawing with notations to illustrate the construction 16.3 given above.
379
Figure 16.4. The parallel using similar triangles.
Construction 16.4 (The parallel obtained from similar triangles). Choose any two points on the given line and draw circles with these centers through the given point P. They intersect in a second point Q, too, which is the the reflection image of P across the line l. The segment PQ intersects the given line in a point O, around which we draw a circle through points P and Q. It intersects the given line l in points A and B. We draw the rays from point Q through points A and B, and circular arcs through these points through point Q. We get the intersection points R and S . The line RS k l is the parallel to line l through point P. All three points P, R and S lie on the parallel, providing a further check. Problem 16.3. Given is a line l and a point P not lying on the line. We want to construct the parallel to line l through the point P. Provide drawing with notations to illustrate the construction 16.4 given above. Construction 16.5 (The parallel using the harmonic quadrilateral). We draw a circle with center any point B on the line l, and obtain the diameter AC on this line. We choose any point D on the line PC and draw lines DA and DB. Let point E be the intersection of lines AP and DB. Let point F be the intersection of lines EC and DA. The line PF is the parallel to the given line l through point P. Problem 16.4. Given is a line l and a point P not lying on the line. We want to construct the parallel to line l through the point P. Provide drawing with notations to illustrate the construction 16.5 given above. 16.2.
Dividing a segment into any number of congruent parts
Problem 16.5. The subdivision of a segment into three congruent parts, can be done as in the figure on page 380. Give a stepwise description of this construction.
380
Figure 16.5. The parallel using the harmonic quadrilateral.
Figure 16.6. Trisection using parallels.
Construction 16.6 (Subdivision using parallels). Given is the segment AB, to be subdivided into n congruent parts. We draw any ray with vertex A and transfer onto it n congruent segments AA1 , A1 A2 , . . . , An−1 An . We draw the parallels to line An B through all intermediate points A1 . . . An−1 . They intersect the given segment AB in the required division points B1 . . . Bn−1 such that segments AB1 , B1 B2 , . . . , Bn−1 B are the n congruent parts of the given segment AB. Remark. The required parallels are easily constructed by at first dropping the perpendicular p from A onto line An B, and then dropping perpendiculars onto line p from the intermediate points A1 . . . An−1 .
381
Figure 16.7. How to trisect a segment but not an angle.
Take two equilateral triangles with the common side AB. Together they give the rhombus ACBF as shown in the figure on page 381. We draw the long diagonal CF. Let M and N be the midpoints of segments AC and AF. The segment BM intersects the long diagonal CF in point D. The segment BN intersects the long diagonal CF in point E. We see and explain the following facts: • At vertex B we get the angles ∠CBM = 30◦ , ∠MBN = 60◦ and ∠NBF = 30◦ . • The points D and E trisect the long diagonal CF. Construction 16.7 (Trisection using triangles). Given is the segment AB, to be trisected. One constructs two congruent isosceles 4ABC and 4ABD, lying on different sides of AB. Let M and N be the midpoints of segments AC and BC. Let P and Q be the intersections of segment AB with segments DM and DN, respectively. Claim: The segments AP, PQ and QB are congruent. Each one measures one third of AB. Problem 16.6. Complete the following proof of that claim. Indicate is yellow, which part of the argument still works, in green, which part does no longer work in hyperbolic geometry. Proof. We show that angles α = ∠APM, β = ∠QPB, γ = ∠PQD, δ = ∠PQC are all congruent. Indeed α β, because they are vertical angles. Next 4AMD 4BND, by SAS congruence with angles at vertices A and B and the adjacent segments matched. Next 4APD 4BQD by ASA congruence with segments AD and BD and the adjacent angles matched. Hence ∠APD ∠BQD, and β = γ, because supplements of congruent angles are congruent. Next 4APM 4BQN by SAS congruence with angles at vertices A and B and the adjacent segments matched. Hence especially AP BQ.
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Figure 16.8. Trisection of a segment
Next 4AQC 4AQD, by SAS congruence, with angle at vertex A and adjacent segments matched. Hence γ = δ. Those congruences allow to get the claims α = δ and AP BQ. Hence by Euclid’s Proposition I.27, the lines MD and CQ are parallel. The reasoning up to this point is valid in neutral geometry. We can carry the argument further in Euclidean geometry. Indeed Euclid I.29 yields ∠AMP ∠ACQ. Hence 4AMP and 4ACQ have all angles pairwise congruent, and are similar. By Euclid VI.4, the sides of equiangular triangles are proportional. Hence we get the proportion |AQ| |AC| = (16.1) |AM| |AP| Since MC AM and AC = AM + MC by construction, the ratios in (16.1) are both equal to 2. Hence |AQ| = 2|AP| , too. Now AQ = AP + PQ implies that the segments AP and PQ are congruent. Thus all three segments AP, PQ and QB have been shown to be congruent, and hence all three have 31 of the length of AB. Problem 16.7 (Using only equilateral triangles). Draw only two circles of radius AB, and then complete the construction just with a straightedge. Construction 16.8. Begin by constructing two congruent equilateral 4ABC and 4ABD, lying on different sides of AB. Line DA intersects the left circle in a second point E. Line DB intersects the right circle in a second point F. Drawn segment BE, which intersects AC in the midpoint M. Draw segment AF, which intersects BC in the midpoint N. Draw segments MD and ND. They intersect the given segment AB in points P and Q. Claim: The segments AP, PQ and QB are congruent. Each one measures one third of AB. Problem 16.8. For clever students: In hyperbolic geometry, does the middle segment get shorter, or longer than the other two segments?
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Figure 16.9. Completed trisection with equilateral triangles
Answer. In hyperbolic geometry, the middle segment gets shorter than the two other segments. Here is a reasoning for the special case that 4AQC is isosceles. In that case δ δ0 := ∠ACQ, because they are base angles of an isosceles triangle. Next, I use the defect of the angle sum 2R − α − β − γ, which is proportional to the area of a triangle. 4APM is part of 4AQC, and hence has smaller area, and smaller defect. Let α0 := ∠AMP. The defect of 4AMP is less than the defect of the larger 4ACQ. Comparison of the defects leads to α + α0 > δ + δ0 and α0 > δ0 . Euclid I.19 says: "If one angle of a triangle is greater than another, then the side opposite to the angle is greater than the other." Now, we have found that ∠AMP = α0 > δ0 = δ = α = ∠APM, and we can use Euclid I.19 in 4APM. Hence AP > AM. Now AQ AC implies PQ = AQ − AP < AC − AM = MC AM < AP. Remark. In hyperbolic geometry, the problem to trisect a given segment cannot be solved with √3 straightedge and compass. Indeed, the problem leads to the Delian problem to construct 2, which is impossible to do with straightedge and compass. 16.3.
Some triangle constructions
Problem 16.9 (Another special triangle). Describe the construction done in the figure on page 384. Explain how you determine the three angles of the triangle 4ABC. Answer. On an arbitrary segment OA, an equilateral triangle 4OAD is erected. This can be done as described in Euclid I.1. We find the intersection point D of a circle around A through O with a circle around O through A. Let B be the second endpoint of diameter AOB. At point O, we erect the perpendicular onto this diameter. Let E be the intersection of the perpendicular with the right circle, lying on the same −−→ −−→ side of AB as point D. The rays AD and BE intersect in point C.
384
Figure 16.10. What are the angles of 4ABC?
Figure 16.11. Construction of a triangle with angles of 45◦ , 60◦ and 75◦ .
We thus get a triangle 4ABC with the angles 60◦ , 45◦ and 75◦ at its vertices A, B and C. The angle ∠CAB = 60◦ is obtained from the equilateral triangle 4OAD. The angle ∠ABC = 45◦ is obtained from the right isosceles triangle 4OBE. Finally, one calculates the third angle ∠BCA = 180◦ − 60◦ − 45◦ by the angle sum of triangle 4ABC. Problem 16.10 (A construction using an altitude). Using Euclid III.21, construct a 4ABC with the three following pieces given: side c = AB = 6, opposite angle γ = ∠BCA = 30◦ , and altitude
385
Figure 16.12. A triangle construction
hc = 5. (hc is the altitude dropped from vertex C onto the opposite side AB). Question (a). Do the construction and measure your angles α and β. Answer. Question (b). Describe the steps for your construction. Construction 16.9. Draw side AB = 6 and its perpendicular bisector p. Let M be the midpoint of AB. The center O of the circum circle lies on the perpendicular bisector. The center angle is double the circumference angle γ. Hence ∠AOB = 2γ = 60◦ , and ∠AOM = 30◦ . For the example given, the point O is especially easy to find because the 4AOB is equilateral. Next, we draw the circle around O through A and B. This is the circum circle of 4ABC, on which vertex C lies. Secondly, vertex C lies on a parallel q to AB of distance |MD| = hc = 5, because of the given altitude. Hence vertex C is an intersection point of this parallel with the circum circle. One may choose any one of these two intersection points. Remark. One can check that, in Euclidean geometry, the triangle 4ABC is not isosceles: Indeed, the isosceles triangle 4ABC 0 with AB = BC 0 = 6 and γ = 30◦ has the altitude √ 3 √ 0 = 27 > 5 ha = 6 · 2 Hence point D lies between M and O, and α < 30◦ . Problem 16.11 (A further construction using an altitude). Using Euclid III.21 as well as the tangent to a circle, construct a triangle 4ABC with the three following pieces given: side c = AB = 6, opposite angle γ = ∠BCA = 30◦ , and altitude ha = 4. (ha is the altitude dropped from vertex A onto the opposite side BC). Question (a). Do the construction and measure your angles α and β.
386
Figure 16.13. A triangle construction
Answer. The construction is shown in the figure on page 386. The angles are measured to be α = 108.19◦ and β = 41.81◦ . Question (b). Describe the steps for your construction. Construction 16.10. Draw side AB = 6 and its perpendicular bisector p. Let M be the midpoint of AB. The center O of the circum circle lies on the perpendicular bisector. The center angle is double the circumference angle γ. Hence ∠AOB = 2γ = 60◦ , and ∠AOM = 30◦ . For the example given, the point O is especially easy to find because the 4AOB is equilateral. Next, we draw the circle around O through A and B. This is the circum circle C of 4ABC, on which vertex C lies. Next we draw the circle A of radius ha = 4 around point A. The triangle side BC lies on the tangent r to this circle from point B. In the figure on page 386, the tangent r has been constructed using the Thales circle with diameter AB. In the figure on page 387, the tangent r has been constructed by Euclid’s method. Vertex C is an intersection point of this tangent r with the circum circle C. In the upper half plane of AB one gets only one solution. Problem 16.12. Given is a triangle 4ABC with its three altitudes. Find in this figure three pairs of congruent angles at the vertices A, B, C. Hint: Draw the circle with a triangle side AB as diameter. Because of the converse Thales’ theorem, two foot-points of altitudes lie on this circle. Now use Euclid III.21 to find congruent circumference angles. Answer. We draw a semicircle with a triangle side AB as diameter. Because of the converse Thales’ theorem, two foot-points D and E of altitudes lie on this circle. By Euclid III.21 , the arc between
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Figure 16.14. A triangle construction
Figure 16.15. Congruent angles produced by the altitudes of a triangle.
these two foot points has congruent circumference angles at the vertices A and B. From the triangle sides BC and CA, we get two further pairs of congruent angles. Problem 16.13 (A construction using an altitude). Using Euclid III.21, construct a triangle 4ABC with the three following pieces given: side c = AB = 6, opposite angle γ = ∠BCA = 60◦ , and altitude hc = 4. (hc is the altitude dropped from vertex C onto the opposite side AB). Question (a). Do the construction and measure your angles α and β.
388
Figure 16.16. A triangle construction
Answer. The construction is shown in the figure on page 388. The angles are measured to be α = 35.49◦ and β = 84.51◦ . Question (b). Describe the steps for your construction. Answer. Draw side AB = 6 and its perpendicular bisector p. Let M be the midpoint of AB. The center O of the circum circle lies on the perpendicular bisector p. Because the value γ = 60◦ is given in the example, we construct at first the equilateral triangle 4AEB. Next we need the circum circle C of the equilateral triangle 4ABE. As follows from Euclid III.21, the vertex C of the triangle 4ABC to be constructed lies on the circle C, too . To get the center of circle C we draw the circle around E through A and B. It intersects the circle around B through A and E at the points A and R. The line AR is the perpendicular bisector of segment BE, and intersects the perpendicular bisector p of segment AB at point O, which is the center of the circum circle C of the equilateral triangle 4ABE. The circum circle C is now easily drawn. Secondly, vertex C lies on a parallel q to AB of distance |MD| = 4, because of the altitude is given to be hc = 4. The vertex C is an intersection point of this parallel with the circum circle C. In the upper half plane of AB one gets two solutions. One may choose any one of these two intersection points. 16.4.
Quadrilaterals
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Figure 16.17. Six quadrilaterals.
Problem 16.14. In Euclidean geometry, find convex quadrilaterals with the interior angles of 30◦ , 60◦ , 120◦ , 150◦ . (i) Convince yourself that such quadrilaterals all are either trapezoids or circular quadrilaterals. (ii) We assume additionally that two adjacent sides are congruent. I do not count separately the reflexion symmetric instances, nor congruent or similar instances,—how many different shapes can these quadrilaterals have? Convince yourself that there exist five different shapes for trapezoids, and one shape for a circular quadrilateral. (iii) Give accurate drawings, using a protractor and straightedge, or geogebra. Answer. (i) If the angles 30◦ and 150◦ are adjacent, one gets a trapezoid, if these two angle are opposite angles, one gets a circular quadrilateral. (ii) We assume additionally that two adjacent sides are congruent. • • • •
In case the angle of 30◦ includes two congruent sides, the adjacent angle cannot be 60◦ . Hence the adjacent angles are 120◦ and 150◦ . In case the angle of 60◦ includes two congruent sides, the adjacent angle cannot be 30◦ . Hence the adjacent angles are 120◦ and 150◦ . In case the angle of 120◦ includes two congruent sides, the adjacent angle cannot be 30◦ . Hence the adjacent angles are 60◦ and 150◦ . In case the angle of 150◦ includes two congruent sides, all three choices of the adjacent angles among 30◦ , 60◦ or 150◦ are possible.
Problem 16.15. Given are three points A, B and C. On which curve D lies the forth vertex of a quadrilateral ABCD, for which
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Figure 16.18. A quadrilateral with α + γ = 120◦ and given vertices A, B, C.
(a) the opposite angles have sum α + γ = 120◦ , (b) the opposite angles have sum α + γ = 270◦ . Again, the vertices A, B, C, D are named in alphabetic order as they follow each other around the quadrilateral. Actually, the curve D is a circle. In two instances for the examples (a) and (b) above, construct this circle, and especially its center. Answer. Because the angle sum of a quadrilateral is 360◦ , the angle at the forth vertex is δ = 360◦ − α − β − γ. In case (a), we get δ = 360◦ − β − 120◦ . This angle can easily be constructed at vertex B since β = ∠ABC is given. In the figure on page 390, the equilateral triangle 4BEF is used to obtain the angle ∠ABF = 120◦ and hence the angle ∠FBC = δ. The complementary angle −−→ −−→ 90◦ − δ is transferred to the rays AC and CA. The intersection of these two rays is the center O of the circle through A and C on which the forth vertex D lies. Indeed, the center angle is ∠AOC = 2δ and, by Euclid III.21, the circumference angle is ∠ADC = δ as required. In case (b), the angle at the forth vertex is δ = 360◦ − α − β − γ = 90◦ − β. Hence we transfer −−→ −−→ angle β to the rays AC and CA. The intersection of these two rays is the center O of circle to be constructed.
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Figure 16.19. A quadrilateral with α + γ = 270◦ and given vertices A, B, C.
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17. Pappus’ and Pascal’s Theorems 17.1. Pappus’ Theorem in Euclidean geometry In the section on Euclidean Geometry and Ordered Fields, we explain in detail how to construct a field from the congruence classes of segments. The question whether multiplication of segment length is commutative can be translated into a question in geometry. To formulate the relevant geometrical theorem, we look once more at the figure on page 404. There are two pairs of parallel lines, resulting from the construction of the
Figure 17.1. Construction of the products ab and ba, with an empirical error.
products ab and ba: EA0 k BP0 EB0 k AQ0 We notice there is a third pair of parallel lines AA0 k BB0 , because of the isosceles and equiangular triangles 4OAA0 and OBB0 . Are the points P0 are Q0 equal? It turns out that the answer is yes if and only if the Theorem of Pappus is valid. For the construction of segment arithmetic, all we need is a simplified version of Pappus’ Theorem containing two isosceles triangles. The versions of Pappus’ theorem given below are proved to hold for a Pythagorean plane. Theorem 17.1 (Simplified Pappus’ Theorem). Given is a Pythagorean plane. Let A, E, B and A0 , P0 , B0 be triples of points on two intersecting lines, different from the intersection point. Assume the triangles 4OAA0 and 4OBB0 are isosceles. If the lines EA0 and BP0 are parallel, then the lines EB0 and AP0 are parallel, and conversely, too.
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Figure 17.2. Pappus’ theorem in the special case with two isosceles triangles.
Figure 17.3. Producing parallel chords
Proof. Let E 0 be the point on the vertical line such that OE OE 0 . The four points A, B, E 0 and P0 lie on a circle, because the angles at its opposite vertices B and E 0 add up to two right angles. (Explain in detail why!) Let C be the circle through these four points. We have to confirm that the chords EB0 and AP0 are parallel. For this circle, the angles ∠E 0 P0 A and ∠E 0 BA are both circumference angles of the same arc E 0 A. By Euclid III.21 they are congruent. On the other hand, an obvious SAS-congruence implies ∠E 0 BA ∠EB0 A0 .
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By transitivity, ∠OP0 A = ∠E 0 P0 A ∠E 0 BA ∠EB0 A0 = ∠OB0 E and Euclid I.27 implies the lines P0 A and B0 E are parallel.
Theorem 17.2 (Pappus Theorem in parallel setting). Given is a Pythagorean plane. Let A, B, C and A0 , B0 , C 0 be both three points on two intersecting lines, all different from the intersection point. If the lines BC 0 and B0C are parallel, and the lines AC 0 and A0C are parallel, then the lines AB0 and A0 B and parallel, too. Remark. In Hilbert’s foundations [22], this theorem is named after Pascal. Pascal’s name is now generally associated with the theorem about the hexagon in a circle or conic section. Following Stillwell’s book [34], I prefer to use the name Pappus for the theorem which does not involve any circle or conic section.
Figure 17.4. Three circles help to produce the third pair of parallel segments.
Following the third proof of Pappus’ Theorem from Hilbert’s foundations. Define three circles: circle CA through the three points A0 , B and C; circle CB through the three points A, B0 and C; finally circle CA through the three points A, B and C 0 . We keep the notation as in the two-circle Lemma. Let D0 be the intersection point of line OA0 with the circle CA . Let CC∗ be the circle through the three points D0 and A, B. Finally let C ∗ be the intersection point of line OA0 with the circle CC∗ . Recall the two-circle Lemma 15.1: If the endpoints of the common chord of two circles lie on two lines, these lines cut the two circles in two further parallel chords. We shall use the two-circle Lemma 15.1 three times. As a first step, the two-circle Lemma 15.1, yields that the chords A0C and AC ∗ are parallel. Hence we conclude that C ∗ = C 0 , since by assumption, the chords A0C and AC 0 are parallel, too. Hence CC∗ = CC , and the salient point D0 lies on both circles CA and CC . A similar reasoning is now done replacing B 7→ A, B0 7→ A0 and CA 7→ CB . Thus one gets that the salient point D0 lies on both circles CB and CC .
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Figure 17.5. Circles CA and CC intersect the angle in the parallel chords A0 C and AC 0
Figure 17.6. Circles CB and CC intersect the angle in the parallel chords B0 C and BC 0
Question. This fact is indeed just another instance of two-circle Lemma 15.1, and the reasoning above. Use the figure on page 395 and go over all details, once more. Answer. Let C∗B be the circle through the three points D0 and A, C. Finally, let B∗ be the intersection point of line OA0 with this circle. By the lemma, the chords BC 0 and B∗C are parallel. Hence B∗ = B0 , and the four points D0 , B0 , C and A lie on a circle C∗B = CB . Question. Use the Lemma 15.1 a third time, now for the circles CA and CB , and confirm that A0 B k AB0 . Answer. Since point D0 lies on all three circles CA , CB and CC , the circles CA and CB have the
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Figure 17.7. Finally, circles CA and CB help to confirm that AB0 and A0 B are parallel.
common chord CD0 . The remaining intersection points of the two lines ABC and A0 B0C 0 with these two circles yields the parallel chords A0 B k AB0 .
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17.2.
Pascal’s Theorem
Theorem 17.3 (Pascal’s Hexagon Theorem). The three pairs of opposite sides of an arbitrary circular hexagon intersect in three points lying on one line.
Figure 17.8. Pascal’s circular hexagon
Figure 17.9. Pascal’s circular hexagon tangled
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Corollary 47 (Pascal’s Hexagrammum Mysticum). The three pairs of opposite sides of an arbitrary hexagon inscribed into any conic section intersect in three points lying on one line. With points and lines of the projective plane, there are no exceptional cases.
Figure 17.10. Hexagrammum Mysticum
Proof. This theorem follows from Pascal’s circular hexagon theorem by applying a convenient projective transformation to the configuration. Indeed, for any given conic section, there exists a projective transformation mapping the given conic section into a circle. This transformation preserves collinearity and incidence, in the projective sense. Hence the given hexagon inscribed into any conic section is mapped into a circular hexagon. By the inverse transformation, the entire configuration of Pascal’s circular hexagon is mapped back into Pascal’s Hexagrammum Mysticum. Remark. In the setting of projective geoemtry, Pascal’s Hexagrammum Mysticum is really the natural common generalization of several theorems from this section. Note that both the circle, and pairs of parallel or intersecting lines are special cases for conic sections. The corresponding special cases for the Theorem of the Hexagrammum Mysticum are Pascal’s circular hexagon theorem 17.3, and the projective versions of the Pappus and Little Pappus theorems, respectively. Furthermore, in the second case, choosing the Pascal line as improper line, we can further specialize to the affine versions of the Pappus theorem 4.6 and Little Pappus theorem 4.9 as stated above. Proof of Pascal’s circular hexagon. Occurring in counterclockwise order, let the circular hexagon have the vertices A, B0 , C, A0 , B, C 0 . Let Pc := AB0 ∩ A0 B, Pa := B0C ∩ BC 0 and Pa := CA0 ∩ C 0 A be the intersection points of opposite sides, extended.
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Figure 17.11. Proving Pascal’s circular hexagon configuration
For the proof we need a second circle Q through the points B, B0 and Pa . Lemma 15.1 is now used with the originally given circle, and this second circle Q. The common chord of these two circles is BB0 . Through both points B and B0 , the given configuration has two lines drawn. Thus one is led to four instances of Lemma 15.1. One of these does not lead to a result, because lines BC 0 and B0C intersect the circle Q in the same point Pa . The other three instances yield three pairs of parallel chords: AC 0 k Q0 Pa AA0 k Q0 Q (17.1) A0C k QPa where Q is the second intersection point of circle Q with line BA0 , and point Q0 is the second intersection of circle Q with line B0 A. The segments AC 0 and A0C are extended, and intersect in point Pb . Thus one gets two triangle 4AA0 Pb and 4Q0 QPa for which three pairs of corresponding sides are pairwise parallel. By the converse Desargues Theorem, these two triangles are in perspective. Hence the three lines AQ0 = AB0 , A0 Q = A0 B and Pb Pa intersect in one point, which is Pc = AB0 ∩ A0 B. Hence the three point Pa , Pb and Pc lie on one line, as to be shown. The exceptional cases because of parallel lines can be eliminated by using the projective plane.
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Figure 17.12. Hexagrammum Mysticum works on a hyperbola!
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18. Arithmetic of Segments—Hilbert’s Road from Geometry to Algebra In this section, we explain Hilbert’s procedure to construct an arithmetic of segments, also called Streckenrechnung. Hilbert constructs a field from a given model of geometry and thus opens up a general road from geometry to algebra. The approach is explained in Hilbert’s foundations. It works for an arbitrary affine plane for which the theorem of Pappus holds. Since existence and uniqueness of the parallel is assumed, one can take advantage of similar triangles in the construction of the field of segment arithmetic. Neither one of the axioms of continuity shall be assumed. Depending on the given model of geometry, we obtain a variety of different fields, rather than the real numbers exclusively. One can even generalize the approach, and assume only the weaker theorem of Desargues instead of Pappus. In this case, one obtains on the side of algebra only a skew field—thus only the associative law still holds, but the commutative law does no longer hold. These important results have already been stated in the Main Theorems 4 and 3. In the context of projective geometry, the commutative law turns out to be equivalent to the theorem of Pappus and the associative law to the theorem of Desargues. As an alternative, we may avoid the theorem of Pappus and get the segment arithmetic more directly from angles in a circle via Euclid’s III.21. This approach is used in Hartshorne’s book and presented here in an own variant. It works in every Pythagorean plane and breaks any connection with ideas from projective geometry. Recall the definitions: Definition (2.2 Pythagorean plane). A Pythagorean plane is a Hilbert plane for which the Euclidean parallel postulate holds. Definition 18.1 (Pythagorean field). A Pythagorean field is an ordered field with the property √ (Pyth) If a, b ∈ F, then a2 + b2 ∈ F. Main Theorem 19. Any affine plane for which the Theorem of Pappus holds is isomorphic to the Cartesian plane over a field. Conversely, the Theorem of Pappus holds in a Cartesian plane over a field. The first result is elaborated in this section in some detail. The second statement repeats part of the Main Theorem 3, and can be checked directly with some rather tedious calculations which we leave out. Furthermore, this section gives the detailed direct proof of the following result: Theorem 18.1. Any Pythagorean plane is isomorphic to the Cartesian plane F2 over its field F of segment lengths. Furthermore, this is a Pythagorean field. 18.1.
Fields
Definition 18.2 (Field). A field is a system of undefined elements (symbols, numbers), two binary relation symbols + and · and two constants 0 and 1 with the following properties: (1) (i) For any two numbers a and b exists the sum a + b. (ii) (a + b) + c = a + (b + c) for any a, b, c. (iii) a + b = b + a for any a, b. (iv) a + 0 = a for any a. (v) for each number a there exists a number −a such that a + (−a) = 0.
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(2) (i) For any two numbers a and b exists the product a · b. (ii) (a · b) · c = a · (b · c) for any a, b, c. (iv) a · 1 = 1 · a = a for any a. (v) for each number a , 0 there exists a number a−1 such that a · a−1 = a−1 · a = 1. (3) (i) 0 , 1. (ii) (a + b) · c = a · c + b · c for any a, b, c. (iii) c · (a + b) = c · a + c · b for any a, b, c. (4) a · b = b · a for any a, b. A system of elements satisfying only (1)(2)(3) is called a skew field. Any model for the field axioms is simply called a field, too. In this context, we denote by F the set of elements. Clearly 0, 1 ∈ F. 18.2. Construction of the field of segment arithmetic To relate Euclidean geometry to algebra, we need to construct a field from the congruence classes of segments. One can define the sum of any two segments by putting them together along a line. Especially, if point B lies between A and C, we define AC = AB + BC. More generally, the sum of any two segments is defined by Definition 7.2. Because of Hilbert’s postulate III.3, the addition of segments defines even an addition for the congruence classes of segments. As already shown in Proposition 7.7 in the section about congruence of segments, angles and triangles, we get an ordered Abelian group from these congruence classes, their negatives and the zero element. In the theorem 9 about the measurement of segments, we have already used a unit segment. To this unit segment, say OE, we assign the length 1. The construction of the length for any segment as an (finite or infinite) binary fraction done in the measurement theorem 9 depends on the Archimedean axiom. Taking the Archimedean axiom for granted, we could simply define the product and the quotient of segment via the product and quotient of the corresponding binary fractions. Although this "old segment arithmetic" is very practical (and usually tacitly assumed in high school), it depends essentially on the Archimedean axiom, and the possibility to approximate irrational numbers by rational numbers. Only to some very special segments, a finite binary or proper fraction can be assigned as length. Hence the old segment arithmetic is not suitable for a general constructive definition of the multiplication of the field. The essential step for the "new segment arithmetic" claimed by Theorem 18.1 is the correct definition of the product of two segments. This is done using the legs of equiangular right triangles. Thus all dependence on the axioms of continuity can be avoided. Remark. Hilbert even shows in detail how the assumption of any axioms of congruence can be avoided, if instead the theorem of Desargues is assumed. We can leave these considerations aside, and define the sum of segments still via Definition 7.2. To be remembered. For the definition of the product ab, we use two similar right triangles. The first triangle has the legs the first factor a and 1, the second one has as its legs ab and the second factor b.
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Figure 18.1. Construction of the product ab for given segments |OA0 | = a, |OB| = b and |OE| = 1.
Look at the figure on page 403. On the perpendicular ray through O, we have put the segment with prescribed length |OA0 | = a On the horizontal ray emanating from vertex O to the right, there are put points O, E and B such that the segments measured from the origin O have the prescribed lengths: |OE| = 1 , |OB| = b Now we draw the parallel to line EA0 through point B. Why does this parallel does intersect the vertical line? Indeed, by Proclus’ Lemma 3.3, a line intersecting one of two parallel lines intersects the other one, too. Since the vertical line intersects line EA0 , it intersects its parallel through B, too. Hence we get the unique intersection point P0 and define the product ab as length of the segment OP0 : ab := |OP0 | Indeed, we have used the similar right triangles 4OEA0 and 4OBP0 . Question. How does the order of the factors a and b become relevant for the construction? Answer. The first factor a = |OA0 | and the unit segment |OE| = 1 appear together as legs of one triangle 4OEA0 . The second factor b determine the segment OB0 through one end of which one draws the parallels EA0 k BP0 . Indeed, the product of segments a and b turns out to have the property that the rectangle with sides a and b has the same area as the rectangle with sides 1 and ab. Remark. The definition of area does not automatically imply commutativity. On the contrary, the usual simple definition of the area of the rectangle as the product of its sides is only possible because of commutativity. 1 The first obvious question is whether our definition given above makes the product commutative. Furthermore, we need to go on and check whether the remaining axiom required for a field are satisfied, and whether we get an ordered field.
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Figure 18.2. Construction of the products ab and ba, with an empirical error exaggerated.
18.2.1. Commutativity To check the commutative law, we repeat the above construction, with the roles of a and b exchanged. We construct point A on the horizontal line such that |OA| = a and a point on the perpendicular line such that |OB0 | = b Now we draw the parallel to line EB0 through point A, and get the intersection point Q0 . By definition ba is length of the segment OQ0 : ba = |OQ0 | As you can see from the figure on page 404, it is now really a question from geometry, whether ab = ba! Proposition 18.1. The segment arithmetic is commutative in any affine plane where the Theorem of Pappus holds. Proof. The product ab is constructed as in the figure on page 404, by the pair of parallels EA0 k BP0 Because of the isosceles and equiangular triangles 4OAA0 and OBB0 , we get another pair of parallel lines AA0 k BB0 . The hexagon AP0 BB0 EA0 and the horizontal and vertical lines h and v are a Pappus configuration. 1
It would be interesting to find out in a psychological study who really knows that!
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Remark. To get the three pairs of opposite sides of the hexagon, it is convenient to shift the points cyclically by three: A P0 B B0 E A0 B0 E A0 A P0 B The three pairs of opposite sides of the hexagon appear underneath each other: AP0 B0 E
P0 B BB0 EA0 A0 A
The Pappus configuration contains the two pairs of parallel opposite sides EA0 k BP0 AA0 k BB0 Hence, by the theorem of Pappus, there is a third pair of parallels EB0 k AP0 The product ba is constructed as in the figure on page 404, by the pair of parallels EB0 k AQ0 Both EB0 k AQ0 and EB0 k AP0 , and both P0 and Q0 lie on the vertical line v. The uniqueness of the parallels and the uniqueness of intersection point implies P0 = Q0 . Hence ab = |OP0 | = |OQ0 | = ba, as to be shown. Remark. Conversely, the commutative law of multiplication, together with the theorem that equiangular triangles have proportional sides, can be used to get Pappus theorem in the simplified version. Remark. In this and the following proofs, we consider only the case of positive quantities a, b, c. It is left to the reader to check that the corresponding results hold for positive as well as negative quantities.
Figure 18.3. Proof of the distributive law ab + ac = a(b + c).
18.2.2. The distributive law
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Proposition 18.2. The segment arithmetic is distributive in any Pythagorean plane. Proof. Given are the segment lengths a, b, c, which we may assume to be positive. We use two perpendicular rays with the common origin O, and transfer the segment a onto one of them,—the vertical one in the drawing; and the two other segments b and c and the unit segment [O, e] onto the other one,—the horizontal one in the drawing. Next we construct the segment [b, b + c], by transferring the segment [O, c] onto the ray with vertex b on the horizontal axis in order to get segment [O, b + c]. The construction of the products ab, ac, and a(b+c) is shown in the figure on page 405. In all three cases, we use parallels to the line EA0 , where A0 is the point on the vertical axis with a = |OA0 |. These parallels go through points b, c and (b + c) on the horizontal axis, respectively. On the vertical axis, we get the intersection points ab, ac and a(b + c), the distance of which to the origin define the respective products. Finally, we draw the parallel to the vertical axis through point b, and get intersection point P with the line through points b + c and a(b + c). By ASA congruence, the triangles 4Ob(ab) and 4b(b + c)P are congruent, since they are equiangular and have one pair of congruent sides. Hence the segments [O, ac] and [b, P] are congruent. Furthermore, the segments [b, P] and [ab, a(b + c)] are congruent since they are opposite sides of a parallelogram. Hence [O, ac] [ab, a(b + c)]. By definition of segment addition, which is now done on the vertical axis, we know that [O, ac] [ab, ab + ac]. Hence we finally get ab + ac = a(b + c), as to be shown.
Figure 18.4. The proof of the associative law yields at first c(ab) = a(cb).
18.2.3. The associative law Proposition 18.3. The segment arithmetic is associative in any affine plane where the Theorem of Pappus holds.
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Proof. From given segments a and c on the vertical line, and e = 1 and b on the horizontal line, we construct at first segments ab and cb on the vertical line. These two segments are transferred onto the horizontal axis, using 45◦ -lines. Finally, one gets, both on the vertical line, segment c(ab) from the parallels [e, c] k [ab, c(ab)] and segments a(cb) from the parallels [e, a] k [cb, a(cb)]. The three points ab, b and cb on the horizontal axis and the three points ab, c(ab) and cb on the vertical axis give a Pappus configuration with two, and hence three pairs of parallels. We conclude that a(cb) = c(ab). By applying the already proved commutative law twice, we get a(bc) = a(cb) = c(ab) = (ab)c
Figure 18.5. Prove commutativity ab = ba.
18.2.4. A direct proof of commutativity of segment arithmetic Proposition 18.4. In any Pythagorean plane, the segment arithmetic is commutative. Remark. The following proof of commutativity for the segment arithmetic does not presuppose Pappus’ theorem. As a prize, we have to restrict our attention to a Pythagorean plane. We use Euclid III.21 about angles in a circle directly. Thus we get a proof that commutativity of segment arithmetic holds in every Pythagorean plane.
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Proof. Given are the positive lengths a > 0 and b > 0, and a unit segment OE. We take the line −−→ OE as horizontal axis, the ray OE as the positive x-axis. At point O the perpendicular is erected to obtain the vertical axis. The reader should look at the figure on page 407 to see how the three points A− , E−0 and B are obtained: The given positive segment a is transferred to the vertical axis to obtain a point A0 such that |OA0 | = a. The same segment a is transferred to the negative horizontal axis to obtain a point A− such that A− ∗ O ∗ E and |OA− | = a. The unit segment is transferred to the negative vertical axis to obtain a point E−0 such that E−0 ∗ O ∗ A0 and |OE−0 | = 1. The segment b is transferred to the positive horizontal axis to obtain a point B such that |OB| = b. The three points A− , E−0 and B do not lie on a line. By theorem 25.1, there exists a circle through points A− , E−0 and B. Indeed, this is the only circle used for the proof. The origin of coordinates O lies in the segment A− B and hence in the interior of this circle. (Clearly point O need not be the center of the circle.) By construction, the vertical axis intersects the circle at point E−0 and goes through the point O in the interior of the circle. Hence it is easy to prove that the vertical axis intersects the circle at second point P0 for which E−0 ∗ O ∗ P0 . Note that no circle-line postulate is needed to get point P0 . Question. Recall how the product ab has been constructed. Answer. As explained by the figure 403, the construction of the product ab uses two similar right triangles. The first triangle has the legs the first factor a and 1, the second one has as its legs ab and the second factor b. To get these triangles, the given positive segment a has been transferred to the positive vertical axis to obtain a point A0 such that |OA0 | = a. To obtain the product ab, we draw the parallel to line EA0 through point B and get its intersection point Y with the vertical axis. The product is given by the length |OY| = ab of the segment from the origin. Question. Why is the parallel obtained for free, — and indeed Y = P0 ? Answer. We reflect the triangle 4OA0 E across the −45◦ line to produce the triangle 4OA− E−0 . The triangles 4OA0 E 4OA− E−0 are congruent, as one easily checks by SAS-congruence. Hence the angles ∠OA0 E ∠OA− E−0 are congruent, too. The angles ∠BA− E−0 ∠BP0 E−0 are both circumference angles of the arc E−0 B, and both lie on the same side of the chord E−0 B. Hence they are congruent by Euclid III.21. Both steps together show the congruence ∠OA0 E ∠OA− E−0 = ∠BA− E−0 ∠BP0 E−0 = ∠OP0 B By Euclid I.27, "congruent z-angles imply parallels". ( See proposition 7.42. ) In a Pythagorean plane, the parallel is unique. Hence the line P0 B is the unique parallel to line EA0 through point B, which is to be used in the construction of the product ab. The intersection point Y with the vertical axis is equal to point P0 . Question. Explain how the product ba has to be constructed, in the left upper quadrant. Answer. The triangle 4OBE−0 is reflected across the +45◦ line. We obtain the triangle OB0 E− , needed to construct the product ba. We draw the line through A0− parallel to E− B0 , and get its intersection point with the vertical axis at point Z with distance |OZ| = ba from the origin. Question. Explain why, because of these constructions, there appear three congruent angles β. Answer. The angles ∠OB0 E− ∠OZA− are congruent because the vertical axis traverses a pair of parallel lines. The angles ∠OBE−0 ∠OB0 E− are congruent because of SAS congruence of the corresponding triangles. We have marked the three congruent angles by β. Question. Explain why the circle through A− , E−0 and B—already drawn—goes through the point Z.
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Answer. The angles ∠E−0 BA− ∠E−0 P0 A− are circumference angles of the same arc, and lie both on the same side of its chord E−0 A− . Hence they are congruent. From ∠E−0 P0 A− ∠E−0 ZA− we see that point P0 = Z lies on this circle. −−→ Since the circle has only one intersection point P0 with the vertical ray OA, we conclude that Y = Z = P0 . Hence we have obtained that both ab and ba is this intersection point, we conclude that ab = ba. Remark. In the drawing from a former version—which was not from Leibniz’ "best of all worlds"—the construction did not come out right. In such a case, would Leibniz have trusted the proof or the drawing?
Figure 18.6. Commutativity ab = ba follows from both lying on the circle.
As an additional benefit, we state: Corollary 48. The Theorem of Pappus holds in every Pythagorean plane. 18.2.5. Alternative proof of associativity The given positive segments |OA| = a > 0 and |OB| = b > 0 are transferred to the vertical axis. The segments |OE| = 1 and |OC| = c > 0 are transferred to the horizontal axis. The products ac and bc are constructed on the vertical axis. Problem 18.1. One goes on, almost as above, with E, A, B replaced by C, AC and BC. The angles α and β each appear four time. Explain in detail how the figure on page 410 yields a proof of a(bc) = b(ac) with any three positive a, b, c > 0, and thus confirms the associative law.
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Figure 18.7. a(bc) = b(ac) both lie on the circle.
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19. Similar Triangles 19.1.
Basic properties from Euclid
Definition 19.1 (Similar triangles). Two triangles 4ABC and 4A0 B0C 0 for which the ratios of corresponding sides a a0 b b0 c c0 = 0 , = 0 , and = , (19.1) b b c c a a0 are equal, are called similar. Hilbert’s—and already Legendre’s—view about the theory of similarity is different from Euclid’s: In Hilbert’s foundations, the results about similar triangles follow from the segment arithmetic. The segment arithmetic, in turn, is justified by Pappus theorem. Following Hilbert’s approach, the properties of similar triangles shall be justified on the basis of the segment arithmetic developed in the previous section. Proposition 19.1. For any two similar triangles, there exists a scaling factor f by which one needs to multiply the length of a side of the triangle 4ABC to obtain the length of the corresponding side of the equiangular triangle 4A0 B0C 0 . Proof. By ordinary calculation with fractions, a0 a a0 b0 = implies = 0 b b a b b0 c0 Similarly we get b = c . This common ratio is the scaling factor, which I call f and obtain a0 = f a , b0 = f b ,
and c0 = f c
Definition 19.2 (Equiangular triangles). Two triangles 4ABC and 4A0 B0C 0 for which the angles across the corresponding sides are congruent are called equiangular. Theorem 19.1. Two triangles are equiangular if and only if they are similar. To prove this basic theorem, we begin with right triangles. The legs of two equiangular right triangles have the same ratio. Given are two equiangular right triangles. We use SAS congruence to get triangles 4OAB and 4OA0 B0 congruent to the original ones, which have a common vertex O at the right angle, and the legs |OB| = a, |OB0 | = a0 and |OA| = b, |OA0 | = b0 lying on two perpendicular rays from this vertex. For convenience, we take these two rays horizontally to the right and upwards. We draw the unit segment |OE| = 1 on the horizontal ray and the parallel to the hypothenuse AB through point E. We obtain the intersection point Q on the vertical ray. By segment arithmetic, the segment |OQ| = q on the vertical ray satisfies qb = a Because the two hypothenuses AB and A B have congruent angles α with the horizontal axis, the lines EQ and A0 B0 are parallel, too. Hence with the same value of q, segment arithmetic now shows that 0 0
qb0 = a0 and a simple calculation shows that a a0 =q= 0 b b
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Figure 19.1. The legs of equiangular right triangles are proportional.
Figure 19.2. Equiangular triangles are similar.
Since we have shown above that the ratio can can define a trigonometric function:
a b
depends only on the angles of the right triangle,
Definition 19.3 (Tangent function for acute angles). The ratio ab of the leg a across to the leg b adjacent to the angle α of a right triangle is the tangent function of this angle: tan α =
a b
Proposition 19.2 (Euclid VI.4). The corresponding sides of equiangular triangles have equal ratios. Equiangular triangles are similar. In formulas: a a0 b b0 = , = b b0 c c0
and
c c0 = a a0
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Proof. Given are two equiangular triangles 4ABC and 4A0 B0C 0 . By Theorem 11.3 from the section about neutral triangle geometry, every triangle has an in-circle. Its center I is the intersection point of the three interior angular bisectors. We drop the perpendicular from the center I onto the triangle side AB and let F be the foot point. We proceed similarly for the equiangular triangle 4A0 B0C 0 . Thus we get two pairs of equiangular right triangles: 4AIF is equiangular with 4A0 I 0 F 0 , and 4BIF is equiangular with 4B0 I 0 F 0 . We get the proportions for the corresponding legs, which we have to add: |AF| |A0 F 0 | |FB| |F 0 B0 | = 0 0 and = 0 0 , |FI| |F I | |FI| |F I | |AB| |A0 B0 | and hence = 0 0 |FI| |F I | Now |FI| = ρ and |F 0 I 0 | = ρ0 are the radii of the in-circles. For the sides |AB| = c and |A0 B0 | = c0 , as well as the other corresponding sides we obtain
Hence
c c0 a a0 b b0 = 0 , = 0 and = , ρ ρ ρ ρ ρ ρ0 b b0 a a0 and = 0 , and similarly = b b c c0
c c0 = a a0
Remark. By the sin Theorem 43.1 proved below, the ratio of any two sides of a triangle is the ratio of the sinus of the opposite sides: a sin α b sin β = , = , b sin β c sin γ
and
c sin γ = , a sin α
Hence we see once more why these ratios depend only of the angles of the triangle. Proposition 19.3. For any two equiangular triangles, there exists a scaling factor f by which one needs to multiply the length of a side of the triangle 4ABC to obtain the length of the corresponding side of the equiangular triangle 4A0 B0C 0 : a0 = f a , b0 = f b ,
and
c0 = f c
Proof. As explained above, the scaling factor is the common ratio f :=
a0 a
=
b0 b
=
c0 c.
Proposition 19.4 (Euclid VI.5). Similar triangles are equiangular. Proof. Given are two triangles 4ABC and 4A0 B0C 0 the lengths of the sides of which satisfy a a0 b b0 = , = b b0 c c0
and hence
a a0 = c c0
As shown earlier, in a Hilbert plane with unique parallels, there exist equiangular triangles of arbitrary sizes. See the section on the natural axiomatization, Wallis’ postulate and Theorem 12.5. We transfer the angle ∠BAC to vertex A0 and angle ∠ABC to vertex B0 , each of both with one side along line A0 B0 . We obtain two new rays, both in the same half plane as point C 0 . The two new rays intersect at a point X. 1 We obtain the equiangular triangles 4ABC and 4A0 B0 X, with 1
The existence of the intersection point X is shown Proposition 12.2, telling equiangular triangles exist in all given sizes.
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the (possibly new) vertex X,— and both 4A0 B0 X and 4A0 B0C 0 lying in the same half plane of line A0 B0 . By Euclid VI.4 (stated in proposition 19.2), the equiangular triangles 4ABC and 4A0 B0 X are similar. Let |B0 X| = a00 and |A0 X| = b00 . We have obtained the proportions b b00 a a00 = 0 and = 0 c c c c 0 00 0 b b a a00 Hence = 0 and = c0 c c0 c0 Hence b0 = b00 and a0 = a00 We see that triangles 4A0 B0C 0 and 4A0 B0 X have three pairs of congruent sides. Hence the two triangles are congruent by SSS-congruence Theorem 7.22. Indeed, they are even equal by the corresponding Lemma for SSS Congruence 7.2. Hence X = C 0 , and we have already shown that 4ABC and 4A0 B0C 0 are similar. Proposition 19.5 (Euclid VI.6). If two triangles have one pair of congruent angles, and the sides containing these pairs are proportional, then the triangles are similar.
Figure 19.3. Addendum to the extended ASA congruence.
Proof. We use Euler’s notation: angle α lies opposite to side a, angle β lies opposite to side b, and angle γ lies opposite to side c. Given are two triangles 4ABC and 4A0 B0C 0 with congruent angles ∠BAC ∠B0 A0C 0 and two pairs of adjacent sides b, c and b0 , c0 which have proportional lengths: b b0 = c c0 Since segment length are comparable, we may assume AB ≤ A0 B0 , without loss of generality. There exists a point B0 in the segment A0 B0 such that α = α0 and
AB A0 B0
(19.2) −−−−→ There exists a ray B0C0 pointing into the half plane of C 0 and making congruent angles β0 = ∠A0 B0C 0 ∠A0 B0C0 . By congruent z-angles, the lines B0C0 k B0C 0 are parallel. Hence by Pasch’s axiom exists the intersection point C0 in the segment A0C 0 . For a Pythagorean plane, it is assumed the angle sum of any triangle is 180◦ . Hence the triangles 4A0 B0C0 and 4A0 B0C 0 are equiangular. By Euclid VI.4 (stated in proposition 19.2), the equiangular triangles 4A0 B0C0 and 4A0 B0C 0 are similar. We obtain the proportion |A0C0 | |A0C 0 | b0 = = |A0 B0 | |A0 B0 | c0
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Hence the assumption
b0 b |AC| = = c0 c |AB| and A0 B0 AB imply |A0C0 | = |AC| and finally AC A0C0
(19.3)
By equations (19.2) and (19.3) and ∠CAB ∠C0 AB0 , the SAS congruence implies that the triangles 4ABC and 4AB0C0 are congruent. Since the triangles 4A0 B0C0 and 4A0 B0C 0 are similar, we finally conclude the triangles 4ABC and 4A0 B0C 0 are similar, as required. 19.2.
Some exercises
Problem 19.1. Given are two triangles with one pair of congruent angles, and the sides across these angles are proportional to a pair of sides adjacent to the angles. Are the two triangles always similar. Give examples, counterexamples, a general reason. Answer. The two triangles need not always to be similar. Take two triangles for which the assumption holds in the form c c0 γ = γ0 , = a a0 It can happen they are not similar. Actually, this is only possible if the angle γ = γ0 < 90◦ is acute, 0 and ac < 1. Take the example γ = γ0 = 30◦ and ac = ac0 = 35 . There exist two triangles with these pieces which are not congruent, and similar neither. See the figure on page 210. I call this figure a "dangling derrick." The important point is that angle γ and side c are across to each other. Problem 19.2. For the two triangles it is assumed b b0 = c c0 Are the two triangles always similar. Give examples, counterexamples, a general reason. α = α0 ,
Answer. The similarity has just been stated in Proposition 19.5. The sides b and c the ratio of which is given are adjacent to the given angle α. Hence Euclid VI.6 tells that the two triangles are similar. Problem 19.3. Draw two circles of equal radii, the center of one lying on the circumference of the other one. Draw the line through the two centers. Use some of the intersection points you have just obtained, and draw two triangles with angles 30◦ , 30◦ , 120◦ of different sizes, in two different colors. Let the lengths of sides of the larger triangle be a and c, and the lengths of the sides of the smaller one be a0 and c0 . Determine the ratio aa0 = cc0 . Answer. The line through the two centers A and B intersects the two circles in points D and E, too. In the figure on page 416, these four points have the order A ∗ D ∗ E ∗ B. Let C be an intersection point of the two circles. For example, we get two isosceles similar triangles 4ABC ∼ 4ACD. Both have the base angles 30◦ and top angle 120◦ . We determine the ratio aa0 = cc0 of the sizes of the two triangles. Since c0 = a and c = 3a0 , we get a c 3a0 = = a 0 c0 a 2 √ a a = 3 and 0 = 3 02 a a
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Figure 19.4. Two isosceles similar triangles with angles of 30◦ and 120◦ .
Problem 19.4. Given is a triangle. Inscribe a square with one side part of a side of the triangle, and all vertices on the sides of the triangle. Answer. We erect a square onto one side a of the triangle. We connect the two new vertices of the square to the third vertex A of the triangle. The connecting lines intersect the side a in two points, which are the endpoints of one side of the square to be inscribed. Finally, we draw the inscribed square. Problem 19.5. A square is inscribed with one side lying on the side a of the triangle, and the other two vertices lying on the two remaining sides of the triangle. Determine the side x of the square in terms of the side a and the area F of the triangle. Answer. We drop the perpendicular from vertex A onto the opposite side. From the similar triangles 4ADK ∼ 4AFL, we get the proportions |DK| |FL| = |AK| |AL| Once more one gets proportions form the similar triangles 4AEK ∼ 4AGL, and adds |EK| |GL| = |AK| |AL| |DE| |FG| = |AK| |AL| a x = a+h h were h is the altitude. Hence x=
ah 2aF = 2 a + h a + 2F
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Figure 19.5. Inscribing a square into a triangle.
Problem 19.6. Suppose that the endpoints of one side of the square are midpoints of two sides of the triangle. Show by dissection that the area of the square is half the area of the triangle.
Figure 19.6. For h = a, one can fold the triangle to cover the square twice.
Problem 19.7. Is the area of the square from the problem above less, more or equal to half of the area of the triangle. Give a simple folding argument to decide without any calculation.
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Figure 19.7. For a < h, one can fold the triangle to cover the square twice, and triangle 4A0 B0 C 0 left over.
Figure 19.8. For a > h, one can fold the triangle to cover the square twice, and part of triangle 4A0 B0 C 0 is covered twice.
Answer. For the special case that the endpoints of one side of the square are midpoints of two sides of the triangle, the area of the square is half of the area of the triangle. In the figure on page 417 you see this borderline case. In all other cases, the area of the square is less than half of the area of
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Figure 19.9. Constructing a circle through a point and touching two lines.
the triangle.To see this, we fold the triangle along three sides of the square. We get three reflection images A0 , B0 and C 0 of the vertices of the triangle. Because of the right angles of the square, the folded sides touch along the lines A0 B0 and A0C 0 . In the figure on pages 418, you see the case that triangle side a < h is less than the altitude h, and in the figure on page 418, the case that triangle side a > h is longer than the altitude. In the first case, one covers the square and an extra triangle 4A0 B0C 0 is left over in the opposite half plane. In the second case, one again covers the square, and the part of the extra triangle 4A0 B0C 0 inside the square is covered twice. Problem 19.8. Given are two intersecting lines l and m and a point A different from their intersection point. We have to construct a circle through the point A touching both lines. Describe the construction given in the figure on page 419. Answer. • We draw the angular bisector of the the two given lines, bisecting the angle α in the interior of which the given point A lies. • We choose any point O0 on the bisector and draw a circle around it touching both line l and line m. • We draw the ray emanating from the intersection point Z of the two given lines and pointing into the interior of the angle α. This ray intersects the circle in two points C and D. • The parallel to line DO0 through point A intersects the bisector in point O. The circle around O through point A does touch the two given lines and solves the problem. • A second solution is obtained by drawing the parallel to line CO0 through point A. This line intersects the bisector in the center O2 . The circle around O2 through point A does touch the two given lines and yields the second solution of the problem, which has not been drawn.
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Reason for the construction. The two triangles 4ZAO ∼ 4ZDO0 are equiangular because the lines AO and DO0 are parallel by construction. Because the lines FO and F 0 O0 are both perpendicular to the same line m and hence parallel, the triangles 4ZFO ∼ 4ZF 0 O0 are equiangular . Since equiangular triangles are similar, we get the proportions ZO ZO0 ZO ZO0 = and = AO DO0 FO F 0 O0 Division yields AO DO0 = 0 0 =1 FO F O Indeed, the latter quotient equals 1 since both points D and F 0 lie on a circle around O0 . Hence the former quotient equals one, too. Hence both points A and F lie on a circle C around O. This means that the constructed circle C touches the line m . Since the centers O0 and O lie on the angle bisector of the two given lines l and m, the circle O touches the line l, too. Problem 19.9. Given are two points A and B and a line l. Construct a circle through the two points touching the line. First ideas. It is useful to consider these cases: (a) One of the points lies on the line. (b) The lines AB and l are different and parallel. (c) The lines AB and l intersect perpendicularly. (d) The lines AB and l intersect, at any other angle. We have solved case (a) earlier in Problem 14.11 in the section "A simplified axiomatic system of geometry". I leave cases (a) and (b) to the reader. Question. What assumption about the two points and the line is needed that the problem can be solvable is cases (c) and (d). Answer. The two points A and B need to lie on the same side of line l, or one of them on the line. The most interesting case (d) is explained below. Problem 19.10. Given is a line l, and two points A and B on the same side of this line. It is assumed that the lines AB and l intersect, but not perpendicularly. Construct a circle through the two points touching the line. Construction for Problem 19.10. Let p the perpendicular bisector of segment AB. The lines p and l intersect, say at point C. We use point C as a center for similarities. We choose any point O0 , C on the line p which lies on the same side as points A and B. We draw the circle D0 around O0 touching the line l, say in point F 0 . This circle can be mapped by a similarity with center C to the circle to be constructed. −−→ We choose an intersection point A0 of CA ∩ D0 . The parallel of line A0 O0 intersects the perpendicular bisector p in point O. We construct the circle D around O touching the line l in point F. Complete justification for Problem 19.10.
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Figure 19.10. Constructing a circle through two points touching a line.
−−→ −−→ Question. Why do the two rays CA and CB both intersect the circle D0 . Answer. One of the points A and B lies in the interior of the angle ∠O0CF 0 , say point A. By the −−→ Crossbar Theorem 1 the segment O0 F 0 and the ray CA intersect, say in point Q. The point Q is in the interior of circle D0 , hence the line CA intersects the circle D0 . Since all −−→ points of a circle lie on the same side of the tangent as the center, even the ray CA intersects the circle D0 , indeed in two points. These are rather obvious consequences of the circle-line intersection property, see Proposition 10.12 and Proposition 10.6 from the section of neutral geometry of circles and continuity. −−→ −−→ The two rays CA and CB are images obtained from each other by reflection across the line p. −−→ The circle D0 . is symmetric to this line. Hence the ray CB intersects the circle D0 , indeed in two further points. Question. By means of similar triangles, check that circle D goes through point A. Answer. The triangles 4COA and 4CO0 A0 are equiangular by construction, and hence similar: |CO| |C 0 O0 | = |OA| |O0 A0 | The right triangles 4COF and 4CO0 F 0 are equiangular by construction, and hence similar: |CO| |C 0 O0 | = |OF| |O0 F 0 | 1
A segment with endpoints on the two sides of an angle and a ray emanating from its vertex into the interior of the angle intersect. See the section about the axioms of order, Proposition 5.9
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Figure 19.11. Theorem of chords, case of chords intersecting inside the circle
Hence
|OA| |O0 A0 | = =1 |OF| |O0 F 0 |
by calculation, and since A0 and F 0 lie on a circle around O0 . Hence A and F lie on a circle around O, as to be shown. 19.3.
Secants in a circle
Theorem 19.2 (Theorem of chords (Euclid III.35)). If two chords cut each other inside a circle, the product of the segments on one chord equals the product of the segments on the other chord. Assume two chords intersect each other outside a circle. The product of the segments, measured from the point of intersection to the two intersection points with the circle are equal for both chords. Problem 19.11. Provide a drawing, with appropriate notation. Proof using proportions. Let AB and CD be two chords of circle γ intersecting at point P inside the circle. We have to check whether |PA| · |PB| = |PC| · |PD|. Following Legendre, I am using proportions. Equiangular triangles are, by definition, triangles with pairwise congruent angles at their corresponding vertices. Congruent circumference angles, as shown in Euclid III.21, occur at vertices A and D, as well as at vertices C and D. Furthermore, we get congruent vertical angles at vertex P. The vertices of the equiangular triangles have to be listed in such an order that these congruent angles occur at corresponding vertices. Hence we get the equiangular triangles 4PAC ∼ 4PDB By Euclid VI.4
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Figure 19.12. Theorem of chords, case of chords intersecting outside the circle
The sides of equiangular triangles are proportional. Hence the ratios of corresponding sides are the same for two equiangular triangles. |PA| |PD| = |PC| |PB|
(19.4)
Now multiplying with the denominators yields |PA| · |PB| = |PC| · |PD| which is just the claim of Euclid III.35. The proof for the case of the segments intersecting outside the circle is almost identical. But note that the two segments QA and QB now overlap. For the case where the extension of the chords intersect outside the circle, one obtains a further result by considering the tangent as a limit of a small secant. In the figures on page 424, we have illustrated the limit C → T, D → T , where T is the touching point of a tangent drawn from point Q to the circle. Thus one is lead to claim Euclid’s next result. Theorem 19.3 (Theorem of chord and tangent (Euclid III.36)). From a point outside a circle, a tangent and a second are drawn. The square of the tangent segment equals the product of the segments on the chord, measured from the point outside to the two intersection points with the circle. Problem 19.12. Provide a drawing, with appropriate notation. Independent proof, again using proportions. Let Q be a point outside circle γ, let T be the touching point of the tangent to the circle through Q, and let AB be a chord of the circle the extension of which runs through Q. We assume that B lies between Q and A. I have to check whether |QT | 2 = |QA| · |QB|. To this end, one compares the triangles 4QAT and 4QT B. We use Euclid III.32:
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Figure 19.13. Getting the limit C → T, D → T
The angle between a tangent line and a chord is congruent to the circumference angle of the arc corresponding to the chord. Hence the angles at vertices A and T , for the two triangles, respectively, are congruent. With vertices listed in an order that these congruent angles occur at corresponding vertices, we get equiangular triangles 4QAT ∼ 4QT B. By Euclid VI.4: The sides of equiangular triangles are proportional. Hence the ratios of corresponding sides are the same for two equiangular triangles. |QA| |QT | = |QT | |QB|
(19.5)
and multiplying with the denominators yields the result.
19.4.
The lens equation
Proposition 19.6. In a trapezoid ABCD is drawn a third parallel to the two parallel sides through the intersection point S of the diagonals. The trapezoid sides cut congruent segments f = FS and g = S G out of the third parallel. (i) The segments f = FS and g = S G cut out on the third parallel are congruent. (ii) The lengths of the parallel sides a = AB and b = CD and the segment f = FS satisfy the lens equation 1 1 1 + = (19.6) a b f Proof. We introduce the segments p = DS and q = S B. Now we use similar triangles to get three equations among the lengths a, b, f, g, p, q Question. Use two similar triangles with common vertex D to find a proportion among these quantities.
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Figure 19.14. A trapezoid and the lens equation.
Answer. The triangles 4DS F and 4DBA are equiangular and hence similar. proportion p p+q = f a
One gets the (19.7)
Question. Use two similar triangles with common vertex B to find a proportion among these quantities. Answer. The triangles 4BS G and 4BDC are equiangular and hence similar. proportion q p+q = g b
One gets the (19.8)
Question. Use two similar triangles with common vertex S to find a proportion among these quantities. Answer. The triangles 4S AB and 4S CD are equiangular and hence similar. One gets the proportion a b = q p
(19.9)
We need now to eliminate p and q from these three equation. Dividing equation (19.7) by equation (19.8) yields p·g b = f ·q a Together with equation (19.9) one gets
p·g p = f ·q q
and hence f = g as required. Question. Use this result now to get the lens equation out of the two equations (19.7) and (19.8). Answer. We add these two equations and divides by p + q to obtain the lens equation (19.6).
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The trapezoid from the figure on page 425 appears in the context of geometric optics. Recall that the perpendicular to the plane of the lens through its center is called the optical axis. For the light rays through a thin lens and close to the optical axis, one can empirically gets to the following three observations: (a) Light rays parallel to the optical axis coming in,—say from the left—are bundled at the focus G behind the lens. (b) There is a second focus F on the other side of the lens at the same distance. Light rays starting at focus F are bend on the other side of the lens into rays parallel to optical axis. (c) Light rays„ even those coming in from different directions, go straight through the center of the lens. These assertions are in reality only an approximation, but they are exact enough for a thin lens to be useful. These assumption lead to a trapezoid, which has the unusual form shown in the figure on page 426. The third parallel FG from above turns out to be the optical axis. The light source is
Figure 19.15. The trapezoid appearing for an optical lens.
located at vertex A, the lens lies in the plane BD. The light rays emanated point A are focused at vertex C. We can check this to true at least for three rays: • the ray AD through the left focus F, which is bend to the ray CD parallel to the optical axis; • the ray AB parallel to the optical axis, which is bend to the ray BC through the right focus G; • the ray AS C which goes straight through the center S of the lens. Too, Proposition 19.6 now confirms the well-known lens equation (19.6). Problem 19.13. A candle is standing at distance a = 3 from a lens with focus length f = 2. At which distance b behind the lens has one to put a screen to get a sharp image of the candle flame. Give a construction and calculate the distance b.
20. The Pythagorean and related Theorems 20.1.
The leg and the altitude theorems
Problem 20.1. The figure on page 427 explains the common definitions for the lengths a, b, c and p, q, h from a right triangle. Find an example where all six segments have lengths with of integer values. Answer. The smallest example, I can find is a = 15 , b = 20 , c = 25 and p = 9 , q = 16 , h = 12
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Figure 20.1. The common notation for a right triangle
Figure 20.2. The altitude theorem for a right triangle
Problem 20.2. Explain for which special situation Euclid III.35 implies the theorem about the altitude of a right triangle, usually stated as h2 = pq. Answer. Take for segment AB a diameter of circle γ and chose CD to be any segment perpendicular to that diameter. One gets a right 4ABC, with P as foot point of its altitude. Because of PC PD and |PC| = |PD|, Euclid III.35 implies pq = |PA| · |PB| = |PC| · |PD| = h2 . Problem 20.3. Explain how Euclid III.36 implies the leg theorem a2 = pc. Answer. In Euclid III.36, as given in theorem 19.3, we choose points A and T on a diameter of circle γ, and point Q on the tangent to the circle at point T . We see that 4AT Q is a right triangle, because tangent and radius are perpendicular to each other. The segment AQ intersects the circle in a second point B. The triangle 4AT Q has altitude T B, because Thales’ theorem
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Figure 20.3. The leg theorem
Figure 20.4. The Pythagorean Theorem
shows that the angle ∠ABT is right. Now Euclid III.36 tells that the square of one leg QT equals the product of the hypothenuse time the projection of that leg onto the hypothenuse. After renaming T → C, B → P, Q → B, A → A, one gets the statement in its usual form a2 = pc. 20.2.
The Pythagorean theorem
Problem 20.4. Explain how Euclid III.36 implies the Pythagorean Theorem a2 + b2 = c2 .
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Figure 20.5. A pair of triangles 4A0 QB and 4C 0 QD of equal area directly confirms the theorem of chords
Answer. For the given right triangle 4ABC, we draw circle with center A through point C. We draw the tangent at point C and see that point B lies on the tangent since tangent and radius are perpendicular to each other. −−→ We draw the ray BA. Let points E and D be the endpoints of the diameter on this ray. Now Euclid III.36, as given in theorem 19.3, tells that |BC|2 = |BD| · |BE| In terms of the sides of the triangle 4ABC, this shows that a2 = (c + b)(c − b) = c2 − b2
and a2 + b2 = c2
We got the Pythagorean Theorem.
Remark. In the spirit of Euclid, one has to understand the product of segments occurring in the theorems above as areas of rectangles or squares. The equality of their areas can really be shown by obtaining one from the other in a finite sequence of cuts and pastes. In the figure below, I want to visualize this interpretation. Take the case of chords intersecting at point Q outside the circle. We extend the two chords to the other side of their intersection point and transfer the segments QA and QC to these opposite rays to obtain a quadrilateral A0 ACC 0 with axial symmetry. It is not hard to check that this quadrilateral has a circum circle and the triangles 4QAC 4QA0C 0 are congruent. The two circle lemma implies that the chords A0C 0 k BD are parallel. Hence we have obtained a new pair of similar triangles 4QA0C 0 ∼ 4PDB which have the center of similarity Q. The figure also contains two triangles of equal area, which confirm Euclid’s theorem III.35 directly as an equality among areas. Since A0C 0 k BD, the triangles |4A0C 0 B| |4A0C 0 D|
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have equal base A0C 0 and equal height, and hence equal areas. Their intersection is the triangle 4QA0C 0 . We subtract this triangle from both sides, and obtain the pair of triangles |4A0 QB| |4C 0 QD| again of equal areas. Now the claim of Euclid III.35— |QA0 | · |QB| = |QC 0 | · |QD| can easily be obtained as an equality among areas. 20.3.
Some number theory inspired by Pythagoras’ theorem
Problem 20.5. In the figure on page 427 is shown the common notation for the six segments a, b, c and p, q, h which occur in a right triangle. Any two among the six segments a, b, c and p, q, h determine the other four ones. We assume that the hypothenuse c and the altitude h are given. Find explicit formulas for the legs a and b and the projections p and q in terms of the hypothenuse c and the altitude h. To determine the sign of the roots occurring in the formulas, we assume that a ≥ b. Answer. From the Pythagorean theorem a2 + b2 = c2 and the formula ab = ch, one obtains a + b. (a + b)2 = a2 + b2 + 2ab = c2 + 2ch √ a + b = c2 + 2ch From known product ab and sum a + b, one gets a quadratic equation for the unknown a. Thus one determines a and finally b. The sign of the roots is determined by the assumption that a ≥ b. √ a2 − a c2 + 2ch + ch = 0 √ √ √ √ c2 + 2ch + c2 + 2ch − 4ch c2 + 2ch + c2 − 2ch = a= 2 √ √2 2 2 c + 2ch − c − 2ch b= 2 Now p and q can be calculated by the leg theorem. √ 2 √ p c2 + 2ch + c2 − 2ch 2c2 + 2 (c2 + 2ch)(c2 − 2ch) a2 p= = = c 4c 4c √ √ 2 4 2 2 2 2 2c + 2 c − 4c h c + c − 4h = = and 4c 2 √ c − c2 − 4h2 q= 2 Lemma 20.1. For any solution of the equation a2 + b2 = c2 with natural numbers a, b, c it is impossible than a and b are both odd. Proof. Assume towards a contradiction that a and b are both odd and a2 + b2 = c2 . There exists a natural number n ≥ 0 such that a = 2n + 1. One checks that a2 = (2n + 1)2 = 4(n2 + n) + 1 ≡ 1 mod 4. Similarly one obtains b2 ≡ 1 mod 4. Hence c2 ≡ 2 mod 4 and c is even, but c2 is not divisible by 4, which is impossible.
431
Proposition 20.1. All solutions of the equation a2 + b2 = c2 with natural numbers a, b, c, for which b divisible by a higher prime power of 2 than a, are given by the formulas a = g(u2 − v2 ) , b = 2guv , c = g(u2 + v2 )
(20.1)
with integer g ≥ 1 and u > v ≥ 1. Moreover, if u and v are relatively prime and u + v is odd, then g = gcd(a, b, c) = gcd(a, b) = gcd(a, c) = gcd(b, c). Proof. It is enough to deal with the case that gcd(a, b, c) = 1. Hence gcd(a, b) = gcd(a, c) = gcd(b, c) = 1 holds, too. By lemma 20.1 one concludes that b is even, and a and c are odd. Hence b + c and c − b are both odd and gcd(c + b, c − b) = 1. Because of (c + b)(c − b) = c2 − b2 = a2 and the uniqueness of the prime decomposition, we conclude that the numbers c + b and c − b are both perfect squares. We conclude c + b = m2 , c − b = n2 and a = m · n where m > n are both odd natural numbers. Define the natural numbers m−n m+n and v := u := 2 2 Clearly u > v are both natural numbers, they are relatively prime and u + v = m is odd. We easily check that m2 − n2 = 2uv 2 m2 + n2 c= = u2 + v2 , and finally 2 a = m · n = u2 − v2 b=
to confirm that equations (20.1) hold with g = 1, as to be shown.
Corollary 49. The unit circle {(x, y) : x2 = y2 = 1} contains infinitely many points for which both coordinates are rational. Reason. One puts x=
u2 − v2 2uv and y = 2 2 2 u +v u + v2
with any integers u, v ∈ Z.
Remark. Using complex numbers, the equations (20.1) can be put together as a + ib = g(u + iv)2 Hence one gets once more c2 = (a + ib)(a − ib) = g2 (u + iv)2 (u − iv)2 = g2 (u2 + v2 )2 and c = g(u2 + v2 ) Corollary 50. In the ring Z + iZ of Gaussian integers, the number a + ib is a perfect square if and only if a2 + b2 is a perfect square.
432
Problem 20.6. Find all cases for which all six segments a, b, c and p, q, h have integer values, and gcd(a, b, c, p, q, h) = 1. Give formulas for these six segments in terms of integer solutions of a02 + b02 = c02 with gcd(a0 , b0 ) = 1. Proof. These integer solutions are a = a0 · c0 , b = b0 · c0 , c = c02 , p = a02 , q = b02 , h = a0 · b0
(20.2)
Here is a simple reason. From proposition 20.1 we know that a = a0 · g , b = b0 · g , c = c0 · g with g = gcd(a, c) = gcd(b, c). From ab = ch we get h=
ab (a0 b0 )g g = a0 b0 · 0 = 0 c c c
which is assumed to be an integer. Now gcd(a0 , c0 ) = 1 and gcd(b0 , c0 ) = 1 imply gcd(a0 b0 , c0 ) = 1. But c0 divides the product (a0 b0 ) · g. Hence Euclid’s lemma implies than c0 divides g. In other words, cg0 is an integer. The remaining lengths are p=
g g a2 = a02 · 0 and q = b02 · 0 c c c
We see that both are divisible by the integer cg0 , as are the lengths h as well as a and b. Hence the assumption gcd(a, b, c, p, q, h) = 1 implies that g = c0 . Finally, one obtains the formulas (20.2). Corollary 51. In the case that all six segments a, b, c and p, q, h have integer values, and gcd(a, b, c, p, q, h) = 1, all segments c, p, q and c + 2h and c − 2h are perfect squares. Reason. One see directly from the equations (20.2) that c, p, q are perfect squares. Moreover c + 2h = c02 + 2a0 · b0 = a02 + b02 + 2a0 · b0 = (a0 + b0 )2 and c − 2h = (a0 − b0 )2
433
Figure 20.6. Two equally good runners start at O and R. Do they better meet at point S or point T .
20.4.
Some (sandbox) applications
Problem 20.7. Two equally fast runners start at the opposite points O and R of the place shown in the figure on page 433. They are allowed to run across the place, but cannot enter any space outside the place. They want to meet on the boundary. Can they meet quicker at point S or at point T. (i) Calculate the distance |OS | for which |OS | = |S R|. (ii) Calculate the point T such that |OT | = |TC| + |CR|. (iii) Calculate the distance |OT | and decide whether which distance is shorter |OT | or |OS |. Answer. (i) Let S = (x, 0) be the coordinates of the meeting point S . Since the distances |OS | = |S R| are equal, one calculates p x = (10 − x)2 + 42 x2 = 100 − 20x + x2 + 16 x = 5.8 √ 2 (ii) Let the distance |OT | = x. Hence the coordinates of the meeting √ point are T = ( √x − 1, 1). Since the distances |OT | = |TC| + |CR| are equal, |TC| = 6 − x2 − 1 and |CR| = 32 + 42 = 5 gives √ x = 11 − x2 − 1 x2 − 1 = (11 − x)2 = 121 − 22x + x2 122 6 x= =5+ 22 11 (iii) We see that point T is where to meet quicker.
434
Figure 20.7. Proof of the parallelogram equation.
Remark. With T = (t, 1) √
t2 + 1 = 6 − t +
√
32 + 42 = 11 − t
t2 + 1 = 121 − 22t + t2 5 t =5+ 11 We get the same distance |OT | = |TC| + |CR| = (6 − t) + 5 = 5 + 20.5.
6 11
The parallelogram equation
Proposition 20.2. The sum of the squares of the four sides of a parallelogram equals the sum of the squares of the diagonals. Remark. Any real numbers satisfy (a + b)2 + (b − a)2 = 2a2 + 2b2 The corresponding equation holds for vectors in a space with an inner product. For a parallelogram, the vectors along the sides and diagonals are −−→ −−→ a := AB , −a := CD , −−→ −−→ b := BC = AD −−→ −−→ a + b = AC , b − a = BD In a vector space with an inner product, the corresponding identity (a + b)2 + (b − a)2 = 2a2 + 2b2 implies that the sum of the squares of the diagonals of a parallelogram equals twice the sum of the squares of two adjacent sides. In this way, we have an analytic proof of the parallelogram equation.
435
Problem 20.8 (Parallelogram equation). The sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its four sides. Prove the parallelogram equation |AC|2 + |BD|2 = 2|AB|2 + 2|BC|2 using the case and the notation of the figure on page 434. Proof. For the parallelogram ABCD, we may assume a = |AB| > |BC| = b, and the angle at B to be obtuse. We drop the perpendiculars from vertex C onto the extension of side AB, and from vertex B onto side CD. The foot points are F and G, respectively. Let the rectangle BFCG have side lengths |BG| = |CF| = h and |BF| = |CG| = k. Apply Pythagoras’ Theorem three times: |AC|2 = |AF|2 + |FC|2 = (a + k)2 + h2
for triangle 4AFC;
|BD| = |BG| + |GD| = h + (a − k)
for triangle 4BGD;
2
2
2
2
2
h2 + k2 = |BF|2 + |FC|2 = |BC|2 = b2
for triangle 4BFC.
Adding the first two equations yields |AC|2 + |BD|2 = (a + k)2 + (a − k)2 + 2h2 = 2a2 + 2k2 + 2h2 = 2a2 + 2b2 We see that the sum of the squares of the diagonals of parallelogram ABCD equals the sum of the squares of its four sides.
21. Trigonometry 21.1.
The law of sines
Definition 21.1 (sin function for acute and obtuse angles). The ratio ac of the leg a across the acute angle α to the hypothenuse of a right triangle is the sin function of this angle: sin α =
a c
For right and acute angles, one defines sin(180◦ − α) = sin α and
sin 90◦ = 1
Definition 21.2 (cosine function for acute and obtuse angles). The ratio bc of the leg b adjacent the acute angle α to the hypothenuse of a right triangle is the cos function of this angle: cos α =
b c
For right and acute angles, one defines cos(180◦ − α) = − cos α and
cos 90◦ = 0
Problem 21.1. Convince yourself that cos α = sin(90◦ − α) sin α tan α = cos α 1 = tan(90◦ − α) tan α tan(180◦ − α) = − tan α
(21.1) (21.2) (21.3) (21.4)
436
Figure 21.1. Proof of the sin theorem for acute, and for obtuse angles.
Problem 21.2. Convince yourself that sin2 α + cos2 α = 1
(21.5)
is equivalent to the Theorem of Pythagoras Theorem 21.1 (The extended sin Theorem). For any triangle with sides a, b, c and angles α, β, γ, and radius R of the circum-circle a b c = = = 2R sin α sin β sin γ Proof. This is an easy consequence of (Euclid III.20) and (Euclid III.21). As stated in theorem 15.4, the central angle is twice the circumference angle. We put the circum circle around the given triangle. Let its radius be R. We connect the endpoints of side a = BC with the center O, and drop the perpendicular from O this side. Let M be the foot point. The central angle is ∠BOC = 2α, since ∠BAC = α. It is bisected by the −−→ ray OM. The side a is bisected by point M. In the right triangle 4OMC, the definition of the sin function yields |CM| a = sin α = |OC| 2R Thus we obtain the required sin theorem for the acute angles of the triangle. For the obtuse angle, we need to use (Euclid III.22), telling that the opposite angles of a circular quadrilateral ABCA0 add up to two right angles. We can take for A0 any point on the arc opposite to arc BAC, and obtain ∠BA0C = 180◦ − α. This time the central angle ∠BOC = 2(180◦ − α). Now −−→ this angle is bisected by the ray OM and we obtain |CM| a sin(180◦ − α) = = |OC| 2R Hence, in order to obtain a sin theorem valid for both acute and obtuse triangles, one defines sin(180◦ − α) = sin α
437
21.2.
The law of cosines
Theorem 21.2 (The cosine theorem). For any triangle with sides a, b, c and angles α, β, γ c2 = a2 + b2 − 2ab cos γ
Figure 21.2. Proof of the cos theorem for acute, and for obtuse angles.
Proof. We drop the perpendicular from vertex A onto the opposite side a. Let D the foot point. The Pythagorean theorem for triangles 4ADC and 4ADB yields, after subtracting b2 = h2 + |DC|2 c2 = h2 + |DB|2 c2 = b2 − |DC|2 + |DB|2 We need now to distinguish the case of an acute and obtuse triangle. Take an acute triangle. By definition of the cosin function cos γ = |DC| b , and hence |DC| = b cos γ. From the drawing, we see that |DC| + |DB| = a. Hence |DB| = a − |DC| = a − b cos γ |DB|2 = a2 − 2ab cos γ + b2 cos2 γ c2 = b2 − |DC|2 + |DB|2 = b2 − b2 cos2 γ + a2 − 2ab cos γ + b2 cos2 γ c2 = a2 + b2 − 2ab cos γ as required. For the obtuse triangle, we get |DB| − |DC| = a. By definition of the cos function ◦ cos(180◦ − γ) = |DC| b , and hence |DC| = b cos(180 − γ). From the drawing, we see that a = |DB| − |DC|. Hence |DB| = a + |DC| = a + b cos(180◦ − γ) |DB|2 = a2 + 2ab cos(180◦ − γ) + b2 cos2 (180◦ − γ) c2 = b2 − |DC|2 + |DB|2 = b2 − b2 cos2 (180◦ − γ) + a2 + 2ab cos(180◦ − γ) + b2 cos2 (180◦ − γ) c2 = a2 + b2 + 2ab cos(180◦ − γ) Hence, in order to obtain a cos theorem valid for both acute and obtuse triangles, one defines cos(180◦ − α) = − cos α
438
Corollary 52 (Pythagorean comparison). For acute angle γ < 90◦ , the sides of a triangle satisfy c2 < a2 + b2 . For obtuse angle γ > 90◦ , the sides of a triangle satisfy c2 > a2 + b2 . Corollary 53 (The converse Pythagorean Theorem). If the sides of a triangle satisfy c2 = a2 + b2 in the sense of segment arithmetic, then the angle across side c is a right angle. Problem 21.3. For a triangle are given the sides a = 10 and c = 6, and the angle γ = 30◦ across the latter side. Find angle α, side b and finally angle β. How many non-congruent solutions do you get? Answer. The sin Theorem yields sin α =
a sin γ 5 = c 6
Hence we get two supplementary angles α1 = 56.44◦
and α2 = 180◦ − α1 = 123.56◦
as possible solutions. The side b is obtained with the cos Theorem: √
c2 = a2 + b2 − 2ab cos γ
b2 − 10 3b + 64 = 0
√ √ √ √ 10 3 ± 300 − 256 b1,2 = = 5 3 ± 11 2 Question. It is not yet clear how these results combine, and whether there are two or even four solutions. Provide a drawing to decide this question. √ √ 3 − 11 = 5.3436 combines only with the obtuse Answer. One sees that the shorter side b = 5 2 √ √ angle α2 . The longer side b1 = 5 3 + 11 = 11.977 combines only with the acute angle α1 . Hence one obtains two non-congruent solutions. Finally, the angle β can be obtained from the angle sum. One gets β1 = 180◦ − α1 − γ = 93.548◦ β2 = 180◦ − α2 − γ = 26.45◦ Question. Obtain the angle β with the sin Theorem. Which problem does arise? Answer. The sin Theorem yields √ √ b sin γ 5 3 ± 11 sin β = = c 12 One cannot yet know whether the acute or obtuse solution is valid. Indeed β2 = 26.44 is acute and β1 = 93.56 is obtuse. Problem 21.4. Use the cosine theorem to give another proof of the parallelogram equation from proposition 20.2. Another proof of the parallelogram equation. Refer to the figure on page 208. Given is a parallelogram ACBD with two pairs of opposite parallel sides AC k BD and BC k DA. Using congruent z-angles and ASA congruence, one sees that the opposite sides are pairwise congruent. Let M be the intersection point of the diagonals. We know already by problem 7.16 that the diagonals of the parallelogram bisect each other.
439
Figure 21.3. For the given data, there are two non-congruent solutions.
We apply the cosine theorem to the triangles 4ACM and 4DCM: |AC|2 = |AM|2 + |CM|2 − 2|AM| · |CM| · cos θ |AD|2 = |AM|2 + |DM|2 − 2|AM| · |DM| · cos η The angle θ = ∠AMC and η = ∠DMC are supplementary and hence cos θ = − cos η. We know already by problem 7.16 that |CM| = |DM|. Hence adding the relations yields |AC|2 + |AD|2 = 2|AM|2 + 2|CM|2 Similarly, we prove that |BC|2 + |BD|2 = 2|BM|2 + 2|CM|2 Since the diagonals of the parallelogram bisect each other we know that 2|AM|2 + 2|BM|2 = |AB|2 and 4|CM|2 = |CD|2 . Hence adding the two equations yields |AC|2 + |AD|2 + |BC|2 + |BD|2 = 2|AB|2 + 2|CD|2
as to be shown.
Problem 21.5. Get an exact expression for the diameter 2R of the circum circle in terms of the three sides a, b, c of a triangle. Proof. The cos Theorem yields c2 = a2 + b2 − 2ab cos γ a 2 + b 2 − c2 2ab The extended sin theorem gives the diameter of the circum circle c c 2abc 2R = = p = p 2 2 2 sin γ 1 − cos γ 4a b − (2ab cos γ)2 2abc 2abc = p = p p 4a2 b2 − (a2 + b2 − c2 )2 (2ab + a2 + b2 − c2 ) (2ab − a2 − b2 + c2 ) 2abc = p p 2 ((a + b) − c2 ) (c2 − (a − b)2 ) 2abc = √ ((a + b + c)(a + b − c)(c − a + b)(c + a − b)) cos γ =
440
21.3.
The addition theorem for tangent
Problem 21.6 (The tangent addition Theorem). Prove the tangent addition theorem tan(x + y) =
tan x + tan y 1 − tan x tan y
(21.6)
for any acute or obtuse angles x, y with acute or obtuse sum x + y.
Figure 21.4. Proof of the tangent addition theorem.
Proof. We prove the addition theorem at first for any acute or obtuse angles x, y with acute or obtuse sum γ = x + y. We assume the perpendicular |FC| = 1 has unit length and the two right triangles have angles x = ∠ACF and y = ∠FCB. From them together, we get the triangle 4ABC as shown in the drawing on page 440. Question. What are the lengths of the sides of the triangle 4ABC. Answer. a = |BC| =
1 1 , b = |CA| = and c = |AB| = tan x + tan y cos y cos x
Question. Write the cosine Theorem for side c in terms of x and y and solve for cos(x + y). Answer. c2 = a2 + b2 − 2ab cos γ 1 1 2 cos(x + y) (tan x + tan y)2 = + − 2 2 cos x cos y cos x cos y " # 1 cos x cos y 1 2 cos(x + y) = + − (tan x + tan y) 2 cos2 x cos2 y
441
I use the abbreviation X = tan x, Y = tan y and cos2 x = 1/(1 + tan2 x) to eliminate the cosine function. i cos x cos y h cos(x + y) = 2 + X 2 + Y 2 − (X + Y)2 = cos x cos y [1 − XY] 2 (1 + X 2 )(1 + Y 2 ) 1 + tan2 (x + y) = (1 − XY)2 (1 + X 2 )(1 + Y 2 ) − (1 − XY)2 X 2 + Y 2 + 2XY tan2 (x + y) = = (1 − XY)2 (1 − XY)2 X+Y tan(x + y) = ± 1 − XY Question. Which sign has 1 − XY for acute angle x + y, which for obtuse angle x + y. What happens for right angle x + y. Answer. 1 − XY > 0 for acute angle x + y, and 1 − XY < 0 for obtuse angle x + y. If and only if x + y is a right angle do we get 1 − XY = 0 since this is the case where the two right triangles 4ACF ∼ 4CBF are similar. Question. How do we justify tan(180◦ − α) = − tan α (21.7) Answer. From the sin and cosine Theorems we have obtained sin(180◦ − α) = sin α , cos(180◦ − α) = − cos α for both obtuse and acute angles. Because of the definition sin α cos α which we now extend to obtuse angles, we get equation (21.7). Question. Which sign has tan(x + y) for acute angle x + y, which for obtuse angle x + y. What happens for right angle x + y. Answer. tan(x + y) > 0 for acute angle x + y, and tan(x + y) < 0 for obtuse angle x + y. If x + y is a right angle, we get tan(x + y) = ∞ because of sin 90◦ = 1 and cos 90◦ = 0. 1 From these considerations of the signs, we see that the same formula tan x + tan y tan(x + y) = (21.6) 1 − tan x tan y holds for any acute or obtuse angles x, y with acute or obtuse sum x + y. tan α =
21.4. Unwound angles Unwound angles occur in the study of the trigonometric functions, physical oscillations, the theory of hyperbolic area and other contexts. The unwound angles have the structure of an ordered Abelian group. This group actually depends on the Hilbert plane considered. For a precise definition let Acute denote the set of the congruence classes for the zero and the acute angles. Let R denote the congruence class of right angles. Definition 21.3. The group of unwound angles is the Cartesian product Z × Acute with the operation of addition defined by setting if α + β < R (m + n, α + β) (m, α) + (n, β) = (21.8) (m + n + 1, α + β − R) if α + β ≥ R 1
The meromorphic nature of the tangent function makes a definition meaningful.
442
Proposition 21.1. In any Hilbert plane, the unwound angles have the structure of an ordered Abelian group. This group actually depends on the Hilbert plane considered. There is a natural homomorphism from the group of unwound angles onto the group of rotations around the origin. The kernel of this homomorphism is 4Z × {0} = {(4n, 0) : n ∈ Z} in other words the angles which are integer multiples of 360◦ . 21.5. Extension to unwound angles Up to now we have defined the sin, cosine and tangent functions for acute and obtuse angles, including the degenerate cases of 0◦ and 180◦ . For the tangent function, we need to allow the infinite value tan 90◦ = ∞. By allowing ∞ as a value, we extend the tangent function to all unwound angles: Problem 21.7. Convince yourself that tan(90◦ + x) = −
1 tan x
(21.9)
holds for 0◦ ≤ x ≤ 90◦ . Answer. The addition theorem (21.6) for the tangent function implies tan(45◦ + x) =
1 + tan x 1 − tan x
holds for 0◦ ≤ x ≤ 135◦ . Iterating the relation implies tan(90◦ + x) =
1 1 + tan(45◦ + x) 1 − tan x + (1 + tan x) = =− ◦ 1 − tan(45 + x) 1 − tan x − (1 + tan x) tan x
The tangent function is defined for all x by requiring the relation (21.9) for all x. Proposition 21.2. The sin and cosine functions can be extended to all unwound angles by setting cos(180◦ + x) = − cos x , sin(180◦ + x) = − sin x In that way, cos x becomes an even function with period 360◦ , and sin x becomes an odd function with period 360◦ . By allowing ∞ as a value, we extend the tangent function to all unwound angles: In this way, tan x becomes an odd function with period 180◦ . Lemma 21.1. We use the imaginary unit i which satisfies i2 + 1 = 0. (1 + i tan x)2 1 + tan2 x 2 tan x cos 2x = 1 + tan2 x 1 − tan2 x sin 2x = 1 + tan2 x
cos 2x + i sin 2x =
(21.10) (21.11) (21.12)
holds for all unwound angles x. With the obvious interpretation of the fractions with the value ∞ for the tangent function, all values are included.
443
Reason. Convince yourself that 1 + i tan 2x cos 2x + i sin 2x = √ for − 90◦ < 2x < 90◦ (21.13) 2 1 + tan 2x can be checked from the definition. But the positive sign of the root enforces the restriction of the domain. We can use the addition theorem of the tangent and conclude that 2 tan x for 1 − tan2 x Together the equations (21.13) and (21.14) imply tan 2x =
− 180◦ < 2x < 180◦
(21.14)
1 − tan2 x + 2i tan x 1 + i tan 2x (1 + i tan x)2 = p cos 2x + i sin 2x = √ = 1 + tan2 x 1 + tan2 2x (1 − tan2 x)2 + 4 tan2 x for −90◦ < 2x < 90◦ . But we can extend the domain of validity of this equation since it does not contain the root any longer. From the way the extensions of the functions have been defined we get cos 2(90◦ + x) + i sin 2(90◦ + x) = − cos 2x − i sin 2x =−
(1 + i tan x)2 (tan x − i)2 (1 − i/ tan x)2 (1 + i tan(90◦ + x))2 = = = 1 + tan2 x tan2 x + 1 1 + 1/ tan2 x 1 + tan2 (90◦ + x)
Hence equation (21.10) still is valid for −90◦ < 2x < 180◦ . The formula is extended to −180◦ < 2x < 180◦ by the symmetry, and then to all x by the periodicity properties of the trigonometric functions. Proposition 21.3. The extended cosine and sin function satisfy the addition theorem cos(x + y) + i sin(x + y) = (cos x + i sin x)(cos y + i sin y) cos(x + y) = cos x cos y − sin x sin y sin(x + y) = cos x sin y + sin x cos y The extended tangent function satisfied the addition theorem tan x + tan y tan(x + y) = 1 − tan x tan y
(21.15) (21.16) (21.17)
(21.6)
Reason. The addition theorem for the tangent function is extended via formula (21.9). From the original formula valid in the domain 0 ≤ x, y, x + y ≤ 180◦ one gets 1 tan x tan y − 1 = tan(x + y) tan x + tan y tan(90◦ + x) + tan y tan x −1 + tan x tan y = 1 − tan(90◦ + x) tan y tan x tan x + tan y
tan(90◦ + x + y) = −
which are equal. With a similar step for y, one extends the domain of validity of formula (21.6) to the domain 0 ≤ x, y ≤ 180◦ . Now periodicity implies that the formula holds for all x, y. The addition theorems for sin and cosine follow now via formula (21.10): [1 + i tan(x + y)]2 [1 − tan x tan y + i(tan x + tan y)]2 = 1 + tan2 (x + y) (1 − tan x tan y)2 + tan2 (x + y) [(1 + i tan x)(1 + i tan y)]2 (1 + i tan x)2 (1 + i tan y)2 = = · (1 + tan2 x)(1 + tan2 y) 1 + tan2 x 1 + tan2 y = (cos x + i sin x)(cos y + i sin y)
cos2(x + y) + i sin 2(x + y) =
444
21.6. Mollweid’s formulas Given any triangle 4ABC, the radius of the circum-circle is denoted by R and the radius of the in-circle by ρ. The radii of the ex-circles touching sides a,b or c, respectively, are denoted by ρa , ρb and ρc . Proposition 21.4 (The sin Theorem and the in-circle). For any triangle with sides a, b, c and angles α, β, γ across them
2R =
ρ 2 sin
α 2
sin β2 sin γ2
b c a = = = 2R (21.18) sin α sin β sin γ ρa ρb ρc = = = β γ β γ α α α 2 sin 2 cos 2 cos 2 2 cos 2 sin 2 cos 2 2 cos 2 cos β2 sin γ2 (21.19)
Proof. As in the proof of proposition 19.2, one gets cot
α |AF| = 2 |FI|
and
cot
β |FB| = 2 |FI|
and hence
c |AB| α β = = cot + cot ρ |FI| 2 2
Now one uses the addition theorem and α + β + γ = 180◦ to get sin α2 sin β2 sin α2 sin β2 2 sin α2 sin β2 sin γ2 ρ = = = c cos α sin β + sin α cos β sin γ sin α+β 2 2 2 2 2 To deal with the ex-circles, one derives at first the formula c α β = cot − tan ρa 2 2
and proceeds quite similarly.
Theorem 21.3 (Mollweid’s formulas and the tangent theorem). Sides and angles of any triangle satisfy a − b sin(α − β)/2 = c sin(α + β)/2 a + b cos(α − β)/2 = c cos(α + β)/2 a + b tan(α + β)/2 = a − b tan(α − β)/2
(21.20) (21.21) (21.22)
Reason for formula (21.20). Use several addition theorems, the angle sum and the cosine theorem three times: sin(α − β)/2 2 [sin(α − β)/2] · [sin(α + β)/2] cos β − cos α cos β − cos α = = = sin(α + β)/2 1 + cos γ 2 cos2 (γ/2) 2 sin2 (α + β)/2 b(2ac cos β) − a(2bc cos α) b(a2 + c2 − b2 ) − a(b2 + c2 − a2 ) = 2abc + c(2ab cos γ) 2abc + c(a2 + b2 − c2 ) (b − a)(c2 − (a + b)2 ) a − b = = c c((a + b)2 − c2 ) =
445
Reason for formula (21.21). Use the addition theorem, the angle sum, the sin theorem twice and then the cosine theorem: sin(α − β) sin α cos β − cos α sin β a cos β − b cos α 2ac cos β − 2bc cos α = = = sin(α + β) sin γ c 2c2 2 2 2 2 2 2 2 2 a +c −b −b +c +a a −b = = 2 2c c2 Now use the half-angle formula and divide by formula (21.20): 2 [sin(α − β)/2] · [cos(α − β)/2] (a − b)(a + b) = 2 [sin(α + β)/2] · [cos(α + β)/2] c2 cos(α − β)/2 a + b = cos(α + β)/2 c Formula (21.22). —results from dividing formula (21.20) by formula (21.21)
22. Measurement of the Circle
Figure 22.1. Archimedes’ approximation of a circular arc
22.1. Approximate measurement of a circular arc In the interior of the angle θ = ∠AOB, we draw a circular arc of radius r = 1. We want to get lower and upper bounds for the length of
446
this arc. Indeed, we want to construct a convergent sequence of more and more precise lower and upper bounds. As a first lower bound, one can simply use the length a1 = |AB| of the chord. To get an upper bound, we draw the tangents to the circle at the endpoints A and B of the arc. They intersect at point C. The length of the union of the segment AC and CB give the first upper bound b1 = |AC| + |CB|. −−→ Question. Why is OC the bisector of angle ∠AOB. Question. Get the formulas for a1 and b1 . From which right triangles are they obtained. Answer. Since |OA| = |OB| = 1, we get from right triangles 4OMA and 4OAC θ θ and b1 = 2 tan 2 2 To get the second, and better lower and upper bounds, we draw the bisector OC of the given a1 = 2 sin
_
angle ∠AOB, and mark the point D where it intersects the circular arc AB. The construction of the _
_
first lower and upper bounds is now repeated for the two arcs AD and DB. Since we have doubled the polygons for half the angle, we obtain a2 = 4 sin
θ θ and b2 = 4 tan 4 4
Proposition 22.1. For a short arc θ of the unit circle, the lengths an of the inscribed polygon with n sides, and the lengths the bn of the circumscribed polygon with 2n sides are calculated recursively: The initial values are θ θ a1 = 2 sin and b1 = 2 tan 2 2 and the two sequences an and bn with n = 1, 2, . . . satisfy the recursion r r bn − an bn − an n+1 n+1 an+1 := 2 , bn+1 := 2 (22.1) 2 bn bn + an Proof. Question. Give the trigonometric formulas for an and bn . θ θ and bn = 2n tan n 2n 2 To check the recursion, we use the trigonometric formulas r r r 1 + cos α α 1 − cos α α 1 − cos α α cos = , sin = , tan = 2 2 2 2 2 1 + cos α an = 2n sin
which hold for α ∈ [0, π] with positive roots.
(22.2)
Problem 22.1. Convince yourself that lengths of the polygons which approximate the circular arc from inside and outside satisfy a1 < a2 < · · · < an < an+1 < · · · < bn+1 < bn < · · · < b2 < b1
(22.3)
Proof. Question. In which triangles does the triangle inequality yield a1 < b1 , a1 < a2 and b2 < b1 . Answer. From the triangle inequality in triangle 4ABC, we obtain a1 < b1 . From the triangle inequality in triangle 4ABD, we obtain a1 < a2 . From the triangle inequality in triangle 4EFC, we obtain b2 /2 = |EF| < |EC| + |CF| = b1 − (b2 /2) and hence b2 < b1 .
447
Finally, we get a2 < b2 and hence together a1 < a2 < b2 < b1 . Repeated bisections yields two sequences of polygons which approximate the circular arc from inside and outside. Their lengths satisfy a1 < a2 < · · · < an < an+1 < · · · < bn+1 < bn < · · · < b2 < b1
For the angle θ = 60◦ one gets the initial values
a1 := 1 and
√ 2 3 b1 := 3
for the sequences with limit lim an = lim bn =
n→∞
n→∞
π 3
For a calculation of π all an and bn need to be multiplied with three.
Problem 22.2.
An+1
A1 := 3 and r Bn − An n+1 := 3 · 2 , 2Bn
√ B1 := 2 3 r Bn+1 := 3 · 2
n+1
Bn − An An + Bn
Compute numerically and report the values of An and Bn for n = 1, 2, . . . , 15. Find a conjecture about the trend for the errors Bn − π and π − An . Which value of n gives the best result for π. In which interval can you say π lies for sure?
Answer. As a trend for the errors, one sees π − An (Bn − π)/2 and Bn+1 − π (π − An )/2. The trend breaks down for n = 11, where the rounding errors makes it impossible to further improve the result. The result for n = 10 is still reliable. Multiplication with 3 · 210 = 3 072 makes the last three digits random—but one still gets that 3.141592 < π < 3.141594.
448
n 3 · 2n 1 6 2 12 3 24 4 48 5 96 6 192 7 384 8 768 9 1 536 10 3 072 11 6 144 12 12 288 13 24 576 14 49 152 15 98 304
Bn An 3.464101615 3 3.215390309 3.105828541 3.159659942 3.132628613 3.146086215 3.139350203 3.1427146 3.141031951 3.14187305 3.141452472 3.141662747 3.141557608 3.141610177 3.141583892 3.141597036 3.141590465 3.141593778 3.141592135 3.141592836 3.141592425 3.141592234 3.141592131 3.141592458 3.141592432 3.141586207 3.141586201 3.141601567 3.141601565
Bn − π π − An .3225089615 .1415926536 .0737976556 .0357641124 .0180672885 .0089640403 .0044935615 .0022424506 .0011219461 .0005607026872 .0002803964226 .0001401812718 .0000700935935 .0000350455515 .000017523088 .0000087613684 .0000043824205 .0000021886731 .0000011239655 .0000005188065 .0000001824518 .000000228241 −.0000004197604 .0000005224336 −.0000001956592 .0000002213274 −.000006446526 .0000064529797 .000008913278 −.0000089116739
Lemma 22.1. The difference of the upper and lower bounds is estimated from above bn − an <
an 4b1 (b1 − a1 ) (b1 − a1 ) < 4n a1 4n−1 a1
(22.4)
Problem 22.3. Explain how to get the estimate (22.4). Reason. Question. Check the following formula: −2 −n a−2 n − bn = 4
Question. We define the quotients qn := check the recursion formula
bn an .
Use the trigonometric formula cos2
q2n+1 =
2qn 1 + qn
α 2
=
1+cos α 2
to
449
Question. Prove that 0 ≤ qn+1 − 1 < (qn − 1)/4 for all n ≥ 1. Answer. Since 0 < an < bn division implies qn > 1 for all n ≥ 1. 2qn qn − 1 −1= qn + 1 qn + 1 qn − 1 qn − 1 < qn+1 − 1 = (qn + 1)(qn+1 + 1) 4 q2n+1 − 1 =
The estimate of qn implies bn+1 − an+1 = an+1 (qn+1 − 1) <
an+1 an+1 (qn − 1) = (bn − an ) 4 4 an
For all n ≥ 1. Finally one gets the result by induction.
Problem 22.4. Use the estimate (22.4) and explain which continuity axioms are needed to conclude that both sequences bn and an converge to the same limit. Answer. As far as the continuity axioms are concerned, one needs to assume the Archimedean axiom (V.1) and Cantor’s principle of boxed intervals (V.2). I have stated these axioms of continuity 14.1 among a simplified axiomatization of geometry 14.1. Because of the estimates (22.3), Cantor’s principle of boxed intervals implies there exists a length x∗ such that a1 < a2 < · · · < an < x∗ < bn < · · · < b2 < b1 for all natural n.
(22.5)
Moreover, the estimate (22.4) together with the Archimedean axiom (V.1) imply that the number x∗ is unique. Indeed, assume that both x∗ and x satisfy the estimate (22.5). Hence |x∗ − x| ≤ bn − an <
4b1 (b1 − a1 ) = K · 4−n 4n a1
hold for all natural n. This is only possible if x = x∗ . Reason. One proves by induction that N < 2N ≤ 4N holds for all natural N. Assume towards a contradiction that x∗ , x. By the Archimedean axiom there would exist a natural number N such that K < N|x∗ − x|. Hence |x∗ − x| < N|x∗ − x|4−n and 4n < N would hold for all natural n. This is false even just for n = N. Remark. The axioms of continuity used above are my own favorites since they can be used directly in the proofs. Moreover they are the axioms of continuity contained in the very first edition of Hilbert’s Foundations of Geometry. 1 22.2.
Arc length and area of a circular sector
Definition 22.1 (Pi). The arc length of a semicircle of radius one is defined to be the number π. Definition 22.2 (Measurement of angle by arc length). The arc length of a circular arc of radius one and central angle θ is defined as the arc measurement of the central angle. Proposition 22.2. Assume that the Archimedean Axiom holds. The measurement of angles by degrees and arc length are proportional. They are related by θ[arc] = 1
π θ[◦ ] 180◦
and
θ[◦ ] =
180◦ θ[arc] π
In later editions of Hilbert’s foundations, his (rather convoluted) axiom of completeness is postulated.
(22.6)
450
Measurement of angle by arc length. The arc length of the circular arc with radius one and central angle θ = ∠AOB is obtained from the common limit of the sequences bn and an θ[arc] = lim an = lim bn n→∞
n→∞
Consider at first the special case that the angle θ = p2−n · 90◦ with integer p and integer n ≥ 0. In other words, one begins with a right angle, and uses repeated bisection and takes any integer multiples of the angles one has thus obtained. In all these cases, the simple proportionality given by equation (22.6) follows directly from the definition of the number π. The general case is obtained using the process for measurement of angles explained in theorem 10. Such a measurement process presupposes the Archimedean property for angles, stated in theorem 8.1. Indeed any angle may be approximated, with arbitrary good accuracy by a sequence of angles θ = p2−n · 90◦ . Hence the proportionality given by equation (22.6) remains valid for any arbitrary angle. Theorem 22.1 (Circumference and area of a circular arc). A circular arc with radius r and 2 central angle θ, to be measured in arc length, has the arc length θ r and the area θ r2 . Theorem 22.2 (Circumference and area of a circle). The circle of radius r has the circumference 2πr and the area πr2 . Corollary 54 ("pizza theorem"). The area of a circle is the product of the radius with half of the circumference. Problem 22.5. Explain the nickname "pizza theorem". The arc length of a circular sector. The approximate measurement of the arc length of a circular arc of radius r is performed as explained in the previous section. One obtains the arclength from the common limit lim an · r = lim bn · r = θ[arc] · r n→∞
n→∞
because the measurement of angles by arc length from definition 22.2.
Figure 22.2. Measuring the area of a regular polygon.
Lemma 22.2. The area of a regular polygon is the product of the radius ρ of the in-circle with half perimeter s. The area of a regular polygon circumscribed around a circular sector equals the radius times half the length of the polygon.
451
Proof. The reasoning is explained for a 7-gon by the figure on page 450. The 7-gon is partitioned into 7 congruent isosceles triangles 4ABO through 4GAO, where point O is the center of the circum-circle. These triangle are rearranged into a trapezoid A0 E 0 O00 O0 . One gets a trapezoid from an n-gon with an odd number n of sides, and one gets a parallelogram from an n-gon with an even number of sides. In both cases, in order to obtain a rectangle, one still needs to bisect the first triangle 4ABO into two right triangles, and shift one of them to the opposite side E 0 O00 . The rectangle M 0 E 0 LO0 that one has obtained, has the side lengths |M 0 O0 | = ρ which is the radius of the in-circle of the original polygon, and |M 0 E 0 | = s which is half of the perimeter of the original polygon. The area of a circular sector. To obtain upper- and lower bounds for the area of a circular sector, one uses the areas of the regular polygons circumscribed and inscribed into the circular sector. By lemma 22.2, the area of a regular polygon circumscribed around a circular sector equals the radius r times half the length of the polygon. Hence one gets the area bn r 2 as an upper bound of the area of the circular sector. To obtain a lower bound, we use the area of a regular polygon inscribed into the circular sector. The arc length of the inscribed polygon is an · r. The radius of the circle inscribed into the inscribed polygon is abnn · r. Hence one gets the area r·
an r an r · bn 2 as an lower bound of the area of the circular sector. Using the formulas (22.2) and the identity sin α = 2 sin α2 cos α2 , one can simplify an r an r a2n r2 θ θ r2 · = · = 2n sin n cos n · bn 2 bn 2 2 2 2 2 2 θ an−1 r r = 2n−1 sin n−1 · = 2 2 2 Altogether, we have obtained the bounds an−1 r2 bn r 2 < area of circular sector < 2 2 and in the limit n → ∞, we obtain area of circular sector = lim an · n→∞
r2 r2 θr2 = lim bn · = 2 n→∞ 2 2
with the central angle θ measured in arc length. Remark. A particular simple "proof" of the pizza theorem is as follows: We know that for a regular n-gon, the area is one-half the product of the radius r of the in-circle to the circumference. "Now image that we make the number of sides of the polygon infinite. Each triangle has height r and an infinitesimal base dr. Since the polygon fills the circle, their areas are equal. Also, the perimeter of the polygon equals the circumference of the circle. Thus the formula in the pizza theorem follows from the corresponding formula for the regular polygons with any number n of sides."
452
This idea is usually attributed to Johannes Kepler, but similar arguments were already used by Democritus and other early Greek mathematicians. Because of the ancient Greeks "horror of the infinite" it should come as no surprise that no important mathematician in ancient times,—even those who devised such proofs,—did considered them to be rigorous. The same can be said of all mathematicians since then, including contemporary ones. Currently, we make the proof acceptable as done above: We give lower and upper estimates for the circumference and area. We use a "squeezing argument" which takes advantage of the fact that the lower and upper estimates have the same limit for n → ∞. In this way the result is obtained from a limit n → ∞ for which the number of sides of the polygon approach infinity. To be even more exact, one has to point out,—as we have done above,—which axioms of continuity are involved to confirm existence and uniqueness of such a limit. The Greeks did not use these terms. But they had their method of exhaustion that was quite similar to the modern rigorous use of limits. This method was first clearly articulated by Eudoxus around 370 B.C. and extensively used by Archimedes. However, without the modern calculus, the method of exhaustion gives no indications how to discover formulas for areas and volumes in the first place. It just needed the genius of Archimedes for every such result! But the mystery about the way Archimedes worked was (at least partly) resolved when a document by Archimedes called "The Method" was discovered around 1903. This document describes an ingenuous procedure in which one or more solids could be mentally sliced into infinitesimally thin pieces. By achieving a balance piece by piece, and using the basic physical law of levers, Archimedes was able to find the volume of the sphere, and similar results. This method of equilibrium is one of the cleverest integration tricks ever devised. As yet, Archimedes did not even mention,—let alone use,—the method in any of his other works, because he did not consider it to be rigorous.
453
23. Area in Euclidean Geometry 23.1.
Equidecomposable and equicomplementable figures
Problem 23.1. Draw two parallelograms on the same base with congruent altitudes, and explain why they are equidecomposable . Clearly draw the dissections you use.
Figure 23.1. A square and a parallelogram that are easily seen to be equidecomposable .
U Answer. In the figure on page 453, U the square is decomposed as 1 2. The parallelogram on the same base is decomposed as 2 3. Since the triangles 1 3 are congruent, the square and the parallelogram as equidecomposable . Problem 23.2. Draw two parallelograms on the same base with congruent altitudes that are not easily seen to be equidecomposable . Explain why they are equicomplementable . Clearly draw the dissections and mark the additional figures used.
Figure 23.2. A square and a parallelogram that are only shown to be equicomplementable .
U Answer. In the figure on page U 453, the square is decomposed as 1 2. The parallelogram on the same base is decomposed as 2 3. But this time 1 and 3 are not congruent. To see that the square and the parallelogram are equicomplementable , we use the triangle 4 as additional figure. Since the unions ] ] 1 43 4 are indeed congruent, we conclude ] ] ] ] ] ] ] square 4∼ 1 2 4∼2 1 4 ∼2 3 4 ] ] ] ∼ 2 3 4 ∼ parallelogram 4 Hence the square and the parallelogram—with the same base and congruent altitudes—are equicomplementable . Similarly, we prove that any two parallelograms with the same base and congruent altitudes are equicomplementable .
454
We can now give the exact modern version for the propositions about area in Euclid’s book one. Proposition 23.1 (Instead of Euclid I.35). Two parallelograms on the same base with congruent altitudes are equicomplementable . Proposition 23.2 (Instead of Euclid I.36). Two parallelograms with congruent bases and congruent altitudes are equicomplementable . Problem 23.3. Explain why two parallelograms with the same base and congruent altitudes are always equicomplementable . Proposition 23.3 (Instead of Euclid I.37). Two triangles with the same base and congruent altitudes are equicomplementable . Remark. Euclid talks about triangles "in the same parallels". Euclid I.37 says literally that "Triangles on the same base in the same parallels are equal." Proposition 23.4 (Instead of Euclid I.38). Two triangles with congruent bases and congruent altitudes are equicomplementable . Problem 23.4. Convince yourself by an informal argument that two triangles with the same base and congruent altitudes are equicomplementable . Problem 23.5. Use the neutral version of Euclid I.37 given by proposition 13.1 above to prove in Euclidean geometry that two triangles with the same base and congruent altitudes are equicomplementable . Answer. In Euclidean geometry, but only in Euclidean geometry, two triangles with the same base and congruent altitudes have the same midline. (Work out further details!) Hence by proposition 13.1 above they are equicomplementable , too. 23.2. The Theorem of Pythagoras and related results As a first step towards the Pythagorean Theorem, Euclid proves the following version of the leg theorem: Proposition 23.5 (Euclid’s leg theorem). The square on a leg of a right triangle is equicomplementable to the rectangle one side of which is the hypothenuse and the other side is the projection of this leg onto the hypothenuse. Modern proof following Euclid. For a given right triangle 4ABC, we have constructed the square ACED over the leg AC and the square ABIH over the hypothenuse AB. We have dropped the altitude from vertex C onto the hypothenuse and got the projection AF of the leg AC. Finally we have obtained the rectangle AFGH, one side of which is the hypothenuse and the other side is the projection AF of the leg AC onto the hypothenuse AB. To show that the square ACDE and the rectangle AFGH are equicomplementable , we use as intermediate steps the two parallelograms ABJD and AHKC. Here point J can be obtained by transferring the leg AD BJ, and point K can be obtained by transferring the altitude FC GK. Question. Convince yourself that both quadrilaterals ABJD and AHKC are indeed parallelograms. Question. Convince yourself that these two parallelograms are congruent. How can one easily obtain parallelogram AHKC from parallelogram ABJD. Question. Why is the square ACED equicomplementable to the parallelogram ABJD.
455
Figure 23.3. Euclid’s leg theorem tells the square over the leg is equicomplementable to the rectangle as indicated.
Figure 23.4. Euclid’s proof of the leg theorem.
Answer. These two parallelograms have the same base AD and congruent altitudes. Hence they are equicomplementable by Euclid I.37 (see proposition 23.3). Question. Why is the parallelogram ACKH equicomplementable to the rectangle AFGH. Answer. These two parallelograms have the same base AH and congruent altitudes. Hence they are equicomplementable by Euclid I.37 (see proposition 23.3).
456
By these steps, we have obtained the following chain of equicomplementable figures: ACED , ABJD ACKH , AFGH By Theorem 13.1, any two congruent figures are equicomplementable . Furthermore, the relation "equicomplementable " is an equivalence relation among figures. Hence we conclude that ACED , AFGH
as to be shown.
Proposition 23.6 (Euclid’s (weaker) Pythagorean theorem). The sum of the squares on the legs of a right triangle is equicomplementable to the square over the hypothenuse.
Figure 23.5. Euclid’s proof of the Pythagorean Theorem.
Modern proof following Euclid. We use Euclid’s leg theorem 23.5 twice, for both legs: ACED , AFGH
and BCK J , BFGI
Since these figures do not overlapping we conclude ] ] ACED BCK J , AFGH BFGI , ABIH as to be shown.
Theorem 23.1 (The Pythagorean Theorem). The sum of the squares on the legs of a right triangle is equidecomposable to the square over the hypothenuse. Problem 23.6. We want to prove the Theorem of Pythagoras by dissection. What went wrong in the above figure on page 457 to get an appropriate dissection. Draw a figure with a more useful dissection.
457
Figure 23.6. What went wrong?
Figure 23.7. A dissection proof of the Theorem of Pythagoras.
Problem 23.7. Prove the Theorem of Pythagoras using the figure on page 457. For a proof by dissection, we need to • give a construction of this diagram which explains in which order it was obtained. • Based on this construction, we need to prove the congruence of the pieces to which the square on the hypothenuse is dissected to the pieces of which the two squares erected on the legs are made up.
458
Proof. Here is a way to do these steps: We construct the center J of the square BCFE on the larger leg of the given right triangle 4ABC. Drop the perpendicular from J onto the hypothenuse AB and erect the "double perpendicular" at point J. The square BCFE cuts the segments N M and KL out of these two perpendicular lines. We need to determine the segment lengths |CK| = x and |BL| = y. We see that y + x = |BL| + |LE| = a. The parallelogram ABLK has two pair of opposite congruent sides. Hence |KL| = |AB| = c and b + x = |AC| + |CK| = |AK| = |BL| = y. Since |K J| = |JL| = 2c , the four quadrilaterals into which the square BCFE is dissected, are congruent. They are transferred by several different parallel shifts to the square ABIH on the hypothenuse. We do this in such a manner that the four right angles at vertex J are transferred to the four vertices A, B, I, H. Hence the transferred quadrilaterals have vertices at the midpoints P, Q, W, Z of the sides of the square ABIH. Furthermore, they have each one vertex A0 , C 0 , G0 or D0 lying inside the square. We get right angles at these four points. Since A0C 0 C 0G0 G0 D0 D0 A0 = y − x = b, we see that the square A0C 0G0 D0 is congruent to the square ACGD erected over the shorter leg AC. Hence the square over the shorter leg, and the four congruent quadrilaterals into which the square over the longer leg has been dissected are put together without overlapping to form the square over the hypothenuse.
Figure 23.8. Dissection proof of the Pythagorean theorem in a special case
Problem 23.8. Given is a right triangle 4ABC with angle α = 60◦ and side a = 2. Squares are erected on its three sides. As done in the figure on page 457, one can construct dissections of them which prove the Theorem of Pythagoras. What kind of quadrilateral is ALKB. Determine the segment lengths x = |LE| and y = |BL|.
459
Answer. The quadrilateral ABLK is a parallelogram. Its opposite sides are congruent. Hence y = |AK| = |AC| + |CK| = |AC| + |LE| = b + x. Secondly, a = |BE| = |BL| + |LE| = y + x. From the equations x + y = a , y − x = b one concludes √ √ a−b a+b 3 3 x= =1− and y = =1+ 2 3 2 3 23.3. Dudeney’s dissection problem How can one dissect a unit square into a few polygons and rearrange them into an equilateral triangle? This is a problem from "The Canterbury Puzzles" by H.E. Dudeney [?], first published in 1908. Dudeney gave the solution indicated in the drawing below, which uses just four polygons. A less ingenious solution is given in the figure on page 460.
Figure 23.9. Dudeney’s dissection of an equilateral triangle and a square of the same area.
It uses five pieces some of which occur as mirror images. Problem 23.9. We assume the side of the square to be equal to one. Calculate segments a and x + y and the angle α in the accompanying drawing for Dudeney’s dissection to confirm its existence. Answer. The side a is obtained from the total area 1 =
√ 3 4
s a = 2 (3
−1/4
)=2
a2 to be √ 3 3
The two congruent quadrilaterals on the left side HAY Q HCY 0 Q
460
Figure 23.10. Another dissection of an equilateral triangle and a square of the same area.
are obtained rotating by 180◦ about H. Hence y = |AY| = |CY 0 |. The two congruent quadrilaterals on the right side GBXL GCX 0 P are obtained rotating by 180◦ about G. Hence x = |BX| = |CX 0 |. To complete the dissection, the triangles 4XLY 4X 0 P0 Y 0 have to be congruent. Hence a − x − y = |XY| = |X 0 Y 0 | = x + y and x+y=
a 2
Since |P0 X 0 | = |LX| = |X 0 P| = 12 , we get from the right triangle 4X 0 P0 Y 0 √ |P0 X 0 | 1 31/4 2 cos α = 0 0 = = < and α ≈ 48.85◦ > 45◦ |X Y | a 2 2 Problem 23.10. Find four segments of the length z = |QY 0 |. Calculate the length |Q0 L| in terms of z. Problem 23.11. Give a Euclidean construction for Dudeney’s dissection beginning with a unit square. Answer. The three figures below on page 461, page 461, and page 462 give stages of such a construction.
461
Figure 23.11. A construction of Dudeney’s dissection starting with the square I.
Figure 23.12. A construction of Dudeney’s dissection starting with the square II.
462
Figure 23.13. A construction of Dudeney’s dissection starting with the square III.
463
Problem 23.12. Calculate the segment x and the ratios the triangle 4YGB.
x a
and ay . You can use the cos Theorem for
Answer. The sides of the triangle 4YGB have the lengths |YG| = 1 , |GB| =
a a , |BY| = x + 2 2
Using the 60◦ angle at vertex B, the cos Theorem yields a 2 a a2 a a a2 a x+ 12 = + x+ − 2 x + cos 60◦ = x2 + xa + − 4 2 2 2 2 2 2 2 ax a x2 + + −1=0 2 4 √ −a + 16 − 3a2 x= q4 √ x −1 + 4 3 − 3 = a 4 q √ y 3 − 4 3−3 = a 4 Hence x ≈ .24549
and
y ≈ .25451
Problem 23.13. Give a Euclidean construction for Dudeney’s dissection beginning with a given equilateral triangle. Answer. The figure on page 464 shows such a construction.
464
Figure 23.14. A construction of Dudeney’s dissection starting with the triangle.
465
23.4. Euclidean area As the first step, we define the area for any triangle 4ABC. In Euclidean geometry, the area of a triangle is defined to be half of base times height. area(4ABC) =
1 1 1 aha = bhb = chc 2 2 2
Problem 23.14. Prove that it does not matter in this definition, which side of the triangle is chosen as base. 1
Figure 23.15. For the calculation of its area, any side of a triangle may be used as its base.
Answer. For the triangle 4ABC, we can take side BC as base. The corresponding altitude is AD, were D is the foot-point of the perpendicular dropped from vertex A onto side BC. As a second possibility, we can take side AC as base. The corresponding altitude is BE, were E is the foot-point of the perpendicular dropped from vertex B onto side AC. The two triangles 4CAD and 4CBE are equiangular, and hence similar. We get the proportion ha |AD| |BE| hb = = = = sin γ b |AC| |BC| a By multiplication with the denominators, we obtain aha = |AD| · |BC| = |BE| · |AC| = bhb = sin γ · a · b which is both the double area. We see that the area of a triangle well defined to be half of base times height. It does not matter which side one chooses as base. Remark. We see from this definition that the area takes values the field of segment arithmetic. This is indeed an Abelian group, as required. Note that this step uses the similarly of triangles with different orientation. 1
If one wants to define the area of rectangle as width time length, one get a similar difficulty—which again needs to be resolved by means of similar triangles.
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Proposition 23.7. The area of the triangle is given by one half product of two sides time and the sin of the angle between them: area(4ABC) =
1 1 1 ab sin γ = bc sin α = ca sin β 2 2 2
Proposition 23.8 (Instead of Euclid I.39). Two equicomplementable triangles with the same base have congruent altitudes.
Figure 23.16. Two triangles with same base and different altitudes are not equicomplementable .
Proof inspired by Euclid. We assume that 4ABC and 4ABD are equicomplementable . Let their altitudes be CF ⊥ AB and DG ⊥ AB. To check whether they are congruent, we transfer GD −−→ onto the ray FC and get the congruent segment FE GD. The triangles 4ABD and 4ABE have the same base and congruent altitudes. By the proposition 23.3, which Euclid has already proved, they are equicomplementable . Hence by theorem 13.1, the triangles 4ABC and 4ABE are equicomplementable . Too, their altitudes dropped from C and E have the same foot point F, and lie on the same line. The question is now, whether C = E. If not, there are two possible cases left. Either E lies between F and C, or C lies between F and E. It is enough to rule out the first case, the argument in the second case is similar. Assume—towards a contradiction—that F ∗ E ∗ C. In that case the triangle 4ABE is a subset of triangle 4ABC, to which it is equicomplementable . Too, the triangle 4ECB lies inside 4ABC, but does not overlap the smaller triangle 4ABE. Hence, by deZolt’s postulate 13.4, the triangles 4ABE and 4ABC cannot be equicomplementable . This contradiction rules out the case F ∗ E ∗ C. The case F ∗ C ∗ E can be ruled out similarly. Hence E = C, and the originally given triangles 4ABC and 4ABD have congruent altitudes CF = EF DG ⊥ AB, as to be shown. Euclid does not succeed to prove deZolt’s postulate in his treatment of area. But he did see that an additional assumptions is needed to get besides the simpler Euclid I.37 its converse Euclid I.39. To this end, he suggests an even stronger very general postulate about any magnitudes: "The whole is more than its parts". Such a general principle is clearly not justifiable. Indeed, it is not valid if dissection constructed from infinite sets are allowed. This is exemplified by the Banach-Tarski paradox. As we have seen from the proof above, deZolt’s postulate would have been a workable version for a new axiom that Euclid could have used to complete his theory of area.
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Short simple proof of Proposition 23.8. Let 4ABC and 4ABD be equicomplementable . By theorem 13.3, they have same area. The area is given by half the product of base times height. Let CF ⊥ AB and DG ⊥ AB be their altitudes. Hence we know that 1 1 |AB| · |CF| = |AB| · |DG| 2 2 Since segment arithmetic is a field, we can multiply both sides by 2 and divide by |AB| , 0 and get |CF| = |DG|
from which follows CF DG, as to be shown.
Already in the segment arithmetic, it was necessary to introduce a unit segment. Areas are naturally measured by taking the square over the unit segment as unit for area. Lemma 23.1. A right triangle T with legs of lengths a and b is equicomplementable to a right triangle with legs 1 and ab, and to a rectangle with legs 1 and area(T ). Problem 23.15. Prove lemma 23.1. Proposition 23.9. Every figure P is equicomplementable to a rectangle one side of which is the unit segment. The second side of this rectangle has the length area(P). Proof. The given figure P can be dissected into triangles. Each of them is either a right triangle or can be dissected into two right triangles. Hence the figure P can be dissected into finitely many right triangles T i for i = 1...n. Each of them is equicomplementable to a rectangle Ri with sides 1 and area(T i ). By proposition 13.3, P=
n ]
Ti ,
i=1
n ]
Ri = R
i=1
were the latter is a rectangle R with sides 1 and n X
area(T i ) = area(P)
i=1
as to be shown. Theorem 23.2. Any two figures with same area are equicomplementable .
Proof. Let P and Q be two figures with same area area(P) = area(Q). By proposition 23.9, figure P , R, where R is a rectangle, one side of which is the unit segment, and the other side has the length area(P). Similarly, figure Q , R0 , where R0 is a rectangle with lengths of sides 1 and area(Q). But two rectangles with the same lengths of sides are congruent, and hence clearly R , R0 . By theorem 13.1, "equicomplementable " is an equivalence relation. Hence P , R , R0 , Q implies the originally given figures P and Q are equicomplementable , as to be shown. Corollary 55. Any two figures are equicomplementable if and only if they have equal areas.
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23.5.
The role of the Archimedean axiom
Theorem 23.3. Assuming the Archimedean axiom (V.1) holds, any two figures with same area are even equidecomposable . Problem 23.16. We assume that the Archimedean axiom holds. Explain why a right triangle T is legs of lengths a and b is even equidecomposable to a rectangle with legs 1 and area(T ). Proof of Theorem 23.4. As explained in the problem 23.16 above, with the assumption of the Archimedean axiom, a right triangle T with legs of lengths a and b and a rectangle with legs 1 and area(T ) are shown even to be equidecomposable . In the same way as in proposition 23.9, we show that any figure P is even equidecomposable to a rectangle one side of which is the unit segment. The second side of this rectangle has the length area(P). Finally, as in the proof of theorem 23.2 that any two figures P and Q with same area are equidecomposable . (We do not use the Archimedean axiom in the two last steps, but the reasoning is done now with the stronger equivalence relation "equidecomposable ".) Problem 23.17. Between two parallels AB k CD of distance 5 units, there are given two parallelograms ABCD and ABEF. Let the lengths of side be |AB| = 1, |AC| = 10 and |AE| = 20. Construct dissections into triangles to shown that the two parallelograms are equidecomposable . How many triangles do you need? Problem 23.18. Let n be a natural number. What is the number t of triangles which are needed to get congruent dissections of a square of side n and a rectangle of sides 1 and n2 ? √ (a) Prove that at least p 2 nq triangles are needed. (b) Give a dissection using 2n triangles. Can one improve the lower or upper bound? Answer. Suppose we have congruent dissections of the square of side n and rectangle of sides 1 and n2 into t triangles. Let L be the length of the longest side of these triangles. The two opposite long sides of the rectangle are dissected into sides of different triangles. Since √ each triangle fits into the square of side n, the longest side L can have length at most 2 n. Hence √ 2n2 ≤ t · L ≤ t 2 n √ which implies t ≥ p 2 nq as claimed. Proposition 23.10. Assume the Archimedean axiom (V.1) does not hold. Then there exist a square and a rectangle that are equicomplementable but not equidecomposable . Proof. We assume the Archimedean axiom (V.1) does not hold. Hence there exist segments OB and OA such that OA > n · OB for all natural numbers n Now OB takes the role of the unit segment. We put the right triangle 4OAB and the right isosceles triangle 4OAA0 on the same side of their common leg OA. We draw the parallel to the hypothenuse −−→ AB through point A0 and get an intersection point P with the ray OA. In this way, we have constructed a segment OP such that |OP| · |OB| = |OA|2
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The right triangle 4OBP and the right isosceles triangle 4OAA0 are equicomplementable . By doubling these figures, we get a rectangle with sides OB and OP, as well as a square with side OA, which, too, are equicomplementable . We claim these two figures are not equidecomposable . Suppose—towards a contradiction— we have congruent dissections of the square and the rectangle into t triangles. Let L be the length of the longest side of these triangles. The two opposite long sides of the rectangle together have total length 2|OP|. They are dissected into sides of different triangles. Since each triangle fits into the square, the longest side L √ can have length at most 2 |OA|. Hence √ 2|OP| ≤ t · L ≤ t 2 |OA| √ 2|OP| · |OA| ≤ t|OA|2 = t|OP| · |OB| √ 2|OA| ≤ t|OB| On the other hand, we have assumed OA > n · OB for all natural numbers n, and hence with n := t we get t|OB| < |OA| √ Thus we arrive at the contradiction 2|OA| < |OA|. This confirms the claim the square and the rectangle are not equidecomposable . The logical relations between the different concepts of content gained in this section are summed up in the following diagram: clear
−−−−−→
P and Q are equidecomposable yTh. 13.3 area(P) yArchimedes and Th. 23.4
P and Q are equicomplementable Th. 13.3 y area(Q)
Th. 23.2 y
P and Q are equidecomposable ←−−−−−−−−−−−−−− P and Q are equicomplementable Archimedean axiom
Corollary 56. Assuming the Archimedean axiom (V.1) holds, any two equicomplementable figures are even equidecomposable . Assuming the Archimedean axiom (V.1) does not hold, there exist two equicomplementable figures that are not equidecomposable . Problem 23.19. Convince yourself once more that two parallelograms with the same base and congruent altitudes are always equicomplementable . Why can one not claim they are always equidecomposable ? 23.6. Some uniqueness results for justification Given is a Pythagorean plane. Let UQ denote a square the sides of unit length. Does assigning the area 1 to this figure determine the area function uniquely? Proposition 23.11. Take any Pythagorean plane. Let G be any ordered Abelian group. Suppose that µ is an measure function having the following properties: (1) the measure µ(P) ≥ 0 of any figure P is a nonnegative value in the ordered Abelian group G; (2) µ(T ) = µ(S ) for any two congruent triangles T S ;
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(3) µ(P
U
Q) = µ(P) + µ(Q) for any non-overlapping figures P and Q.
Then there exists an order preserving homomorphism φ : F 7→ G from field F of segment arithmetic to the group G such that µ = φ ◦ area. In other word, each measure is a homomorphic image of the area: µ(P) = φ(area(P)) for all figures P (23.1) and we get the diagram figure P area y
ymeasure µ
area(P) −−−−−→ measure µ(P) φ
Proposition 23.12. Under the additional assumption that (1*) the measure µ(T ) > 0 is strictly positive for any triangle T . the mapping φ is injective, hence an endomorphism. Proof of Proposition 23.11. By the same reasoning as used in the proof of Theorem 13.3, we derive from the assumptions (2) and (3) the following more general properties of the measure µ:
(4) Any two equidecomposable figures P ∼ Q have the same measure µ(P) = µ(Q). (5) Any two equicomplementable figures P , Q have the same measure µ(P) = µ(Q). We go on to construct the homomorphism φ : F 7→ G with the property (23.1). For any segment length r, we define φ(r) = µ(R) to be the measure of the rectangle R, one side of which is the unit segment and the second side has the length r. Given is any figure P. By Proposition 23.9, the figure P is equicomplementable to a rectangle R , P, one side of which is the unit segment and the second side has the length area(P). Hence we get both area(R) = area(P), as well as µ(R) = µ(P) from property (5). In the end that means µ(P) = µ(R) = φ(r) = φ(area(R)) = φ(area(P)) holds for any figure P, as required. It remains to check that the mapping φ preserves the addition and the order. Take two rectangles R with sides r × 1, and S with sides s × 1. We may assume they are non overlapping and have a U common side of unit length. In that case R S is a rectangle with sides r + s and 1, and thus ] φ(r + s) = µ(P R) = µ(R) + µ(S ) = φ(r) + φ(s) confirms the additivity. Now assume that r ≤ s, and both R and S lie on the same side of their common side of unit length. In this case R ⊆ S , and hence φ(r) = µ(R) ≤ µ(S ) = φ(s) confirms that φ preserves the order.
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Proof of Proposition 23.12. In order to check whether the homomorphism φ is injective, we suppose that r and s are two segment lengths and φ(r) = φ(s). The rectangles R with sides r × 1 and S with sides s × 1 have the same measure since µ(R) = φ(r) = φ(s) = µ(S ). Since any two segments are comparable, we may assume r ≤ s and hence R ⊆ S . We use the equality µ(R) = µ(S ) to rule out the R is strictly less that S . In that case, the difference S \ R would be the disjoint union of two congruent right triangles. By assumption (1*), they would have a positive measure and hence additivity would imply µ(R) < µ(S ). Thus we have seen that φ(r) = φ(s) implies µ(R) = µ(S ), secondly R = S and finally r = s, as to be checked for injectivity of φ. Theorem 23.4. In a Pythagorean and Archimedean plane, any measure function µ is equal to the common known area, provided it has the following properties: (0) the measure µ(UQ) = 1 for a unit square; (1) the measure µ(P) ≥ 0 of any figure P is a nonnegative value in the segment arithmetic; (2) µ(T ) = µ(S ) for any two congruent triangles T S ; U (3) µ(P Q) = µ(P) + µ(Q) for any non-overlapping figures P and Q. It is interesting to note that this uniqueness proof uses really Eudoxus idea to get equality from rational upper and lower bounds. This principle appear in Euclid’s book V as a definition for the equality of ratios of magnitudes. We reformulate this result as a proposition about Archimedean fields. Proposition 23.13 (Eudoxus principle). Given are any four elements a, b, c, d > 0 in an Archimedean field. The ratios a : b and c : d are equal, if and only if for any whole numbers m and n either one of these three cases occurs: both
na > mb and nc > md,
both
na = mb and nc = md,
both
na < mb and nc < md.
Proof. By the the Archimedean axiom there exists a natural number n such that (n + 1) · a > m · b. By the basic properties of the natural numbers, there exists a smallest such n. In this case both n · a ≤ m · b as well as (n + 1) · a > m · b hold. By assumption, the same inequalities hold for a and b replaced by c and d. Hence n · a ≤ m · b < (n + 1) · a and n · c ≤ m · d < (n + 1) · c Division implies b n+1 n c n+1 n ≤ < and ≤ < m a m m d m We have squeezed both fractions b/a and c/d between the same upper and lower rational bounds. By subtraction we conclude 1 b c 1 ≤ − ≤ m a d m Since m is an arbitrary natural number, and the Archimedean axiom has been assumed to hold, we conclude ba = dc . This last step also called the principle of exhaustion. −
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Remark. Eudoxus and Archimedes’ principles are used by Euclid in book XII.2 to prove that the "The areas of circles are in the same ratio as the squares on their diameters". Too, they are essential for Euclid’s theory of volume. Lateron, the same prinicples appear again in Archimedean proof of the law of levels, and many times more in mathematical physics. Proof of Proposition 23.4 . By Proposition 23.11, there exists a homomorphism φ : F 7→ F from the field of segment arithmetic to itself such that µ(P) = φ(area(P)) for all figures P
(23.1)
Moreover normalization for the unit square implies φ(1) = φ(area(UQ)) = µ(1) = 1 The additivity of the area and the measure of any n non overlapping congruent rectangles R of dimensions r × 1 imply, by means of an easy induction on n that φ(n · r) = φ(area(n · R)) = µ(n · R) = n · µ(R) = n · φ(area(R)) = n · φ(r) For any natural n and m we conclude m·φ
r
m 1 φ m n φ m
r =φ m· = φ(r) m ! 1 · r = · φ(r) m n · r = · φ(r) m
proving φ(qr) = q φ(r) for any rational q. We need the corresponding property for any even irrational segment length s. Given is any segment s > 0 and denominator m , 0, by the Archimedean axiom there exists a natural number n such that (n + 1) · 1 > m · s. By the basic properties of the natural numbers, there exists a smallest such n. In this case, we get n · 1 ≤ m · s < (n + 1) · 1 and n · 1 ≤ m · φ(s) < (n + 1) · 1 The second inequalities hold since φ(qr) = qφ(r) for any rational q and r = 1, and φ(1) = 1, and finally the homomorphism φ preserves the order. We have squeezed both s and φ(s) between the same upper and lower rational bounds, and conclude by subtraction −
1 1 · 1 ≤ φ(s) − s ≤ · 1 m m
Since m is an arbitrary natural number, and the Archimedean axiom has been assumed to hold, we conclude φ(s) = s. Once more, we have used Archimedes’ principle of exhaustion. Hence it turns out that φ is the identity and the only additive positive measure µ is the common area. 23.7. About the volume of polyhedra Already Gauss brought it to the attention of mathematicians, whether a corresponding justification of the three-dimensional volume based on dissections is possible. In a letter to Gerling in 1844, Gauss says that it is too bad that the equality of volume of two symmetrical, but not congruent figures, can be proved only using the method of exhaustion. Gerling could solve the dissection problem for this example. In his reply, Gerling gives a direct proof that any triangular pyramid can be dissected into twelve pieces that are congruent to those of
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a dissection of its mirror image. Gauss responds that this is a nice result. But it is still unfortunate that the proof of Euclid’s proposition (XII.5) "Pyramids of the same height on triangular bases are in the same ratio as their bases" needs the method of exhaustion. Hilbert conjectured that the method of exhaustion is really necessary. At the International Congress of Mathematicians in Paris 1900, he presented his famous list of the 23 most important problems facing mathematics in the twentieth century. The third problem of this list addresses the present question: Give two tetrahedra with equal bases and equal heights, which in no way can be dissected into congruent tetrahedra, and which, even by adding congruent tetrahedra, cannot be complemented into polyhedra, which in turn can be dissected into congruent tetrahedra. In the same year, Max Dehn has given such a counterexample. Max Dehn proves that it is not possible to dissect a regular tetrahedron into a finite number of pieces that can be reassembled into a cube, even after possibly adding on other figures that are equivalent by dissection. The important consequence is that the method of exhaustion is really necessary in Euclid’s proof of (XII.5).
Figure 23.17. Herons’ formula for the area of triangle.
23.8.
Heron’s formula for the area of a triangle
Problem 23.20. Derive Heron’s formula 1p area(4ABC) = (a + b + c)(a + b − c)(a − b + c)(b + c − a) 4 for the area a triangle in terms of its three sides a, b and c. Answer. Take the triangle 4ABC with vertices at the centers of the two circles and their intersection point A. We begin with the basic fact that the area of a triangle is half the product of base and altitude. We use the base CB and the altitude AP and get Area 4ABC =
1 |CB| · |AP| 2
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Let |CP| = p and |BP| = q. The height is calculated with the Pythagorean theorem in the right triangle 4CPA h2 = |AP|2 = |CA|2 − |CP|2 = b2 − p2 = (b − p)(b + p) Similarly, the Pythagorean theorem in the right triangle 4BPA yields h2 = |AP|2 = |BA|2 − |BP|2 = c2 − q2 Subtracting yields b2 − c2 = p2 − q2 = p2 − (a − p)2 a2 + b2 − c2 = a2 + p2 − (a − p)2 = 2ap 2a(b + p) = (a + b)2 − c2 = (a + b + c)(a + b − c) 2a(b − p) = −(a − b)2 + c2 = (−a + b + c)(a − b + c) Hence we get p 1 1 p |CB| · |AP| = 2a(b + p) 2a(b − p) 2 4 1 p = (a + b + c)(a + b − c)(−a + b + c)(a − b + c) 4 Algebraic relations for the pieces of a triangle Area 4ABC =
23.9.
Problem 23.21. Let r be the radius of the in-circle of triangle 4ABC. Prove the area is (a + b + c) r area(4ABC) = 2 Answer. It is explained in an earlier problem 11.3 in the context of neutral triangle geometry, how to construct the in-circle of a triangle. The perpendiculars are dropped from its center I onto the
Figure 23.18. An arbitrary triangle is partitioned into six right triangles.
three sides. The foot points F, G, H are the touching points of the in-circle to the three sides. The six segments from the center of the in-circle to the vertices and the touching points partition the given triangle into six right triangles. All of them have the radius of the in-circle as one of their legs. The other legs have lengths x = |AH|, y = |BF|, and z = |CG|, each one occurring twice for two congruent triangles. Since the circumference is a + b + c = 2x + 2y + 2z, the total area of the given triangle 4ABC is xr yr zr (a + b + c) r area(4ABC) = 2 · +2· +2· = (x + y + z) r = 2 2 2 2
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Problem 23.22. Let R is the radius of the circum-circle of triangle 4ABC. Prove the area is area(4ABC) =
abc 4R
Answer. As already stated in the Proposition 23.7, the area of a triangle is one half product of two sides time and the sin of the angle between them: area(4ABC) =
1 ab sin γ 2
On the other hand, the extended sin theorem tells that c = 2R sin γ Hence sin γ =
c 2R .
Plugging into the first formula yields the required formula area(4ABC) =
abc 4R
Problem 23.23. Can the three sides of a triangle be determined from the radius of the circumcircle, the radius of the in-circle and the circumference of the triangle? Answer. Yes, this is possible, but one needs to solve a third order equation. Hence the problem cannot be solved by a Euclidean construction— one needs to use Cardanos’ formula. We assume as given: the radius r of the in-circle, the radius R of the circum-circle, and the circumference 2s. From the two previous problems, we get area(4ABC) =
abc = sr 4R
Hence we can calculate the sum a + b + c = 2s and the product abc = 4Rrs. We define the polynomial P(x) := (x − a)(x − b)(x − c) = x3 − (a + b + c)x2 + (ab + ac + bc)x − abc To get all coefficients of this polynomial, we have still to use Heron’s formula, given in Problem 23.20 below in the form 1p (a + b + c)(a + b − c)(a − b + c)(b + c − a) area(4ABC) = 4 This can also be written with the semi-perimeter s as p area(4ABC) = s(s − a)(s − b)(s − c) Hence we get one further value of the polynomial P: P(s) =
area(4ABC)2 = rs2 s
One now can calculate the polynomial and obtain P(x) = x3 − 2sx2 + (rs + s2 + 4Rr)x − 4Rrs From this information, it is possible to calculate the three sides, for example by Cardanos’ formula or by some numerical method.
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24. Coordinate Planes—Descartes’ Road from Algebra to Geometry The Cartesian plane over a field F takes as coordinates the elements of an arbitrary finite or infinite field F— in place of the real number coordinates for the real Cartesian plane. We recall from definition 3.14: • The "points" of the Cartesian plane are ordered pairs (x, y) of elements x, y ∈ F. • The "lines" of the Cartesian plane of the field F are equations ax + by + c = 0 with coefficients a, b, c from the field, of which a and b are not both zero. A "point lies on a line" if and only if the coordinate pair (x, y) satisfies the equation of the line. One defines the slope of a line in the usual way. Lines are parallel if and only if they have the same slope. To find the parallel to a given line through a given point, one uses the point slope equation of a straight line. This procedure shows that the Euclidean parallel property holds. Leaving the details aside, we repeat the results: •
(Theorem 3.2 ). In a Cartesian plane over any field, there exist a unique line between any two points. There exists a unique parallel to a line through a given point. Hilbert’s axioms (I.1)(I.2)(I.3a)(I.3b) and the Euclidean parallel postulate hold. Hence a Cartesian plane over a field is an affine plane. The Main Theorem 3 implies the converse of the elementary Theorem 3.2. Together, there is the sharp result: (Theorem 4.5). Any affine plane is isomorphic to a plane over a field if and only if the Theorem of Pappus holds in its projective completion. 24.1.
A Hierarchy of Cartesian planes
Definition 24.1 (Ordered field). An ordered field is a field with an order relation < having the following properties: (i) For any two numbers a and b, one of the three relations holds: either a < b or b < a or a = b. (ii) For any two numbers a and b, no more than one of the three relations a < b or b < a or a = b hold. (iii) a < b and b < c imply a < c, for any a, b, c. (iv) a < b implies a + c < b + c, for any a, b, c. (iv) a < b and 0 < c imply a · c < b · c and c · a < c · b, for any a, b, c. Every ordered field contains the rational numbers as a subfield: Q ⊆ F. Hence every ordered field has the characteristic zero, and especially it is infinite. The points of a plane over an ordered field have a natural order relation. It turns out that Hilbert’s relevant axioms of order hold, too. Definition 24.2. A point B = (b1 , b2 ) lies between a point A = (a1 , a2 ) and a point C = (c1 , c2 ) if and only if bi = λai + (1 − λ)ci for i = 1, 2 with 0 < λ < 1. (24.1)
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Proposition 24.1. In a Cartesian plane over an ordered field, Hilbert’s axioms of order (II.1)(II.2)(II.3)(II.4) hold. Proposition 24.2. Conversely, assume that the axioms of order (II.1)(II.2)(II.3)(II.4) hold in any Cartesian plane constructed over some field. Then one can introduce into this field an order compatible with the order of the plane. Remark. There exists an affine incidence plane for which the Euclidean parallel property (IV*), the axioms of order (II.1) through (II.4) all hold, but the Theorems of Desargues and Theorem of Pappus are both violated. In 1902, F.R. Moulton [15] has described such an example in his article "A simple non-desarguesian plane geometry". Because of Theorem 4.5 this is an example of an ordered Euclidean incidence plane that is not a Cartesian plane over a field. We are now ready to introduce congruence. From a Pythagorean field, we reconstruct a Pythagorean plane as the analytic geometry over the field. This construction works for an arbitrary Pythagorean field, and thus we continue our road from algebra to geometry. The introduction of Cartesian coordinates turns out to be a perfect way to produce many different models for Pythagorean planes. Recall the definitions: Definition (2.2 Pythagorean plane). A Pythagorean plane is a Hilbert plane for which the Euclidean parallel postulate holds. Definition (18.1 Pythagorean field). A Pythagorean field is an ordered field with the property √ (Pyth) If a, b ∈ F, then a2 + b2 ∈ F. This section gives the detailed direct proof of the following result: Theorem 24.1. The Cartesian plane over a Pythagorean field is a Pythagorean plane. The converse result have already been obtained in the section on Hilbert’s arithmetic of segments. By Theorem 18.1, any Pythagorean plane is isomorphic to the Cartesian plane over its field of segment lengths, which is an Pythagorean field. Taking both results together, we state the important result: Main Theorem 20 (Fundamental Theorem about Pythagorean Geometry). The Cartesian plane F2 over a Pythagorean field F is a Pythagorean plane. Conversely, any Pythagorean plane is isomorphic to the Cartesian plane over its field of segment arithmetic, which is a Pythagorean field. Hence there is a bijection between the Pythagorean planes and the Pythagorean fields. The proof of the first part uses the fact that enough rigid motions exist in a Cartesian plane over a Pythagorean field. For the proof of the second part, one constructs the field with Hilbert’s new arithmetic of segments (Streckenrechnung) as explained in the section "Arithmetic of Segments— Hilbert’s Road from Geometry to Algebra". Our next step towards more and more familiar geometry is to add some axiom justifying the classical use of ruler and compass. Definition 24.3 (Euclidean plane). A Euclidean plane is a Hilbert plane for which both the Euclidean parallel postulate and the circle-circle intersection property hold.
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Definition 24.4 (Euclidean field). A Euclidean field is an ordered field with the following property: (3) If a ∈ F and a ≥ 0, then
√ a ∈ F.
Putting results together we get what I would be willing to name the fundamental theorem of Euclidean geometry: Main Theorem 21 (Fundamental Theorem about Euclidean Geometry). The Cartesian plane F2 over a Euclidean field F is a Euclidean plane. Conversely, any Euclidean plane is isomorphic to the Cartesian plane over its field of segment arithmetic, which is a Euclidean field. Hence there is a bijection between the Euclidean planes and the Euclidean fields. Definition 24.5 (real closed field). A field is called real closed if every polynomial of odd order has at least one zero. In a fourth step, we use Archimedes’ axiom in order to justify measurements. Finally, we introduce a suitable axiom of completeness. Definition 24.6 (Archimedean plane). A Archimedean plane is a Hilbert plane for which the Archimedean Axiom hold. Definition 24.7 (Archimedean field). A field or skew field is called Archimedean if for any a, c > 0 there exists a natural number n such that a < n · c. Theorem 24.2. The Cartesian plane F2 is Archimedean if and only if the field F is Archimedean. Definition 24.8 (Real Euclidean plane). A real Euclidean plane is a Hilbert plane for which both the Euclidean parallel postulate and the Dedekind axiom hold. Putting results together we get what I would be willing to name the fundamental theorem of geometry: Main Theorem 22 (Fundamental Theorem of Geometry). The real Cartesian plane R2 is a complete, Archimedean Euclidean plane. Conversely, any complete, Archimedean Euclidean plane is isomorphic to the real Cartesian plane. The real numbers are obtained as its field of segment arithmetic. We thus get a chain of models of strictly decreasing familiarity: • The real Euclidean plane is a Euclidean plane. • Any Euclidean plane is a Pythagorean plane. • Any Pythagorean plane is a semi-Euclidean plane. • Any semi-Euclidean plane is a Hilbert plane. • Any Hilbert plane is an affine plane. • Any affine plane is an incidence plane. The converse does not hold in any of these steps.
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24.2.
About Archimedean fields
Proposition 24.3 (Theorem 59 of Hilbert). Every Archimedean skew field is commutative. Proof. We get by an easy induction that n·a=a·n for any natural number n and any a of the skew field. Take two arbitrary elements a and b, for which we need to check the commutative law. It is easy to see that it is enough to consider—towards a contradiction—the case a > 0 , b > 0 , ab − ba > 0 (24.2) From the existence of the inverse, we conclude that there exists an element c > 0 such that (a + b + 1) · c = ab − ba
(24.3)
There exists a number d such that 0 < d < 1, d < c Recall that each nonempty set of natural numbers has a smallest one. Hence it is easy to see that there exist natural numbers n, m ≥ 0 such that m · d < a ≤ (m + 1) · d
and
n · d < b ≤ (n + 1) · d
Multiplication and subtraction yield mn · d2 < ba < ab ≤ mn · d2 + (m + n + 1) · d2 ab − ba < (m + n + 1) · d2 (m + n + 1) · d2 < (a + b + 1) · d < (a + b + 1)c ab − ba < (a + b + 1)c contradicting relation (40.11). Hence the assumption (24.2) cannot be true, and the commutative law is confirmed. 24.3. Congruence in a Pythagorean plane For any two points A = (a1 , a2 ) and B = (b1 , b2 ), we introduce the distance p d(A, B) = (a1 − b1 )2 + (a2 − b2 )2 and define the congruence of any two segments AB and A0 B0 by postulating AB A0 B0 ⇔ d2 (A, B) = d2 (A0 , B0 ) From this definition one easily checks that congruence of segment is an equivalence relation and hence Hilbert’s axiom (III.2) holds. Problem 24.1 (The rational plane is not Euclidean). The rational Cartesian plane Q2 is an incidence plane, but not a Hilbert plane. Which are the two among Hilbert’s axioms of incidence (I.1)(I.2)(I.3), and the axioms of order, congruence, parallelism, and continuity that do not hold in this case? √ We may √ assume to be known that 2 is not rational. Explain with a simple counterexample involving 2 that one of the axioms of congruence does not hold.
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Answer. Only the axiom of congruence (III.1), and the completeness axiom (V.2) are not satisfied in the rational Cartesian plane Q2 . To see that the axiom of congruence (III.1) is violated, we take the unit square with vertices −−−−−−−−−→ (0, 0), (1, 0), (0, 1), (1, 1) and try to transfer its diagonal onto the √ positive x-axis, the ray (0, 0)(1, 0). p This is impossible because a segment (0, 0)( q , 0) with length 2 does not exist. The completeness axiom (V.2) is violated, since it is possible to adjoin additional points into the rational plane in such a way to obtain a Pythagorean plane. Problem 24.2. Check that Hilbert’s axiom (III.1) about the transfer of segments holds in a plane over a Pythagorean field. Proof. In axiom (III.1) are given arbitrary points A, B and A0 and an arbitrary ray with vertex A0 . −−→ We may suppose this ray is a0 = A0C with C , A0 . To check the claim made by the axiom, we need to prove existence of a point B0 on this ray for which the segments AB A0 B0 are congruent. With the coordinates A = (a1 , a2 ) , B = (b1 , b2 ) , A0 = (a01 , a02 ) , C = (c1 , c2 ) the equation of the ray is x = a01 + λ(c1 − a01 ) and y = a02 + λ(c2 − a02 ) with λ ≥ 0 To get the required congruence AB A0 B0 , we need to satisfy the equation (b1 − a1 )2 + (b2 − a2 )2 = (x − a01 )2 + (y − a02 )2 Plugging in the equation for the ray yields (b1 − a1 )2 + (b2 − a2 )2 = λ2 [(c1 − a01 )2 + (c2 − a02 )2 ] It is possible to solve for the parameter λ > 0 since C , A0 . We plug into the equation of the ray and get the coordinates of the required point B0 = (b01 , b02 ) to be s (b1 − a1 )2 + (b2 − a2 )2 b01 = a01 + λ(c1 − a01 ) and b02 = a02 + λ(c2 − a02 ) with λ = (c1 − a01 )2 + (c2 − a02 )2 Hence we have proved existence of the point B0 = (b01 , b02 ).
Problem 24.3 (adding two adjacent segments on a line). Check that Hilbert’s axiom (III.3) holds in a plane over a Pythagorean field. Proof. In axiom (III.3) are given two segments AB and BC lying on a line with B as the only point in common. In the same manner, two segments A0 B0 and B0C 0 are given on a line with only point B0 in common. In other words, we assume that point B = (b1 , b2 ) lies between point A = (a1 , a2 ) and point C = (c1 , c2 ), and similarly for the primed items. Hence by definition 24.2 bi = λai + (1 − λ)ci for i = 1, 2 with 0 < λ < 1. = λ0 a0i + (1 − λ0 )c0i for i = 1, 2 with 0 < λ0 < 1.
b0i The axiom (III.3) takes
AB A0 B0 and BC B0C
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as assumption for the given segments. By definition, this means (b1 − a1 )2 + (b2 − a2 )2 = (b01 − a01 )2 + (b02 − a02 )2 (b1 − c1 )2 + (b2 − c2 )2 = (b01 − c01 )2 + (b02 − c02 )2 We plug in equation (24.3) and get (1 − λ)2 [(a1 − c1 )2 + (a2 − c2 )2 ] = (1 − λ0 )2 [(a01 − c01 )2 + (a02 − c02 )2 ] λ2 [(a1 − c1 )2 + (a2 − c2 )2 ] = λ0 2 [(a01 − c01 )2 + (a02 − c02 )2 ] If either A = C or A0 = C 0 , both equalities hold, and nothing is to be proved. We can assume A , C and A0 , C 0 , and divide the two relations to get (1 − λ)2 (1 − λ0 )2 = λ2 λ0 2 Since 0 < λ < 1 and 0 < λ0 < 1, we obtain via the root 1 − λ 1 − λ0 = λ λ0 0 λ=λ (a1 − c1 )2 + (a2 − c2 )2 = (a01 − c01 )2 + (a02 − c02 )2 AC A0C 0 Hence we have proved if AB A0 B0 and BC B0C 0 , then AC A0C 0
as was claimed.
24.4. Transfer of an angle Recall the definition 7.14 of an orientated angle as an ordered pair (h, k) of two rays h and k with a common vertex . In every ordered incidence plane, it is possible to define and fix an orientation. We establish an orientation in the coordinate plane. Definition 24.9 (Slope of a line). For a line AB through points A = (a1 , a2 ) and B = (b1 , b2 ), we define the slope b2 − a2 m= b1 − a1 Opposite rays on the same line have the same slope. A vertical ray gets the slope ∞ < F, hence the slope of a line is an element in F ∪ {∞}. Definition 24.10. We define the tangent of the orientated angle between two lines p and q with slopes m p and mq in F ∪ {∞} by tan x (p, q) =
mq − m p 1 + m p mq
as an element of F ∪ {∞}. The tangent of the orientated angle depends only on the two lines and does not distinguish opposite rays. The tangent of the orientated angle between two rays is defined by the tangent between the lines of the two rays.
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Remark. In the case that one of the slopes is m p , 0 and the other one mq = ∞, the formula makes sense by using the rule ∞ − mp 1 = 1 + mp · ∞ mp In the case that one of the slopes is m p , 0 and 1 + m p mq = 0, the formula makes sense by using the rule ±1/0 = ∞. Finally there remains the special case of the slopes m p = 0 and mq = ∞, for which we need to agree that tan x (p, q) = ∞. Remark. The tangent takes the value infinity if and only if the angle ∠(p, q) is a right angle. This occurs if the slopes are either 0 and ∞, or the denominator 1 + m p mq = 0. Recall the definition 7.17 of congruence of orientated angles Definition 24.11 (Congruence of angles). Two angles ∠(h, k) and ∠(h0 , k0 ) are called congruent if and only if either • the orientated angles x (h, k) and x (h0 , k0 ) are congruent, or • the orientated angles x (k, h) and x (h0 , k0 ) are congruent. Question. Convince yourself that two orientated angles x (h, k) and x (h0 , k0 ) are congruent if and only if the angles ∠(h, k) and ∠(h0 , k0 ) are both congruent and they have the same orientation. Question. Let the angles ∠(h, k) and ∠(h0 , k0 ) have the same orientation. Convince yourself that tan x (h, k) = tan x (h0 , k0 ) if and only if the angles ∠(h, k) and ∠(h0 , k0 ) are congruent. Question. Let the angles ∠(h, k) and ∠(h0 , k0 ) have opposite orientations. Convince yourself that tan x (h, k) = − tan x (h0 , k0 ) if and only if the angles ∠(h, k) and ∠(h0 , k0 ) are congruent. Lemma 24.1. If an angle is congruent to its supplement, the slopes mh and mk of its two sides are either 0 and ∞, or satisfy 1 + mh mk = 0. Proof. Let h0 be the ray opposite to h, and assume that the angle ∠(h, k) is congruent to its supplement ∠(h0 , k). By definition we conclude congruence of the orientated angles x (h, k) and x (k, h0 ) since they have the same orientation. Because h and h0 have the same slope, we conclude tan x (h, k) = tan x (k, h0 ) mk − mh mh − mk = 1 + mk mh 1 + mk mh We conclude that either the slopes mh = mk are equal—impossible for a non degenerate angle—or the slopes are 0 and ∞, or 1 + mh mk = 0. Problem 24.4. Explain why unique angle transfer according to axiom (III.4) is possible in the rational plane Q2 , and indeed axiom (III.4) does hold. Indication of the solution. Given is an angle ∠(h, k) and a ray h0 . Assume that ray k lies in the left half-plane of ray h, and the angle ∠(h, k) has to be transferred to the left half-plane of h0 . In this case, the two orientated angles x (h, k) and x (h0 , k0 ) have the same orientation, and need to be congruent. Hence tan x (h, k) = tan x (h0 , k0 ) m0 − m0 mk − mh t := = k 0 h0 1 + mk mh 1 + mk mh
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We can solve the last equation for the slope m0k and get m0k =
m0h + t 1 − m0h t
This determines the slope of the line of k0 . Since the ray k0 lies in the left half-plane of h0 , the ray k0 has been determined, too. Now we assume that ray k lies in the left half-plane of ray h, and the angle ∠(h, k) has to be transferred to the right half-plane of h0 . In this case, the two orientated angles x (k, h) and x (h0 , k0 ) with h and k switched have the same orientation, and need to be congruent, too. Hence tan x (h, k) = − tan x (h0 , k0 ) m0 − m0 mk − mh t := = − k 0 h0 1 + mk mh 1 + mk mh We can solve the last equation for the slope m0k and get m0k =
m0h − t 1 + m0h t
This determines the slope of the line of k0 . Since the ray k0 lies in the right half-plane of h0 , the ray k0 has been determined, too. Well, I see clear enough that all remaining details can be worked out. We thus have checked that the requested rays h0 and h00 have rational slope and hence exist in the rational plane Q2 . Lemma 24.2. Hilbert’s axiom (III.4) about the transfer of angles holds in a plane over a Pythagorean field. 24.5.
Enough rigid motions exist in a Pythagorean plane
Definition 24.12 (Rigid motion). A rigid motion in a Pythagorean plane is a bijection φ from the set of points to itself and a bijection of the set of lines to itself which preserves incidence and order, and maps any segment AB or angle ∠(h, k) to a congruent segment φ(A)φ(B) AB or angle ∠(φ(h), φ(k)) ∠(h, k). Definition 24.13 (Similarity). A similarity in a Pythagorean plane is a bijection φ from the set of points to itself and a bijection of the set of lines to itself which preserves incidence and order, and maps any angle ∠(h, k) to a congruent angle ∠(φ(h), φ(k)) ∠(h, k), and any pair of congruent segments AB A0 B0 to a pair of congruent segments φ(A)φ(B) φ(A0 )φ(B0 ). Definition 24.14 (Preserving versus reversing orientation). A rigid motion φ of any ordered affine plane preserves orientation if and only if any triangle 4ABC and its image 4φ(A)φ(B)φ(C) always have the same orientation. The rigid motion φ reverses orientation if and only if any triangle 4ABC and its image 4φ(A)φ(B)φ(C) always have the opposite orientation. −−→ Equivalently, we can state that a rigid motion preserves orientation if any rayAB and point C −−−−−−−→ in the left half plane of this ray is mapped to a ray φ(A)φ(B) and point φ(C) in the left half plane of the image ray. Theorem 24.3 (There are enough rigid motions). In a Pythagorean plane are given are two points P and Q, and rays rP and rQ with these vertices. There exist exactly two rigid motions φ and ψ, which map P to Q and rP to rQ . The first mapping φ preserves the orientation, and the second mapping ψ reverses the orientation.
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Lemma 24.3. For any point A = (a1 , a2 ), there exists a rigid translation which maps A to O, and which is orientation preserving. Lemma 24.4. The reflection P at the origin is a rigid motion which preserves the orientation. Lemma 24.5. The rotation by a right angle around the origin is a rigid motion which preserves the orientation. Problem 24.5. Check that the linear equations x0 = cx + sy y0 = −sx + cy with c2 + s2 = 1 define an isometry in the Pythagorean plane. Problem 24.6. Given the orientated angle ∠EOA with E = (1, 0), O = (0, 0), A = (x, y) and y > 0. Check that the linear equations x0 = cx + sy y0 = −sx + cy with c2 + s2 = 1, c > 0, s > 0 map the orientated angle ∠EOA to a congruent angle. −−→ −−→ Answer. The sides OE and OA of the angle ∠EOA have the slopes 0 = m p and y/x = mq . Hence tan x (p, q) =
mq − m p y = 1 + m p mq x
−−−→ Point E = (1, 0) is mapped to E 0 = (c, −s). Hence the ray OE 0 has the slope s m0p = − c −−→ The other image ray OA0 has the slope m0q =
y0 ) −sx + cy = x0 cx + sy
Hence tan x (p0 , q0 ) =
m0q − m0p 1+
m0p m0q
=
(−sx + cy)c + s(cx + sy) y = c(cx + sy) − s(−sx + cy) x
We can also check that both the angle ∠EOA and its image ∠E 0 OA0 have the same orientation. This is easy enough for the special case A = (0, 1), from which the general case follows. Hence the angle ∠EOA and its image ∠E 0 OA0 are congruent. Lemma 24.6. The linear equations x0 = cx + sy y0 = −sx + cy with c2 + s2 = 1 define a rigid rotation in the Pythagorean plane. Lemma 24.7. The reflection I x across the x-axis is a rigid motion which reverses the orientation.
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−−→ Lemma 24.8. Given is a Pythagorean plane. For any ray OA pointing into the right half plane, there exists a rigid rotation which maps this ray into the positive x-axis and preserves orientation. Proof. Let A = (a1 , a2 ) be the coordinates of A, point E = (1, 0) and a2 m = tan ∠EOA = a1 −−→ be the slope of ray OA. In a Pythagorean field, we can get the cosine and sinus of the rotation angle ∠EOA as a1 a2 1 m = 2 = 2 c= √ and s = √ 2 2 1 + m2 a1 + a2 1 + m2 a1 + a2 √ These quantities exist: Indeed, the √ square root 1 + m2 exists in a Pythagorean field. Moreover, 1 + m2 ≥ 1 cannot be zero, hence 1 + m2 , 0 and the division by this quantity is possible. The linear equations x0 = cx + sy y0 = −sx + cy −−→ −−→ can be checked to be the rigid rotation mapping the ray OA to the positive x-axis OE.
Lemma 24.9. Any composition of rigid motions, and the inverse of a rigid motion is again a rigid motion. In other words, the rigid motion are a group. The orientation preserving motions are a subgroup. Proof of the Theorem about enough rigid motions. By composition of a translation and a rotation, a ray towards right can be mapped to the positive x-axis. For an arbitrary ray, the same result can be achieved by use of a point reflection or a rotation by a right angle as third factor. We thus obtain an orientation preserving rigid motion with the required property. Composition with a reflection across the x-axis yields an orientation reversing rigid motion with the required property. Corollary 57. Every rigid motion can be realized by a composition a translation, a rotation about the origin, and possibly a reflection across the x-axis. Proposition 24.4. In a Pythagorean coordinate plane are given two triangles 4ABC and 4A0 B0C 0 . If the congruences AB A0 B0 , AC A0C 0 , ∠BAC ∠B0 A0C 0 hold, then the two triangles are congruent. Especially Hilbert’s SAS-axiom (III.5) holds in a plane over a Pythagorean field. Proof. I restrict myself to the case that angle ∠BAC is acute. There exists a rigid motion φ mapping −−→ −−→ A to O, the ray AB to the positive x-axis OE, and vertex C to a point in the upper half plane. Hence φ(A) = O = (0, 0) , φ(B) = (0, c) and φ(C) = (x, y) with y > 0 The rigidity of the mapping φ implies c = d(A, B) =
p (a1 − b1 )2 + (a2 − b2 )2 ,
x2 + y2 = d2 (A, C) = (a1 − c1 )2 + (a2 − c2 )2 and y = tan ∠BAC x
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All corresponding claims holds for a mapping φ0 related to the second triangle 4A0 B0C 0 . Comparing the relations obtained for both triangles, and using the congruences assumed, we get c = d(A, B) = d(A0 , B0 ) = c0 x2 + y2 = d2 (A, C) = d2 (A0 , C 0 ) = x02 + y02 y y0 = tan ∠BAC = tan ∠B0 A0C 0 = 0 x x 0 y > 0 and y > 0 From the assumption that angle ∠BAC ∠B0 A0C 0 is acute, we get x > 0 and y > 0 and similarly x0 > 0 and y0 > 0. Hence we conclude x = x0 , y = y0 and c = c0 . Hence the two "moved triangles" 4φ(A)φ(B)φ(C) with the vertices φ(A)(0, 0) , φ(B) = (0, c) and φ(C) = (x, y) and the corresponding primed one are equal. Hence they are congruent, too! Since a rigid motions maps a triangle to a congruent one, we conclude that the original triangles 4ABC 4A0 B0C 0 are congruent, too, as to be shown. In the case of an obtuse angle ∠BAC ∠B0 A0C 0 , we get x > 0 and y < 0 and similarly x0 > 0 and y0 < 0. Thus x = x0 , y = y0 follows similarly to above. The proof is even easier for the case that angle ∠BAC ∠B0 A0C 0 is right. End of the proof of Theorem 24.1. We finish the proof that a Cartesian plane over a Pythagorean field is a Pythagorean plane. • By Theorem 3.2, a Cartesian plane over a field is an affine plane. This means that Hilbert’s axioms (I.1)(I.2)(I.3a)(I.3b) and the Euclidean parallel property hold. • By Proposition 24.1, in a Cartesian plane over an ordered field, Hilbert’s axioms of order (II.1)(II.2)(II.3)(II.4) hold. • In the solution of Problem 24.2, we have checked that Hilbert’s axiom (III.1) about the transfer of segments holds in a plane over a Pythagorean field. • From the definition of congruence of segments, one sees that congruence is an equivalence relation, and hence Hilbert’s axiom (III.2) holds. • In the solution of Problem 24.3, we have checked that Hilbert’s axiom (III.3) about segment addition holds in a plane over a Pythagorean field. • By Lemma 24.2, Hilbert’s axiom (III.4) about the transfer of angles holds in a plane over a Pythagorean field. • By Proposition 24.4, Hilbert’s SAS-axiom (III.5) holds in a plane over a Pythagorean field. Thus all five axioms of congruence (III.1) through (III.5) are satisfied in a Cartesian plane over a Pythagorean field. 24.6.
Euclidean fields and intersection properties of circles
Theorem 24.4. The following three properties are equivalent in any Pythagorean coordinate plane F2 : (1) (2) (3)
The field F is Euclidean. The circle-line intersection property holds in the plane F2 . The circle-circle intersection property holds in the plane F2 .
Lemma 24.10. A Euclidean field is Pythagorean.
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Lemma 24.11 ((1) ⇒ (2)). The circle-line intersection property holds in the plane over a Euclidean field. Proof of (1) ⇒ (2). Given is a circle and a line which contains an interior point of the circle. Let P be the foot point of the perpendicular p dropped from the center of the circle onto the line. Since a Euclidean field is Pythagorean, there exist enough rigid motions in the plane F2 . Hence there exists −−→ a rigid motion mapping the ray OP onto the positive x-axis. In the special case O = P any one of the two rays with vertex perpendicular to the given line will be mapped onto the positive x-axis. Thus we have put the circle with radius c > 0 and the line into a Cartesian coordinate system, with coordinates C = (0, 0) for the center and P = (p, 0) for the foot point. Question. What are the equations for the circle and the line. Answer. x2 + y2 = c2 (24.4) x=p Since the line contains a point (p, q) inside the circle, we know that p2 ≤ p2 + q2 < c2 and hence c2 − p2 > 0. Question. What are the coordinates (x, y) of the intersection points of the circle and the line. Answer. Plugging x = p into the equation of the circle yields y2 = c2 − p2 . We have seen above that c2 − p2 > 0 follows since the line contains a point inside the circle. The field is assumed to be Euclidean, hence we can take the square root of a positive quantity and get the ptwo solutions p y = ±p c2 − p2 . We have obtained two intersection points with the coordinates (p, c2 − p2 ) and (p, − c2 − p2 ). We have checked that the circle-line intersection property holds in the coordinate plane over a Euclidean field. Lemma 24.12 ((2) ⇒ (1)). If the circle-line intersection property holds for the Pythagorean plane F2 , then the field F is Euclidean. Proof of (2) ⇒ (1). Given is f > 0, a positive element of the field. We construct in the plane F2 the circle and the line with the equations 24.4 with c=
1+ f 2
and p =
f −1 2
The assumption f > 0 implies −c < p < c. Hence the point (p, 0) lies inside the circle. By assumption, the circle-line intersection property holds. Hence circle and line intersect in some point (p, y). From the equation of the circle and the assumption we get y2 = c2 − p2 = f Hence y =
p
f is a square root of f , the existence of which was to show.
Lemma 24.13 ((1) ⇒ (3)). The circle-circle intersection property holds in the plane over a Euclidean field. Proof of (1) ⇒ (3). The two circles C with center C and radius c, and D with center D and radius d are given. We assume that one of the circles has points both inside and outside the other one. Question. Which inequalities hold between the two radius c, d > 0 and the distance a = |CD| between the two centers. Which case occurs in Proposition 10.10.
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Answer. If a ≥ c + d, the interiors of the two circles are disjoint, which is case (i) of Proposition 10.10. If a ≤ |c − d|, the interior of the smaller circle lies entirely in the interior of the larger one. These are the cases (ii) and (iii) of Proposition 10.10. These cases are excluded by the assumption that one circle has points both inside and outside the other one. Hence |c − d| < a < c + d These inequalities correspond in Proposition 10.10 to only remaining possibility (iv). Since a Euclidean field is Pythagorean, there exist enough rigid motions in the plane F2 . Hence −−→ there exists a rigid motion mapping the ray CD between the two centers onto the positive x-axis. Question. What are the coordinates of the two centers of the circles. Answer. We have put the two circle in a Cartesian coordinate system, with coordinates of the centers C = (0, 0) and D = (a, 0), where a = |CD| is the distance of the two centers. Question. What are the equations for the two circles. Answer. x2 + y2 = c2 (x − a)2 + y2 = d2 Question. What can one conclude from subtracting the two equations about the possible intersection of the two circles? Answer. Substraction of the two equations yields a linear equation with one unknown x. One can even solve for x. a2 + c2 − d2 x= (24.5) 2a This is the equation of a line. Hence the intersection points of the two circles lie on this line l. Conversely, the intersection points of the line l with either circle lie on the other circle, too. Hence it is now enough to look for the intersection points of circle C and line l. We may assume that c ≥ d, and use the abbreviation a2 + c2 − d2 p= 2a The point P with coordinates (p, 0) lies on the line l. Question. Why do the circle C and the line l intersect? Answer. The point P lies inside the circle C since −c < p < c The right inequality can be checked as follows: a2 + c2 − d2 < c 2a a2 + c2 − d2 < 2ac a2 + c2 − 2ac < d2 (a − c)2 < d2 −d < a − c < d c−d < a < c+d The last of these inequalities is true by assumption, and they are all equivalent. Checking that c + p > 0 is a quite similar calculation left to the reader. Since the line l has the point P inside circle C, the circle-line intersection property guarantees that they intersect.
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Question. What are the coordinates (x, y) of the intersection points of the circle and the line. Answer. Plugging x = p into the equation of the first circle yields y2 = c2 − p2 . We have seen above that c2 − p2 = (c + p)(c − p) > 0. The field is assumed to be Euclidean, hence we can take the p square 2 root of a positive quantity y and get the two intersection points with the coordinates (p, c2 − p2 ) p 2 and (p, − c − p2 ). Thus we have that the circle-circle intersection property holds for a Euclidean plane. End of the proof of Theorem 24.4. We know from Proposition 10.15 that the circle-circle intersection property implies the line-circle intersection property. This holds even more generally in every Hilbert plane. Thus we know that (3) ⇒ (2) for the items of Theorem 24.4. We have proved that (2) ⇒ (1)— the circle-line intersection property implies that field is Euclidean— in Lemma 24.12 Moreover, Lemma 24.13 states that (1) ⇒ (3)—The circle-circle intersection property holds in the plane over a Euclidean field. Thus all three statements (1)(2)(3) of Theorem 24.4 are equivalent, as to be shown. Problem 24.7 ((3) ⇒ (1)). Give a direct proof of the statement that a Pythagorean plane F2 for which the circle-circle intersection property holds has a Euclidean field F. Proof of (3) ⇒ (1). Given is f > 0, a positive element of the field. We construct in the plane F2 two circles with the same radius c and the distance a between there centers. We choose 1+ f c= and a = | f − 1| 2 From the equations of the two circles x2 + y2 = c2 (x − a)2 + y2 = c2 we get the coordinates (x, y) of an intersection point to be x = 2a and s !2 !2 q p 1+ f 1− f y = y2 = − = f 2 2 Since the intersection point is assumed to exist, we see that the square to be shown.
p
f exists for any f > 0, as
From the results above, we get a proof of Proposition 10.16 as promised in the section on the neutral geometry of circles. Definition (Proposition 10.16). In any Hilbert plane, the line-circle intersection property together with the parallel axiom imply the circle-circle intersection property. Reason. Assuming the parallel axiom, we may invoke theorem 18.1: any Pythagorean plane is isomorphic to the Cartesian plane over its field of segment lengths, which is an Pythagorean field F. Because the circle-line intersection property is assumed, we know from Lemma 24.12 above, that the field F is even Euclidean. Hence by Lemma 24.13, the circle-circle intersection holds, too. Corollary 58. If the distance a = |CD| between the centers of two circles C with center C and radius c, and D with center D and radius d satisfy the inequalities |c − d| < a < c + d they intersect in two points.
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Corollary 59 (Heron’s formula). In Euclidean geometry, the area of a triangle with the sides a, b and c is 1p area(4ABC) = (a + b + c)(a + b − c)(a − b + c)(b + c − a) 4 Proof. I replace b by d as always above and consider the triangle with the vertices C, D and an intersection point X = (p, y) of the circles C and D. We get the Euclidean area q ay a area(4CDX) = = c2 − p2 2 2 The definition of p implies 2ac − a2 − c2 + d2 d2 − (a − c)2 (d − a + c)(d + a − c) = = 2a 2a 2a 2ac + a2 + c2 − d2 (a + c)2 − d2 (a + c + d)(a + c − d) c+ p= = = 2a 2a 2a p 2 2 The coordinates of the intersection points are (x, y) = (p, ± c − p ) with c− p=
a2 + c2 − d2 2a (a + c + d)(a + c − d)(a − c + d)(c − a + d) 2 2 c −p = 4a2 ] Hence the area of the triangle with sides a, c and d is q a 1p area(4CDX) = c2 − p2 = (a + c + d)(a + c − d)(a − c + d)(c − a + d) 2 4
(24.6)
p=
(24.7)
Problem 24.8. Confirm that the line point P and the line l are constructible with Hilbert tools and unique parallels, even if no circle intersection properties are assumed. Proof. From the expression 24.6 for c − p and c + p, one gets the proportions c− p c+d−a = (24.8) a+c+d 2a c+ p a+c+d = (24.9) a+c−d 2a From these proportions, construction of either c − p or c + p can be implemented with Hilbert tools, using parallels and similar triangles. Since the line l has the point P inside circle C, the circle-line intersection property guarantees that they intersect. By the remarks above, there intersection points are the same as the intersection points of the two circles C and D. Here is a "fairly clean" solution for the construction of point P. Let AB and FE be the diameters of the two given circles C and D lying on the line CD. Question. We assume the order A ∗ F ∗ B ∗ E. Calculate the distances |AF|, |FB|, |BE| and |AE|. Answer. |AF| = a + c − d |FB| = c + d − a |BE| = a − c + d |AE| = a + c + d
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Figure 24.1. The intersections C ∩ D = C ∩ l = D ∩ l are equal—point P and line l can be constructed with Hilbert tools.
We construct a rectangle with one side AB, and the perpendicular sides congruent to 2a. The coordinates of its vertices are A = (−c, 0), B = (c, 0), H = (c, 2a), G = (−c, 2a) We draw a line g with slope −1 through point F = (a − d, 0). It intersects the vertical sides of the rectangle, or their extensions, in the points I = (−c, a + c − d) and J = (c, −(c + d − a)) The parallel to line GE through point I intersects the x-axis in point P = (p, 0), because of the proportion (24.9) and the similar triangles 4API ∼ 4AEG. With a similar reasoning, we get that the parallel to line HE through point J intersects the x-axis in the same point P = (p, 0), too.
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25. Standard Euclidean Triangle Geometry 25.1.
The circum-center
Lemma 25.1. In a Pythagorean plane are given any angle ∠(h, k) with vertex O, and two lines p and q which intersect the sides of the angle at points P = h ∩ p and Q = k ∩ q. We assume that P , O and Q , O. If p ⊥ line h and q ⊥ line k, then the lines p and q intersect at a unique point.
Figure 25.1. Do the perpendiculars p and q intersect?
Proof. Let O = h ∩ k be the vertex of angle ∠(h, k). Let P = h ∩ p and Q = k ∩ q be the points where the perpendiculars are erected on the sides of the angle. We prove: "p , q": By assumption O , P and O , Q and hence triangle 4OPQ exists. The case p = q leads to a contradiction as follows. We would obtain a triangle 4OPQ with two right angles at vertices P and Q. By the exterior angle theorem 7.40 (indeed already Lemma 7.5) such a triangle cannot exist. "p ∦ q": We know already that p , q. Because of the assumptions P , O and Q , O, neither −−→ −−→ lines p ⊥ OP nor q ⊥ OQ can go through the vertex O. We show that the assumption p k q leads to a contradiction as follows. With assumption p k q we may use Proclus’ lemma 3.3. Because the line of h intersects p, Proclus’ lemma yields that the line of h intersects the parallel q k p, too. Let R = line h ∩ q be their intersection point. Since line q does not go through point O, we conclude R , O. Hence RO is the line of ray h. Since Q , O has been assumed, we conclude that QO is the line of ray k. Since ∠(h, k) is assumed to be an angle, the points O, Q and R do not lie on a line. Hence the triangle 4OQR exists.
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It has been shown in proposition 7.44 that in a Pythagorean plane, any two parallel lines form congruent z-angles with any transversal. The line of ray h is transversal to the parallels p k q. Too, the line of h is perpendicular to p by assumption. Hence the line of h is perpendicular to q. By assumption, the line of k is perpendicular to q, too. We have obtained a triangle 4OQR with two right angles at vertices Q and R. By the exterior angle theorem such a triangle cannot exist.
Figure 25.2. No, the triangle 4OQR cannot exist.
By Hilbert’s proposition 1 (see 3.1), the lines p and q intersect at a unique point, as to be shown.
Theorem 25.1 (The Circum-center). In a Pythagorean plane, the bisectors of all three sides of any triangle intersect at one point,—which is the center of the circum-circle. Proof. Because of lemma 25.1, any two perpendicular side-bisectors of the triangle 4ABC intersect. Let the bisector p of side AB intersect the bisector q of side BC at the point O. By corollary 9, a point lies on the perpendicular bisector a segment if and only if its distances to the endpoints of the segment are congruent. Since point O lies on the bisector of AB, we conclude AO BO. Since point O lies on the bisector of BC, we conclude BO CO. Hence point O has congruent distances to all three vertices of triangle 4ABC. Hence point O is the center of the circum circle. Too, we conclude that O lies on the bisector of the third side CA.
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Figure 25.3. Constructing a triangle of double size.
25.2.
Double and half size triangles
Proposition 25.1 (Double ASA). Given is a triangle 4A0 B0C 0 and a segment AB = 2A0 B0 , of double length than A0 B0 . Then there exists a triangle 4ABC, equiangular with 4A0 B0C 0 . All three sides have length double as the corresponding sides of the original triangle. Reason. Let M be the midpoint of segment AB, and hence A0 B0 AM MB. We stay now in one −−→ −−→ half plane of line AB, and transfer the angle β = ∠A0 B0C 0 onto both rays BA and MA. Similarly, we −−→ −−→ transfer the angle α = ∠B0 A0C 0 onto the two rays AB and MB. By the extended ASA congruence one gets two intersection points Mb and Ma . The two new triangles and the one to begin with are all congruent. 4A0 B0C 0 4AMMb 4MBMa (25.1) Since we deal with Euclidean geometry, the angle sum of a triangle is 2R (Euclid I.32). Hence angle addition at vertex M implies ∠Ma MMb γ. Now SAS congruence implies 4A0 B0C 0 4Ma Mb M
(25.2)
Hence congruent z-angles (Euclid I.27) yield that the lines AB and Ma Mb are parallel. The −−−→ extensions of the already produced ray AMb intersect both parallel segments AB and Ma Mb at congruent angles α. (Here we need to use Euclid 1.28 and hence the Euclidean parallel property!) −−−→ Similarly, the extensions of the already produced ray BMa intersect both parallel segments AB −−−→ −−−→ and Ma Mb at angles β. By the extended ASA congruence, the rays AMb and BMa do intersect, say at point C, and 4A0 B0C 0 4Mb MaC (25.3) Altogether, the congruences (40.11),(40.11) and (46.7) have produced four new congruent triangles. Furthermore Ma , Mb and M are the midpoints of the sides of the larger new triangle 4ABC. Proposition 25.2 (Half ASA). Given is a triangle 4ABC, and a segment A0 B0 of half length as AB. Then there exists a triangle 4A0 B0C 0 , equiangular with 4ABC, and all three sides of it have half length as the sides of the original triangle. Reason. The half segment A0 B0 is smaller than the original segment AB. By my earlier remarks about extended ASA and Pasch’s axiom in the section on Neutral Geometry ??, A0 B0 < AB implies the existence of a triangle 4A0 B0C 0 such that ∠ABC ∠A0 B0C 0
and ∠BAC ∠B0 A0C 0
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By Euclid I.32 both triangles 4ABC and 4A0 B0C 0 have the same angle sum 2R. Hence two pairs of congruent angles are enough to show that they are equiangular. Now we are back to the situation of Proposition 25.1, which is applied to the smaller triangle 4A0 B0C 0 and the (larger) segment AB. Based on the segment AB, we construct a triangle 4ABD, which is equiangular with triangle 4A0 B0C 0 and hence with the original triangle 4ABC. Because of 4ABC 4ABD, uniqueness of angle transfer yields D = C. As expected, we got back the original triangle. By Proposition 25.1, the sides of triangle 4ABD = 4ABC are double the sides of 4A0 B0C 0 . Of course, this means that the sides of 4A0 B0C 0 are half the sides of 4ABC.
Figure 25.4. Construction and properties of the midpoint triangle.
Proposition 25.3 (The midpoint triangle). The midpoints of the sides of a triangle and the segments between them produce four smaller congruent triangles. They are equiangular to the original triangle, and have sides half as long as the corresponding sides of the original triangle. "Halving then doubling gives back the original triangle". We draw the parallel to side BC through point Mc . By Pasch’s axiom, the parallel intersects segment AC, say in point Pb . Furthermore, the "halved" triangle 4AMc Pb is equiangular to the original triangle 4ABC. By Proposition 25.2, all three sides of 4AMc Pb are half of the corresponding sides of 4ABC. Hence AC = 2APb and Pb = Mb is the midpoint of side AC. This means that the parallel to BC through point Mc does intersect triangle side AC in its midpoint. Similarly, one sees that the other two sides of the midpoint-triangle 4Ma Mb Mc are pairwise parallel to the sides of the original triangle. By Euclid I.28, a transversal crossing two parallel lines produces congruent z-angles. (This is a strong statement, valid only in Euclidean geometry!). Hence the midpoint triangle partitions the original triangle into four smaller, half-size triangles, which are all equiangular to the original triangle. Proposition 25.4 (Double SAS). Assume that the two triangles 4ABC and 4A0 B0C 0 have congruent angles ∠BAC 4B0 A0C 0 , and the corresponding adjacent sides satisfy AB 2A0 B0 and AC 2A0C 0 . Then they are equiangular, and BC 2B0C 0 . Reason. This follows easily from Double ASA:—as the reader should check—: By Double ASA, there exists a triangle 4ABC∗ , which is equiangular to the given triangle and all three sides are of double length than the original triangle 4A0 B0C 0 . Hence AC∗ 2A0C 0 . On the other hand, AC 2A0C 0 was assumed. Hence axiom III.2 yields AC∗ AC. Reproducing this −−→ segment on the ray AC yields a unique point C = C∗ . 1
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Figure 25.5. The three medians intersect in the centroid.
25.3.
The centroid
Theorem 25.2 (The Centroid). The three medians of a triangle intersect in one point, called the centroid. The centroid divides the medians in the ratio 2 : 1. Reason, given for Euclidean geometry. Apply the Double SAS Proposition 25.4 to the two triangles, 4AMb Mc and 4ABC. We see that they are equiangular, and that the side BC 2Mb Mc . Hence the segment Mb Mc is parallel to the side BC. At first, we define point S as intersection of the two medians BMb and CMc . The triangles 4S Mb Mc and 4S BC are equiangular. We can apply the Double ASA Proposition 25.1 to the triangle 4S Mb Mc and the (longer) segment BC. Hence we conclude that S B = 2S Mb and S C = 2S Mc . By the fact of dividing the median BMb as 2 : 1, the point S on median BMb is uniquely defined. Now define point T as the intersection point of medians BMb and AMa . The same argument as above shows that point T divides the median BMb as 2 : 1, too. Hence S = T is the intersection of all three medians. 25.4.
The orthocenter
Theorem 25.3 (The Orthocenter). In Euclidean geometry, the three altitudes of a triangle intersect in one point—which is called the orthocenter. Concise proof. We can double one side of the given triangle. By Double ASA, Proposition 25.1, one constructs an equiangular triangle 4A0 B0C 0 with all three sides doubled. The original triangle 4ABC is congruent to the midpoint triangle 4Am BmCm of 4A0 B0C 0 . 4ABC 4Am BmCm The altitudes of the triangle 4Am BmCm are the perpendicular bisectors of the sides of the larger triangle 4A0 B0C 0 . Hence they intersect in one point. This point is the circum-center of 4A0 B0C 0 , as well as the orthocenter of 4Am BmCm . Proposition 25.5. In Euclidean geometry, every triangle 4ABC is the midpoint triangle of a larger triangle 4A0 B0C 0 . 1
If you still need a bottle of wine from Hilbert: Applying the SAS axiom III.5 to the two triangles 4ABC and 4ABC∗ , we conclude ∠ABC ∠ABC∗ . Hence the −−→ −−−→ uniqueness of angle transfer stated in axiom III.3 implies r = BC = BC∗ . Hence C = C∗ is the unique intersection point −−→ of the two rays r and AC.
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Figure 25.6. Every triangle 4ABC is the midpoint triangle of a larger triangle 4A0 B0 C 0 .
Proof I: Use extended ASA congruence three times. Use extended ASA congruence three times, and get three new triangles: 4ABC 4CB0 A lying on opposite sides of AC. Similarly, 4CB0 A 4ABC are lying on opposite sides of AC. Finally 4A0CB 4ABC are lying on opposite sides of BC. Because the angle sum is α + β + γ = 2R, point C lies on segment A0 B0 . Furthermore B0C AB CA0 . Hence C is the midpoint of segment A0 B0 . Similarly, we see that A is the midpoint of segment B0C0 and B is the midpoint of segment C0 A0 . hence the given triangle 4ABC is the midpoint-triangle of the larger triangle 4A0 B0C 0 . Proof II: Erect the perpendiculars on the altitudes. One constructs the doubled triangle 4A0 B0C 0 by erecting the perpendiculars on its altitudes at all three vertices of triangle 4ABC. Let pa , pb , pb be these three perpendiculars. Using angle sums, one can check that −−→ ∠(pa , AC) γ
−−→ and ∠(pc , CA) α
By the extended ASA congruence, lines pa and pc intersect at a point B0 , and 4ABC 4CB0 A These two triangles lie on opposite sides of AC. Similarly, one get a new triangle 4CB0 A 4ABC These two triangles lie on opposite sides of AC. Similarly, one get a new triangle 4A0CB 4ABC
498
on opposite side of BC and 4BAC 0 4ABC on opposite side of AB. In this way, the intersection points of the three lines pa , pb , pb yield a larger triangle 4A0 B0C 0 . Question. What happens if one does the same two constructions in hyperbolic geometry? Answer. Can one construct the three new congruent triangles outside of the given triangle by the extended ASA theorem. One ends up a figure with four congruent triangles, but its outer boundary is a convex hexagon, not a triangle. Alternatively, one can use the double perpendiculars, erected at the vertices on the altitudes of the given triangle. These three lines need not even intersect! In case they do intersect, the smaller triangle is indeed the midpoint-triangle of the larger triangle. But the larger triangle 4A0 B0C0 has angles smaller than the angles of the given triangle 4ABC. Constructive proof for the orthocenter—using midpoints. By Proposition 25.5, the given triangle 4ABC is the midpoint-triangle of a larger triangle 4A0 B0C0 . By Proposition 11.1, the altitudes of the triangle 4ABC are the perpendicular bisectors of the sides of the larger triangle 4A0 B0C 0 . By Proposition 25.1, the perpendicular bisectors intersect in one point. This point is the circum-center of the larger triangle 4A0 B0C 0 , as well as the the orthocenter of the original triangle 4ABC. Corollary 60. The circum-center of a triangle is the orthocenter of the midpoint triangle. 25.5.
The in-circle and the three ex-circles
Theorem 25.4 (The in-circle and three ex-circles). In Euclidean geometry, a triangle has an incircle and three ex-circles. All four circles have the three sides of the triangle, or their extensions, as tangents. The in-circle touches the three sides from inside the triangle. The ex-circles touch one side from outside, and the extensions of the two other sides. Reason. The existence of the in-circle holds already in neutral geometry. To get the ex-circle touching side AB form outside, consider the two exterior bisectors eA and eB at vertices A and B. They both form acute angles with segment AB, and both lie on the side opposite to vertex C. By Euclid’s fifth postulate these exterior bisectors do intersect, say in point Ic . 1 Now one argues similarly as in the case of the in-circle. Point Ic has congruent distances to all three sides of the triangles, and hence lies on the interior bisector of angle ∠BCA, too. Point Ic is the center of the ex-circle touching side AB from outside, as well as the extensions of the two other sides. Remark. The bisectors of the exterior angles of the triangle are the perpendiculars to the angular bisectors, erected at the vertices. 25.6.
The road to the orthocenter via the orthic triangle
Definition 25.1 (The orthic triangle). Let Fa , Fb and Fc denote the foot points of the three altitudes. The triangle 4Fa Fb Fc is called the orthic triangle. 1
The unique parallel to eA through point B is different to eB .
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Figure 25.7. The altitudes are the angular bisectors of the orthic triangle.
Theorem 25.5. For an acute triangle, the altitudes are the inner angular bisectors of the orthic triangle. For an obtuse triangle, the only the altitude dropped from the obtuse angle is an inner angular bisector. The other two altitudes are exterior angular bisectors. Corollary 61. The three altitudes of a triangle intersect in one point. The case of an acute triangle. We shall repeatedly use the congruence of circumference angle, stated by Euclid III.21. Both angle ∠CAFa ∠CBFb R − γ because of the angle sum and the right angles. By the converse Thales theorem, the four points C, Fb , Fc and B lie on a semicircle with diameter BC. Hence the congruence of circumference angles (Euclid III.21) implies that ∠CBFb ∠CFc Fb . Similarly, the four points C, Fa , Fc and A lie on a semicircle with diameter AC, and the congruence of circumference angles implies ∠CAFa ∠CFc Fa . Hence ∠CFc Fa ∠CFc Fb , and the altitude CFc bisects the angle ∠Fa Fc Fb of the orthic triangle. Similarly, we can show that all three altitudes are angular bisectors of the orthic triangle. The Corollary: Constructive proof for the orthocenter—using bisectors. Since we did not need the intersection point of any two altitudes, but we know from Proposition 11.3 that the three angular bisectors intersect in one point, one gets an additional proof that the three altitude intersect in one point. Proposition 25.6. In Euclidean geometry, every acute triangle 4ABC is the orthic triangle of two non congruent larger triangle, one acute 4A1 B1C 1 , and one obtuse 4A2 B2C 2 . Proof. The first good exercise. Do it!
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Figure 25.8. The orthic triangle for an obtuse triangle.
25.7.
The Euler line
Theorem 25.6 (The Euler line). The orthocenter, the centroid, and the circum-center of a triangle lie on one line, called the Euler line. The centroid trisects the segment joining the orthocenter and the circum center. Remark. In the exceptional case of an equilateral triangle, all three centers are equal, but the Euler line is not defined. Definition 25.2. A central dilation with center C and ratio k , 0 is a mapping that maps point A to a point A0 such that the three points A, A0 and C lie on a line and A0C CA. If k > 0, the center C lies outside the segment AA0 . If k < 0, the center C lies inside the segment AA0 . Proposition 25.7. A central dilation maps a triangle 4XYZ to an equiangular triangle 4X 0 Y 0 Z 0 such that X 0 Y 0 kXY, Y 0 Z 0 kYZ and Z 0 X 0 kZX and these corresponding segments are parallel. Partial reason. We shall need only the special case k = −2, which can be covered with congruence theorems and Double congruence theorems from above. Proof of Euler’s Theorem. Given is a triangle 4ABC. We use a central dilation with the centroid S as center and ratio k = −2. The Centroid Theorem 25.2—together with the definition of a central dilation—implies that the midpoints Ma , Mb and Mc are mapped to the vertices A, B and C. Because of conservation of angles, the altitudes of the midpoint triangle are mapped to the altitudes of the original triangle.
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Furthermore, the orthocenter H2 of the midpoint triangle is mapped to the orthocenter H of triangle 4ABC. By Proposition 11.1, the orthocenter H2 of the midpoint triangle is the circum-center O of triangle 4ABC. Hence the dilation maps O to H and, of cause, center S to itself. Directly from the definition of a central dilation, we see that the three centers O, S and H lie on one line and 2 OS S H, with the centroid S between the circum-center and the orthocenter.
Figure 25.9. The Euler line comes from a dilation with center S .
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Figure 26.1. Menelaus’ Theorem
26. Harmonic Points 26.1.
The Theorems and Menelaus and Cevá
Theorem 26.1 (Theorem of Menelaus). Let a transversal cut the three sides of 4ABC, extended if necessary, in the three points X, Y, Z. One orders the six segments as they occur on a closed curve circumventing the triangle: AX, XB, BY, YC, CZ and finally ZA. Note this curve can be drawn without lifting the pencil. That ordering understood, one gets AX
·
BY
·
CZ
XB YC ZA
= −1
Proof of Menelaus’ Theorem. Draw the parallel to triangle side AB through vertex C. Let G be its intersection point with the transversal. Similar triangles with common vertex Y and opposite parallel side GC and XB yield the proportion |GC| |YC| = |XB| |Y B|
(26.1)
Similar triangles with common vertex Z and opposite parallel side GC and XA yield the proportion |GC| |ZC| = |XA| |ZB|
(26.2)
Dividing (26.1) over (26.2) yields |XA| |YC| |ZB| = · |XB| |Y B| |ZC| Now rearranging yields the assertion.
Problem 26.1. All three sides of an equilateral triangle are trisected. Three transversals are drawn from the vertices to one of the partition points of the opposite side, in such a way to obtain a smaller equilateral triangle in the middle. Assume that the area of the original triangle is 21. Find the area of the triangle in the middle. Which fraction of the total area is covered by the middle triangle?
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Figure 26.2. The area of the middle triangle is one seventh of the total area.
Figure 26.3. What is the area of the triangle in the middle?
Answer. Since the total area is given to be 21, the areas of the three triangles adjacent to the left side add up to 7. Hence the areas x of the middle triangle, y of the quadrilaterals, and z of the small triangles in the corners satisfy the two equations x + 3y + 3z = 21 y + 2z = 7 A third equation among the three unknowns x, y, z can be obtained from Menelaus’ theorem. We use the left triangle 4BCD and the transversal AU. Starting at vertex B, we enumerate the six
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Figure 26.4. The Theorem of Cevá
segments along the sides of the triangle cut by the transversal and obtain BU
·
CV
UC V D
·
DA AB
= −1
The first and third ratio are given. The second ratio is y+z z because of the areas of the triangles left to the side CD. Hence we obtain 1 y + z −1 · · = −1 2 z 3 and hence y + z = 6z. Plugging y = 5z into the first two equations, we obtain z = 1, y = 5 and finally x = 3. Hence the area of the middle triangle is one seventh of the total area. Definition 26.1. Four points A, B, X, Y on a line are called harmonic, if the two points X and Y divide segment AB from inside and outside in the same ratio. In other words, the cross ratio (AB, XY) =
AX · BY BX · AY
= −1
for the directed segments. Remark. To remember the correct appearance of the points on top and bottom of the fraction, you can use the diagram A B X→Y ↓ ↑ B A X→Y Theorem 26.2 (Theorem of the harmonic quadrilateral). Take any quadrilateral ABCD. Let a pair of opposite sides AB and CD intersect in point X, and the other pair of opposite sides in point Y. Let the diagonals intersect in point M. Let the line Y M intersect side AB in point X 0 . Then the four points A, B, X, X 0 are harmonic points.
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Figure 26.5. The harmonic quadrilateral
Theorem 26.3 (Theorem of Cevá). Let M be a point inside (or outside) of triangle 4ABC. Let the three lines through M and the vertices of the triangle cut the opposite sides in the three points X, Y, Z. One orders the six segments XA, XB, Y B, YC, ZC and ZA such that their midpoints occur in that order on a 360◦ around the triangle. That ordering understood, one gets |XA| |Y B| |ZC| · · =1 |XB| |YC| |ZA| Proof. take triangle 4ABC from the figure of the theorem of Cevá, and let the two line AY and BZ intersect in point M. Now draw line ZY and use it as a transversal for the theorem of Menelaus. Let X 0 be the intersection of the transversal with the extended side AB. The theorem of Menelaus yields |X 0 A| |Y B| |ZC| · · = −1 |X 0 B| |YC| |ZA| One needs still to draw line CM. Let X 0 be its intersection point with line AB. One gets harmonic quadrilateral based on ABYZ. Hence A, B, X, X 0 are four harmonic points. (AB, XX 0 ) =
|AX| · |BX 0 | = −1 |BX| · |AX 0 |
The two equations together imply |XA| |Y B| |ZC| · · =1 |XB| |YC| |ZA| as to be shown.
Problem 26.2. Use the Theorem of Cévà to prove: The three segments from the vertices of a triangle to the touching points of its in-circle intersect at one point. Answer.
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Figure 26.6. The Gergonne point.
Figure 26.7. Three angles of 45◦ .
26.2.
The circle of Apollonius
Proposition 26.1. Given are two circles which have the diameters AB and XY lying on one line, say in the order A ∗ X ∗ B ∗ Y. Let point P be an intersection point of the two circles. The three angles ∠APX, ∠XPB and ∠BPY are 45◦ if and only if the two circles intersect orthogonally.
507
Proof. In the figure on page 507, the angles at vertices A and O are α and 2α, the latter being an exterior angle at the top of an isosceles triangle. Similarly, the angles at vertices Y and Q are y and 2y, respectively. The angle sum in the triangles 4OPQ and 4APY yield ∠OPQ = 180◦ − 2α − 2y ∠APY = 180◦ − α − y ∠OPQ − 90◦ = 2(∠APY − 135◦ ) Hence ∠OPQ is a right angle if and only if α + y = 45◦ if and only if ∠APY = 135◦ . In this case, the two circles intersect perpendicularly. Too, all three angles ∠APX, ∠XPB and ∠BPY are 45◦ since by Thales’ theorem ∠XPY = ∠APB = 90◦ .
Figure 26.8. How three angles of 45◦ are obtained.
−−→ Problem 26.3. Explain why the ray T Y bisects the angle ∠UT V. Find and mark in the figure above: angle α in red, angles 90◦ − α in blue, angles 90◦ − 2α in green.
508
Figure 26.9. Find angles α, 90◦ − α and 90◦ − 2α.
509
Figure 26.10. The points on Apollonius circle have all same ratio of distances from points C and G.
Figure 26.11. The points on Apollonius circle have all same ratio of distances from A and B, and the segments AX and BX appear under equal angles.
For any four harmonic points A ∗ X ∗ B ∗ Y, a circle with the harmonic division points X and Y as diameter XY, is called an Apollonius circle. Theorem 26.4 (Apollonius’ Theorem). All points on an Apollonius circle over the harmonic division points X, Y of segment AB have the same ratio of distance from A and B. From any point R on the Apollonius circle, the segments AX and BX appear under congruent angles ∠ARX ∠XRB.
510
Let now C be any point not on line AB. We mark the three adjacent angles u, v, v0 with vertex C and points A, X, B, Y on their sides, and finally the angle u0 , such that the four angles add up to two right angles. The following statements are equivalent: (1) u v: the two segments AX and XB are seen from C appearing under congruent angles; (1’) u0 v0 : the two segments AY and BY are seen from C appearing under supplementary angles;
1
(2) |AC| |AX| = |BC| |BX| (2’) |AC| |AY| = |BC| |BY| (3) Point C lies on the Thales’ circle over the harmonic division points X and Y. Lemma 26.1. |AC| sin v |AX| = |BC| sin u |BX| |AC| sin v0 |AY| = |BC| sin u0 |BY| AX · BY sin u · sin v0 (AB, XY) = =− sin v · sin u0 BX · AY
(26.3) (26.4) (26.5)
Proof. Let x = ∠AXC and y = ∠AYC. We use the sin theorem several times and get |AC| sin x |BC| sin x |AC| sin y |BC| sin y
|AX| sin u |BX| = sin v |AY| = sin u0 |BY| = sin v0 =
(26.6) (26.7) (26.8) (26.9)
Note we have used that the sin of supplementary angles are equal. By division we get formulas as claimed. "(1) → (2)": Under the assumption u = v, formula (26.3) yields, as to be shown |AC| |AX| = |BC| |BX| "(2) → (1)": Under the latter assumption, formula (26.3) yields, sin u = sin v. Since u + v < 2R, the two angles u and v cannot be supplementary. Hence we get u = v, as to be shown. 1
more exactly, one has to say "angles adding up to two right"
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"(2) ↔ (20 )": Follows at once from the assumption that A, B, X, Y are harmonic points.
We see with a similar argument that (1) ↔ (20 ). Hence all four statements (1),(1’),(2) and (2’) are equivalent. "(1) → (3)": Assume u = v. By the reasoning above, we conclude that u0 = v0 . Hence the rays −−→ −−→ CX and CY are interior and exterior angular bisectors of the angle ∠ACB. As seen earlier (see section 11.6 about the in-circle and ex-circles), these bisectors are perpendicular to each other. By the converse Thales’ theorem 39, the vertex C lies on the circle with diameter XY, which is an Apollonius circle. "(3) → (1)": Assume that point C lies on an Apollonius circle with diameter XY. By Thales’ theorem, angle ∠XCY is a right angle, and hence u + u0 and v + v0 are right angles. From formulas 26.3 and 26.4 we conclude sin v sin v0 cos v = = sin u sin u0 cos u tan u = tan v
(26.10) (26.11)
Both u and v are acute angles. Hence we conclude u = v, as to be shown.
Problem 26.4. Given is the segment AB, for simplicity we assume it to have length |AB| = 3. Provide a drawing and short explanations for the following: • find the point X in the segment AB for which |AX| = 2|BX|; −−→ • find the point Y on the ray AB but outside the segment AB for which |AY| = 2|BY|; • construct the Apollonius circle of all points Z for which |AZ| = 2|BZ|.
Figure 26.12. The points on this Apollonius circle have all the ratio 2 : 1 of distances from points A and B.
Answer.
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Figure 26.13. Harmonic points XY, UV from Apollonius circle.
Problem 26.5. Explain an easy construction of the forth harmonic point with three points A ∗ X ∗ B given. Use a Thales circle over diameter AB. Confirm that the construction is valid. Construction. Let the given segment XY be a diameter of circle δ. At the given point U inside the segment XY, we erect the perpendicular, and let S and T be the intersection points of the perpendicular with this circle. The tangent at T intersect the line XY at the forth harmonic point V. −−→ Reason of validity. To check that X, Y, U, V are harmonic points we prove that the ray T Y bisects the angle ∠UT V. Indeed, by Euclid III.32, the angle between a chord and the tangent at its endpoint is congruent to the circumference angle of this chord. Hence ∠S T Y ∠YT V. The converse argument is easily checked, too. Remark. As follows easily from the leg theorem, U and V are inverse images of each other. The circle with diameter UV contains a pair of inverse points. As well known (from the basics for Poincaré’s disk model) it is thus orthogonal to the circle δ. Furthermore, the 4VT S has Y as center of its in-circle, and X as center of its ex-circle. Once more, this implies that (XY, UV) = −1. Proposition 26.2. Given are two circles which have the diameters AB and XY lying on one line, say in the order A ∗ X ∗ B ∗ Y. Let point P be an intersection point of the two circles. (AB, XY) = − tan2 ∠APX
(26.12)
Corollary 62 (Thales’ second donkey). Equivalent are: (a) Two circles intersect perpendicularly. −−→ −−→ −−→ −−→ (b) the four rays PY, PA, PX, PB from an intersection point P of the two circle make three angles ◦ of 45 . (c) Their two diameters on the line through their center are four harmonic points. Problem 26.6. Prove this corollary, ready to be engraved on the tomb of my former high school math teacher.
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Figure 26.14.
Figure 26.15. The attraction by a conducting sphere is equivalent to the attraction by a mirror charge.
26.3.
An application to electrostatics
Problem 26.7. Near a conducting sphere of diameter AB = 3 m, a point charge of Q Coulomb is placed, say at point Y with distance BY = 1 m from the conducting sphere. We want to calculate the attraction force between the point charge and the sphere. On any conducting boundary, the total potential has to be constant. To assure the potential to vanish on the surface of the conducting sphere, we determine an appropriate mirror charge. The electric field is equal to the field produced by the given point charge together with the field of the mirror charge Q0 . Thus outside of the conducting sphere, its effect is equivalent to the effect of the mirror charge. The location and strength of the mirror charge is determined by means of the boundary condition Q Q0 + =0 |CY| |CX| which needs to hold for any point C on the surface of the sphere. Convince yourself that, because of Apollonius theorem, the mirror charge Q0 is located at the forth harmonic point X.
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(a) Calculate the distances |AX| and |BX| for the forth harmonic point X. (b) Use the boundary condition to determine the mirror charge. (c) Calculate the attraction force F from Coulomb’s law F=
Q0 Q |XY|2
Answer. (a) The forth harmonic point satisfies |AX| |AY| = =4 |BX| |BY| since |AY| = |AB| + |BY| = 4 m and |BY| = 1 m are given. With |AB| = 3 m, we get |AX| =
4 3m 5
and
|BX| =
1 3m 5
(b) The total potential from the two charges needs to satisfy the boundary condition for any point C on the conducting surface. This turns out to be achievable because of Apollonius theorem. Indeed, with any point C on the conducting sphere, we get −
Q0 |CX| |BX| 3 = = = Q |CY| |BY| 5
(c) The distance of the charge to the mirror charge is |XY| =
! 3 8 +1 m= m 5 5
We get the attraction force F from Coulomb’s law F=
3 Q2 15 Q2 Q0 Q = − = − Coulomb m−2 5 (8/5)2 64 |XY|2
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Figure 26.16. Albrecht Dürer 1530
26.4. The perspective view We want to construct the perspective view of objects, lying mainly in a horizontal xy-plane as they appear in a vertical xz-plane. I call the horizontal xy-plane the object plane and the vertical xz-plan the picture plane. Problem 26.8. Suppose you know that the given quadrilateral is a perspective image of a square in a horizontal plane. We have to • reconstruct the horizon • reconstruct the position of the viewer’s eye. To this end, we use the half-plane above the horizon and assume it to be rotated by a right angle around the horizon towards the viewer, out of the vertical picture plane. Answer. • The extensions of the opposite sides AB and CD meet at point G, the opposite sides BC and DA meet at point H. The line GH is the horizon. • The position of the viewer’s eye is reconstructed in the half-plane above the horizon. The diagonals BD and AC intersect the horizon in the points I and J, respectively. The viewer’s eye lies at the intersection of the Thales’ circles above the segments GH and I J. The light rays from the points G and H on the horizon (or points nearby, in reality) enter the viewer’s eye at right angles. Hence the viewer’s eye lies on a Thales circle with diameter GH. Similarly, light rays from the points I and J on the horizon enter the viewer’s eye at right angles. Hence the viewer’s eye lies on a Thales circle with diameter I J, too.
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Figure 26.17. From were does a square look like this?
Figure 26.18. Reconstruction of the position of the eye.
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Figure 26.19. Reconstruction of the eye position and the object square from its perspective image.
Problem 26.9. As in problem 26.8, we reconstruct a square in a horizontal object plane from its perspective image in the picture plane. Reconstruct the horizon, the position of the viewer’s eye, and finally the object square in the object plane.
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Figure 26.20. Tiling of a plane.
Problem 26.10. Explain how the tiling of the plane with translations of the original square is constructed in the perspective view. Why can there no contradictions arise? Give an example of the little Theorem of Desargues applied to two triangles, which shows that an incidence holds as expected.
Figure 26.21. The mapping by perspective in coordinates.
The following problem gives the mapping of perspective in three-dimensional Cartesian coordinates. The eye of the viewer is positioned behind the picture plane at point E, say with coordinates (0, − f, h). The length f is the distance of the eye from the picture plane. In the set up of photography it is the focus length. The coordinate h is determined by the height of the eye above the object plane. For any given point P = (x, y, z), we want to determine its image Q = (x0 , 0, z0 ) on the picture plane. The three points E, P and Q lie on the line given by the light ray from the object to the eye
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of the viewer. From the parametric representation of a line in three dimensional space, we get Q = tE + (1 − t)P 0 0 x 0 = t − f + (1 − t) h z0
x y z
with a real parameter t varying along the line. Problem 26.11. Determine parameter t as a function of y. Get the coordinates x0 and z0 as a function of x, y, z. Answer. The second component of the system gives 0 = −t f + (1 − t)y. Hence t = 1−t =
f f +y .
y f +y
and
We plug this solution into the equations for the first and third component to get fx f +y hy + f z z0 = th + (1 − t)z = f +y x0 = (1 − t)x =
Problem 26.12. Use the result fx f +y hy + fz z0 = f +y
x0 =
to determine the image of a ray y = mx, z = 0 with x ≥ 0. Which point Q∞ on the horizon do you get in the limit x → ∞. Answer. The image of the given ray, and the point on the horizon are fx f + mx hmx z0 = f + mx
x0 =
Hence the line z0 = h is the horizon.
lim x0 =
x→∞
f m
lim z0 = h
x→∞
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27. Advanced Euclidean Geometry
Figure 27.1. How the Euclidean egg should look.
27.1.
A Euclidean egg
Problem 27.1. Use several circular arcs to compose a continuously differentiable curve of the shape of an egg. On each arc, the two endpoints and the midpoint should be easy to construct. Construction 27.1 (A Euclidean egg). Take a segment AB and draw the two circles C1 and C2 with center A through point B, and with center B through point A. Draw the common chord c of the two circles, and let O be its intersection point with segment AB. Draw a circle C3 around O with diameter AB and let point C be one of its intersection points with the common chord c. Let point −−→ D be the intersection of ray AC with the first circle around A. Draw a circle C4 around C through point D. Claim: Arcs from the four circles can be used to compose the egg. Problem 27.2. Prove that an angular bisectors for triangle 4ABC intersects both the leg of this triangle and the common chord c on the smallest circle C4 . − → − → Let the rays AI and BI be angular bisectors for triangle 4ABC, and point I be the center of the in circle. Let G be the intersection point of the side BC with the opposite bisector. One can use Apollonius’ Lemma 40.2 about the angular bisector. Lemma 27.1. The angular bisector of any triangle cuts the opposite side in the ratio of the lengths of the two adjacent sides.
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Figure 27.2. A Euclidean egg—how it is done.
Answer. Apollonius’ Lemma and Pythagoras’ Theorem for triangle 4ACO imply |IC| |AC| √ = = 2 |IO| |AO| With segment length |AB| = 1,
implies
|IC| = |IO|
1 2
√ |IC| = 2 − |IC|
√ 2 |IC| = √ = 2(1 + 2)
√ √ 2( 2 − 1) 2 √
√
On the other hand, the smallest circle C4 has the radius 1 − 22 = 2−2 2 , which is the same. Hence in circle center I lies on the circle C4 . One can use a similar argument, this time for triangle 4ABC. Apollonius’ Lemma and Pythagoras’ Theorem yield |GB| |AB| √ = = 2 |GC| |AC| With segment length |AB| = 1, √ 2 − |GC| √ |GB| = 2 = 2 |GC| |GC| implies again √ √ √ 2 2( 2 − 1) |GC| = √ = 2 2(1 + 2) which is again the radius of the circle C4 . Hence point G lies on that circle, too.
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Problem 27.3. Construct the midpoints of the four arcs of the egg. Answer. Two of the midpoints lie on the common chord c. To get the other two midpoints, let I be the intersection point of the common chord with the smallest circle, lying inside triangle 4ABC. − → The ray BI intersects the first circle around B in the midpoint of the arc on that circle. Similarly, − → the ray AI intersects the second circle around A in the midpoint of the other symmetric arc on the second circle.
Figure 27.3. This is a beginning.
27.2.
The egg built from inside
Problem 27.4. Find the angles marked in the figure on page 522. c are congruent, we get ∠FBC ∠FAC = γ. Answer. Because circumference angles for the arc FC The angle sum in the right triangle 4ACD yields ∠ADB = 90◦ − γ. Congruence of the base angles of the isosceles triangle 4ACD implies 90◦ − γ = 45◦ + γ Hence γ = 22.5◦ , and one gets ∠FBC = 22.5◦ , ∠ADB = 67.5◦ . Problem 27.5. Find the angles marked in the figures 27.2 and 27.2 in terms of α. Answer. Because triangle 4ABC is both right and isosceles, one gets β = 45◦ − α from angle subtraction. Because of congruence of the base angles of the isosceles triangle 4ABI, one gets ∠ABI = α. Because the exterior angle of triangle 4AIE 0 is the sum of two nonadjacent interior angles, one gets ∠AIE 0 = 2α. Next ∠E 0 IC = 90◦ − α, because it is the vertical angle to the congruent angle in the right triangle 4OIB. Now we can draw the egg with the triangle 4ABC, and its angular bisectors inside. Problem 27.6. Set α := 22.5◦ in a figure similar to the figures "too small" or "too large" on pages 523 or 523. Prove that the angular bisectors for the right triangle 4ABC produce three or more lines intersecting at points I, E, F, G—surprising coincidences!
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Figure 27.4. Too small.
Figure 27.5. Too large.
Answer. Because of the assumption α := 22.5◦ , the triangle 4ICE 0 gets congruent base angles. Hence it is isosceles and C = G is the center of the circle through the three points I, E and E 0 . The line AF is both angular bisector and altitude of the isosceles triangle 4AD0 B. By the converse
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Figure 27.6. The egg built from inside—just right.
Thales theorem, the foot point F lies on the circle with diameter AB. 27.3.
A triangle construction using the sum of two sides
Problem 27.7 (Using a sum of two sides). Construct a triangle with side AB = c = 6, ∠BCA = γ = 60◦ and the sum of the other two sides a + b = 10. Show that the two solutions 4ABC and 4A0 BC 0 obtained are symmetric to each other. Construction 27.2. At first, one constructs a triangle 4ABD with sides AB = 6 and DB = 10 and angle ∠BDA = 30◦ . (a) To construct this triangle, begin by drawing any segment BD of length 10 and a ray DE forming the ∠BDE = 30◦ . Then one draws a circle of radius 6 around point B. It intersects the ray DE in two points, say A and A0 . (b) We need still to get vertex C. One uses the isosceles triangle 4ACD. Point C is the intersection of segment BD with the perpendicular bisector of DA. Measuring yields a = 6.9, b = 3.1, approximately. (c) Step (b) has to repeated with point A0 in place of A. Measuring yields a0 = 3.1, b0 = 6.9, approximately. Proof of congruence of the two solutions. The two isosceles 4CAD and 4C 0 A0 D both have two base angles of 30◦ . The two base angles φ of the isosceles 4A0 BA are congruent, too. Hence angle addition at vertex A0 yields φ + α0 + 30◦ = 180◦
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Figure 27.7. Triangle construction from given side, sum of the two other sides, and the angle across.
The angle sum of 4ABD yields φ + β + 30◦ = 180◦ Comparing the two formulas yields α0 = β. Since both triangles 4ABC and 4A0 B0C 0 have congruent angles γ = 30◦ at vertices C and C 0 , and the angle sums of the two triangles 4ABC and 4A0 B0C 0 are both 180◦ , we conclude β0 = α. We see that the vertices of the two triangle 4ABC and 4BA0C 0 correspond in this order. They are congruent, as one checks with the ASA-congruence theorem. Remark. It turns out that problem (27.7) is solvable for given values of c, a + b and γ, if and only if there exists a right triangle with hypothenuse a + b and leg c, and the angle 2δ across that leg is γ δ 2 ≥ 2 . For the side lengths a + b = 10, c = 6 given above, the borderline case occurs for a = b = 5, c = 6 and 2δ = 36.84◦ . The problem is solvable if and only if γ ≤ 73.68◦ . Open Problem. Consider the same triangle construction is hyperbolic geometry. How many solutions does one get? Remark. I can answer that question via hyperbolic trigonometry: It turns out that the problem is c δ solvable if and only if there exists a right triangle with hypothenuse a+b 2 and leg 2 , and the angle 2
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across that leg is α, β.
δ 2
≥ γ2 . Again, the solution turns out to be unique, up to interchange of a, b and
Figure 27.8. In Euclidean geometry, an isosceles solution exists for a + b = 10, c = 6 and
γ 2
≤ 36.84◦
Open Problem. The Euclidean construction given above does not work in hyperbolic geometry, because the angle sum of triangle is less than two right. Find a construction which works in the hyperbolic case! 27.4.
Archimedes’ Theorem of the broken chord
Theorem 27.1 (Archimedes’ Theorem of the broken chord). Let two segments AC and CB make up a broken chord in a circle. Let M be the midpoint of that one of the two circular arcs AB, which contains point C. Drop the perpendicular from M onto the longer one of chords AC and CB, and let F be the foot point. We claim that F is the midpoint of the broken chord AC ∪ CB. In the case AC < CB as drawn in figure on page 527, one gets AC + CF FB and |AC| + |CB| = |FB| 2
(A)
Proof. Point M is the intersection of the perpendicular bisector of AB with the circle arc AB which contains point C. −−→ We extend segment CB beyond point C to the ray BC, and let D be the point on that ray such that MA MD.
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Figure 27.9. Archimedes’ broken chord
Next, Euclid III.21 implies ∠MBD = ∠MBC ∠MAC, 1 since these are circumference angles of the same arc MC. Together with congruence of the base angles of the isosceles triangle 4DBM, transitivity implies congruence of three angles, all denoted by α in the figure on page 527. ∠MDC = ∠MDB ∠MBD = ∠MBC ∠MAC
(1)
DM BM AM
(2)
By construction The goal is to prove the congruence of 4ACM and 4DCM with S S S congruence. To this end, we need to compare the two segments AC and DC. The triangle 4ADM is isosceles by construction, and hence has congruent base angles ∠ADM ∠DAM. Above we have shown that ∠MDC ∠MAC. Hence either angle addition or substraction implies ∠ADC ∠DAC. Finally, the converse isosceles triangle proposition (Euclid I.6) implies AC DC. Since obviously AM DM and CM CM, we have now matched three pairs of sides, and the SSS-congruence implies the triangle congruence 4ACM 4DCM. The hypothenuse-leg theorem yields the triangle congruence 4MBF 4MDF, and hence BF FD. In other words, the altitude of an isosceles triangle bisects its base side. Hence |BF| = |FD| = |FC| + |CD| = |FC| + |CA| as to be shown. Remark. Indeed 4ACM 4DCM, but the triangles 4BCM and 4DCM are not congruent. For both pairs, we can match SSA, because of (1)(2), and the side CM is common of the two triangles, and congruent to itself. The difficulty is that matching SSA do not immediately imply congruence. The triangles 4BCM and 4DCM are not congruent, because their angles at vertex C are supplementary, and not congruent. 1
∠MBD = ∠MBC, because points C and D, and hence all three points M, C and D always lie on the same side of AB.
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In this situation, it is just as easy to check that the angles ∠ACM ∠DCM are congruent (I call this the SSAA congruence). Question. Show that angles ∠ACM ∠DCM are indeed congruent. Answer. Since point C lies on the arc AM, the angle ∠ACM is supplementary to the circumference angle of arc AM, which is congruent to arc MB. Hence angle ∠ACM is congruent to the supplement of angle ∠BCM, which is again the supplement of angle ∠DCM. Question. What happen if one drops the perpendicular from M onto the shorter one of chords AC and CB. Question. What happens if points C and M are chosen to lie on different sides of line AB? Proof using trigonometry. Let P be any point on the circle, not on the arc ACB. Let α = ∠APC and γ = ∠CPB be the circumference angles of the arcs AC and CB, respectively. We assume that the circle has radius 1. The corresponding chords have lengths and average |AC| = 2 sin α , |CB| = 2 sin γ |AC| + |CB| = sin α + sin γ (27.1) 2 The chords to the midpoint M have circumference angle α+γ 2 and length α+γ |AM| = |MB| = 2 sin 2 The arc CM has the circumference angle ∠CBM = γ−α . Hence using the definition of the cosin 2 function in the right 4FBM yields γ−α α+γ γ−α |FB| = |MB| cos = 2 sin cos (27.2) 2 2 2 Because of (27.1) and (27.2), Archimedes’ assertion that |AC| + |CB| = |FB| (A) 2 is equivalent to the trigonometric formula γ−α α+γ cos sin α + sin γ = 2 sin 2 2 which is less well known, but correct. Problem 27.8 (Archimedes meets Thales for a triangle construction). In the triangle 4ACB, the following pieces are given: side AB = c, opposite angle ∠ACB = γ, and sum of its adjacent sides AC + CB. Construct the triangle using Archimedes’ broken chord. Is there more than one solution? Construction 27.3. We use Archimedes theorem of the broken chord. One constructs segment AB, and a circle Arch with center O, through its endpoints such that the circumference angle of arc AB is γ. The hard part is now to locate vertex C on the circle. Again look at the drawing from Archimedes broken chord. We can readily get point M, but still need to get foot point F. Here is how to get two coordinates of F. Clearly, point F lies on a circle around B of given radius |BF| = |AC+CB| . We draw this circle, 2 named Circ. What else can be done to find F? Apply the tangents from point M to circle Circ. Thales needs to draw his circle T h, with diameter MB! Because of the right angle ∠MFB, the point F is a intersection of Thales circle T h with circle Circ. Finally point C is the intersection of line BF with circle Arch.
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Figure 27.10. Triangle construction (27.8)
Figure 27.11. The second symmetric solution for construction (27.8)
530
Question. Explain how one gets the circle Arch. Answer. The center O lies on the perpendicular bisector of segment AB. The isosceles triangle 4ABO has two congruent base angles 90◦ − γ. Or, we can use the fact that segment AB and the tangent to circle Arch at point A form the angle γ. Next we get the radius of circle Arch, since it is perpendicular to the tangent. Now the center lies where this radius intersects the perpendicular bisector of AB. Question (a). Compare with the solution with the one we have found earlier in another way. Use Archimedes theorem of the broken chord and Thales’ circle to construct a 4ACB with given side AB = 6, opposite angle γ = 60◦ and sum of adjacent sides AC + CB = 10. Measure angle α, and compare with the other construction from above. Question (b). Use Archimedes theorem of the broken chord and Thales’ circle to construct a triangle 4ACB with side AB = 6, opposite angle γ = 30◦ and sum of adjacent sides AC + CB = 14.
Figure 27.12. The two symmetric solutions of (b)
Construction 27.4 (Still another variant for the triangle construction (27.7)). One constructs segment AB of the given length, its perpendicular bisector, and the point O on the bisector such that angle ABO = 90◦ − γ. Let Arch be the circle with center O through A and B. On this circle, the circumference angle of arc AB is γ. Let M be the intersection of the perpendicular bisector of segment AB with circle Arch lying on the same side of AB as center O. Let Hal f be the circle with center M, through A and B. On this circle, the arc AB has the circumference angle γ2 . Let D be the intersection point of circle Hal f with the circle around B of radius a + b. Finally point C is the intersection of line BD with circle Arch.
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Figure 27.13. Still another variant of triangle construction (27.8)
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27.5.
The Theorem of Collignon
Theorem 27.2 (The Theorem of Collignon). On the outside of the four sides of an arbitrary quadrilateral are erected isosceles right triangles. The two pairs of additional vertices of the triangles erected on opposite sides of the quadrilateral are connected. Then the two connecting segments are congruent and perpendicular.
Figure 27.14. The Theorem of Collignon yields two congruent segments PR and QS orthogonal to each other.
Problem 27.9. To prove the Theorem of Collignon, it is helpful to use vectors, represented by complex numbers. Any directed segment is given by the difference of endpoint minus start point, e.g. the segment PS is given by s − p. Let a, b, c, d ∈ C be the complex numbers representing the vertices of the quadrilateral—as one follows its sides in counterclockwise order. Let p, q, r, s be the extra vertices at the four right angles of the isosceles right triangles 4APB, 4BQC, 4CRD and 4DS A. (i) Find the complex numbers corresponding to the vertices P, Q, R, S . (ii) Find the complex numbers for the segments PR and QS . (iii) Prove the claim. (iv) Formulate the theorem for a self intersecting quadrilateral. (v) Is there an easy relation between the diagonals of the original quadrilateral ABCD and the segment PR?
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Solution. (i) Let M be the midpoint of segment AB. The difference AM = m − a = b−a 2 . The ◦ difference MP is obtained by rotating AM by 90 clockwise, hence MP = −iAM and P = M + MP =
b−a 1+i 1−i a+b −i = a+ b 2 2 2 2
Together with the other three cases, we get 1+i 1−i a+ b 2 2 1+i 1−i Q= b+ c 2 2 1+i 1−i R= c+ d 2 2 1+i 1−i S = d+ a 2 2 P=
(ii) PR = r − p =
−1 − i −1 + i 1+i 1−i a+ b+ c+ d 2 2 2 2
and QS is calculated similarly. QS =
−1 − i −1 + i 1+i 1−i b+ c+ d+ a 2 2 2 2
(iii) Comparison of the two results yields −1 − i [a − ib − c + id] 2 1−i [a − ib − c + id] = i PR QS = 2 PR =
The equation QS = i PR means that QS is obtained by rotating PR by an angle of 90◦ counterclockwise. This implies the claim. (iv) The theorem remains valid for a self intersecting quadrilateral, if one erects the triangles always on the same side—say at right hand—as one cycles through its sides. Problem 27.10. Draw the figure for the Theorem of Collignon with a generic parallelogram, and for a self intersecting parallelogram. Explain what is remarkable. Proof. In case of a parallelogram, the two congruent orthogonal segments PR and QS bisect each other at its midpoint. The case of a self intersecting "quasi-parallelogram", looks surprising. The two congruent orthogonal segments PR and QS need not even intersect. 27.6.
Vectors and special quadrilaterals
Problem 27.11. Check: any four points A, B, C, D on a line satisfy AB · CD + BC · AD = AC · BD as an identity for the directed segments.
(27.3)
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Figure 27.15. In case of a parallelogram, the two congruent orthogonal segments PR and QS bisect each other at its midpoint.
Conclude that for any quadrilateral ABCD, the corresponding vector equation with dot products is valid: −−→ −−→ −−→ −−→ −−→ −−→ AB · CD + BC · AD = AC · BD (27.4) You can put −−→ −−→ −−→ a := AB , b := BC , c := CD
(27.5)
−−→ −−→ −−→ AC = a + b , BD = b + c , AD = a + b + c
(27.6)
and get
Answer. This exercise is left to the reader. Problem 27.12. Only for a part out of the 24 possible arrangements of the four different points A, B, C, D on a line does one get an equation |AB| · |CD| + |BC| · |DA| = |AC| · |BD|
(27.7)
for the absolute values. Find all the possible orders of the four points for which this equation holds. Answer. Any permutation of A, B, C, D brings one of the four letters to the first place. Equation (27.7) holds iff its terms are either all three equal to the terms of the equation (27.4) for the directed segments, or all three terms are the negatives of those in equation (27.4). This requirement leads to two possible orders of the remaining three letters. For example, in case B is the first letter, the order B ∗ C ∗ D ∗ A leads to the terms of equations (27.7) and (27.4) being the same, whereas order B ∗ A ∗ D ∗ C leads to the terms of equations (27.7) and (27.4) being all three negative to each other. Thus one ends up with the eight possible arrangements. Equation (27.7) holds exactly for the following orders: A∗ B∗C ∗D , D∗C ∗ B∗A ,
B∗C ∗D∗A , C∗B∗A∗D ,
C∗D∗A∗B, B∗A∗D∗C ,
D∗A∗ B∗C A∗D∗C ∗ B
and
(27.8) (27.9)
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In other words, the four letters can be permuted cyclically and reversed in order. Still another formulation: the letters A, C are alternating with the letters B, D. 1 27.7. The Theorem of Ptolemy A simple closed polygon is a polygon the sides of which do not intersect, except for the endpoints of adjacent sides. A polygon with four vertices is called a quadrilateral. Let the vertices A, B, C, D of a simple quadrilateral occur in alphabetic order, as one moves around its boundary. The segments AC and BD are called diagonals. The segments AB and CD are one pair of opposite sides. The segments BC and AD are the second pair of opposite sides. Proposition 27.1 (Ptolemy’s Theorem). A quadrilateral has a circum circle, if and only if the product of its diagonals equals the sum of the products of the two pairs of opposite sides. Corollary 63. For any four vertices A, B, C, D |AC| · |BD| ≤ |AB| · |CD| + |BC| · |DA|
(27.10)
|AC| · |BD| = |AB| · |CD| + |BC| · |DA|
(27.11)
The equation holds if and only either the four point lie on a circle, and are ordered clockwise or counterclockwise— or they lie on a line, in a way to be obtained by orthogonal projection from the above points on a circle. Especially, for any four vertices A, B, C, D lying neither on a a circle, nor a line the strict inequality holds in (27.10). Remark. If all four vertices lie on a line, the equation AC · BD = AB · CD + BC · DA
(27.12)
does hold for the directed segments. Proof of the Corollary. It is left to the reader to consider the case of four points lying on a line. Let the quadrilateral be ABCD, and assume that the three points B, C and D do not lie on a line. On the segment AB, we construct a triangle 4ABE ∼ 4DBC, by transferring the angle of 4DBC at it vertices D and B. We put point E and C on the same side of line DB. By Euclid VI.4, the sides of equiangular triangles are proportional. Hence |BA| |BD| = and |AE| |DC| |BA| |BE| = |BD| |BC|
(27.13) (27.14)
(i) If E = C then A = D. If E = A then A = B. −−→ −−→ (ii) The rays AE = AC are equal if and only if the the four points A, B, C, D lie on a circle and points A and D lie on the same side of BC. −−→ −−→ Reason for (ii). Assume the rays AE = AC are equal. Then the angles ∠BAC ∠BAE are congruent. But ∠BDC ∠BAE because of the similar triangles 4DBC ∼ 4ABE. Hence ∠BAC ∠BDC. Now Euclid III.21 implies that the four points A, B, C, D lie on a circle, and points A and D lie on the same side of BC. The converse is as easy to check. 1
The permutations obtained in this way are not a subgroup of S 4 .
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Figure 27.16. Two pairs of similar triangles: 4DBC ∼ 4ABE and 4ABD ∼ 4EBC.
The latter proportion (27.14), and the congruent angles ∠ABD ∠EBC show that the two triangles 4ABD and 4EBC have their angles at vertex B congruent, and the sides containing that angle proportional. Hence, by Euclid VI.6, they are similar, too: 4ABD ∼ 4EBC and hence |BD| |BC| = |AD| |EC|
(27.15)
|BD| · |AE| = |AB| · |CD| |BD| · |EC| = |BC| · |AD| |BD| · (|AE| + |EC|) = |AB| · |CD| + |BC| · |DA|
(27.16) (27.17) (27.18)
The proportions (27.13) and (27.15) imply
The triangle inequality |AC| ≤ |AE| + |EC| and equation (45.2) immediately yield the first assertion (27.10). Now we show that the points A, B, C, D lie on a circle if and only if |AC| · |BD| = ±|AB| · |CD| ± |BC| · |DA| (iii) If |AC| · |BD| = |AB| · |CD| + |BC| · |DA|, then point E lies between A and C. The four points A, B, C, D lie on a circle and point A lies on the arc BD opposite to C. (iv) If |AC| · |BD| = |AB| · |CD| − |BC| · |DA|, then point A lies between E ad C. The four points A, B, C, D lie on a circle and point A lies on the arc BC opposite to D. (v) If |AC| · |BD| = −|AB| · |CD| + |BC| · |DA|, then point C lies between A ad E. The four points A, B, C, D lie on a circle and point A lies on the arc CD opposite to B.
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Figure 27.17. If E lies between A and C, the point A lies on the arc BD opposite to C.
Figure 27.18. If point A lies between E and C, the point A lies on the same arc BD as point C.
Reason for (iii). The equations (45.2) and (27.11) imply |AE|+|EC| = |AC|. Hence E lies between −−→ −−→ A and C and the rays AE = AC are equal. As explained in item (ii), this implies that the four points A, B, C, D lie on a circle and points A and D lie on the same side of BC. The same argument with B and D interchanged, yields that points A and B lie on the same side
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Figure 27.19. If C lies between A and E, the point A lies on the arc CD opposite to B.
of DC. Hence point A lies on the arc BD opposite to C.
Reason for (iv). Interchanging B and C bring one back to case (iii).
Reason for (v). Interchanging B and D bring one back to case (iv).
We can attack to get the converse of item (iii): Assume that the four points A, B, C, D lie on a circle and point A lies on the arc BD opposite to C. Clearly, the points A and D lie on the same side of BC. Hence we can use item (ii), and conclude −−→ −−→ that the rays AE = AC are equal. Hence either E lies between A and C— or C lies between A and E. To rule out the second case, note that it would imply |AC| · |BD| = |AB| · |CD| − |BC| · |DA|, and hence, by item (iv), point A would lie on the arc CD opposite to B–contrary to the assumption. We are left with the first case, which implies equation (27.11), as expected. Similarly, we get (iii*) The equation (27.11) holds if and only if point E lies between A and C. which happens if and only if the four points A, B, C, D lie on a circle, and point A lies on the arc BD opposite to C. (iv*) The equation |AC| · |BD| = |AB| · |CD| − |BC| · |DA| holds if and only if point A lies between E and C, which happens if and only if the four points A, B, C, D lie on a circle, and point A lies on the arc BC opposite to D. (v*) The equation |AC| · |BD| = −|AB| · |CD| + |BC| · |DA| holds if and only if point C lies between A and E, which happens if and only if the four points A, B, C, D lie on a circle, and point A lies on the arc CD opposite to B. Item (iii*) yields the proof—except the case of four points lying on a line, which I leave to the reader.
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27.8.
The quadrilateral of Hjelmslev
Definition 27.1. A quadrilateral with two right angles at opposite vertices is called a Hjelmslev quadrilateral. Recall that a quadrilateral with two right angles at opposite vertices is called a Hjelmslev quadrilateral. In a Hjelmslev quadrilateral quadrilateral, we draw the diagonal between the two right angles and drop the perpendiculars from the two other vertices. Their occur some remarkable congruences.
Figure 27.20. The Theorem of Hjelmslev.
Proposition 27.2 (Hjelmslev’s Theorem). In a Hjelmslev quadrilateral, (a) at each of the two vertices with arbitrary angle, there is a pair of congruent angles between the adjacent sides, and the second diagonal and the perpendicular dropped onto the first diagonal, respectively; (b) there is a pair of congruent segments on the diagonal between the right vertices, measured between these vertices and the foot points of the perpendiculars. Problem 27.13. Prove Hjelmslev’s conjecture about angles in Euclidean geometry. Proof of the angle congruence in Euclidean geometry. Let ABCD be the Hjelmslev quadrilateral, with right angles at the opposite vertices A and C. The midpoint of the segment BD is the center of the circum-circle of the quadrilateral ABCD, as follows from the converse Thales’ Theorem, Corollary 39. At vertex B, we have to check congruence of the angles β := ∠ABD and β0 = ∠GBC. Since the triangle 4ABD has a right angle at vertex A, the angle α := ∠ADB is complementary to β. Since
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Figure 27.21. Euclidean proof of the angle congruence of Hjelmslev.
the triangle 4GBC has a right angle at vertex G, the angle α0 := ∠GCB is complementary to β0 . The angles α and α0 are circumference angles of the same arc AB and hence congruent by Euclid III.21. Hence their complementary angles β β0 are congruent, too. Lemma 27.2. If two chords in a circle have congruent or supplementary circumference angles, the two chords are congruent. Proof. The proof can be left as an exercise.
Proof of the segment congruence in Euclidean geometry. We extend the perpendicular DH and let F be the intersection point of the extension with the circum circle. We claim the triangle congruence 4CBG 4AFH from which the claimed congruence CG AH follows immediately. The triangle congruence is now shown by the SAA congruence theorem. Besides the right angles at vertices G and H, a further pair of congruent angles exists at vertices B and F. Indeed ∠CBG = β0 β = ∠DBA ∠DFA = ∠HFA The congruence of the hypothenuses is shown with the converse of Euclid III.21, given in lemma 27.2 above. The circumference angles of the segments CB and AF at the vertex D are ∠CDB = γ0 γ2 = ∠ADF Their congruence was shown in the first part of Hjelmslev’s Theorem. Hence the triangle and segment congruences stated above are proved.
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Figure 27.22. Euclidean proof of the segment congruence of Hjelmslev.
Problem 27.14. Convince yourself that the theorem holds for a self-intersecting (not simple) quadrilateral, too. Give a drawing for that case. Check how the proof has to be modified for the case of a self-intersecting quadrilateral, and provide a drawing for that case. For the case of intersecting sides AB and CD, the segments CG and AH shown to be congruent will not overlap.
Figure 27.23. The theorem of Hjelmslev for a self-intersecting quadrilateral.
Indication of proof . I choose a quadrilateral were the sides AB and CD intersect. In that case, the segments and angles, congruence of which is stated, do not overlap, if they did not overlap for convex quadrilateral neither. Some details are different. For example ∠CBG = β0 β = ∠DBA 2R − ∠DFA = ∠HFA
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Figure 27.24. The proof of Hjelmslev’s theorem for an overlapping quadrilateral.
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28. The Regular Pentagon 28.1. The Euclidean construction with the Golden Ratio The figure on page 545 shows an easy Euclidean construction of a regular pentagon. The justification of the construction begins by first considering the regular 10-gon. Furthermore—as customary since Legendre—we take a shortcut by solving a quadratic equation. By a "slice of a regular n-gon", I mean an isosceles triangle with base of which is one side of the polygon, and the third vertex of which is the center of the circum circle. Problem 28.1. Find two equiangular triangles within the slice of the regular 10-gon. Set up the relevant proportion to determine the side length x of the 10-gon within a circum circle of radius 2. Get a quadratic equation for x and solve it.
Figure 28.1. Two similar triangles appear within the slice of the regular 10-gon.
Similar triangles appear within the slice of the regular 10-gon. Let O be the center of the circum circle and AB a side of the 10-gon. The isosceles triangle 4OAB has base angles 72◦ . The top angle is 36◦ —which is just half of the base angles. Hence it is natural to bisect one of them. Let BC be the bisector. The triangles 4OAB and 4BAC are equiangular. By Euclid VI.4, the sides of equiangular triangles are proportional. Hence we get a proportion |AB| |AC| = |AO| |AB|
(28.1)
Both triangles have a pair of congruent base angles. By Euclid I.9, they are isosceles, and hence OC BC BA. By assumption we set |AO| = 2 and |AB| = x, and hence get
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|AC| = |AO| − |OC| = 2 − x. Thus the proportion (28.1) yields x 2−x = 2 x x2 = 2(2 − x) The quadratic equation for x has the two solutions x1,2 = −1 ±
√ 5.
Definition 28.1 (Golden ratio). The ratio of the longer to the shorter side of the isosceles triangle with angles 36◦ , 72◦ , 72◦ is called the golden ratio. Its value is √ 5+1 φ= = 1.61803 . . . 2 Problem 28.2. Check that φ = 2x and 1 − φ = − 2x , and that these are the two solutions of the quadratic equation φ2 − φ − 1 = 0. Problem 28.3. Do and describe, and justify he Euclidean construction of the regular pentagon. For simplicity, choose the radius of the circum circle equal to 2 units. Answer. One begins the construction with a circle around O of radius |OA| = 2. Next one bisects the segment OO0 perpendicular to OA. Let D be the midpoint. √ The right triangle 4OAD has legs |OA| = 2, |OD| = 1 and hence hypothenuse |AD| = 5, by the Theorem of Pythagoras. The circle around D √ through point O intersects the hypothenuse in point E. Segment subtraction yields |AB| = |AE| = 5 − 1 = x, which is just the side of the regular 10-gon. Finally, a circle around A through point E intersects the circum circle drawn in the beginning in two adjacent vertices B and B5 of a pentagon. Finally, one gets all vertices B B1 B2 B3 B4 B5 of the regular pentagon since its sides are congruent, and its vertices all lie on the circum circle. The construction is done in the figure on page 545. Problem 28.4. Alternatively, the equation for the side x of the 10-gon can be justified with Euclid III.36, the theorem about secants and tangents to a circle. This is done in the figure on page 545. Explain which circle you use, and how to get a tangent and a pair of secants from point B outside this circle. Justification using secants in a circle. Let AF and BC be the angular bisectors in triangle 4OAB. We use the circle through points O, C and F. The center H of this circle lies on the third angular bisector of the angle with vertex O. Since ∠OCH ∠COH = 18◦ and ∠ACB ∠CAB = 72◦ , angle subtraction at vertex C confirms that the angle ∠BCH = 180◦ − 18◦ − 72◦ = 90◦ is a right angle. We use the circle through points O, C and F. The tangent from point B to this circle touches at point C. Finally, we use Euclid III.36—about secants and tangents to a circle—and get |BC|2 = |BO| · |BF| x2 = 2(2 − x) as above.
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Figure 28.2. A simple Euclidean construction of the regular pentagon.
Figure 28.3. Euclid’s theorem of secants gives another derivation of the golden mean.
28.2.
Relation between the sides of pentagon and 10-gon
Problem 28.5. Let M be the midpoint of the rhombus ABCB5. Use the Theorem of Pythagoras
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for the right triangle 4OMB and calculate the side y = |BB5| of the pentagon. The length of a side for the pentagon. The right triangle 4OMB has the legs |OM| = |MB| = 2y . The hypothenuse is |OB| = 2. Hence the Theorem of Pythagoras yields
2+x 2
and
(2 + x)2 y2 + =4 4 4 from which we get We can use x =
√
y2 = 16 − (2 + x)2 5 − 1 and get the side of the pentagon explicitly: √ √ y2 = 16 − (2 + x)2 = 16 − ( 5 + 1)2 = 10 − 2 5 q √ y = 10 − 2 5
Theorem 28.1 (A Relation between the Sides of Pentagon and 10-gon (Euclid XIII.10)). The sides of the regular pentagon, hexagon, and 10-gon inscribed into the same circle are the sides of a right triangle. Problem 28.6. Use the information gathered so far to give a short algebraic proof of Euclid XIII.10. Algebraic proof of Euclid XIII.10. Since x2 = 4 − 2x we get y2 = 16 − 4 − 4x − x2 = 12 + (2x2 − 8) − x2 = 4 + x2 By the converse Pythagorean Theorem, y is the hypothenuse of a right triangle with legs 2 and x, as claimed by Euclid XIII.10. A more traditional proof of Euclid XIII.10. In the regular pentagon BCDEF, bisecting the angle ∠FOB produces the vertex A of a regular 10-gon with the same circum circle around O. We drop the perpendicular from O onto the side AB of the 10-gon, and let N be its intersection point with the side FB of the pentagon. Euclid’s theorem of tangent and secants (Euclid III.36) gives rectangles for the squares of segments FO and BA. −−→ Question. Why is the ray FO a tangent to the circle through points O, N and B and touches at point O. What does Euclid III.36 imply. Answer. The angles ∠FON ∠FBO = 54◦ are congruent. Hence they are angle between tangent and secant and circumference angle for the same chord ON. Hence we conclude that |FO|2 = |FN| · |FB| Of course, you get the same result from the observation that the triangles 4FON ∼ 4FBO are equiangular and hence similar. −−→ Question. Why is the ray BA a tangent to the circle through points A, N and F and touches at point A. What does Euclid III.36 imply. Answer. The angles ∠BFA ∠BAN = 18◦ are congruent. Hence they are circumference angle and angle between tangent and secant for the same chord NA. Hence we conclude that |BA|2 = |BN| · |BF|
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Figure 28.4. Proving Euclid XIII.10.
Of course, you get the same result because the triangles 4BFA ∼ 4BAN are equiangular and hence similar. Adding the two equations results in |FO|2 + |BA|2 = |FN| · |FB| + |BN| · |BF| = (|FN| + |BN|) · |BF| = |BF|2 Indeed, as to be shown, the sum of the squares of the hexagon side |FO| and the 10-gon side |BA| is the square of the pentagon side |BF|. By a "10-gon three-side-diagonal", I mean a diagonal spanning over three adjacent sides of the 10-gon, hence corresponding to a central angle three times the central angle of a side. Theorem 28.2 (A Relation between the diagonals of pentagon and the 10-gon). The diagonal of the regular pentagon is the hypothenuse of a right triangle with the side of the hexagon, and the three-side-diagonal of the 10-gon as legs. Problem 28.7. Use the figure on page 548 to prove Theorem 28.2. Solution of Problem 28.7. In the regular pentagon BCDEF, bisecting the angle ∠FOB produces the vertex A of a regular 10-gon with the same circum circle around O. 1 We drop the perpendicular from O onto the side AF of the 10-gon, and let N be its intersection point with the diagonal FC of the pentagon. Euclid’s theorem of tangent and secants (Euclid III.36) gives rectangles for the squares of segments FO and CA. 1
We substitute AB 7→ AF, FB 7→ FC.
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Figure 28.5. Proving Euclid XIII.10.
−−→ Question. Why is the ray FO a tangent to the circle through points O, N and C and touches at point O. What does Euclid III.36 imply. Answer. The angles ∠FON ∠FCO = 18◦ are congruent. Hence they are angle between tangent and secant and circumference angle for the same chord ON. Hence we conclude that |FO|2 = |FN| · |FC| Of course, you get the same result from the observation that the triangles 4FON ∼ 4FCO are equiangular and hence similar. −−→ Question. Why is the ray CA a tangent to the circle through points A, N and F and touches at point A. What does Euclid III.36 imply. Answer. The angles ∠CFA ∠CAN = 54◦ are congruent. Hence they are circumference angle and angle between tangent and secant for the same chord NA. Hence we conclude that |CA|2 = |CN| · |CF| Of course, you get the same result because the triangles 4CFA ∼ 4CAN are equiangular and hence similar. Adding the two equations results in |FO|2 + |CA|2 = |FN| · |FC| + |CN| · |CF| = (|FN| + |CN|) · |CF| = |CF|2 Indeed, as to be shown, the sum of the squares of the hexagon side |FO| and the 10-gon threediagonal |CA| is the square of the pentagon diagonal |CF|.
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Problem 28.8. We use the notation as in the figure on page 548. Check that the side and three-side diagonal of the 10-gon, and the side and diagonal of the pentagon satisfy √ √ 5−1 5+1 |AC| |AF| = and = |OA| 2 |OA| 2 s s √ √ |FC| |FB| 5− 5 5+ 5 = and = |OA| 2 |OA| 2 Give a more simple algebraic proof of Theorem 28.2. 28.3.
The construction with Hilbert tools √
Problem 28.9. Find the irreducible monic polynomial with zero gates, and show that the number is an algebraic integer.
5−1 2 .
Find all algebraic conju-
q √ 5− 5 Problem 28.10. Find the irreducible polynomial with zero z1 = 2 . Find all its algebraic conjugates, and show that the number is a totally real algebraic integer. Solution of Problem 34.7 . Simple arithmetic shows that (2z2 − 5)2 − 5 = 0 and hence P(z) := z4 − 5z2 + 5 = 0 is the monic polynomial in the ring Z[z] with zero z1 . Hence z1 is an algebraic integer. The Eisenstein criterium tells: For a polynomial to be irreducible in the ring Z[z], it is sufficient that all its coefficients except the leading one are divisible by the same prime number p, but the constant coefficient is not divisible by p2 . The polynomial P(z) is irreducible since it satisfies this condition for p = 5. Hence its zeros are exactly the algebraic conjugates of z. Obviously, these zeros are s s s s √ √ √ √ 5+ 5 5+ 5 5− 5 5− 5 , − , , − 2 2 2 2 which all four turn out to be real.
The set of all lengths constructible by Hilbert tools is called the Hilbert field and denoted by Ω. By Proposition 34.13, all lengths constructible by Hilbert tools are totally real. Hence the domain T of all totally real Euclidean numbers satisfies Ω ⊆ T ⊆ K, where K denotes the Euclidean field. 1
To come back to the pentagon, we have seen in Problem 34.7 by means of algebraic arguments that z1 ∈ T . From Artin’s Theorem, we conclude that even z1 ∈ Ω. Here is an even easier direct proof leading to the same conclusion: By Proposition 31.1, the Hilbert field Ω is the smallest field with properties (1) 1 ∈ Ω. (2) If a, b ∈ Ω, then a + b, a − b, ab ∈ Ω. 1
By a theorem of Emil Artin, the Hilbert field Ω is exactly the field T of the totally real algebraic Euclidian numbers.
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(2a) If a, b ∈ Ω and b , 0, then (3) If a, b ∈ Ω, then
√
a b
∈ Ω.
a2 + b2 ∈ Ω.
For the side and diagonal of the pentagon, we get v s t √ 2 √ 1 ± 5 5± 5 + 1 = 2 2 which we can now see to be obtained recursively by the steps (1)(2)(3) specified above. Thus we conclude that its side and its diagonal, and hence the pentagon is constructible with Hilbert tools. It is now time to really do a construction by Hilbert tools! Problem 28.11. Use the three right triangles from above, and find a construction of the regular pentagon by Hilbert tools. For simplicity, I have chosen 4 as radius of the circum circle.
Figure 28.6. A construction of the regular pentagon by Hilbert tools.
Construction of the pentagon by Hilbert tools. We can use the three right triangles which came up above: √ (i) Triangle 4ODA has legs |OA| = 2, |OD| = 1 and hypothenuse |AD| = 5. (ii) Euclid XIII.10 yields a right triangle with legs 2, x and hypothenuse y. (iii) Triangle 4OMB has legs |OM| =
2+x 2 , |MB|
= 2y , and hypothenuse |OM| = 2.
551
We begin with a segment |OA| = 2 and erect the perpendiculars at its endpoints O and A. Transfer a segment |OD| = 1 onto the first perpendicular, as well as to the hypothenuse of the resulting triangle 4ODA. √ We get a segment |ED| = 5 − 1 = x. and transfer this segment onto the second perpendicular −−→ and the ray opposite AO to get segments |AE| = |AF| = x. The hypothenuse of the right triangle 4OAE is |OE| = y, congruent to the side of the pentagon in a circle of radius 2. We transfer this segment onto the first perpendicular on both sides of O, and get segments |OG| = |OH| = y. The isosceles triangle 4FGH is one of the slices of a pentagon with circum circle around F with radius |FG| = |FH| = 4. The four remaining slices are obtained by angle transfer at center F.
Figure 28.7. Still another construction of the pentagon with Hilbert tools.
Problem 28.12. Again, we assume that |OA| = 2. Check that the construction done in the figure on page 551 yields the segments √ √ 5−1 5+1 |OC| = and |OB| = 2 s s2 √ √ 5− 5 5+ 5 |DC| = and |DB| = 2 2 Problem 28.13. Use the information gathered so far justify the construction of the regular pentagon with Hilbert tools—as done in the figure on page 551.
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Validity of the construction . Here are the major steps of the construction done in the figure on page 551: (i) On the radius OA, a perpendicular is erected. Its midpoint is D. −−→ −−→ (ii) Construction the inner and outer angular bisectors DC and DB of the angle ∠ODA. Let C and B be the intersection points of the bisectors with the line OA. (iii) Transfer segment DC at point B onto the perpendicular erected on line OA. On both sides of this line, one gets the congruent segments BE BF DC. (iv) Transfer segment DB at point C onto the perpendicular erected on line OA. On both sides of this line, one gets the congruent segments CG CH DB. (v) Finally connects the vertices AHFEG of a regular pentagon. The side and diagonal of the pentagon were obtained in Problem 28.8. By Problem 28.12, they agree with the segments we have just constructed.
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Figure 28.8. This is a Euclidean construction for the regular pentagon.
28.4.
Variants of the Euclidean construction
Problem 28.14 (Another Euclidean construction of the regular pentagon). From the construction done in the figure on page 553, and |OA| = 2 given, calculate exact expressions for the segments (i) |OC| and |OD|. (ii) |BC| and |BD|. Answer. We follow the steps of the construction to get exact expressions for the segments. √ √ (i) |OC| = 5 − 1 and |OD| = √5 + 1. From |OB| = 2 and |MB|2 = |MO|2 + |OB|2 = 5 one gets |MC| = |MD| = |MB| = 5 and √ |OC| = |MC| − |MO| = 5 − 1 √ |OD| = |MD| + |MO| = 5 + 1 (ii) Two more applications of the Theorem of Pythagoras: √ √ |BC|2 = |BO|2 + |OC|2 = 22 + ( 5 − 1)2 = 10 − 2 5 √ √ |BD|2 = |BO|2 + |OD|2 = 22 + ( 5 + 1)2 = 10 + 2 5 q q √ √ |BC| = 10 − 2 5 and |BD| = 10 + 2 5 By comparison with the results of Problems ?? and ??, we get a proof that the construction does yield a regular pentagon. Problem 28.15. Get a formula for sin 18◦ . Guess and check the exact values of sin kφ, cos kφ and tan kφ for multiples of φ = 18◦ and k = 1, 2, 3, 4. This is just an exercise in clever use of the calculator!
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Answer. We can use the figure on page 543. The slice of the regular 10-gon is the isosceles triangle 4OAB with top angle 36◦ . We drop the perpendicular from center O onto the side AB. The foot point G is its midpoint. The right triangle 4OGA yields √ |GA| 5−1 sin 18◦ = = |OA| 4 As a further hint, we use the expressions obtained in Problem 28.14: √ √ |OD| |OC| 5−1 5+1 = and = |OB| 2 |OB| 2 s s √ √ |BC| 5− 5 |BD| 5+ 5 = and = |OB| 2 |OB| 2 But some of them are larger than one, so you still need to divide. Here is the result: kφ
sin kφ √
18◦ 36◦ 54◦ 72◦ 90◦
5−1 q 4√ 5− 5 8 √ 5+1 q 4√ 5+ 5 8
1
cos kφ q √ 5+ 5 8 √ 5+1 q 4√ 5− 5 8 √ 5−1 4
0
tan kφ √ √ 3/2 2 5 20 5 − √ √ 3/2 10 5 20 5 − √ 3/2 √ 2 5 20 5 + √ √ 3/2 10 5 20 5 + ∞
Figure 28.9. This construction does not yield a regular pentagon.
28.5.
A false pentagon
Problem 28.16 (A false pentagon construction). Into the construction done in the figure on page 554, put a Cartesian coordinate system with O = (0, 0), A = (2, 0) and B = (0, 2). (i) Calculate exact expressions for the coordinates of B1 .
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(ii) Calculate exact expressions for the coordinates of B2 . (iii) Determine exactly the ratio |B3 B2 | |B2 B1 | Check of the false pentagon construction . We need to calculate the coordinates for intersections points of circles. √
Point B1 is an intersection point of the circles Circ1 and (i) The coordinates of B1 are ( 455 , 34 ). Circ2 . Since |OB| = 2 and |BM|2 = |BO|2 + |OM|2 = 5, they have the equations: Circ1 : x 2 + y2 = 4 Circ2 :
x2 + (y − 2)2 = 5
Subtraction yields (y − 2)2 − y2 = 1,
and hence y =
With the equation for Circ1 we get 9 x + = 4, 16 2
√
3 . 4
√ and hence x =
55 . 4
(ii) The coordinates of B2 are ( 3 1655 , − 23 Point B2 is an intersection point of the circles Circ1 16 ). and Circ3 . They have the equations: Circ1 : x 2 + y2 = 4 √ 2 !2 3 55 + y − =5 Circ3 : x − 4 4 √ 2 !2 55 3 2 − y2 = 1 Subtraction yields x − − x + y − 4 4 √ 55 55 3 9 − x+ − y+ =1 2 16 2 16 √ 55x + 3y = 6 Plugging into the equation for Circ1 yields √ 2 55 2 x + 2 − x = 4 3 √ 64x2 4 55x − =0 9 3 √ 3 55 x1 = 0 and x2 = 16 The solution x = 0 leads back to point B. The new intersection point B2 corresponds to the solution x2 . Plugging into the linear relation of x and y yields √ 55x + 3y = 6 √ √ 3 55 55 + 3y = 6 16 55 23 y=2− =− 16 16
556
(iii) The exact ratio of the longer to the shorter side is √ √ 3 55 |B3 B2 | 3 11 8 = √ = |B2 B1 | 8 5
557
29. Circles, Tangents, Power and Inversion 29.1.
The equipower line of two circles
Definition 29.1. The power of a point P with respect to a circle C is defined by POW(P, C) = ±|PA| · |PB| Here A and B are the two intersection points of any line l through P with the circle C. The power is negative for points inside the circle—and positive for points outside the circle. We denote the power of the point P with respect to a circle C by POW(P, C). Soundness of definition. Let k be any other line through P intersecting the circle C, now in the points A0 and B0 . We need to confirm that we get the same value for the power, using line k. Indeed |PA| · |PB| = |PA0 | · |PB0 | by Euclid III.35 and III.36. Hence the defining product does not depend on the choice of the line through point P. Moreover, we can choose a line through the center of circle in which case the intersection points A and B are known to exist. Therefore the power is well defined. Given are any two circle CA and CB with different centers A , B. Let RA = b and RB = a be their radii and |AB| = c be the distance of their centers. Theorem 29.1 (The equipower line). Given is any Pythagorean plane, neither the circle-circle intersection property nor the circle-line intersection are assumed. Given are two circles with different centers. All points which have equal powers with respect to the two circles lie on a line which is perpendicular to the line AB through the centers of the two circles. The equipower line is unique and constructible with Hilbert tools.
Existence of the equipower line. We choose any two different points P and Q on circle CA which are not mirror images across the line AB, and a third point R on CB which does not lie on line PQ. Let C be the circle through the three points P, Q and R. The circles C and CB either intersect in a second point S , or they have a common tangent r touching at point R. In the first case, we define the line r = RS . Lemma 29.1. The lines p = PQ and r = RS intersect at a unique point Z. Reason for the Lemma. Indeed, assume towards a contradiction that lines p and r are equal or parallel. The center C of circle C is the intersection of the perpendicular bisectors b of PQ and c of RS . These bisectors are perpendicular to p and r, respectively. Under the assumption that p k r, they are parallel or equal, too. Since b and c intersect at center C, they are equal. Hence there would exist a line b = c such that P, Q and R, S are two pairs of reflection images across c. Furthermore, the three centers A, B and C all lie on line c. This contradicts the choice of points P and Q. The point Z lies both on the common chord PQ of circles C and CA , and the common chord or tangent r = RS of circles C and CB . Hence by the definition of the power of a circle POW(Z, CA ) = |ZP| · |ZQ| = POW(Z, C) = |ZR| · |ZS | = POW(Z, CB )
558
Figure 29.1. Construction of the equipower line of two circles.
We see that point Z lies on the equipower line of the two given circles CA and CB . We construct the perpendicular e to line AB through point Z and let F be the foot point on line AB. For any point E on the line e POW(E, CA ) = (|EA| + b)(|EA| − b) = (|FA| + b)(|FA| − b) + |EF|2 = POW(F, CA ) + |EF|2 follows easily from the Pyrthagoras |EA|2 = |FA|2 + |EF|2 for the right triangle 4AFE. Here b is the radius of circle CA . Similarly, we get POW(E, CB ) = POW(F, CB ) + |EF|2 We may choose E = Z and conclude from POW(Z, CA ) = POW(Z, CB ) that POW(F, CA ) = POW(F, CB ) POW(E, CA ) = POW(E, CB ) for all points E of the line e. Thus we have shown existence of the equipower line. All steps of the above construction can be done by Hilbert tools. Uniqueness of the equipower line. For the points on the line AB, use coordinates A = (0, 0), B =
559
(c, 0), F = ( f, 0). Assume that F is any point on the line AB for which equal powers. One calculates POW(F, CA ) = POW(F, CB ) |FA|2 − b2 = |FB|2 − a2 f 2 − ( f − c)2 = b2 − a2 f =
b2 − a2 + c2 2c
to conclude that point F is uniquely determined. The argument from above shows that every equipower line is perpendicular to the line AB connecting the two centers. Hence the entire equipower line is uniquely determined. Corollary 64. In any Pythagorean plane, the equipower lines of any three circles with three different centers either intersect in one point, or they are all three parallel. The second case occurs if and only if the three centers lie on a line. Reason. Assume any two of the three equipower lines of the three circles intersect in a point Z. Then the point Z has equal power with respect to all three circles, and hence is the intersection of all three equipower lines. On the other hand, assume that two of the three equipower lines are parallel. Then the lines AB and BC connecting the centers are both perpendicular to these equipower lines and hence equal. Hence the three centers A, B, C lie on a line. The possible different cases for the mutual positions of any two circles in a Hilbert plane have been explained in Proposition 10.10. In a Pythagorean plane, we get now additional statements about the their equipower line e, and the intersection point F of the equipower line with the line AB. Let RA = b > 0 and RB = a > 0 be their radii and |AB| = c > 0 be the distance of their centers. Thus we exclude the case of concentric circles, for which no equipower line exists. Lemma 29.2. Equivalent statements are (i) Point F lies inside circle CA ; (ii) f 2 < b2 ; (iii) (a + b − c)(b + c − a)(c + a − b) > 0; (iv) a < b + c, b < a + c, c < a + b; (v) Point F lies inside circle CB ; (vi) ( f − c)2 < a2 . Proof. From 2c f = b2 − a2 + c2 one gets after squaring 4c2 ( f 2 − b2 ) = (b2 − a2 + c2 )2 − 4b2 c2 = (b2 − a2 + c2 + 2bc)(b2 − a2 + c2 − 2bc) = (b + c + a)(b + c − a)(b − c + a)(b − c − a)
560
We see that (i) is equivalent to (ii) which is equivalent to (iii). In the product (iii) all three factors are positive. Indeed that a, b, c are positive implies that no two of the three factors can be zero or negative. Hence (iii) and (iv) are equivalent. From the formula 4c2 (( f − c)2 − a2 ) = (a2 − b2 + c2 )2 − 4a2 c2 = (a2 − b2 + c2 + 2ac)(a2 − b2 + c2 − 2ac) = (b + c + a)(b + c − a)(b − c + a)(b − c − a) we see that (iii) and (vi) are equivalent. Since (v) and (vi) are equivalent, we are ready.
Proposition 29.1. In a Pythagorean plane are given two circles with different centers. The following list exhausts all possible cases for the circles, their equipower line e, and the intersection point F of the equipower line with the line connecting the two centers. (i) a + b < c: The interiors of the two circles are disjoint. The equipower line lies exterior to both circles and A ∗ F ∗ B: the point F lies between the two centers A and B. (i0) a + b = c: The interiors of the two circles are disjoint. The two circles touch from outside. The equipower line is their common tangent and A ∗ F ∗ B. (ii) b + c < a: The interior of circle CA is a subset of the interior of circle CB . The equipower line lies exterior to both circles and F ∗ A ∗ B. (ii0) b + c = a: The interior of circle CA is a subset of the interior of circle CB . The two circles touch from inside. The equipower line is their common tangent, F ∗ A ∗ B and F is the touching point of the common tangent. (iii) a + c < b: The interior of circle CB is a subset of the interior of circle CA . The equipower line lies exterior to both circles and F ∗ B ∗ A. (iii0) b + c = a: The interior of circle CB is a subset of the interior of circle CA . The two circles touch from inside. The equipower line is their common tangent, F ∗ B ∗ A and F is the touching point of the common tangent. (iv) a < b + c, b < a + c, c < a + b: The interiors of the two circles intersect, but neither one is a subset of the other one. In that case, each circle intersects both the interior and exterior of the other one. The point F lies in the intersection of the two interiors, and A ∗ F ∗ B. The equipower line intersects the interior of both circles. This Proposition holds without assuming neither the circle-circle intersection property nor the circle-line intersection. Lemma 29.3. For any two circles CA and CB with different centers and their equipower line e CA ∩ e = C B ∩ e = CA ∩ C B Two circles intersect if and only if any one of them intersects their equipower line. Proof. Assume that any of the three intersections from the lemma is nonempty. Assuming CA ∩ e = {C, D}, we conclude that points C and D have power zero with respect to both circles, and all three intersections are the same set {C, D}. Assuming CA ∩ CB = {C, D}, we conclude that points C and D have power zero with respect to both circles, and hence lie on the equipower line By the uniqueness of the equipower line, all three intersections are the same set {C, D}.
561
Theorem 29.2. Given is any Pythagorean plane, and the circle-line intersection property is assumed. (a) For two circles touching from inside or outside, the equipower line is their common tangent. (b) For two circles which do not intersect and have different centers, the equipower line is the locations of all points around which there exists a circle perpendicular to the two given circles. In other words, the equipower line is the locations of all points which have congruent tangent segments to both circles. (c) For circles intersecting in two points, the equipower line is their common chord. Proof. We use Proposition 29.1 and its notation. Item (a) corresponds to the cases (i0) or (ii0) or (iii0). The lengths a, b, c satisfy either a + b = c or b + c = a or b + c = a. Further details are left to the reader. Item (b) corresponds to the cases (i) or (ii) or (iii). The lengths a, b, c satisfy either a + c < b or b + c < a or a + b < c. Here the two circles do not intersect and have different centers. The line-circle intersection property implies that the tangent from any point outside to a circle is constructible by Construction 10.1. (This construction even works in every Hilbert plane.) Take any point E on the equipower line. As we know from Proposition 29.1, point E lies outside both circles CA and CB . Let S , T be the touching points of a tangent from E to circles CA and CB , respectively. From Euclid III.36 and the definition of the power, we get POW(E, CA ) = |ES |2
and
POW(E, CB ) = |ET |2
For a point on the equipower line, we conclude that |ES | = |ET |. Too, the circle around E through S and T intersects both circles CA and CB perpendicularly. Item (c) corresponds to the case (iv) The lengths a, b, c satisfy a < b + c and b < a + c and c < a + b. By Lemma 29.2, each circle intersects both the interior and exterior of the other one. Moreover, the point F lies in the interior of both circles. We show that the two circles intersect. Since point F lies on the equipower line e, the circle-line intersection property yields the existence of the intersection points {C, D} = CA ∩ e = CB ∩ e = CA ∩ CB Hence two circles intersect provided that one of them intersects both the interior and exterior of the other one. Too, we see that the equipower line is the common chord, as to be expected. Corollary 65. In any Pythagorean plane, the circle-line intersection property implies the circlecircle intersection property. Moreover, the tangent from a point outside to a circle is constructible.
562
Figure 29.2. Construction of an outer common tangent by similar triangles.
Figure 29.3. Construction of an inner common tangent by similar triangles.
29.2. Common tangents of two circles The drawing on page 562 constructs an outer common tangent using similar triangles. The second drawing on page 562 shows the construction of an inner common tangents, again using similar triangles. Construction 29.1 (Construction of the common tangents to two circles by means of similar triangles). To get the two outer common tangents, we construct at first their intersection point I. We draw an arbitrary pair of parallel radii OA and O0 A0 , and get point I as intersection of line AA0 with the line OO0 connecting the two centers. A special case occurs for two circles of equal radii—in that case the lines AA0 , OO0 and the common tangent S S 0 , are all three parallel. To get the two inner common tangents, we draw a pair of any two anti-parallel radii OB and 0 0 O B . The lines BB0 and OO0 intersect in point J, which is the intersection of the two inner tangents, too.
563
Finally we construct the tangents from points I and J to anyone of the two circles, and obtain their common tangents as well as the touching points. Reason of validity of the construction. Because of Thales’ (second) Theorem, the equiangular triangles 4IOA and 4IO0 A0 have proportional sides: |OA| |IO| = 0 0 0 |IO | |O A | Because of symmetry, the intersection point X of the outer tangents lies on the line OO0 through the two centers. We get point I as the intersection point of line AA0 with the line OO0 connecting the two centers. We confirm that X = I. Let us assume we have constructed a tangent from point I to the circle around O0 , and we have obtained the touching point S 0 . Let F be the footpoint of the perpendicular dropped from the center O of the other circle onto this tangent. We get a second pair of equiangular triangles 4IOF and 4IO0 S 0 . Again by Thales’ (second) Theorem, they have proportional sides: |IO| |OF| = 0 0 0 |IO | |O S | Both proportions together imply |OA| |OF| = 0 0 0 0 |O A | |O S | 0 0 0 0 Since |O A | = |O S | are two congruent radius, we conclude |OA| = |OF|. Hence footpoint F lies on the circle around O, too, and we have obtained a common tangent. Remark. Alternatively, we can base the justification on the Converse Desargues Theorem 4.2: We assume the line S S 0 to be a common tangent. As before, we draw an arbitrary pair of parallel radii OA and O0 A0 . Indeed, the two triangles 4OS A and 4O0 S 0 A0 have pairwise parallel sides. By the Converse Desargues Theorem, they are in perspective. Hence the three lines OO0 , S S 0 and AA0 either intersect in one point I, or are all three parallel. 29.3. Definition and construction of the inverted point Let D be a open circular disk of radius R and center O, and denote its boundary circle by ∂D. Definition 29.2 (Inversion by a circle). The inversion by the circle ∂D is defined to be the mapping from the plane plus one point ∞ at infinity to itself, which maps an arbitrary point P , O, to its −−→ inverse point P0 —defined to be the point on the ray OP such that |OP| · |OP0 | = R2 . Hence, especially, all the points of ∂D are mapped to themselves. The inversion maps the origin O to ∞, and ∞ to O. We denote the images by inversion with primes. Problem 29.1. Do an example for the construction of the inverse point. Use the theorems related to the Pythagorean theorem. Construction 29.2 (Inversion of a given point). Let P be the given point. One erects the −−→ perpendicular onto ray OP at point P. Let C be an intersection point of the perpendicular with ∂D. Next one erects the perpendicular on radius OC at point C, and gets a tangent to circle ∂D. The inverse point P0 is the intersection of that tangent with the ray OP. Reason. Indeed, by the leg theorem, | OP| · | OP0 | = |OC|2 = R2 .
Remark. Here is an alternative justification of the construction. Put in Thales’ circle with diameter CP0 . By the converse Thales’ theorem, P lies on that circle. Then use the chord-tangent theorem Euclid III.36. One concludes | OP| · | OP0 | = |OC|2 = R2 , once more.
564
Figure 29.4. Construction of the inverse point
Figure 29.5. The gear of Peaucollier.
29.4. point.
The gear of Peaucollier The gear of Peaucellier allows a construction of the inverse
Problem 29.2 (The gear of Peaucellier). Six stiff rods are linked and can be turned flexibly by each other within a plane. Two rods of length a are linked to each other at point O, the remaining endpoints are A and C. The four remaining rods have a different length b—they are linked to a rhombus, and the points A and C on one diagonal are linked to the first two rods. (i) Assume that a > b. Prove that the endpoints of the other diagonal of the rhombus are inverse points by a suitable circle ∂D with center O. Find the radius of this circle. You can use the Theorem of chords Euclid III.35 for a second circle L with center A.
565
(ii) Check that the assumption a > b implies that point O lies outside circle L. Let OT be a tangent to the same circle. Use Euclid III.36 to get the length of OT . (iii) Check the Theorem of Pythagoras for the right triangle 4OT A. (iv) Prove that the circles ∂D and L intersect each other perpendicularly. (v) What happens in the case a < b?
Figure 29.6. The Theorem of chords explains the gear of Peaucollier.
Answer. We draw the circle L around A through points P and P0 . This circle intersects line OA in two endpoints E and F of a diameter. Too, the three points O, P and P0 lie on a line. (i) The assumption a > b implies that P and P0 lie on the same side of O. We use the Theorem of chords Euclid III.35 (see 19.3) to conclude |OP| · |OP0 | = |OE| · |OF| = (a + b)(a − b) Hence P and P0 are inverse points by a circle of radius R, and R2 = (a + b)(a − b) = a2 − b2 (ii) By the Theorem of chord and tangent Euclid III.36 (see 19.3), the length of the tangent from point O to the circle L satisfies |OT |2 = |OP| · |OP0 | Together, we get |OT |2 = a2 − b2 = R2 , and hence |OT | = R. (iii) The right triangle 4OT A has legs |OT | = R, |T A| = b, and hypothenuse |OA| = a. Hence the Theorem of Pythagoras tells that a2 = R2 + b2 , as we have already seen above.
566
(iv) The segment AT is a radius of circle L. The perpendicular segment OT is a radius of circle ∂D, and at the same time a tangent of circle L. Hence AT is a tangent of circle ∂D, and the two circles intersect perpendicularly. (v) In the case b > a, the point O lies between P and P0 . Now P and P0 are antipodal points by a circle ∂D of radius Ra , but R2a = b2 − a2 . Point O lies now inside this circle. The drawing on page 566 provides an example.
Figure 29.7. In case b > a, the gear of Peaucollier constructs the antipodal point.
29.5. Invariance properties of inversion After two further definitions, we can state the main result of this section. Definition 29.3. A generalized circle is defined to be either a circle or a straight line. Definition 29.4. The cross ratio of four point A, B, C, D is defined as (AC, BD) =
|AB| · |CD| |CB| · |AD|
A→C C←A
B→D B→D
Remark. Remember:
Main Theorem 23. The inversion by a circle maps generalized circles to generalized circles, conserves angles, and conserves the cross ratio. Proposition 29.2. The inversion by a circle maps generalized circles to generalized circles.
567
Figure 29.8. A circle, not going through O is mapped to a circle.
Circles not through O are mapped to circles. Given is a circle C which does not go through center O. We prove that its inverse image is a circle, too. Take any two lines l and k through O. Let A, B and P, Q be their intersection points with C. By definition of inversion, |OA| · |OA0 | = |OB| · |OB0 | = R2 . And hence R4 R4 = |OA| · |OB| p where p is the power of point O relative to circle C. For the points P and Q on the second line k, one calculates again |OA0 | · |OB0 | =
|OP0 | · |OQ0 | =
R4 R4 = |OP| · |OQ| p
Since this is the same value |OP0 | · |OQ0 | = |OA0 | · |OB0 | Euclid III.35 and III.36 imply that the four points A0 , B0 , P0 and Q0 lie on a circle C0 . Indeed, since the line k is arbitrary, we have shown that the images of all points of C lie on that circle C0 . Question. What is the power of point O relative to the inverted circle C0 . Answer. The calculation above show that C0 .
R4 p
is the power of point O relative to the inverted circle
In the case that point O lies outside of circle C, one can conclude even more. In the limiting case that P moves to Q, line k becomes a tangent from point O to circle C. In the same process, P0 −−→ −−→ moves to Q0 . Hence the ray OP = OP0 becomes a common tangent of the two circles C and C0 . Hence both common tangents of circles C and C0 intersect at point O. This construction suggests that circles C and C0 are preimage and image for a central dilation with center O. We now prove this claim in both cases that O lies outside or inside of C.
568
Figure 29.9. The common tangent of a circle and its inverse image intersect at O.
Lemma 29.4. A central dilation z with center O and ratio k=
R2 p
maps the circle C to the circle C0 . The intersection points P and Q of circle C with any central ray k are mapped as P 7→ Q0 and Q 7→ P0 —in a different way as inversion by circle ∂D maps them. The central dilation z maps the center Z of circle C to the center Z∗ of circle C0 , and the touching point C of a common tangent to touching point C2 . Hence |OC2 | |OZ∗ | = =k |OC| |OZ| Proof. We use proportions. Indeed |OQ0 | |OP0 | |OP0 | · |OP| R2 = = = =k |OP| |OQ| |OQ| · |OP| p Hence the dilation z maps P 7→ Q0 and Q 7→ P0 . Too, the center Z of circle C is mapped to he center Z∗ of circle C0 , and the touching point C of a common tangent to touching point C2 . A circle through O is mapped to a line. We now consider the exceptional case that the point O lies on C, and prove that its image by inversion is a line. Let OA be a diameter of circle C and P be −−→ an arbitrary point on that circle. We erect the perpendicular c∗ on ray OA at the inverse point A0 . −−→ Let P∗ be the intersection of c∗ with the ray OP. The right triangles 4OAP and 4OP∗ A are similar. Hence by Euclid VI.6, corresponding sides have the same ratio: |OP| |OA0 | = |OA| |OP∗ |
569
Figure 29.10. A circle through O is mapped to a line.
and hence |OP| · |OP∗ | = |OA| · |OA0 | = R2 . Thus P∗ = P0 is the inverse image of point P, and the line c∗ = C0 is the inverse image of C. Proposition 29.3. The inversion by a circle conserves angles. The angle between a radial ray and a circle not through O is conserved. To show angles are conserved, we need to map further objects by the dilation z. Let t be the tangent to circle C at point P. The dilation z maps the tangent t to the tangent t∗ to the circle C0 at point Q0 . By simple facts about central dilations, the tangents t and t∗ are parallel. Hence by Euclid I.29, the lines t and t∗ −−→ intersect the ray OP in congruent angles α = α00 . Now let t2 be the tangent to circle C0 at the point P0 . (Why is t2 not the image of t under the inversion? This question distracts a bid, but see: the image of tangent t is a circle though O, and has a common tangent with C0 at point P0 .) In general, the tangents t∗ and t2 intersect, say at point S . We get an isosceles 4S P0 Q0 with two congruent base angles α = ∠S Q0 P0 ∠S P0 Q0 Thus the tangent t to circle C at P, and the tangent t2 to circle C0 at P0 both intersect the projection −−→ ray OP at angle α. −−→ In the exceptional case that the tangents t∗ and t2 are parallel, they both intersect the ray OP at right angles. Since the tangents t∗ and t are always parallel, both the tangent t to circle C at P, and −−→ the tangent t2 to circle C0 at P0 intersect the projection ray OP at right angle. Problem 29.3. Show that the angle between a central ray and a circle through O is conserved by inversion. To this end, prove that the three angles α, α0 and α00 in the figure on page 569 are congruent. Answer.
570
Figure 29.11. The angle between a circle and the radial ray is conserved.
The angle between generalized circles is conserved. It is now easy to see that angles between circles are conserved. One maps two circles C1 , C2 intersecting at point P into two generalized −−→ circles C01 , C02 . They intersect at point P0 . One puts in the common projection ray OP and uses angle addition. The angle between the two generalized circles C1 and C2 is congruent to the angle between C01 and C02 . It is left to the reader to check the remaining cases—involving generalized circles.
Figure 29.12. Relation of the ratio of three points, and ratio of their inverted images.
571
Proposition 29.4 (Conservation of the Cross Ratio). The inversion by a circle conserves the cross ratio of any four points A, B, C, D. (The four points need not lie on a circle.) Reason. In figure on page 570, the circle through points A, B, C is mapped by inversion to the circle through the inverted points A0 , B0 , C 0 . On the inverted circle C0 , both the inverted points A0 , B0 , C 0 , and the dilated points A2 , B2 , C2 are marked. The first goal is to show that |A0 B0 | |AB| |OC2 | = · ((*)) |C 0 B0 | |CB| |OA2 | The easy part is to use the central dilation z. Because central dilations conserve ratios, we get |AB| |A2 B2 | = |CB| |C2 B2 |
(1)
We need now to relate the distances |A2 B2 | and |C2 B2 | to the distances |A0 B0 | and |C 0 B0 |. To this end, we use angles in circle C0 to find similar triangles 4OA2 B2 ∼ 4OB0 A0 Clearly 4OA2 B2 and 4OB0 A0 have a common angle at vertex O. By Euclid III.35, we get supplementary ∠B0 B2 A2 and ∠B0 A0 A2 , because these angles subtend the two disjoint arcs from B0 to A2 on circle C0 . Because the angles ∠OB2 A2 and ∠B0 B2 A2 are supplementary, we get congruent angles ∠OB2 A2 ∠B0 A0 A2 ∠B0 A0 O. Hence the two triangles have two pairs of congruent angles. Because the angle sum in any triangle is two right angles (Euclid I.32), all three angles of the two triangles are pairwise congruent. Hence, by Euclid VI.4, these triangles are similar. Hence |A0 B0 | |A2 B2 | = |OB0 | |OA2 | This ratio is actually
sin α0 sin ω .
Similarly one gets that |C 0 B0 | |C2 B2 | = |OB0 | |OC2 |
and dividing the two ratios yields |A0 B0 | |A2 B2 | |OC2 | = · |C 0 B0 | |C2 B2 | |OA2 |
(2)
Now (1) and (2) imply the claim (*). Now one can use a similar relation for the three points A, D and C, just replacing point B by point D: |A0 D0 | |AD| |OC2 | = · |C 0 D0 | |CD| |OA2 |
((**))
(It is not required that all four points A0 , B0 , C 0 , D0 lie on one circle, so that second relationship uses possibly a different pair of circles.) Division of the two relations (*) and (**) cancels out the last 2| fraction |OC |OA2 | on the left, and leads to an equation between of cross ratios: (A0C 0 , B0 D0 ) = (AC, BD)
572
30. A Glimpse at Elliptic Geometry 30.1.
Elliptic geometry is derived from spherical geometry
Problem 30.1. This is a quite interesting interpretation of the undefined terms of incidence geometry: A "point" is interpreted as a pair {P, Pa } of antipodal points on the unit sphere S 2 in Euclidean three dimensional space. A "line" is to be a great circle C on the sphere. A "A point {P, Pa } lies on a line C" means that the great circle C contains both antipodal points P and Pa . Question. Which of the axioms of incidence geometry are satisfied, and which not? Answer. All three axioms (I.1),(I.2) and (I.3) are satisfied.
Figure 30.1. A spherical line
Reason for (I.1). Given two "points" {P, Pa } and {Q, Qa }, they determine in the ambient Euclidean three space two different lines PPa and QQa , which both pass through the origin (0, 0, 0). The two lines determine a plane , which intersects S 2 in a great circle . This great circle is the "line" connecting the two given points. Reason for (I.2). Given two "points" {P, Pa } and {Q, Qa }, they determine the plane uniquely, and hence the great circle from above uniquely . Hence the "line" connecting the two given points is unique.
573
Reason for (I.3). Every great circle contains infinitely many points, and hence more than one antipodal pair of points. The three points (±1, 0, 0), (0, ±1, 0) and (0, 0, ±1) do not lie on a great circle. Question. Which parallel property (elliptic, Euclidean, hyperbolic, or neither) does hold? Answer. The elliptic parallel property holds. Any two great circles intersect. Hence there do not exist parallel lines. 30.2.
The conformal model
Definition 30.1 (The stereographic model of elliptic geometry). Consider the following model of elliptic geometry, built in the Euclidean plane as ambient underlying reality. Throughout, we denote the open unit disk by D = {(x, y) : x, y real, x2 + y2 < 1} Its boundary is the unit circle ∂D = {(x, y) : x, y real, x2 + y2 = 1} The center of D is denoted by O. • The "points" are the points in the interior of the unit disk D as well as pairs of antipodal points on its boundary ∂D. • The "lines" are closed circular segments that pass through a pair of antipodes on ∂D. Problem 30.2 (Construction of an elliptic line). Construct the elliptic line through any two given points P and Q. The solution uses pairs of antipodes and the Lemma below. Definition 30.2 (The antipode). For any given point P, we define the antipode Pa to be the point −−→ on the ray opposite to OP such that OP · OPa = 1. For the center O itself, the antipodal point is the point at infinity. Lemma. If a circle passes through a pair of antipodes, it consists entirely of pairs of antipodes. Hence a circle is an elliptic line if and only if it passes through a pair of antipodes. Proof of the Lemma. Assume that a circle C passes through a pair of antipodes P and Pa . Let Q be a third point on the circle. We draw the chords POPa and QO. Let Q2 be the second intersection point of the (Euclidean) line OQ with circle C. Because of the theorem of chords (Euclid III.35), we know that OP · OPa = OQ · OQ2 By definition of antipodes OP · OPa = 1 and hence OQ · OQ2 = 1 The point O lies inside the circle C, and hence Q and Q2 lie on different sides of O, (O lies between Q and Q2 .) This confirms that the second intersection point Q2 = Qa is the antipode of Q . Question. Provide the drawing for the Lemma.
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Figure 30.2. An elliptic line consists of pairs of antipodes
Figure 30.3. Construction of the antipode
Construction 30.1 (The antipode). To construct the antipode of a given point P, one uses the theorems related to the Pythagorean Theorem. One erects the perpendicular h to radius OP at point O . Let C and Ca be the intersections of h with ∂D. Next one erects the perpendicular on CP at point C. The antipodal point Pa is the intersection at that perpendicular with the line OP. Indeed, by Thales’ theorem, the four points P, C, Pa and Ca lie on a circle. Hence Euclid III.35 implies that OP · OPa = OC · OCa = 1. Question. Provide the drawing for this construction.
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Construction 30.2 (Elliptic line through two given points). Let two points P and Q be given. To find the elliptic line through P and Q, we construct the antipode Pa of one of the two points, say P. In the special case that P lies on ∂D, the construction is easy, since PPa is a diameter of D. If P lies in the interior of D, and is not the center O itself, we proceed as explained above. Finally, we need a circle C through the three points P, Pa and Q. Its center is the intersection of the perpendicular bisectors of segments PQ, PPa and QPa . Reason for validity of the construction. Thus the circle C consists entirely of pairs of antipodes. Hence the two circles C and ∂D intersect in the pair of antipodes A and Aa . Hence C is an elliptic line. Problem 30.3. Do the construction for an example. Put in—as a check—the common chord of the two circles. It should come out to be a diameter of ∂D.
Figure 30.4. Construction of an elliptic line
30.3.
Falsehood of the exterior angle theorem
Proposition 30.1 (No Exterior Angle Theorem). The Exterior Angle Theorem does not hold in spherical geometry. Neither does it hold in elliptic geometry. The following construction gives a counterexample. For simplicity, take antipodes B and D = Ba on the equator ∂D. We connect them by two different half great circles a and c, lying symmetric to the south pole O. Next draw a segment through O that cuts c at point A and a at point C, but not at a right angle. Nevertheless, you get a figure which is point symmetric by point O. Hence the triangle 4ABC has the interior angle γ = ∠ACB congruent to the nonadjacent exterior angle δ = ∠DAC. Question. Draw the figure, as explained, in the conformal model, just the southern hemisphere. Measure and report the three angles of your 4ABC. What do you observe about segments AB and CD, and what about segments BC and AD.
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Figure 30.5. The exterior angle can be congruent or smaller than a nonadjacent interior angle.
Answer. The segments AB CD are congruent, and segments BC AD are congruent.
Proposition 30.2 (No SAA Congruence). S AA congruence does not hold in spherical geometry. Neither does it hold in elliptic geometry. Proof. The following construction gives an example to confirm this claim. Choose any angle γ < 45◦ . We construct two non-congruent triangles with AB = 45◦ , α = 90◦ and γ as given. Take two pairs of antipodes A, Aa and B, Ba on the equator with AB = 45◦ . We chose for C any point on the diameter AOAa , thus getting a right angle α = 90◦ . Let C2 be the point such that segment CC2 has midpoint O. Finally, one needs to get the great semi-circles BCBa and BC2 Ba . They are point symmetric with respect to center O. Question (a). Draw the figure as explained into the southern hemisphere of the conformal model. Stress the triangles 4ACB and 4AC2 B with two different colors. Question (b). Measure and report the angles γ = ∠ACB and γ2 = ∠AC2 B. Answer. Question (c). Which pieces of the two triangles are pairwise congruent? Answer. Obviously the common side AB is congruent to itself. Both triangles have a right angle at vertex A. Because of point symmetry of the entire figure by O, they are congruent. Angles γ1 = ∠ACB γ2 = ∠AC2 B are congruent, too. Question (d). For which congruence theorem are these just the matched pieces? Nevertheless, the two triangles 4ACB and 4AC2 B are not congruent!
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Figure 30.6. Two non congruent triangles matching in two angles and the side across one of them.
Figure 30.7. SAA congruence in neutral geometry
Answer. The matching pieces are those required for SAA congruence (Theorem 25 in Hilbert’s Foundations). Question (e). What do you observe about the two (corresponding but not congruent) sides of triangles 4ACB and 4AC2 B, which lie across the second given angle, and were not prescribed initially? Answer. These are the two sides BC and BC2 . They turn out to be supplementary. 30.4.
Area of a spherical triangle
Proposition 30.3 (The area of a spherical triangle). The area of a spherical triangle is proportional to the excess of its sum of angles over two right angles. With its angles measured in radians, the area of a spherical 4ABC on a sphere of radius R is A = (α + β + γ − π) R2
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Figure 30.8. Another example of two non congruent triangles matching in two angles and the side across one of them.
By a lune LA is denoted the region lying between two half great circle. Question (a). Let α be the angle at vertex A between the two great half circles bounding the lune, measured in radians. Find a formula for the area the area |LA | of this lune. Answer. The area |LA | is proportional to α. For α = π, one gets a half sphere which has area 2π R2 . Hence the area of the lune between to half great circle intersecting at angle α is |LA | = 2αR2 Now we want to find the area of 4ABC. To get an easy drawing, I use the conformal model. We can assume that vertices A and B lie on the equator ∂D, and vertex C on the southern hemisphere, I mean inside disk D. Question (b). Draw an example. I find the case with acute angles α and β easier to draw. Extend the side AC to the antipode Aa on the equator ∂D, and furthermore to the antipode Ca on the northern hemisphere outside the disk D. Similarly, extend the side BC to the antipode Ba on the equator ∂D, and furthermore to the antipode Ca on the northern hemisphere, outside the disk D. Check whether the three points C, Ca and O lie on a Euclidean line. In the figure, we use three lunes: (a) the lune LA with tips A and Aa , bounded by the extended sides ABAa and ACAa of 4ABC. (b) the lune LB with tips B and Ba , bounded by the extended sides BABa and BCBa of 4ABC. (c) the lune LC with tips C and Ca , bounded by the extended sides CAaCa and CBaCa .
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Figure 30.9. A spherical triangle
The first two lunes lie totally on the southern hemisphere (interior of disk D). Only the third lune wraps around the equator to the northern hemisphere (exterior of disk D). Question (c). You may color the three lunes with three different colors. (I have used different textures instead only for photocopying.) What are the areas of the three lunes? Answer. |LA | = 2α R2 , |LB | = 2β R2 , |LC | = 2α R2
(1)
We now derive two equations for areas. The two lunes LA and LB overlap in the 4ABC. They cover the entire southern hemisphere except of the 4CAa Ba . Since a hemisphere has area 2π R2 , one gets for the areas |LA | + |LB | = |LA ∩ LB | + |LA ∪ LB | = |4ABC| + 2π R2 − |4CAa Ba |
(2)
The lune LC consists just of the two non-overlapping triangles 4Aa BaCa and 4CAa Ba . Hence |LC | = |4CAa Ba | + |4Aa BaCa |
(3)
Question (d). Add the two equations (2)(3) and use (1). Get the area of 4ABC. Answer. Adding (2) and (3) yields |LA | + |LB | + |LC | = |4ABC| + 2π R2 − |4CAa Ba | + |4CAa Ba | + |4Aa BaCa | = |4ABC| + 2π R2 + |4Aa BaCa | = 2|4ABC| + 2π R2 Because the areas of the lunes are |LA | = 2α R2 , |LB | = 2β R2 , |LC| = 2α R2 , we finally get |4ABC| = (α + β + γ − π) R2
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30.5. Does Pythagoras’ imply the parallel postulate? A purely geometric proof would have to struggle with some obvious difficulties. At first the squares of the sides of a triangle cannot be interpreted as area of any figure in neither spherical nor hyperbolic geometry. In these geometries, there do not exist any similar figures other than congruent ones, and squares do not exist at all. So the squares of the sides can only be interpreted as numbers. I am for now considering these difficulties as not crucial. Here are possible approaches to address the question posed: (1) One can look at the special Pythagorean triples of integers such that a2 +b2 = c2 . The smallest one is the triple 32 + 42 = 52 . So a meaningful question is whether there exist right triangles with the sides 3, 4 and 5 in either spherical or hyperbolic geometry. (2) Take any right triangle 4ABC. In spherical geometry its sides satisfy cos c = cos a cos b (3)
Does this imply that Pythagoras’ Theorem can never hold? Take any right triangle 4ABC. In hyperbolic geometry—with Gaussian curvature K = −1— the sides of a right triangle satisfy cosh c = cosh a cosh b
Does this imply that Pythagoras’ Theorem can never hold? I do no know about any approach to answer at least part of these questions in the framework of neutral geometry—where they would naturally fit, similarly to Legendre’s Theorems about the angle sum. My approach problems (2) and (3) uses substantial tools from trigonometry and calculus. 30.5.1.
Main results obtained by calculus
Theorem 30.1. If a spherical right triangle has the short arc as its hypothenuse c, this side is strictly shorter than for the Euclidean flat triangle with legs of the same lengths. Hence in single elliptic geometry, the hypothenuse c of any right triangle is strictly shorter than for the corresponding Euclidean flat triangle with legs of the same lengths. Theorem 30.2 ("The lunes of Pythagoras"). For spherical right triangles where the hypothenuse c is the longer arc, this side may be either shorter or longer, or of equal length as for the Euclidean flat triangle with legs of the same lengths. Assume the length of the hypothenuse is restricted to the interval c ∈ (π, 2π). The legs satisfy a2 + b2 = c2 —as in the Euclidean case— if and only if (a, b) lies on one curve C ⊂ E, which connects the boundary points (s∗ , 0) and (0, s∗ ) inside the quarter circle E = {(a, b) : 0 < a < 2π , 0 < b < 2π , a2 + b2 < 4π2 } Here the number s∗ ≈ 257◦ is the unique solution of tan s∗ = s∗ ∈ (π, 3π 2 ). Theorem 30.3. In hyperbolic geometry, the hypothenuse c is longer than for the Euclidean flat triangle with legs of the same lengths. 30.5.2.
Examples for the "lunes of Pythagoras"
Definition 30.3 ("The lunes of Pythagoras"). The lunes of Pythagoras are right spherical triangles, the sides a, b, c ∈ (0, 2π) of which satisfy both cos c = cos a cos b c =a +b 2
2
2
and
(30.1)
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On the pages 581 through 585, one can see several examples—obtained by solving the equation √ cos a2 + x2 − cos a cos x := g(a, x) = 0 x2 numerically for several values of a. I show the flat triangle, its sides in radian measure, together with the spherical triangle in stereographic projection, with side b on the equator (red), and side a extended through the south pole (blue). Sides c and b intersect in both vertex A and its antipode. The lengths of the sides are given in degree measure. Problem 30.4. Give your own comments about these figures.
Figure 30.10. The Pythagoras lune with a = 45◦ .
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Figure 30.11. The Pythagoras lune with a = 70◦ .
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Figure 30.12. The Pythagoras lune with a = 90◦ is highly degenerate.
Figure 30.13. The Pythagoras lune with a = 120◦ .
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Figure 30.14. The Pythagoras lune with a = 150◦ .
Figure 30.15. The Pythagoras lune with a = 180◦ corresponds to a flat right triangle with sides 3, 4 and 5—it is degenerate, too.
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Figure 30.16. An approximately isosceles lune.
30.6.
The stereographic projection
Definition 30.4 (The stereographic projection). In Euclidean 3-space, take a sphere S 2 of diameter 1. The stereographic projection is the central projection which maps S 2 onto the tangent plane T to the sphere at the south pole S , with the north pole N as center of projection. The north pole N itself is mapped to a single point ∞ at infinity. The equator is mapped to the unit circle ∂D. The southern hemisphere is mapped to the disk D, and the northern hemisphere is mapped to the exterior R2 ∪ { ∞ } \ D of the disk. Main Theorem 24 (Properties of the stereographic projection). (1) Circles are mapped to circles or Euclidean lines. (2) Angles are preserved. 30.6.1. The stereographic projection is an inversion To master this tricky three-dimensional problem, it is helpful to use three-dimensional inversion, defined by the straightforward generalization of definition 30.5. Thus the preservation of circles by the stereographic projections can be derived from the preservation of circles by inversion. Let B be the open interior of a three dimensional ball of radius R and center Z, and denote its boundary sphere by ∂B. Definition 30.5 (Inversion by a sphere). The inversion by the sphere ∂B is defined to be the mapping from the Euclidean 3-space plus one point ∞ at infinity to itself, which maps an −−→ arbitrary point P , Z, to its inverse point P0 —defined to be the point on the ray OP such that |OP| · |OP0 | = R2 . Hence, especially, all the points of ∂B are mapped to themselves. The inversion maps the origin Z to ∞, and ∞ to Z.
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Proposition 30.4 (The stereographic projection as an inversion). Let ∂B be the sphere with center Z = N at the north pole through the south pole S . Since the principal sphere S 2 has diameter 1, the new sphere ∂B has radius 1. Inversion by ∂B maps the sphere S 2 to the the tangent plane T at the south pole S . The stereographic projection is the restriction of the inversion by ∂B to the principal sphere S 2 . Proof. Again the drawing plane is determined by the meridian S 1 through the north pole N, south pole S and the generic point A. Let ∂B be the intersection of ∂B with the drawing plane—which is the circle around the north pole N through the south pole S . Inversion by ∂B maps the north pole N to infinity, and the south pole S to itself. The diameter NS is mapped to itself. The circle with diameter NS and the tangent t at the south pole are both orthogonal to this diameter NS . Hence preservation of angles implies that the circle with diameter NS is mapped to the tangent t. Rotation of the entire figure around the axis NS yields a three dimensional figure, from which one sees that inversion by ∂B maps the sphere S 2 to the the tangent plane T at the south pole S . −−→ This mapping takes any point A ∈ S 2 along the projection ray NA to the image point P ∈ T . Hence it is just the stereographic projection. Direct check using triangle geometry. Take any generic point A on the median S 1 and let P be the image of A by the stereographic projection. From the similar right triangles 4NS A and 4NPS , one gets the proportion |NA| |NS | = = cos(∠ANS ) |NS | |NP| and hence |NA| · |NP| = |NS |2 = 1 which confirms that point P is the inverse image of A by a circle of radius 1 around north pole N, as to be shown.
Figure 30.17. The stereographic projections maps circles to circles.
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Reason for item (1) of the Main Theorem. We show that the image of any circle on the sphere S 2 is a circle or line in the tangent plane T through the south pole S . Take any circle C on the sphere S 2 . Denote its center by A. In the special case that the center A is north pole N or south pole S , one can immediately show that C is mapped to a circle. Hence we exclude this special case, and let B and C be the two points of the circle lying on the meridian NA. Suppose the stereographic projections maps the points A, B and C to the image points P, Q and R in the tangent plane T . We use the plane through the north pole N, south pole S and point A as our drawing plane. Points B, C, P, Q, R lie in this plane, too. Let U be the circle through the points B, C and Q. Because the pairs A, P, B, Q and C, R are inverse points by circle ∂B, |NB| · |NQ| = |NC| · |NR| = 1 Now Euclid’s theorem of chords implies that circle U consists entirely of pairs of inverse points. Hence all four points B, C, Q, R lies on circle U. Furthermore circle U is orthogonal to ∂B. Because of the orthogonality, inversion by the circle ∂B maps the circle U to itself. Rotation around the common diameter of the orthogonal circles U ⊥ ∂B produces the two orthogonal spheres U ⊥ ∂B. Again because of the orthogonality, inversion by the sphere ∂B— which is just the stereographic projection—maps U to itself. Since the stereographic projection maps the sphere S 2 to the tangent plane T , it maps the intersection S 2 ∩ U to the intersection U ∩ T . But U ∩ S 2 clearly is a circle. Because the centers of both spheres U and S 2 lie in the drawing plane, this circle has diameter BC, and hence U ∩ S 2 = C is the originally given circle. Its stereographic image is U ∩ T , which is a circle in the tangent plane T , as to be shown. Again, because the center of sphere U lies in the drawing plane, the image circle U ∩ T has diameter QR. Reason for item (2) of the Main Theorem. Let any three points A, B, C on the sphere be mapped to the points P, Q, R on the tangent plane t. The two lines PQ and PR have as preimages two circles ABN and ACN on the sphere S 2 . It is enough to show that the angle ∠QPR between the two lines is congruent to the angle between the two circles. We can measure the angle between these two circles as the angle between their tangent at the second intersection point N. The tangent t1 to the circle ABN at the north pole N is a line in the plane spanned by the −−→ −−→ projection rays NA and NB. It is the parallel to line PQ through the north pole N. Similarly, the tangent t2 to the circle ACN at the north pole N is the parallel to line PR through the north pole N. Because parallel shift leaves angles invariant, the angle between the lines PQ and PR is congruent to the angle between the tangents t1 and t2 , as to be shown. 30.6.2. Many congruent angles Proposition 30.5 (Antipodes). (1) Antipodes A and Aa on S 2 are mapped to antipodal points P and Pa with respect to the circle ∂D. (2) Great circles (elliptic lines) are mapped to circles or Euclidean lines through a pair of antipodes. (3) Points on the northern and southern hemispheres of S 2 which are symmetrical to the equatorial plan, are mapped to inverse points with respect to the circle ∂D.
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Figure 30.18. Antipodes on the sphere are mapped to antipodes in the conformal model.
Reason for item (1). Draw a section in the plane with points N, S and the generic point A on the sphere S 2 . Let P be the image of A. The points O, P, Aa , Pa lie in the same plane. Angle ∠ANAa = 90◦ by Thales’ theorem, and angle ∠NS P = 90◦ , because the radius is orthogonal to the tangent by Euclid III.16. The altitude theorem for the right triangle 4Pa NP yields |S P| · |S Pa | = |NS |2 = 1 Hence P and Pa are antipodal points with respect to the unit circle ∂D.
Let S 1 be the meridian through point A. This is just the great circle lying in our drawing plane. Let a be the tangent line to the circle S 1 at point A. Let R be the intersection point of that tangent with line t = S P. Let β be the congruent base angles of the isosceles triangle 4NOA. We claim that triangle 4RAP is isosceles, too. Indeed, angle addition at vertex A shows that β + 90◦ + ∠RAP = 180◦ and the angle sum of the triangle 4NPS yields β + 90◦ + ∠RPA = 180◦ Hence the triangle 4RAP has congruent base angles α = ∠RAP ∠RPA and hence it is isosceles. Question. Here is a drawing of what has been explained so far. In this figure, find and mark with same color angle α = 90◦ −β at vertices N, P, Pa , A, Aa five times. Find and mark with another color of your choice angle β at these vertices as often as possible. (I do not take into account vertical angles.) Reason for item (3). Take any generic point B on the median S 1 and let Q be the image by the stereographic projection. The antipode Ba has the image point Qa such that |S Q| · |S Qa | = 1
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Figure 30.19. Many congruent angles
Figure 30.20. Points symmetric to the equatorial plane are mapped to inverse points with respect to the equator circle ∂D.
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and the south pole S lies between Q and Qa . Rotating the antipode Ba by 180◦ around the axis NS yields the point C, such that B and C are reflection images by the equatorial plane. The stereographic projection maps point C to point R, which is obtained from point Qa by a rotation of 180◦ around the south pole S . Hence |S Q| · |S R| = |S Q| · |S Qa | = 1 and the south pole S lies outside the segment QR. Clearly this confirms that R is the inversion image of Q by the equator circle ∂D. Direct check of item (3) with triangle geometry. Take any generic point B on the median S 1 and let Q be the image by the stereographic projection. From the similar right triangles 4NS B and 4NQS , one gets the proportion |QS | |BS | = = tan(∠BNS ) |NS | |NB| Now let point C be the reflection image of B by the equatorial plan, and R be the stereographic image of C. From the similar right triangles 4NS C and 4NRS , one gets the proportion |RS | |CS | = = tan(∠CNS ) |NS | |NC| Since segments BC and NS are parallel, the triangles 4NS B 4S NC are congruent. Hence |BS | = |NC| and |NB| = |S C|, and finally |QS | |RS | |BS | |CS | · = · =1 |NS | |NS | |NB| |NC| and hence |QS | · |RS | = |NS |2 = 1 which confirms that the stereographic images Q and R of the symmetric points B and C are inverse images by the equator circle ∂D.
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Figure 30.21. The diameter in the drawing plane is projected symmetrically.
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31. Euclidean Constructions with Restricted Means 31.1. Constructions by straightedge and unit measure Given is a segment UV, which we call the unit segment. The axiom (III.1) about transfer of segments is restricted to the transfer of the unit segment. Transfer of angles is not postulated at all. We thus consider the weakened axioms: III.1 unit If A0 is a point on the line a0 , then it is always possible to find a point B0 on a given side of the line a0 through A0 such that the unit segment UV is congruent to the segment A0 B0 . In symbols UV A0 B0 . III.4 unit Let the angle ∠(h, k) and the ray a0 be given. Then there exists at most one ray k0 such that the angle ∠(h, k) is congruent to the angle ∠(h0 , k0 ) and at the same time all interior points of the angle ∠(h0 , k0 ) lie on the given side of a0 . Every angle is congruent to itself, thus it always holds that ∠(h, k) ∠(h, k) Definition 31.1 (Hilbert plane with restricted transfer). A geometry with the axioms of incidence (I.1),(I.2),(I.3), the axioms of order (II.1) through (II.5), but the axioms of congruence (III.1) and (III.4) replaced by the weaker axioms (III.1 unit) and (III.4 unit) is called a Hilbert plane with restricted transfer. Main Theorem 25 (Straightedge and unit measure are equivalent to Hilbert tools (Theorem of Hilbert and Kürschàk)). 1 For a Hilbert plane with parallel axiom (IV.1), all constructions that can be done with Hilbert tools, can be done using only straightedge and unit measure. In other words, for Euclidean geometry, the axioms for a Hilbert plane with restricted transfer are equivalent to the Hilbert’s original axioms. In a Hilbert plane with restricted transfer, the Hilbert tools originally postulated by axioms (III.1) and (III.4) have been forbidden. In order to emulate these Hilbert tools, we are going to solve the following construction problems with straightedge and unit measure: We need the parallel axiom (IV.1) to achieve that. Construction 31.1. Transfer a given segment onto a given ray, starting the new congruent segment at the vertex of the ray. Construction 31.2. Transfer a given angle onto a given ray, with the vertex of the new congruent angle at the vertex of the ray, one side of the new angle on the given ray, and the other side in a prescribed half plane. Construction 31.3. Construct the parallel to a given line through a given point. Construction 31.4. Construct any perpendicular to a given line. Construction 31.5. Construct the perpendicular to a given line through a given point. Construction (31.3). By transfer of the unit segment UV to any point M of the given line l, one gets a point A such that UV AM. A second transfer yields a point B such that UV MB, and M becomes the midpoint of segment AB. Let P be the given point through which we have to get the parallel. Draw line AP and choose on it any point C such that P lies between A and C. Draw segments CM and PB. They intersect in −−→ some point S . Draw the segment BC and the ray AS . They intersect in some point Q. Line PQ is the required parallel to AB through point P. Reason using the harmonic quadrilateral. The construction uses a special case of the complete quadrilateral from projective geometry. 1
This Theorem is Hilbert’s Proposition 63—who gives credit to J. Kürschàk [26].
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Figure 31.1.
Another construction of the Euclidean parallel— the lines are numbered in the order they are constructed.
Figure 31.2. The harmonic quadrilateral.
Theorem 31.1 (Theorem of the harmonic quadrilateral). Take any quadrilateral ABQP. Let a pair of opposite sides AB and PQ intersect in point Y, and the other pair of opposite sides in point C. Let the diagonals intersect in point S . Let the line CS intersect side AB in point X. Then the four points A, B, X, Y are harmonic points. In other words, the cross ratio (AB, XY) =
AX · BY BX · AY
= −1
(31.1)
Via the Theorem 31.1 of the harmonic quadrilateral, we show the lines AB and PQ cannot intersect for special case occurring in the construction. Since X = M is the midpoint of segment AB, with lines AB and PQ intersecting in any point Y, the cross ratio would be (AB, MY) =
AM · BY BM · AY
=−
BY AY
, −1
(31.2)
Hence the harmonic property (31.1) could not be satisfied. The only remaining possibility is that the lines AB and PQ are parallel.
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Figure 31.3. Transfer of a given segment AB onto line l at point P.
Construction (31.1). Given is a segment AB to be transferred onto the ray l at point P. Draw the parallel to AB through point P, and the parallel to AP through point B. The two parallels intersect −−→ in point Q. Use the unit measure UV to get the segment PC on the parallel ray PQ and the segment PD on the given ray l such that UV PC and UV PD. Draw the parallel to line CD through point Q. It intersects the given ray l in point E such that AB PE. Question. There are some special cases: (1) The given ray l is parallel to AB, but does not lie on AB. (2) The given point P lies on line AB, but the ray l does not lie on AB. (3) The given point P and the given ray l both lie on line AB. Modify the construction to cover these special cases. Construction (31.4). Choose an arbitrary point M on the given line l, and transfer the unit segment UV onto both opposite rays of l with vertex M. One gets the congruent segments UV BM MC. Now choose two further rays with vertex M in the same half plane of line l, and transfer the unit segment UV onto them to produce the congruent segments UV MD and UV ME. The lines BD and CE intersect in a point F, and the lines BE and CD intersect in a point H. Now the line FH is perpendicular to the given line l. Reason for validity. The four points B, D, E, C lie on a circle with diameter BC. Hence by Thales’ theorem, ∠BDC and ∠BEC are right angles. The two altitudes BE and CD of triangle 4BFC intersect in H. Because all three altitudes of a triangle intersect in one point, FH is the third altitude and hence perpendicular to BC.
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Figure 31.4. Constructing a perpendicular line, using only straightedge and unit measure.
The construction problem 31.5 can now be solved using the just established construction 31.4 of a perpendicular, and construction 31.3 of the parallel through a given point. Before proceeding to construction 31.2, we solve a special case of that problem—just rotating a given angle. Construction 31.6. For a given β = ∠D0 AB0 and line l = AC 0 1 through its vertex, on both sides of line l, construct the congruent angles ∠C 0 AE1 ∠C 0 AE2 β.
Figure 31.5. Rotating a given angle about its vertex.
Construction (31.6). Let D be the foot point of the perpendicular dropped from point B onto AD0 . Similarly, let C be the foot point of the perpendicular dropped from point B onto line AC 0 . 1
Primes have been used where new points will be needed below, producing the same rays and angles.
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Next we drop the perpendicular from vertex A onto line DC, and let E be its foot point. Finally, −−→ ∠CAE β is the angle to be constructed. Its legs are the given ray AC 0 and the perpendicular dropped from A onto line DC. Remark. Both points C and D lie on a circle with diameter AB. Hence the construction of the perpendiculars can easily be done with rusty compass, by choosing AB twice the radius of the rusty compass and drawing a circle with diameter AB.
Figure 31.6. Validity of the construction to rotate an angle.
Reason for validity. The four points A, B, D, E lie on a circle C with diameter AB. Hence ∠DAB ∠DCB because they are angles in circle C over the same arc DB. Finally, ∠DCB ∠EAC because their sides are pairwise perpendicular. (Let F be the intersection lines BC and AE. One gets three similar right triangles 4AFC ∼ 4ACE ∼ 4CFE.) Remark. To transfer the angle to the other side of line l, one interchanges points B0 and D0 on the sides of the given angle β. Then one proceeds as above. We are now in the position to solve the construction problem 31.2. −−−→ Construction (31.2). Given is an angle β = ∠DAB, and a ray A0 B0 . At first we construct the parallel to the line A0 B0 through vertex A. Next we rotate the given angle around its vertex A to produce a congruent angle with one side on this parallel. Finally, we need to construct the parallel to the other side of the rotated angle through point A0 . Now we have the required angle at vertex A0 .
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Figure 31.7. Transferring a given angle to both sides of the given ray.
31.2.
Tools equivalent to straightedge and compass
Definition 31.2 (Traditional Euclidean tools). Constructions done using straightedge and compass only for the performance of the following steps: draw a line through two different points draw a circle with a given center through a given point intersect two lines intersect a circle and a line intersect two circles choose an arbitrary point choose a point on a given line choose a point on a given circle are called constructions by traditional Euclidean tools. Main Theorem 26 (Constructions by compass only (Theorem of C. Mohr and Lorenzo Mascheroni)). For the Euclidean geometry with parallel axiom (IV.1) and circle-circle intersection property, all constructions that can be done with straightedge and compass, can be done using only a (collapsible) compass.
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Figure 31.8. Construction of the midpoint with compass only.
Example 31.1 (Construction of the midpoint with compass only). Given is the segment AC to be bisected. We draw a circle A with center A through point C, and a second circle C with center C through point A. By the circle-circle intersection property, these two circles intersect in two points X and Y. Now we find the second endpoint of diameter AA0 of circle C. One uses simply the regular hexagon inscribed into this circle. This step is a Euclidean construction. We need the the intersection points Z and Z 0 of circle A with the circle around A0 through point A. Finally, the the two circles around Z and Z 0 through point A intersect in the midpoint M. This step, too, is a Euclidean construction. Reason for the construction. The isosceles triangles 4A0 AZ and 4ZAM are equiangular. By Euclid VI.4, their sides are proportional. Hence |A0 A| |ZA| = =2 |AZ| |AM| which confirms that |AM| = 21 |ZA| = 12 |AC|.
Main Theorem 27 (Constructions with one circle (Theorem of Poncelet and Steiner)). For the Euclidean geometry with parallel axiom (IV.1) and circle-circle intersection property, all constructions that can be done with straightedge and compass, can be done using only the straightedge, and only one circle with its center given. The Arabian mathematician Abul-Wefa suggested to use straightedge and a rusty compass This is a compass with fixed radius. From the Theorem of Poncelet and Steiner, we can deduct
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that rusty compass and straightedge are equivalent to the traditional Euclidean tools. But, since the actual constructions with fixed circle are rather awkward, I prefer to pose some problems using the rusty compass. Problem 31.1. Find the intersection of a given line l with a circle, of which the center O and one point B are given, using straightedge and rusty compass. Here is the idea how to proceed. We build two images, which are mapped to each other by a central dilation with center O. One image contains the line l and the circle C, the other one contains a parallel line l0 and a circle C 0 around O of radius one unit which can be drawn using the given compass. You see that this is a problem of Euclidean geometry. It makes ample use of parallels!
Figure 31.9. Intersection of line and circle with the rusty compass.
Answer (Description of the construction). Begin by drawing a circle C 0 of the given unit radius −−→ around point O. Let B0 be the intersection point of C 0 with the radial ray OB. Next choose an arbitrary point A on the line l. Then construct two similar triangles: 4OAB and 4OA0 B0 . This is −−→ easy, because A0 is just the intersection of OA with the parallel to line AB through point B0 . Next we get line l0 . It is the parallel to line l through point A0 . Let H10 and H20 be the intersections of circle C 0 with line l0 . (If they do not exist, the circle C and line l do not intersect, neither.) The intersections H1 and H2 of the circle C and line l are now easy to get, because they lie on the rays −−−→0 −−−→ OH1 and OH20 . Reason for validity. Here is the reason why Hi , for i = 1 or 2, lie on the circle C: Because of the similar triangles 4ABO ∼ 4A0 B0 O one gets the proportion |OA| |OB| = 0 |OA | |OB0 |
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Because of the similar triangles 4AHi O ∼ 4A0 Hi0 O one gets the proportion |OHi | |OA| = |OHi0 | |OA0 | By construction OHi0 = OB0 = 1 is the radius of the rusty compass. Hence the proportions imply OHi = OB, again for i = 1, 2.
Figure 31.10. What the teacher wanted.
Problem 31.2 (David at school 1). David Hilbert sees that the geometry teacher has draw the figure on page 600 on the board. The teacher tells that the square ABCD have sides 1 unit, the two curved lines are circular arcs, and that G is the center of the semicircle. Then the teacher poses the following problems: (a) Calculate the length of segment AH. (b) Use compass and straightedge to reproduce the drawing for your notebook. 2
Answer. AH = AF · AB can be seen using q √the right 4FBH. Because of AB = 1 and AF = √ 2 be get AH = 2 − 1 and hence AH = 2 − 1.
√ 2 − 1,
Problem 31.3 (David at school 2). At that point, David realizes that he has lost his compass. Luckily enough, he still has a straightedge and finds a rusty compass, with opening just one unit in his back pocket. David manages to get points A through G with a score of circles of his rusty compass, and the straightedge. Really do similar constructions, even if you need more (or less!) circles. Report what you have done, best by leaving some obvious details aside. But count how many circles you did need.
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Answer. The unit square ABCD can be drawn with a rusty compass. using 7 circles, to get a stack of equilateral triangles. To get point F, one can bisect the 45◦ angle ∠ABD, and then drop the perpendicular from point D onto the bisector. One needs 5 extra circles for that purpose. To get point G, note that it is the midpoint of segment AF. Hence one drops the perpendicular from the midpoint M of the square ABCD onto the same bisector. One needs only 3 circles. Problem 31.4 (David at school 3). Use an extra drawing to explain how to get point H with the same rusty compass. Count how many circles you need. One can use a central dilation with center B, mapping G 7→ A.
Figure 31.11. What Hilbert—at school 3—could have done.
Answer. Images for the same central dilation are A 7→ F and H 7→ H 0 , since |BG| |BA| = |BA| |BF| Erect the perpendicular at point F onto line AB. Get H 0 is the intersection of this perpendicular with the circle around A through point B. Point H is the intersection of lines BH 0 and AD. 31.3. Hilbert tools and Euclidean tools differ in strength Finally, we explain, why Hilbert tools are different, and indeed weaker than the traditional Euclidean tools. In the following discussion, it is agreed that some unit segment AB of length |AB| = 1 is given. Recall that by definition 34.1, the Hilbert field Ω is the smallest real field with the that a, b ∈ Ω implies √ a2 + b2 ∈ Ω. Proposition 31.1 (The Hilbert field are the segment lengths constructible with Hilbert tools ). The set of all lengths constructible with straightedge and unit measure are exactly those in the Hilbert field.
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Indication of reason. Let Ω be the Hilbert field, which is just the smallest field with properties (1)(2)(3). Let Ωconstr be the set of lengths constructible with Hilbert tools. All constructions with Hilbert tools can only produce lengths which are obtained via the algebraic operations mentioned above. Hence Ωconstr ⊆ Ω. Too, it is easy to check that Ωconstr is a field. Since the domain Ω is the smallest domain with the properties (1)(2)(3), we get Ωconstr = Ω. By Proposition 31.1 and Proposition 34.13, all lengths constructible by Hilbert tools are totally real. The domain T of all totally real Euclidean numbers has the properties (1),(2),(2a) and (3), too. Hence the minimality of Ω, shown in Proposition 31.1 implies Ω ⊆ T ⊆ K, where K is the Euclidean field. 1 q√ 2 − 1 is not totally real. A segment of Proposition 31.2 (Counterexample). The number z = that length cannot be constructed by Hilbert tools, just starting from a given unit segment. In the figure on page 600, points A through G are constructible with Hilbert tools, but not point H is not constructible. q√ Reason for the counterexample. The number z = 2 − 1 is a root of P(z) = (z2 + 1)2 − 2 = 0. This polynomial is irreducible over the rational numbers. Hence z is not root of any integer polynomial of degree two or three. But the polynomial P(z) has two complex roots. Hence z < T and z < Ω. Let the segment AB in the figure on page 600 be the unit segment. All segments constructible starting from AB —of length |AB| = 1—with q √ straightedge and unit measure are totally real algebraic numbers. But the number z = 2 − 1 is not totally real. Hence a segment of that length cannot be constructed by Hilbert tools, just starting from a given unit segment. Finally, here is the algebraic field of numbers constructible by the traditional Euclidean tools. Definition 31.3 (constructible field). The constructible field K is the smallest real field with properties (1) 1 ∈ K. (2) If a, b ∈ K, then a + b, a − b, ab ∈ K. (2a) If a, b ∈ K and b , 0, then ab ∈ K. √ (3) If a ∈ K and a ≥ 0, then a ∈ K. Main Theorem 28. The lengths constructible with straightedge and compass are exactly the numbers in the constructible field K. The lengths constructible with Hilbert tools are exactly the numbers in the Hilbert field Ω ⊂ K. q√ The Hilbert field Ω ⊂ K is a proper subset of the constructible field K. Especially, 2−1∈ K \ Ω. Proof. The algebraic operations in the definition of the constructible field can be emulated by the traditional Euclidean tools— straightedge and compass. No other algebraic operations can be emulated by these construction tools. Similarly, the algebraic operations in the definition of the Hilbert field are exactly those which can be emulated with Hilbert tools. 1
Emil Artin has proved that even Ω = T , but we do not use that much stronger result here.
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Straightedge and rusty compass are stronger than straightedge and unit measure. What is possible, what is impossible, concerning the construction of point H in the figure on page 600 from given points A and B? q√ Point H in this figure —and a segment of length z = 2 − 1— can be constructed with straightedge and rusty compass. The reason is that these tools are equivalent to the traditional Euclidean q √ tools, as follows from the Theorem of Poncelet and Steiner. Clearly ,the length z = 2 − 1 can be constructed with traditional Euclidean tools. It would even be possible to restrict the use of the compass to drawing only a single circle. On q √the other hand, nor point H in the figure on page 600 —and neither the segment of length 2 − 1— can be constructed with straightedge and unit measure. Indeed, by the Theorem of z= Hilbert and Kürschàk, tools. But a segment q √ straightedge and unit measure are equivalent to Hilbert q√ of length z = 2 − 1 cannot be constructed with Hilbert tools, since 2 − 1 is not totally real.
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32. Trisection of an Angle and the Delian Problem
Figure 32.1. Trisection of an acute angle
32.1.
Trisection by Archimedes
Construction 32.1 (Trisection of an angle by Archimedes). Suppose that on a straightedge two points C and D of distance d have been marked. Given is the angle α = ∠AOB to be trisected. We draw a circle C of radius d centered at the vertex O of the angle. This circle cuts the sides of the angle at points A and B. One places the marked ruler such that the following three requirements are met: −−→ 1. One of the marks on the ruler yields point C on the ray opposite to OA. 2. The other mark yields point D on the circle C. 3. The point B lies on the ruler. Result: With this construction, the angle γ = ∠OCD is one third of the given angle α = ∠AOB. Problem 32.1. Provide a drawing for this construction in the following cases (a) for a given acute angle α < 90◦ . (b) for a given angle 90◦ < α < 135◦ . (c) for a given angle α > 135◦ . Problem 32.2. In all cases, measure your angles α and γ.
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Figure 32.2. Trisection of an obtuse angle less than 135◦ .
Figure 32.3. Trisection of an obtuse angle larger than 135◦ .
Answer (Measurements). For my examples, I got In example (a): α = 54◦ , γ = 18◦ In example (b): α = 108◦ , γ = 36◦ In example (c): α = 153◦ , γ = 51◦ Problem 32.3. Let A0 be the second end of diameter A0 A. Since the base angles of the isosceles
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triangle 4COD are congruent, the trisected angle appears both at vertex C and the vertex O, too. How can one check the accuracy of the construction without using numbers or a protractor? Answer. It is easy to use a compass to measure the chord A0 D of the angle γ = ∠COD = ∠A0 OD. Then one can check whether this chord fits three times on the arc AB of the original angle α. Remark. In general, it is not possible to trisect an angle just with the tradition Euclidean tools— compass and unmarked straightedge. But, once the two-marked straightedge is put into the position such that the three requirements are satisfied, the construction is exact—as shown below. Problem 32.4. Give the reason, why Archimedes’ trisection construction is exact. Justification of Archimedes trisection. Since the marked segment CD has a length equal to the radius of the circle, and all points of a circle have the same distance from the center, we know that the three segments OB OD CD are congruent. Hence the triangle 4COD is isosceles. Its congruent base angles γ = ∠OCD ∠COD appears both at vertex C and at vertex O. The exterior angle at the top of this triangle is β = ∠ODB. By Euclid I.32, the exterior angle of a triangle is the sum of the two nonadjacent interior angles. Hence β = ∠ODB = ∠OCD + ∠COD = 2γ In the case of an angle α < 135◦ , the bases angle of the second isosceles triangle 4ODB are β = ∠ODB ∠OBD
(32.1)
In case (c), the supplements of the base angles of triangle 4OBD are β = 2γ. Finally, we use Euclid I.32 for the triangle 4OBC. Since the given angle α is an exterior angle for this triangle, we conclude α = ∠AOB = ∠OCB + ∠CBO = γ + β = γ + 2γ = 3γ . Hence γ = α3 , as to be shown.
Problem 32.5. Explain the difference of the drawings for the cases (a), (b) and (c). Answer (Difference between the cases). In cases (a) and (b), point D in between C and B. Hence D does not lie in the interior of the given angle, ∠AOB. The base angles of the isosceles 4OBD are β = 2γ. This is the same situation as did occur for an acute angle. In case (c), point B is between C and D and point D lies in the interior of angle ∠AOB. The supplements of the base angles of triangle 4OBD are β = 2γ. 32.2.
Trisection by Nicomedes
Construction 32.2 (Trisection by Nicomedes, 240 B.C.). Given is the angle α = ∠COB to be trisected. On the ruler is marked a segment PQ with length double as OB. One drops the perpendicular p from B onto line OC. Next one erects at point B a second perpendicular q on p. Indeed, line q is the parallel to OC through point B. Finally, one places the marked ruler such that the following three requirements are met: 1. The ruler line goes through the vertex O of the given angle.
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Figure 32.4. Trisection of an acute angle
2. The point marked P on the ruler lies on the perpendicular p. 3. The point marked Q on the ruler lies on the parallel q. Result: We claim that the angle γ = ∠QOC is one third of the given angle α = ∠AOB. Remark. In general, it is not possible to trisect an angle just with the tradition Euclidean tools— compass and unmarked straightedge. But, once the two-marked straightedge is put into the position such that the three requirements are satisfied, the construction is exact—as we prove below. Problem 32.6. Provide a drawing for this construction for several different angles: (a) an angle 90◦ < α < 135◦ . (b) an angle α > 135◦ . Tell in a few words, what made you stumble at first, and how this second drawing for case (b) differs from the drawing in case (a), or the drawing for an acute angle given above. Measure and report your angles α and the trisected angles γ. Answer. One needs to turn the ruler such that the point P on the perpendicular lies below the line OC, on the other side than point B. Too, there exists a second fake solution with M = O which does not yield angle trisection. To get appropriate angle trisection, the order of the points P, O, M and Q on the ruler is different in cases (a) and (b). In case (a), points Q and M lie on the same side, and point P on the other side of point O. In case (b), points P and M lie on the same side, and point Q on the other side of point O.
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Figure 32.5. Trisection of an obtuse angle between 90◦ and 135◦ .
Figure 32.6. Trisection of an obtuse angle larger than 135◦ .
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Problem 32.7. Question. Suppose that M is the midpoint of the segment PQ. Give a reason why the three segments MP, MQ and MB are congruent. Answer. The angle ∠PBQ is a right angle by construction. Hence the converse Thales’ Theorem 39 implies that point B lies on the circle with the hypothenuse of triangle 4PQB as diameter PQ, and hence the three segments PM, MB and MQ are congruent. Problem 32.8. Prove that the construction done in part (a) of problem (32.6) produces an exact trisection. Validity of the construction. We know that the three segments PM, MB and MQ are congruent, where M is the midpoint of segment PQ. By construction, segment OB has half the length of segment PQ. Hence all four segments OB, PM, MB and MQ are congruent. We get the isosceles triangles 4MQB and 4OMB. Question. Let γ = ∠MBQ ∠MQB measure the base angles of the first isosceles 4MQB. Find congruent z-angles of size γ. Mark all three angles congruent to γ with matching color. Answer. By construction, lines OC and BQ are parallel. With the transversal OQ, they form zangles ∠COQ ∠OQB = γ, which are congruent by Euclid I.28. The base angle ∠OMB of the second triangle is an exterior angle at the top vertex of the first 4MQB. By Euclid I.32, the exterior angle of a triangle is the sum of the two nonadjacent interior angles. Hence β = ∠OMB = ∠MBQ + ∠MQB = 2γ The bases angle of the second isosceles triangle 4OMB are β = ∠OMB ∠BOM
(32.2)
Finally, angle addition at vertex O implies α = ∠COB = ∠COQ + ∠QOB = 2γ + γ = 3γ Hence γ = α3 , as to be shown.
Question. The meaning of the angles β from equation 32.2 will change in the case (b) from problem 32.6. What is the the meaning of the angles β in case one trisects an angle α > 135◦ . Why is equation 32.2 still true. Answer. The angles β become the supplements of the base angles of the isosceles triangle 4OMB. But, of course, the supplements of the two base angles of an isosceles triangle are congruent to each other. Hence equation 32.2 is still true. 32.3.
Trisection with Nicolson’s angle, and by origami
Construction 32.3 (Trisection with Nicolson’s angle (1883)). Nicolson’s angle consists of a strip of two parallels with distance AB to which an extra square of congruent side BC is attached at point B. Thus B becomes the midpoint of segment AC. The points A, B and C are marked on the Nicolson angle. −−→ To trisect any given angle ∠POR, we first draw the parallel to side OP at distance AB which cuts the interior of the given angle. Next we place the Nicolson angle in such a way that −−→ • on the parallel to side OP lies point A; −−→ • on the other side OR lies point C;
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Figure 32.7. Trisection with Nicolson’s angle.
• The vertex O lies on the edge of the Nicolson angle with extension through point B. −−→ −−→ The rays OA and OB trisect the given angle ∠POR. Construction 32.4 (Trisection of an angle by paper folding). Take a rectangular sheet of paper. One needs the right angle at the lower left corner, which is point A. Choose an acute angle θ to be trisected. Place the angle you have chosen with its vertex at A, and one side along the lower edge a of the paper. Draw the other side t of angle θ in blue across the paper. The trisection construction is begun by drawing, on the front side of the paper, a parallel b to the lower edge a. Next one draws a parallel c to b which has the double distance from the lower edge a. This can conveniently be done by folding the paper along b. Let B and C be the intersections of the parallels b and c with the vertical left margin l of the paper. Mark point C on the back side of the paper. Now the paper is folded such that (i) the lower left corner A is flipped over to a point A0 on the line b, (ii) point C is flipped over onto point C 0 on the side t of angle θ. −−→ Claim: The lower edges a and the ray AA0 form the angle θ/3.
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Figure 32.8. Trisection by paperfolding
Problem 32.9. (a) Actually do the construction on a separate sheet of paper. Mark the quantities −−→ as explained above. Let α := ∠S AA0 be the angle between the lower edges a and the ray AA0 . Measure and report your angles θ and α. What about accuracy of the trisection? Answer. My choice came out to be θ = 39.5◦ . Measurement yields ∠A0 Aa = 13◦ . Thus the relative error is about 1%. Question (b). Find and mark the trisected angle at five or more places. Answer. The angle 3θ appears three times with vertex A, and twice with vertex W, where the paper crest intersects the left paper margin. Validity of the construction. The paper folding generates a reflection across the paper crest s as symmetry axis. Let W be the intersection of the paper crest s with the left margin. Let S , T and U be the points where the paper crest s intersects lines a, b and c. Let A0 , B0 and C 0 be the reflection images of A, B and C.
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Figure 32.9. Reflection at the folding line
Question (c). Draw the lines S A0 , T B0 and UC 0 and the line WA0 B0C 0 . What do you observe about these lines? Answer. The three lines S A0 , T B0 and UC 0 are all perpendicular to line WA0 B0C 0 . Hence they are parallel. Question (d). Compare the segments A0 B0 and B0C 0 . Answer. The two segments A0 B0 and B0C 0 are congruent. Here is the reason: A segment is congruent to its reflected segment. Hence AB A0 B0 and BC B0C 0 . The two segments AB and BC are congruent by construction. Hence, because congruence is transitive A0 B0 B0C 0 . We need to compare the three angles α := ∠S AA0 , β := ∠A0 AT and γ := ∠B0 AC 0 . Question (e). How do the two triangles 4AB0 A0 and 4AB0C 0 compare? How do the two angles β and γ and compare? Answer. The two triangles 4AB0 A0 and 4AB0C 0 are congruent. Indeed, they have a common side AB0 , both have a right angle at vertex B0 , furthermore the other pair of legs A0 B0 B0C 0 , is congruent, too. Hence they are congruent by SAS congruence. Hence, the two angles β and γ are congruent, too.
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Question (f). Give a direct reason that the angles α and β are congruent. Answer. The angles α := ∠S AA0 and β0 := ∠AA0 T are z-angles (alternate interior angles) between the parallels a and b. By Euclid I.29, z-angles between parallels are congruent. The angles β0 and β are congruent, because they are images by reflection. By transitivity, we see that α and β are congruent. Question (g). Draw the diagonals of quadrilateral AS A0 T . What kind of quadrilateral is this? Answer. The two diagonals are perpendicular, because the segment from a point A and its reflection image A0 is perpendicular to the reflection axis S T . The four sides are congruent. The diagonals are perpendicular. Hence the quadrilateral AS A0 T is a rhombus. Question (h). What can you say about the angles that these diagonals form with the sides of the quadrilateral? Answer. These two angles are congruent, as shown above. Too, it is well known that the diagonals in the rhombus bisect its angles. To conclude the argument recapitulate: we have shown that the three angles α := ∠S AA0 , β := ∠A0 AT and γ := ∠B0 AC 0 are congruent. Since by angle addition θ = α + β + γ, we get θ = 3α and α = 3θ , as to be shown. Problem 32.10 (Some experimentation). To better understand the trisection construction and the proof of its validity from above, do some paper folding where one or another of the the assumptions are violated. Question (a). Choose the three parallel lines a, b and c such that AB < BC. How do the three angles α, β and γ turn out? Answer. The two angles α and β are still congruent, but α < γ. Question (b). Choose the three parallel lines a, b and c such that AB > BC. How do the three angles α, β and γ turn out? Answer. The two angles α and β are still congruent, but α > γ. Question (c). Choose only the two lines a and c parallel, but not line b. How do the three angles α, β and γ turn out? Answer. 32.4. Construction of the cubic root by two-marked ruler The construction given here is a slight simplification of the version reported by Arthur Baragar in the recent article [5]. Let the ruler have two marks C and D with distance |CD| = 1. Given is the segment PQ of length a < 4. I √3 use the two-marked ruler for the construction of a segment of length 2a. Construction 32.5 (Nicomedes’ Construction of a Cubic Root). Let T be the midpoint of the given segment PQ. Construct the perpendicular bisector of the half segment T Q. Let A be the midpoint of T Q and O be a point on the bisector such that |OT | = 1. Extend line OT and let B the point such that T is the midpoint of segment OB. Finally, one places the two-marked ruler. We turn it around point O and put the two marks on the lines c = BQ and d = PQ, such that the following three requirements are met: 1. The ruler line goes through point O. 2. The point marked C on the ruler lies on the line c = BQ.
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Figure 32.10. Nicomedes’ construction of
√3
2a.
3. The point marked D on the ruler lies on the line d = PQ. Claim: The segment QD has length
√3
q 2a, and the segment OC has the length
3
a2 2.
Problem 32.11. Provide a drawing and actually do the construction for a = 1. Measure your results and check for accuracy. What kind of triangle is 4OT Q?
Simple modern proof. The proof uses only the two most important facts of Euclidean geometry: similar triangles, and the Theorem of Pythagoras.
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Let u = |QD| and v = |OC| denote the lengths of the segments obtained by the construction. The Theorem of Pythagoras for the two right triangles 4OAQ and 4OAD, and substraction yields a2 |OA|2 + = |OA|2 + |AQ|2 16 a 2 |OA|2 + u + = |OA|2 + |AD|2 4 a 2 a2 u+ − 4 16 au 2 u + 2
= |OQ|2 = 1 = |OD|2 = (1 + v)2
(32.3)
= (1 + v)2 − 1 = v2 + 2v
(32.4)
To eliminate the quantity v, we use proportions. Point T is the midpoint of both segments PQ and OB. Hence congruent z angles easily imply that the lines OP and QB are parallel. With center D, and the parallels CQ = QB k OP produce the equiangular triangles 4DCQ and 4DOP. By Euclid (VI.4), the sides of equiangular triangles are proportional, and hence v |OC| |PQ| a = = = 1 |CD| |QD| u Now we eliminate of the quantity v from equation (32.4) and get au 2a a2 u2 + = + 2 2 u u a 3 u u+ = a(2u + a) 2 One solution of this equation is u4 = − a2 , corresponding to v4 = −2. In the drawing, this solution corresponds to putting the ruler marks at C4 = B and D4 = T . Cancelling the factor (u + 2a ) on both sides yields the equation u3 = 2a q √3 2 3 We get the solution |QD| = u1 = 2a and |OC| = v1 = a2 . which is the solution we have been looking for. Remark. This is another version to derive equation 32.4 from Hartshorn’s book [?], p. 263. Draw a circle around point O of unit radius. It goes through points Q and T . Let EF be its diameter on the line OD, with point E between D and O. Since |DC| = |EO| = 1, we get |DE| = |CO| = v. Now equation 32.4 follows from Euclid’s theorem of chords III.36 since a u u+ = |DQ| · |DT | = |DE| · |DF| = v(v + 2) 2 32.5. Definition and equations of the conchoid The conchoid of Nicomedes with focus O, guiding line AB, and marked segment p, is the location of all points Q away from the guiding line −−→ by distance p. as measured on a radial ray. Thus the radial ray OQ cuts the guiding line AB in point P, and the segment PQ has length |PQ| = p. Problem 32.12 (The Conchoid in Cartesian coordinates). From this definition, find the equation of the conchoid of Nicomedes in Cartesian coordinates, with given segment length p, the focus at the origin, and vertical guiding line x = d.
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Figure 32.11. The conchoid of Nicomedes
Answer. Drop the perpendicular from the generic point Q of the conchoid onto the x axis, and let F be the foot point. Similar triangles 4OFQ and 4OAP yield the proportion x |OF| |OA| |OF| − |OA| x − d = = = (32.5) cos θ = = r |OQ| |OP| |OQ| − |OP| p Hence multiplying, Pythagoras’ theorem and solving for y yield x x−d = r p r · (x − d) = p · x (32.6) 2 2 (x + y )(x − d)2 = p2 x2 q x y=± p2 − (x − d)2 x−d Problem 32.13 (The conchoid in polar coordinates). Find the equation of the conchoid of Nicomedes in polar coordinates. Answer. Since x = r cos θ and y = r sin θ, the equation (32.6) above implies x x−d d cos θ = = = r p r−p d r−p= cos θ d r = p+ cos θ
(32.7)
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32.6. The conchoid and the construction of the cube root We now use the conchoid for a better understanding of Nicomedes’ construction 32.5 of a cubic root. The point O around which the ruler is turned is the focus. We place this point at the origin of the Cartesian coordinate system. Any one of the two lines c = BQ and d = PQ can be used as guiding line. I shall use PQ as guiding line. The guiding line is put parallel to the y-axis. Since A is the foot point of the perpendicular dropped from the focus q onto the guiding line, the x-axis is the line OA. With |OA| = d = 1 − has length p = |CD| = 1.
a2 16 ,
the guiding line has again the equation x = d. The marked segment
Question. Give the equation of the conchoid corresponding to the construction 32.5 in Cartesian coordinates, implicitly and explicitly. q a2 Answer. Since p = 1 and d = |OA| = 1 − 16 , the equation 32.6 becomes (x2 + y2 )(x − d)2 = x2 y=±
√ x 32dx + a2 4(x − d)
(32.8)
It is now just an exercise of endurance—left to the reader—to calculate the equation of the line c = QB, get the coordinates √of points C and D, and finally the segment length u = |QD|. A graph with a = |PQ| = 1 and d = 415 is given in the figure on page 618. One can see that segments OQ and OT are tangent to the loop of this conchoid at the origin, and the conchoid passes through the two real solutions at points B and C. 32.7. Duplication of the cube by two-marked ruler Indeed, the duplication of the cube— also known as the Delian problem—is a special case of Nicomedes’ more general construction of the third root explained earlier. Nevertheless, I include this section, because of the historic importance of the problem, and because this special case turns out to be remarkably pretty. Problem 32.14 (Duplication of the Cube √ with Two-Marked Ruler). In the drawing of page 619, 3 Nicomedes’ construction of the third root 2a is done in the special case a = 2. We have assumed that both the radius of the circle as well as the segment marked on the straightedge have unit length |T B| = |CD| = 1. (a) Describe the construction. What kind of quadrilateral is OPBQ? Find two equilateral triangles inside of it. How is it constructed most easily? How has the the two-marked ruler to be positioned? (b) The construction yields two √3 √3 segments OC and QD, which we claim to have the lengths |OC| = 2 and |QD| = 4. To confirm this claim, do the following: • •
find two similar triangles and set up a proportion for the segments above. Calculate the lengths of all three sides of the right triangle 4OAD exactly and check the theorem of Pythagoras.
Construction 32.6 (Nicomedes’ Solution of the Delian Problem). Let the ruler have two marks C and D with distance |CD| = 1. Draw a circle with center T and the same radius 1. Into the circle, we inscribe the regular hexagon OQEBPF by means of six cords of length 1. Since OT B and QT P are diameters, we get the rectangle OPBQ with these diameters as its diagonals. Too, the diagonals of this rectangle form two equilateral triangles 4OT Q and 4PT B. The lines QP and QB need to be extended to the opposite side of Q. Finally, one places the marked ruler such that the following three requirements are met:
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Figure 32.12. Another gift for Apollo.
1. The ruler line goes through point O. 2. The point marked C on the ruler lies on the line QB. 3. The point marked D on the ruler lies on the line QP. √3 √3 Claim: The segment OC has length 2, and the segment QD has length 4. Answer. The quadrilateral OPBQ is indeed a rectangle. Hence the lines QC and PO are parallel. Thus the figure on page 619 contains the similar triangles 4CDQ ∼ 4ODP. We get the proportions √3 |CD| |OC| 1 2 = √3 = |QD| |PQ| 2 4 The right triangle 4OAD has the sides √ √3 3 1 √3 |OA| = , |AD| = + 4 , |OD| = 1 + 2 2 2
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Figure 32.13. Nicomedes’ Solution of the Delian Problem.
Hence !2 √3 √3 √3 2 √3 √3 3 1 √3 + 4 = 1 + 4 + 16 , |OD|2 = 1 + 2 = 1 + 2 2 + 4 |OA| + |AD| = + 4 2 2
2
which turn out to be the same value. The lengths of segments OC and QD give both middle proportions between |T P| = 1 and |PQ| = 2. One can simplify this drawing a bid further, drawing neither the rectangle nor its circum circle, but adding a unit circle around Q. This leads to the following construction of a segment of length √3 2. Construction 32.7 (Simplified Nicomedes’ Construction). Let the ruler have two marks C and D with distance |CD| = 1. Draw a circle with center Q of radius 1. Let OQ be the horizontal diameter and let QM be perpendicular to OQ. Next we need on the circle point T , such that triangle 4OQT is equilateral. Finally, one places the marked ruler such that the following three requirements are met:
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Figure 32.14. My final gift to Apollo.
1. The ruler line goes through point O. 2. The point marked C on the ruler lies on the perpendicular QM. 3. The point marked D on the ruler lies on the line QT . √3 Claim: The segment OC has length 2. Problem 32.15. Provide a drawing and actually do the construction. Measure your results and check for accuracy. Simple modern proof. Being unfaithful to the old Greeks, I use trigonometry. Let α = ∠QOC. (It turns out to be about 37◦ .) Let x = |OC| be the length of the segment obtained by the construction. From the right 4OQC, we get cos α =
|OQ| 1 = |OC| x
and
x=
1 cos α
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Figure 32.15. The Delian problem, once more
The sin theorem in 4OQD implies |OD| |OC| + |CD| |QD| = = = √ ◦ 1 sin α sin 120 2 3
1 cos α + 1 √ 1 2 3
2 + 2 cos α = √ 3 cos α
The sin theorem in 4QCD implies |QD| |CD| = sin(90◦ + α) sin 30◦ |QD| = |CD| · 2 cos α = 2 cos α One can eliminate |QD| and get from (32.9) and (46.7) 2 cos α 2 + 2 cos α = √ sin α 3 cos α √ 2 3 cos α = (1 + cos α) sin α We use the variable x =
1 cos α
(32.9)
(32.10)
(32.11)
and square 3 1 = 1+ x x4
!2
1 1− 2 x
3 = (x + 1) (x − 1) 2
2
! (32.12)
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One solution of this equation of forth order is x4 = −2. Hence one can factor (x + 1)2 (x2 − 1) − 3 = x4 + 2x3 − 2x − 4 = (x + 2)(x3 − 2) √3 Hence another solution of equation 46.9 is x1 = 2. Geometrically, the solution x4 is obtained by putting C4 := B and D4 := T . The ruler line is AT B, where ∠QOT = −60◦ , and |OT | = |T B| = 1. The solution x1 is the one√we were looking for. It corresponds to the ruler line OCD, where 3 ∠QOC = α ≈ 37◦ , and OC = 2. Duplication of cube using just geometry. To keep the reasoning in the style of the ancient geometers, I need the Theorem of Menelaus 26.1. We use this theorem for the triangle 4OCB and transversal T Q. One gets |DC| |T O| |QB| · · =1 (32.13) |DO| |T B| |QC| Now use that by construction |DC| = |T O| = |T B| = 1 and |DO| = |DC|+|CO| = 1+ x. Furthermore, Pythagoras theorem yields |QB|2 = 22 − 12 = 3 and |QC|2 = |OC|2 − |QO|2 = x2 − 1. Hence I get √ 1 3 1 =1 (32.14) · · √ x+1 1 x2 − 1 Now, we go on as in the first version above. Squaring and factoring yields (x + 1)2 (x2 − 1) − 3 = x4 + 2x3 − 2x − 4 = (x + 2)(x3 − 2)
√3 This is an equation of forth order. One solution is x1 = −2. The second solution is x2 = 2, which ◦ is the one √3 we are looking for. It corresponds to the ruler line OCD, where ∠OAC = α ≈ 37 , and |OC| = 2. 32.8. Duplication of the cube and the curve of Agnesi For the duplication of cube construction, we shall now use the line d = QB as guiding line of a conchoid. In that special case, the distance from focus to guiding line is exactly equal to the length of the marked segment. This special conchoid has p = d and is also known as the curve of Agnesi. Problem 32.16. Put the simplified Nicomedes’ construction 32.7 in a coordinate system with origin at focus O = (0, 0), and center of the circle Q = (1, 0). Find the equation of the conchoid with focus O and guiding line QM ⊥ QO. Find the equation of line QT . Calculate the coordinates of the points C, D and C4 := B, D4 := T and the distance |OC|. Produce a graph and the segments OC and OB and check their lengths. Answer. The equation of line QT is y=
√ 3(x − 1)
The conchoid has parameters p = d = 1, hence its equation is x p y= 1 − (x − 1)2 x−1 Intersection points have the x-coordinate satisfying √ x p 3 (x − 1) = 1 − (x − 1)2 x−1 3(x − 1)4 = x2 (−x2 + 2x) 4x4 − 14x3 + 18x2 − 12x + 3 = 0 (2x − 1)(2x3 − 6x2 + 6x − 3) = 0 (2x − 1)[2(x − 1)3 − 1] = 0
(ON)
(Conchoid)
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from which we get the coordinates of the two intersection points to be √ √ 1 1 3 3 D4 = (x4 , y4 ) = , − and D = (x, y) = 1 + √3 , √3 2 2 2 2 and hence the points on the guiding line are √ ! y √ y4 3 C4 = 1, = 1, − 3 and C = 1, = 1, √3 x4 x 2+1 A further exercise in arithmetic yields √3 x2 + y2 ( 2 + 1)2 + 3 √3 2 |OC| = = = 4 √3 x2 ( 2 + 1)2
Figure 32.16. The Delian problem and the curve of Agnesi.
32.9.
A close look at Nicomedes’ trisection
Problem 32.17. Repeat Nicomedes’ trisection construction 32.2. Extend the ruler line, the perpendicular p and the parallel q as long as needed, to both sides. Use a small angle of about 30◦ . One needs to turn the ruler an entire 360◦ , such that the point P covers the entire perpendicular line p, and the other marked point slides along the entire parallel q. Find out how many solutions you get.
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Figure 32.17. Nicomedes’ trisection construction has four solutions
Problem 32.18. For all four solutions Pi Qi with i = 1, 2, 3, 4, mark the midpoints Mi of these segments. What is special about one of these midpoints M4 . What figure do you get from the segments OM1 , OM2 , OM3 . Answer. There exists a fake solution with M4 = O which does not yield angle trisection. The three other segments OM1 , OM2 , OM3 are a star with three congruent angles of 120◦ . 32.10. Trisection and the conchoid Let α = ∠AOB be the angle to be trisected, with one side −−→ OA on the positive x-axis. I put Nicomedes’ construction in a Cartesian coordinate system. As above, let p = |PQ| be the segment marked on the ruler. The origin O is the focus, and as guiding line, I choose line AB. The point A = (d, 0) is the intersection of the guiding line with the x-axis. Hence the distance from the focus to the guiding line is d = |OA| = |OB| cos α =
|PQ| cos α p cos α = 2 2
since Nicomedes’ construction requires |OB| = |PQ| 2 . The Cartesian coordinates of points A and B are ! p cos α p cos α p sin α A= , 0 and B = , 2 2 2 Question. What are the equations of the lines p and q used in the trisection construction?
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Answer. Line p is the parallel to the y-axis through point B, and has the equation x=
p cos α 2
(p)
Line q is the parallel to the x-axis through point B, and has the equation y=
p sin α 2
(q)
The ruler in the trisection construction is put onto a radial ray which intersection guiding line p at point P, and line q at point Q such that |PQ| = p. Hence point Q is the intersection of the horizontal line q with the conchoid. Problem 32.19 (Analytic justification of Nicomedes’ angle trisection). Nicomedes construction can be understood as intersecting the conchoid with a line. Use polar coordinates (r, θ), eliminate distance d. Find the angles θ1 , . . . , θ4 of the four intersection points of the conchoid with the line q. Answer. In polar coordinates, the equations of the conchoid, and of line q are r = p+
d p cos α = p+ cos θ 2 cos θ p sin α r sin θ = 2
(32.15)
After division by p, one uses some trigonometric identities and gets: p cos α p sin α = 2 cos θ 2 sin θ 2 sin θ cos θ + cos α sin θ = sin α cos θ 2 sin θ cos θ = sin α cos θ − cos α sin θ sin 2θ = sin(α − θ) p+
It is not difficult to see that this equation has the four solutions θ1 =
α − 360◦ α + 360◦ α , θ2 = = θ1 − 120◦ , θ3 = = θ1 + 120◦ , θ4 = 180◦ − α 3 3 3
The equation q yields the radii corresponding to these intersection points. ri =
p sin α 2 sin θi
for i = 1 . . . 4
Unfortunately, this formula makes r < 0 if 180◦ < θ < 360◦ . I remedy this by substituting (r2 , θ2 ) 7→ (−r2 , θ2 + 180◦ ) = (|r2 |, θ1 + 60◦ ) Question (b). Find the angle τ between the positive x-axis and the tangent to the conchoid at its double point. Answer. On that tangent, the segment from origin O to the intersection point T with the guiding line AB has length p, too. Hence cos τ =
|OA| d = = 2 cos α |OT | p
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Figure 32.18. The conchoid gives all four solutions
Question (c). Use the following program LINE on the TI84 to produce some examples for the trisection. Put the calculator to mode: angle in degrees, graph polar coordinates. Graph the function \r1 = P (1+.5cos(A)/cos(θ)) and run program LINE for a few different angles A. Explain your observations.
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PROGRAM: LINE :PROMPT A :2 7→ P :Polar :-P7→Xmin: 1.5P7→Xmax :Xmin7→Ymin: Xmax7→Ymax :Zsquare :Horizontal(.5Psin(A)) :Vertical(.5Pcos(A)) :(cos−1 (.5cos(A)))+1807→T :{ 0,A,A/3,A/3+60,A/3+120,180-A,T } 7→ L1 :For(I,1,7) Line(0,0,3cos(L1 (I)), 3sin(L1 (I))) :End 32.11.
Archimedes trisection yields more solutions
Problem 32.20. Follow the same procedure as given for Archimedes’ trisection, with one small difference in item 1: Put the point marked C on the ruler anywhere on the line OA. Answer. If one follows this modified procedure of construction carefully, one finds six solutions with Ci Di OB for i = 1, . . . , 6. Three of these solution C1 , C2 , C3 are related to trisection. −−→ For the first solution, C1 lies on the ray opposite to OA and outside the circle. This is the solution, one usually considers first. For the second solution C2 lies inside the circle, but in general not at the center O. −−→ For the third solution, C3 lies on the ray OA and outside the circle. The other three C4 , C5 , C6 are obvious "fake" solutions. For the forth solution, the ruler is a diameter of the circle, hence let C4 = O and BD4 be a diameter. Two further solutions correspond to D5 = D6 = B, and the two points C5 and C6 on the OA with distance OA from B. Proposition 32.1 (A close look at Archimedes’ trisection). A careful analysis shows that the procedure leads to six solutions. Problem 32.21. Provide six drawings for this construction with different angles. Use a different paper, and report your results below—ignore the fake solutions. In each example, measure and report your chosen angle α, and the three angles γi = ∠OCi Di for i = 1, 2, 3. (a) Chose α < 45◦ . (b) Exactly α = 45◦ . (c) Chose 45◦ < α < 90◦ . (d) Exactly α = 90◦ . (e) Chose 90◦ < α < 135◦ . (f) Exactly α = 120◦ . Answer. (a) Chose α < 45◦ .
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Figure 32.19. Archimedes’ star of Mercedes
Answer. For this example, I got α =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
, γ1 =
, γ2 =
, γ3 =
(b) Exactly α = 45◦ . Answer. For this example, I got α = (c) Chose 45◦ < α < 90◦ . Answer. For this example, I got α = (d) Exactly α = 90 . ◦
Answer. For this example, I got α = (e) Chose 90◦ < α < 135◦ . Answer. For this example, I got α = (f) Exactly α = 120◦ . Answer. For this example, I got α =
Proposition 32.2 (The star of Mercedes in Archimedes’ trisection). The three angles γi = ∠OCi Di for i = 1, 2, 3 are related to the chosen angle α = ∠AOB via α α α , γ2 = 60◦ + , γ3 = 60◦ − 3 3 3 −−−→ −−−→ −−−→ The triangle 4D1 D2 D3 is equilateral and the three rays OD1 , OD2 , OD3 form a star of Mercedes. The three ruler lines form at vertex B two congruent adjacent angles γ1 =
∠C1 BC2 ∠C2 BC3 60◦
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Proof. The claim
α 3 was shown earlier. Recall that the isosceles 4OC1 D1 has two congruent base angles γ1 , and the isosceles 4OD1 B has two congruent base angles 2γ1 . Finally α = γ1 + 2γ1 , because α is an exterior angle in triangle 4OC1 B, and γ1 , 2γ1 are the two nonadjacent interior angles of that triangle. The claim α γ3 = 60◦ − 3 follows by the same reasoning applied to trisecting the supplementary angle ∠C1 OB 180◦ − α. The claim α γ2 = 60◦ + 3 needs a bid more care to be checked. The isosceles triangle 4OC2 D2 has two congruent base angles γ2 . The isosceles triangle 4OD2 B has two congruent exterior base angles 2γ2 . Finally γ1 =
2γ2 = α + (180◦ − γ2 )
(32.16)
because 2γ2 is an exterior angle in triangle 4OC2 B, and α, 180◦ −γ2 are the two nonadjacent interior angles. Solving (32.16) for γ2 yields the claim. The angles between the ruler lines at vertex B can now be calculated via the angle sum in 4C1 BC2 and triangle 4C2 BC3 . One gets ∠C1 BC2 ∠C2 BC3 60◦ . −−−→ −−−→ −−−→ To get the angles between the three rays OD1 , OD2 and OD3 , we can use Euclid III.20 "the central angle is twice the circum angle". Hence ∠D1 OD2 = 2∠D1 BD2 = 2∠C1 BC2 = 2 · 60◦ = 120◦ and similarly, ∠D2 OD3 = 2∠D2 BD3 = 2 · 60◦ = 120◦ . As I imagine, my dear friends, this is all the old guy Archimedes could have told his students, would he still be alive. 32.12. Archimedes’ trisection and the conchoid Archimedes construction can be interpreted as intersecting a conchoid and a circle. I keep the drawing from above. The Cartesian coordinate system get the origin at point B, the x-axis pointing to the right, and the y-axis upwards. The marked segment has length p = 1, the guiding line in the horizontal line with the Cartesian equation y = −d, and the distance from focus to guiding line is d = sin α. In polar coordinates, this conchoid has the equation d sin α r = p− =1− (32.17) sin θ sin θ Problem 32.22. (a) Find the equation of the circle in polar coordinates (r, θ). (b) Use your TI84 and graph the two curves for α = 40◦ . (c) Point B is a double intersection point of the self crossing conchoid and the circle. It occurs at r = 0 and is now discarted. Find the four angles for which the conchoid and the circle intersect with r , 0. Answer (a). Let R be a generic point on the circle, and M be the midpoint of segment BR. From r the right triangle 4BOM one gets cos(θ − α) = |BM| |BO| = 2 . The circle has the equation r = −2 cos(θ − α)
(32.18)
Answer (b). To use the TI84 and graph the two curves for α = 40◦ . Choose a graphing window with Zquare and Ymax = 1.666, Ymin = -1.666.
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Answer (c). To find the four angles θ1 , . . . , θ4 for which the conchoid and the circle intersect. Putting equal the radial coordinate r from the equations of the conchoid (32.17) to that of the circle (32.18), and multiplying by sin θ yields the equation 1−
sin α + 2 cos(θ − α) = 0 sin θ
for angle θ. One easy solution corresponds to putting the ruler through the midpoint of the circle. Hence θ4 = α + 180◦ is a solution, as one can check. To cancel that zero, I factor out cos θ−α 2 and get, after some calculations sin θ − sin α + 2 cos(θ − α) sin θ = 2 cos
θ−α 3θ − α sin 2 2
(32.19)
Question (d). Check this formula. Answer. Put x = (32.19) gets
θ−α 2
and y =
3θ−α 2 .
Hence θ = y − x and α = y − 3x, and the left hand side of
sin θ − sin α + 2 cos(θ − α) sin θ = sin(y − x) − sin(y − 3x) + 2 cos 2x sin(y − x) = sin(y − x) − sin(y − x) cos 2x + cos(y − x) sin 2x + 2 cos 2x sin(y − x) = sin(y − x) + sin(y − x) cos 2x + cos(y − x) sin 2x = sin(y − x) + sin(y − x) cos 2x + cos(y − x) sin 2x = sin(y − x) + sin(y + x) = 2 sin y cos x θ−α 3θ − α = 2 cos sin 2 2 Question. Find the four angles for which the conchoid and the circle intersect. Answer. The four zeros are θ1 =
α α + 360◦ α − 360◦ , θ2 = = θ1 + 120◦ , θ3 = = θ1 − 120◦ , θ4 = 180◦ + α 3 3 3
Unfortunately, the circle equation (32.18) yields r < 0 for some angles. One can again substitute (r, θ3 ) 7→ (−r, θ + 180◦ ) to remedy that awkward situation. The circle is passed twice, but the conchoid only once, if the angle takes a 360◦ turn. Hence one gets four solutions, which are all different. Question. How does one loose the two fake solutions D5 and D6 ? Answer. One can loose the origin in polar equations, because it should be checked for all angles θ— but we did not take this into account. For the conchoid, these two fake solutions should appear for θ5 = α and θ6 = α − 180◦ . Indeed at these two angles one get the self crossing loop for the conchoid at r = 0. But the equation (32.18) of the circle has r = ±2 and yields both times the point D4 .
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Figure 32.20. Archimedes’ trisection and the conchoid
Figure 32.21. Archimedes’ trisection and the conchoid
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33. The Heptagon 33.1. The construction We inscribe a regular 7-gon inside a circle. The construction uses a marked ruler to trisect an angle. Probably, it was already known to Archimedes. Construction 33.1 (Construction of a Regular Heptagon). Given is a circle C of radius 3 with horizontal diameter BOA, and point F on the radius OA with OF = 1. Let 4OAC be equilateral. Draw a second circle A around F through point C. Next use Archimedes’ construction to trisect the angle 3α = ∠AFC. One gets a point G where the segment HC intersects the circle A just drawn, and three congruent segments HG GF FC. The 4HFC has ∠FHC = α, ∠FCH = 2α. Finally, draw circular arcs of radius 3 around H. Let I, K be its intersection points with the original circle of the same radius around O. The isosceles 4HOI has base angles β = ∠BOI ∠IHO. The assertion is that 360◦ β= 7 Now you can get a regular heptagon with K, B, I three of its seven vertices.
Figure 33.1. The heptagon
Problem 33.1. (a) Actually do the construction, using an entire sheet. Draw the 7-gon and check for accuracy. (b) Use trigonometry and a numerical computation to check the validity of the construction.
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Figure 33.2. How I begin
33.2.
Trigonometric calculations
Answer. These are the quantities for which exact expressions are needed: Question (i). Find the radius |FC| of circle A. Answer. Let √ D be the midpoint of segment OA. The altitude of the equilateral 4OAC is 3 3 |DC| = . Hence the Pythagorean theorem yields |FC|2 = |FD|2 + |DC|2 = 14 + 27 4 = 7, and 2 √ |FC| = 7 Question (ii). We define angle ∠AFC =: 3α. Calculate cos 3α Answer. cos 3α =
|FD| 1 = √ |FC| 2 7
Question (iii). Calculate x :=
|FH| |OA|
in terms of cos α. Answer. By the sin theorem, the ratio of two sides of a triangle equals the ratio of the opposite angles. For 4HFC, the sin theorem and the double angle formula sin 2α = 2 sin α cos α imply |FH| sin 2α = = 2 cos α |FC| sin α Hence
√ |FH| |FH| |FC| 2 7 cos α x= = · = |OA| |FC| |OA| 3
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Figure 33.3. Use Archimedes’ trisection
Question (iv). Recall that β = ∠BOI ∠IHO are the base angles of the isosceles 4HOI. Calculate cos β in terms of x. Answer. Let M be the midpoint of segment HO. From the right 4HI M one gets cos β =
|HO| |HF| − 1 3x − 1 = = 2|HI| 6 6
These are fairly all quantities needed for the construction. Proposition 33.1. As the construction is done, we define the two angles 3α := ∠AFC and β := ∠BOI. 1 cos 3α = √ 2 7 √ |FH| 2 7 cos α 1 x= = = 2 cos β + |OA| 3 3
(33.1) (33.2)
satisfies the third order equation x3 −
7 7 x− =0 3 27
Proof. Because we have shown that cos 3α =
1 √ , 2 7
(33.3)
the identity cos 3α = 4 cos3 α − 3 cos α yields
4 cos3 α − 3 cos α −
1 √ =0 2 7
√ √ 3 2 7 cos α 7 2 7 cos α 7 − · − =0 3 3 3 27 This means that x =
√ 2 7 cos α 3
satisfies the third order equation (33.3)
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(c) Use formulas (33.1) and (40.11) for a numerical calculation of the quantities 3α, α, x, cos β, β and 7β. Answer.
33.3.
3α = 79.10660535◦ , α = 26.36886845◦ , x = 1.580312937 , cos β = 0.6234898019 , β = 51.42857143◦ , 7β = 360◦
(33.4)
Using complex numbers
On giants shoulders. Easier insight is gained from complex numbers. (That is why we are on "giants shoulders".) Define 2π γ := , z1 = eiγ 7 (33.5) 1 u := 2 cos γ + 3 We prove that u is a root of equation (33.3). The complex number z is a root of the polynomial equation z7 − 1 = 0 Division by z − 1, next z3 , and the binomial formula, next 2 cos γ = z + z−1 , then again the binomial formula lead to the identities z6 + z5 + z4 + z3 + z2 + z + 1 z7 − 1 = (z − 1)z3 z3 3 −3 2 = z + z + z + z−2 + z + z−1 + 1 = (z + z−1 )3 − 3(z + z−1 ) + (z + z−1 )2 − 2 + z + z−1 + 1 = (z + z−1 )3 + (z + z−1 )2 − 2(z + z−1 ) − 1 = 8 cos3 γ + 4 cos2 γ − 4 cos γ − 1 !3 1 2 cos γ 1 = 2 cos γ + − − − 4 cos γ − 1 3 3 27 !3 ! 1 7 1 7 = 2 cos γ + − 2 cos γ + − 3 3 3 27 This means that u := 2 cos γ +
1 3
satisfies the third order equation (33.3), as to be shown.
(33.6)
Question (d). A cubic equation has three real or complex roots, counting multiplicity. What are the two other roots of equation (33.3)? Answer. The other two solutions of equation (33.3) are u2 := 2 cos 2γ +
1 , 3
u3 := 2 cos 3γ +
1 3
(33.7)
Indeed, because z21 = e2iγ and z31 = e3iγ are two further solutions of z7 − 1 = 0, the calculation from above shows that u2 and u3 satisfy equation (33.3), too. Question (e). Calculate u2 and u3 from (33.7) numerically and check that they satisfy equation (33.3) Answer. u2 = −0.1117085346 , u3 = −1.468604402
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Question (f). We have shown above that x = 2 cos β + 31 and u = 2 cos γ + 13 both satisfy the same equation (33.3). What is the actual value of β? Is the construction working exactly, or is it only an approximation? Answer. One can conclude that x = u or x = u2 or x = u3 . A reason why only the first case is possible, can be obtained from geometry directly. The second case x = u2 would mean that x = 2 cos β + 13 = 2 cos 2γ + 31 . Hence we would get β = 2γ! We can see from the construction that this does not happen. Indeed the angle β is acute, but the ◦ angle 2γ = 2 · 360 7 is obtuse. Similarly, one rules out the third case. Still a different reason for the first case is direct numerical evidence. ◦ Thus we have finally confirmed that x = 2 cos β + 13 = 2 cos γ + 31 and β = γ = 360 7 . Hence the construction is exactly valid. 33.4.
Geometric proof The following geometric proof is credited to François Vièta.
Proposition 33.2. The lines QO and IA are parallel. Proof. By means of Euclid V.6, proportions can be used to show that lines QO and IA are parallel. To this end, we need to check whether |HQ| |HO| = (33.8) |HI| |HA| To get (33.8), we use as first step chord in a circle, Euclid III.36. This proposition implies that |HQ| · |HI| = |HB| · |HA|. Hence |HQ| |HA| |HQ| · |HI| |HA| |HA| · |HB| |HA| |HB| · |HA|2 · · · = = = |HI| |HO| |HO| |HO| |HO| · |OB|2 |HI|2 |OB|2 That √ seems stuck, but we need to use the construction yet. From the construction |HO| = 2 7 cos α − 1 = 3x − 1, |HA| = 3x + 2, |HB| = 3x − 4, |OB| = 3 and hence |HB| · |HA|2 (3x − 4)(3x + 2)2 = (3x − 1) · 9 |HO| · |OB|2 Now equation (33.3) mysteriously implies that this last fraction equals one. Hence we get indeed equation (33.8), Hence by Euclid V.6, the lines QO and IA are parallel. Proposition 33.3 (The angle sum of a triangle comes in seven parts). Let Q be the second intersection point of line HI and the main circle around O. Lines QO and IA are parallel. The angles ε, 2ε = β, 3ε comes up seven times with vertices H, Q, I, O, A as shown in the drawing. Hence the angle sum of 4HIA implies that 7ε = 180◦ and 360◦ β= 7 Question (g). Mark the angles ε, 2ε, 3ε seven times in the drawing below! Proof. Now we use Euclid I.29 to produce congruent z-angles from the parallel lines, together with congruence of base angles of isosceles triangles. One gets that ε = ∠HOQ ∠HAI ∠AIO ∠IOQ From the congruent base angles of isosceles 4HIO, one gets β := ∠OHI ∠HOI = 2ε Form the exterior angle in 4HQO, and finally the bases angles of isosceles 4QOI, one gets β + ε = ∠OQI ∠OIQ = 3ε Hence the angle sum of 4HIA comes in seven parts and 7ε = 180◦ , as to be shown.
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Figure 33.4. Angle sum in seven parts
Remark. ε = β2 = 25.71428571 < α. Hence lines OQ and FG are not parallel, but intersect in the β α lower half plane. I conjecture that 360 ◦ is irrational, and only 360◦ is rational. 33.5.
A false but almost regular heptagon
Problem 33.2 (A false heptagon). A false construction of a regular heptagon is shown in the figure on page 33.5. We use a circum circle of radius |OA| = 2. 1. Calculate the false side |AM|. 2. Calculate the true side of the regular heptagon. 3. How much too short is the false side. 4. Calculate the accumulated center angle ∠WOA by which the false heptagon does not close. Answer. 1.
|OA| = 2 is assumed, hence |OM| = 1. Pythagoras for the triangle 4OAM gives |AM|2 = |OA|2 − |OM|2 = 3 √ |AS | = |AM| = 3 ≈ 1.732050808
2.
for the constructed side. The central angle for the regular heptagon is 360◦ /7. The true side is 2|OA| sin
3. 4.
360◦ ≈ 1.735534956 14
The true side is about 0.0035 longer. More instructive is to know the angle ∠AOW by which the false 7-gon fails to close. Let α = ∠S OA. √ α |AS | 3 sin = = hence 2 2|OA| 4 α ≈ 51.3178◦ and ∠AOW = 7α − 360◦ ≈ −0.775◦
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Figure 33.5. A false heptagon.
33.6.
Plemelj’s construction
Construction 33.2 (Plemelj’s construction of a regular heptagon). Draw the circle with center O passing through A. Find on this circle the point M so that AM OA. Bisect segment OM at N, and trisect it at P. Find point T on the segment NP so that ∠NAT = ∠NAP/3. The segment AT is the needed side of the regular heptagon with the circum circle drawn in the beginning. For the justification of the construction, we check that both the constructed side, and the true side of the regular heptagon inscribed into the unit circle satisfy the cubic equation √ √ y3 + 7y2 − 7 = 0 But this cubic equation has a unique positive solution. Hence the construction yields the regular heptagon. Calculation of the constructed side. The tangent and cosine of the angle α = ∠NAP to be trisected are √ √ 1 27 tan α = 1/(3 3) and cos α = √ = √ 2 28 1 + tan α
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Figure 33.6. A regular heptagon.
For the trisected angle, u = 2 cos(α/3) is a root of the polynomial √ 27 3 3 u − 3u − 2 cos α = u − 3u − √ = 0 7 For the constructed side length √ √ 3 3 y := |NA| = = 2 cos(α/3) u we get the polynomial equation √ √ √ 3 3 3 27 − √ =0 − 3 y y 7 √ 2 √ 3 y + 7y − 7 = 0 This equation has exactly one positive solution. You can use Descartes’ rule of signs and elementary calculus to check. Hence it really determines the constructed piece uniquely. Reduction of the true side to a cubic root. We have seen in equation (33.6) that 8 cos3 γ + 4 cos2 γ − 4 cos γ − 1 = 0
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has the solutions γ = 2πk/7 with k = 1, 2, 3. Hence the polynomial R(v) := v3 + v2 − 2v − 1 has the zeros 2 cos(2πk/7). The actual side is xk = 2 sin
πk 7
with k = 1 and the diagonals are obtained for k = 2, 3. They are related to half the angle. We use 2 − xk2 = 2 − 4 sin2
πk 2πk = 2 cos 7 7
and see that these are zeros of the polynomial R7 (v). Now we substitute: R(2 − x2 ) = (2 − x2 )3 + (2 − x2 )2 − 2(2 − x2 ) − 1 = −x6 + 7x4 − 14x2 + 7 √ The polynomial is still reducible over Q( 7) and hence over the constructible field. Indeed √ √ √ √ x6 − 7x4 + 14x2 − 7 = (x3 + 7x2 − 7)(x3 − 7x2 + 7) From Descartes rule of signs, and by direct calculation, we see that the first factor has the zeros 4πk π 2πk 4πk 2 sin π7 , −2 sin 2πk 7 , −2 sin 7 , whereas the second factor has zeros −2 sin 7 , 2 sin 7 , 2 sin 7 . Question. How can we now see that the regular 7-gon is not constructible with ruler and compass? Answer. The polynomial x6 − 7x4 + 14x2 − 7 has as zeros the side and diagonals of the regular heptagon inscribed into the unit circle. By the Eisenstein criterium with p = 7, we see that this polynomial is irreducible over the integers. Hence any of its roots r generate a field extension [Q(r) : Q] = 6 of dimension six. But this is not a power of two. Hence r is not in the constructible field.
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34. Geometric Constructions and Field Extensions We shall consider the possibility and impossibility of constructions in a Pythagorean plane. The means of constructions may vary. But to even speak of any proof of possibility or impossibility of a construction, we need to impose the following minimal requirements: 1. 2. 3. 4. 5.
It has to be clearly specified which means of construction are allowed. We need to postulate or be able to prove that any construction step has a unique result. When any arbitrary elements are chosen, it needs to be indicated. It needs to be explained how to check whether the construction process is finished and stops. A construction is only allowed to use finitely many steps.
Specifically, we consider the following means of construction, in the order of increasing strength: • Hilbert tools as specified by Definition 7.10 from the section about congruence of segments, angles and triangles. • Traditional Euclidean tools as specified by Definition 14.7 from the section about a simplified axiomatic system of geometry. • A neusis or sliding ruler. This is a two-marked straightedge. At first we impose the restriction that the marked points can only be adjusted to lie on a straight line, but it is not possible or allowed to adjust a marked point in a way that is lies on a circle. The trisection by Nicomedes described in the Construction 32.2 and Nicomedes’ solution of the Delian problem described in the construction 32.7 are two important examples. • Traditional Euclidean tools, and additionally a two-marked straightedge. Now the marked points can be adjusted to lie either a straight line or a circle. An important example for such a construction is Archimedes’ trisection of an angle described in the Construction 32.1. • Paper folding. As a example, we have described in the Construction 32.4 the trisection of an angle by paper folding. There exist further options, some of which are classics, some are a part of mathematical physics— others are more or less entertainment. For example, we may consider: • The "lazy-boy" construction of the tangent from the point to the circle, discussed in Problem15.14 from the section on Euclidean geometry of circles. • Intersection of conic sections with straight lines. • Mutual intersection of conics. • Building three-dimensional models of polyhedra. • Intersections of further curves to be specified. • A wheel rolling without friction, on a straight line or even a circle. • Balancing of a level as a mechanical tool, used in the Method by Archimedes for the calculation of the volume of a ball. • Other mechanical tools. • Optical tools. In this section, we need a lot of material from modern algebra. My main sources are Michael Artin’s book [2] on Algebra and the relevant chapter in Robin Hartshorne’s book [19]. 34.1. A reminder about polynomials The integers are denoted by Z. For an indeterminant or variable x, the expressions 1, x, x2 , x3 , . . . are called monomials. An integer combination of monomials an xn + an−1 xn−1 + · · · + a1 x + a0 with integer coefficients an , an−1 , . . . , a0 ∈ Z is called an integer polynomial or polynomial over the integers. The set of all integer polynomials is denoted by Z[x]. One has to put the indeterminant into square brackets. The set Z[x] is a first example for a ring. For any ring R, we can define the polynomial ring R[x] by taking elements an , an−1 , . . . , a0 ∈ R as coefficients in the respective polynomials. Again,
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provided that R is a ring, the set of polynomials R[x] is a ring, too. It is called the polynomial ring over the ring R. A polynomial with the leading monomial an xn and an , 0 is defined to have the degree n. A polynomial with the leading monomial xn is called monic. The polynomials of positive degree are those with n ≥ 1 and hence those in which the indeterminant x really appears. The polynomials with degree 0 are the elements a0 , 0 of the ring R. The zero polynomial has the degree undefined1 —hence deg 0 , 0. A ring with no null divisors can be embedded into a quotient field. For any field F, one can define the polynomial ring F[x]. This is not a field, but a ring with important nice properties. Especially the following algorithms work: • division with remainder; • the Euclidean algorithm; • the extended Euclidean algorithm. • There exists a greatest common divisor for any two nonzero polynomials. Division with remainder is possible for polynomials over a field. For any p, q ∈ F[x] with q , 0, we say p : q is a with remainder r as a short-hand expressing the polynomial equation p = aq + r with deg r < deg q The polynomials a and r are unique up to nonzero factors in the field F. Of course, the remainder r may turn out to be zero. Furthermore, the Euclidean algorithm and even the extended Euclidean algorithm can be performed in the ring F[x], and always stops after finitely many steps. Hence there exists a greatest common divisor gcd(p, q) for any two not both zero polynomials p, q ∈ F[x]. Proposition 34.1. Any two polynomials p, q ∈ F[x] not both of which are zero have a greatest common divisor gcd(p, q) ∈ F[x]. Moreover, there exist polynomials r, s ∈ F[x] such that rp + sq = gcd(p, q) The polynomials gcd(p, q) , 0, r, s are only unique up to nonzero factors in the field F. 34.2. Algebraic numbers Algebraic numbers, algebraic integers, and their conjugates can be defined over any base field F. In case no base field is mentioned, it is assumed, in these notes and in many books by many authors, the base field are the rational numbers. By a field extension is meant a pair of fields F ⊆ K. It is customary to write K/F for a field extension. The most common field extensions are R/Q, C /Q and C /R. A number α is called algebraic or, more accurately, algebraic of a field F, if there exists a nonzero polynomial p ∈ F[x] such that p(α) = 0. In that case α is called a root of the polynomial p. The lowest degree of polynomials with the root α is called the degree of the algebraic number α. The corresponding polynomial is called the minimal polynomial of the algebraic number α. Using division with remainder, it is easy to see that the minimal polynomial is unique up to a factor in F. A number α is called an algebraic integer if it is the root of a nonzero monic minimal polynomial p ∈ Z[x]. 1
To put deg 0 = −1 seems to me no good choice neither.
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The numbers a ∈ F in the base field, including zero 2 , have the degree one. Indeed, p(x) = x−a is a polynomial of degree one with the root α. All other numbers a < F have either degree at least two, or they are never roots for any field extension K/F. The numbers, for which there does not exist any field extension in which they become a root are called transcendental. The set of all real or complex algebraic numbers is denoted by A. This is a countable set. 34.3. The dimension of a field extension For any field extension K/F , the upper field K is a vector space over the base field F. Hence this vector space has a dimension, which can be finite or infinite. We denote the dimension of a field extension by dim K/F or [K : F]. An extension is called finite-dimensional or simply finite if the dimension [K : F] is finite. Obviously [K : F] = 1 if and only if K = F. Proposition 34.2 (Tower Theorem). The dimensions of a chain of two field extensions F ⊆ L ⊆ K are multiplied: [K : F] = [K : L] · [L : F] Indication of reason. Let n = [L : F] and l1 , . . . , ln be a basis of the extension L/F. Let m = [K : L] and k1 , . . . , km be a basis of the extension K/L. Then one can check that the nm products ki l j are linearly independent over F and generate the largest field K. Hence the extension K/F has the dimension nm. A polynomial p ∈ R[x] over any ring is called irreducible if any factorization p = rs has one factor which is a unit of the base ring R. A polynomial p ∈ F[x] over any field F is called irreducible if any factorization p = rs has one factor of degree zero. Equivalently, we can require that p cannot be factored into any polynomials of lower degree. By these definitions, the zero polynomial is not irreducible. Every irreducible polynomial p ∈ F[x] has degree at least one. Every irreducible polynomial p ∈ Z[x] is either a prime number, or has degree at least one. Given is an extension K/F and one number α ∈ K which is algebraic over the base field F. The minimal polynomial of α is the polynomial in F[x] of lowest degree which has the zero α. The one-element extension F(α) is defined to be the smallest field containing both α and all elements of F. Proposition 34.3. The minimal polynomial is irreducible and unique up to nonzero factors in F. Conversely, let α be a root of an irreducible polynomial. This polynomial is the minimal polynomial of the one-element extension F(α) generated by α. Proposition 34.4 (Extension generated by one element). The dimension of the extension F(α)/F generated by one algebraic number α is the degree of the minimal polynomial of the algebraic number α. The powers 1, α, α2 , . . . αn−1 are a basis for the extension F(α)/F. All further roots which the minimal polynomial may have in K are called the algebraic conjugates of α. We see that they have the same minimal polynomial. Proposition 34.5. The extension field F(α) generated by one algebraic number α over the base field F is isomorphic to the ring quotient F[x]/(p) modulo the principal ideal (p) generated by the minimal polynomial p of α. deg 0 = 1 for the number zero, but deg 0 = −1 for the polynomial zero.—That is the only contradiction I have ever found in modern algebra!
2
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The fact that the quotient ring F[x]/(p) is a field guaranties the existence of a field extension corresponding to any irreducible polynomial p. Moreover, any two such extensions are isomorphic, via an isomorphism which fixes all elements of F. For any algebraic conjugate β of α, the extensions F(α) and F(β) are isomorphic, via an isomorphism which fixes all elements of F. Further and larger extensions K/F can be generated by adjoining any set of numbers α, β, · · · ∈ K. The extension F(α, β) is defined to be the smallest field containing both α, β and all elements of F. This definition implies immediately that F(α, β) = F(α)(β) = F(β)(α) = F(β, α). An extension obtained by adjoining finitely many elements is called finitely generated. For a finitely generated extension we obtain a spanning set αi−1 βk−1 with i = 1 . . . deg α and k = 1 . . . deg β. A similar idea is used in the tower theorem, too. But in the present context, these elements may turn out to be linearly dependent over F. We obtain only an upper bound for the dimension of multiple extensions: [F(α, β) : F] = [F(α, β) : F(α)][F(α) : F] ≤ [F(β) : F][F(α) : F] = deg α · deg β [F(α, β, γ) : F] ≤ deg α · deg β · deg γ Remark. We may have inequality in [F(α, β) : F(α)] ≤ [F(β) : F]. This happens if adjoining α splits the irreducible polynomial q ∈ F[x] for β into two polynomials q, r ∈ F(α)[x] of lower degree. In that case q = rs and hence β is either a root of r or s and [F(α, β) : F(α)] ≤ max(deg r, deg s) < deg q = [F(β) : F] A field extension K/F is called algebraic if all numbers in K are algebraic over the base field F. Otherwise, the field extension is called transcendental. Proposition 34.6. An extension K/F is finite if and only if it is algebraic and generated by adjoining finitely many elements to the base field. There exist finite as well as infinite√dimensional algebraic extensions. For example: • The extensions C /R and Q( −1)/Q are two dimensional. • The extensions Ω/Q and K/Q of the Hilbert field and the constructible field over the rational numbers are infinite dimensional. • The extension of all algebraic numbers√A/Q over the rational numbers is infinite dimensional. • The extensions (A ∩ R)/Q and A/Q( −1) are infinite dimensional. Since F[x] has no null divisors (is an integral domain), the set of quotients p/q of polynomials p, q ∈ F[x] is a field. The quotient field is also called the field of rational functions. Any oneelement transcendental extension is isomorphic to the field of rational functions. Hence every transcendental extension is infinite dimensional. Proposition 34.7. Suppose K/F is a field extension and the polynomial p ∈ F[x] of degree n has n roots α1 . . . αn in K. Then the polynomial p factors into n Y p(x) = (x − αi ) i=1
already in the algebraic extension F(α1 , . . . , αn )/F. The dimension d of this extension is at most the factorial n!: d = [F(α1 , . . . , αn ) : F] ≤ 1 · 2 · 3 · · · (n − 1) · n All fields L ⊇ F of dimension d in which the polynomial p splits, are isomorphic with an isomorphism fixing the elements of F. If the polynomial p is irreducible, then n is a divisor of dimension d.
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34.4.
The field of algebraic numbers
Proposition 34.8. The reciprocal of an algebraic number a , 0 over any field F is algebraic, and contained in the simple extension F(a). Indeed, the reciprocal 1/a can be expressed as evaluation of a polynomial s ∈ F[x] of degree deg s < deg a: 1 = s(a) a Especially, the reciprocal of an algebraic number over the rational numbers can be expressed as evaluation of an integer polynomial s ∈ Z[x] with deg s < deg a, divided by a positive integer 0 < n ∈ Z: 1 s(a) = a n Proof. We want to express the reciprocal of an algebraic number. Let a , 0 have the minimal polynomial p ∈ F[x]. We need to obtain the reciprocal 1/a. To handle this case, we choose q(x) = x. As easily, we can obtain the reciprocal 1/q(a) with q(a) , 0 for any polynomial q ∈ F[x] with deg q < deg p. We use the extended Euclidean algorithm for p and q. Their greatest common divisor g = gcd(p, q) is a divisor of p with degree deg g ≤ deg q < deg p. Since p is the minimal polynomial, it is irreducible. Hence the divisor g of p is either a multiple of p or a multiple of 1. Since deg g < deg p and g , 0, we see that deg g = 0. In other words, g ∈ F is a constant polynomial. Since 0 , g ∈ F, we may even choose g = 1. By the extended Euclidean algorithm, the greatest common divisor g = gcd(p, q) can be expressed as rp + sq = g with polynomials r, s ∈ F[x]. Evaluation at x = a yields 1 = g(a) = r(a)p(a) + s(a)q(a) = s(a)q(a) Hence
1 = s(a) q(a)
is a representation of the reciprocal of an algebraic number by evaluation of a polynomial over the base field. Remark. Moreover, in the quotient ring F[x]/(p) we get the identity 1 ≡ s mod p q We have proved once more: Corollary 66. The quotient ring F[x]/(p) of a polynomial ring of a field over a principal ideal (p) of an irreducible polynomial p is a field. Question. Write
with an integer denominator and cancel.
√ 5+4 2 √ 4+2 2
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Answer.
√ √ √ √ √ 5 + 4 2 (5 + 4 2)(4 − 2 2) 20 − 16 + 6 2 2 + 3 2 = √ = √ √ = 8 4 4 + 2 2 (4 + 2 2)(4 − 2 2) 3 Question. Use the extended Euclidean algorithm for the polynomials x − 2 and x2 + 1. Answer. We need two divisions with remainder until 5 turns out to be the last nonzero remainder. (x3 − 2) : (x2 + 1) = x remainder − x − 2 (x2 + 1) : (−x − 2) = −x + 2 remainder 5 Then go backwards to express the last remainder as a combination of x3 − 2 and x2 + 1. 5 = (x2 + 1) − (x + 2)(x − 2) = (x2 + 1) + [(x3 − 2) − x(x2 + 1)](x − 2) = (x3 − 2)(x − 2) + (x2 + 1)(−x2 + 2x + 1) Question. Find the reciprocal 1 2 x +1 in the quotient field Q[x]/(x3 − 2). Answer. 5 ≡ (x2 + 1)(−x2 + 2x + 1) mod (x3 − 2) 1 −x2 + 2x + 1 ≡ mod (x3 − 2) 5 x2 + 1 gives the answer in the quotient field Q[x]/(x3 − 2). Question. Write 1 2/3 2 +1 with an integer denominator. √3 Answer. We use the last answer with x = 2 and get −22/3 + 2 · 21/3 + 1 1 = 5 22/3 + 1 As an alternative, we can go back to the identity 5 = (x3 − 2)(x − 2) + (x2 + 1)(−x2 + 2x + 1) and √3 plug in x = 2. We get the same answer. Proposition 34.9. The sum a + b, the product ab, and the quotient a/b with b , 0 of any two algebraic numbers are algebraic. The sum a + b and the product ab of two algebraic integers is and algebraic integer. Indication of reason. Because of the existence of a splitting field, we may assume that the minimal polynomials p ∈ F[x] with p(a) = 0 and q ∈ F[x] with q(b) = 0 split in the field K. Hence m n Y Y p(x) = (x − ai ) and q(x) = (x − bk ) i=1
k=1
The polynomials S (x) =
m Y n m Y n Y Y (x − ai − bk ) and P(x) = (x − ai bk ) i=1 k=1
i=1 k=1
have coefficients which are totally symmetric in the roots ai as well as bk . By the Symmetric Functions Theorem, 1 the coefficients of P and Q are F-polynomial functions of the elementary 1
Theorem 16.1.6 in Michael Artin’s book [2]
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symmetric functions of ai and bk . The elementary symmetric functions of ai are by Viëta’s Theorem the coefficients of p, and the elementary symmetric functions of bk are the coefficients of q. These elementary symmetric function, and hence the coefficients of P and Q are elements of the base field F. Hence S , P ∈ F[x]. This means that the sums ai + bk and the products ai bk are algebraic over F. Theorem 34.1 (The field of algebraic numbers). The set A of all (real or complex) algebraic numbers is a field (Eisenstein 1850). The set A is countable (Cantor 1874). 34.5.
Minimal fields
Definition 34.1 (Hilbert field). The Hilbert field Ω is the smallest real field with the properties (1) 1 ∈ Ω. (2) If a, b ∈ Ω, then a + b, a − b, ab ∈ Ω. (2a) If a, b ∈ Ω and b , 0, then ab ∈ Ω. √ (3) If a ∈ Ω, then 1 + a2 ∈ Ω. Definition 34.2 (constructible field). The constructible field K is the smallest real field with the properties (1) 1 ∈ K. (2) If a, b ∈ K, then a + b, a − b, ab ∈ K. (2a) If a, b ∈ K and b , 0, then ab ∈ K. √ (4) If a ∈ K and a ≥ 0, then a ∈ K. It is also called the surd field. Remark. There are many very different Pythagorean fields, and many different Euclidean fields, among them the real number field R. There exist non-Archimedean Pythagorean fields, as well as non-Archimedean Euclidean fields, too, but these are not the minimal fields. Only the Hilbert’s field Ω and the surd field K are minimal and hence they are uniquely defined structures. Since the Hilbert field Ω is smallest Pythagorean field, we have the following more direct characterization of Ω: Proposition 34.10. Any number x ∈ Ω in the Hilbert field can be obtained in finitely many steps. We start with a rational number—or even just 0 and 1. Each step constructs a new number using only the field operations and the operation provided √ 2 by the assumption (3): a > 0 7→ 1 + a , applied to the numbers already obtained. Note that the total number of construction steps depends on the number x ∈ Ω and hence has no global bound. The corresponding statement holds for the constructible field K. Proposition 34.11. Any number k ∈ K in the constructible field can be obtained in finitely many steps as follows: We start with a rational number—or even just 0 and 1. Each step constructs a new number using only the field operations and the operation provided by the assumption (4): √ a > 0 7→ a, applied to the numbers already obtained. Corollary 67. Since there are only countably many such processes, the constructible field is countable. The Hilbert field is a subfield of the constructible field and countable, too.
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34.6.
Towers in the constructible field
Corollary 68. For any number k ∈ K in the constructible field there exists a tower of finitely many two dimensional extensions Q = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn = Q(k) Especially the dimension [Q(k) : Q] is a power of two. Proof of the Corollary. Fix any number k ∈ K in the constructible field. The numbers obtained up to the i-th step of the construction process generate a field Fi . Clearly Fi−1 ⊆ Fi for all construction steps. Either the i-th step involve√only field operation in which case Fi−1 = Fi , or the i-th step is an operation of the type a > 0 7→ a. In the latter case √ a ∈ Fi−1 , a > 0 , Fi = Fi−1 ( a) and [Fi : Fi−1 ] = 2 Suppose construction of the number k ∈ K involves N steps of the second kind. We have obtained a tower of N extensions Q = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ FN 3 k In this tower, all extensions are two-dimensional, generated by adjoining a single root of a quadratic equation. Hence [FN : Q] = 2N . We may intersect all fields of the tower with the field Q(k) and discard the step with equality Fi−1 ∩ Q(k) = Fi ∩ Q(k). Thus we obtain the tower in the corollary, with n ≤ N. Actually we can also begin with the tower for which N is minimal. Thus we get FN = Q(k) and hence [Q(k) : Q] is a power of two. Proposition 34.12. Conversely, suppose that any number k ∈ R is obtained in a tower of twodimensional extensions Q = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn 3 k (34.1) Then the number k is in the constructible field. Theorem 34.2 (Towers in the constructible field). The constructible field consist of the real numbers k ∈ R for which a tower of two-dimensional extensions as in equation (34.11) exists. Remark. Every number of the constructible field lies in a finite dimensional extension of the rational numbers. Nevertheless, the entire constructible field is an infinite dimensional extension of the rational numbers. Lemma 34.1. Let the field F have characteristic not equal √ two. Let L/F be any field extension of dimension two. Then there exists a ∈ L such that L = F( a). √ Moreover, if L ⊆ R, then there exists D > 0 in L such that L = F( D). Finally, if the field F is a subfield of the constructible field, then L is a subfield of the constructible field, too. Proof of the Lemma. Suppose that [L : F] = 2 and α ∈ L \ F. The extension is generated by α since no proper intermediate field F ⊂ M ⊂ L can exist. Hence L = F(α). The minimal polynomial p ∈ F[x] of α has degree 2. Let p(x) = x2 + bx + c The quadratic equation p(α) = 0 has the solution √ α1,2 =
−b ±
b2 − 4c 2
649
The division by 2 , 0 is possible in a field of characteristic not two. Hence p(x) = (x − α1 )(x − α2 ). It is impossible that α1 = α2 : that would contradict the fact that p is irreducible. We can produce 2 the same field extension √ by adjoining the square root of the discriminant D = b − 4c. In other words, L = F(α) = F( D). In the case that L ⊆ R, the fields F and L have characteristic zero. Hence the solution formula for √the quadratic√equation works. Since α ∈ L ⊆ R, the quadratic formula implies α1 − α2 = b2 − 4c = D√ ∈ R. Moreover, α1 , α2 as mentioned √ above, and hence D > 0. 0 → 7 D in Proposition 34.11. If Thus the field extension F( D)/F corresponds to a step D > √ the field F is a subfield of the constructible field, then L = F( D) is a subfield of the constructible field, too. Proof of Proposition 34.12. We may intersect all fields of the tower (34.11) with the real number field R, or even R ∩ A and discard the steps with equality Fi−1 ∩ R = Fi ∩ R. We still get a working tower since k ∈ Fn ∩ R. The lemma proves inductively for all i = 1 . . . n that Fi−1 ⊆ K. Hence k ∈ K. 34.7.
Totally real extensions
Proposition 34.13. All numbers in the Hilbert field are totally real algebraic numbers. That means that they are roots of a irreducible polynomial of (possibly high) degree N with integer coefficients, all N roots of which are real. √ Lemma 34.2. If the number x is totally real, then
1 + x2 is totally real, too.
Proof. Assume the number x is totally real. Hence there exists an integer (even irreducible) polynomial P ∈ Z[x] of degree N all roots x = x1 , x2 . . . xN of which are real. We define R(y) = P(y2 − 1), which is again an integer polynomial and has the roots q q q ± 1 + x12 , ± 1 + x22 · · · ± 1 + x2N q These are all real numbers and hence
1 + x12 is totally real, as to be shown.
Remark. The factoring R(y) = Q(y)Q(−y) with q q q Q(y) = y − 1 + x12 y − 1 + x22 · · · y − 1 + x2N need not be possible over the rational numbers. Of course, this factoring works for coefficients in the Hilbert field. Proof of Proposition 34.13. Any number x ∈ Ω in the Hilbert field can be obtained in are finitely many algebraic field extensions Q = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ FN 3 x In this tower, all extensions are two-dimensional and generated by adjoining a single root: q 1 + xi2 ∈ Fi+1 \ Fi with xi ∈ Fi for i = 0 . . . N − 1. By the lemma, we see successively that all intermediate fields Fi are totally real. Hence x is totally real.
650
Theorem 34.3 (Gauss’ Lemma). An integer polynomial which factors over the rational numbers into factors of lower degree, already factored into integer polynomials of lower degree. The latter factoring is obtained from the former one by adjusting integer factors. Especially, a monic integer polynomial that factors over the rational numbers, even factors over the integers into monic integer polynomials. Problem 34.1. Show that the polynomial x4 + 2x2 − 1 is irreducible over the rational numbers. q√ 2 − 1, and show that this algebraic integer is not totally real. Find the algebraic conjugates of Answer. The only rational zeros of the polynomial x4 + 2x2 − 1 could be ±1, but they are obviously not zeros. Any factorization of the polynomial into two quadratics could only be of the form (x2 + ax + 1)(x2 − ax − 1) = x4 − a2 x2 − 2ax − 1 , x4 + 2x2 − 1 The factorization turns q√ q √ out to be impossible. Hence the polynomial is irreducible. Its zeros are ± 2 − 1 and ±i 2 + 1, which are algebraically conjugate to each other, but not all real. q√ Proposition 34.14 (A constructible number not in the Hilbert field). The number z = 2−1 is not totally real. It is a constructible number not in the Hilbert field. Problem 34.2. Can one factor the polynomial x4 + 2x2 − 1 over the Gaussian integers. Answer. The only possibility of factoring would be x4 + 2x2 − 1 = (x2 + ax + i)(x2 − ax + i) = (x2 + i)2 − a2 x2 = x4 + (2i − a2 )x2 − 1 This leads to a2 = 2(1 + i) which has no integer solution. Hence the polynomial is irreducible over the Gaussians, too.
Figure 34.1. Find the length of segment BY, and AY exactly.
Problem 34.3. These are the steps for the construction of the figure on page 650: We draw a unit −−→ square ABCD. Let E be the point on the ray AB such that the diagonal AC is congruent to the segment AE. Let M be the midpoint of the side AB. Draw the circle around M through the point E. This circle intersects the side BC at point Y. Calculate exact simplified root expressions for the lengths of (i) segment ME; (ii) segment AF; (iii) segment AY. Decide whether these segments are constructible with Hilbert tools.
651
Answer. Since the diagonal has the length |AC| =
√ 2, we get
|ME| = |AE| − |AM| =
√ 2−
1 2
and Pythagoras’ Theorem for triangle 4MY B yields BY = MY − MB = q √ |BY| = 2 − 2 2
2
2
√
1 2− 2
!2 −
√ 1 =2− 2 4
once more Pythagoras’ Theorem, now for triangle 4ABY, yields √ √ AY 2 = AB2 + BY 2 = 12 + 2 − 2 = 3 − 2 q √ |AY| = 3 − 2 All three segment lengths are totally real and hence constructible with Hilbert One see this q tools. √ √ 1 since both expressions ± 2 − 2 are real. Similarly, all four expressions ± 2 ± 2 are real, as q √ well as all four expressions ± 3 ± 2 are real.
Figure 34.2. Find the length of segment BF exactly.
−−→ Problem 34.4. Draw a unit square ABCD. Let E be the point on the ray AB such that the diagonal AC is congruent to the segment AE. Let N be the midpoint of the segment AE. Draw the circle with diameter AE. This circle intersects the side BC at point F. Calculate exact simplified root expressions for the lengths of (i) segment BF; (ii) segment AF. Decide whether these segments are constructible with Hilbert tools.
652
Answer. Since the diagonal has the length |AC| = theorem for the right triangle 4AFE, one obtains
√
2, we get |BE| =
√ 2 − 1. From the altitude
√ BF 2 = |AB| · |BE| = 1 · ( 2 − 1) q √ |BF| = 2−1 From Pythagoras’ Theorem, now for triangle 4ABY, one obtains √ √ AF 2 = AB2 + BF 2 = 12 + 2 − 1 = 2 √4 |AF| = 2 Neither one of these two segment lengths is totally real. Hence they are not constructible with Hilbert tools. But nevertheless, they are constructible with straightedgeqand compass. √ Take the example (i). One sees that not all four expressions ± ± 2 − 1 are real. To be completely accurate, one has still to check that these four numbers are algebraic conjugate to each 2 2 other,—by checking q √that the polynomial (x + 1) − 2 is irreducible. This step completes the proof that the number 2 − 1 is not totally real. q √ Take the example (ii). One sees that not all four expressions ± ± 2 are real. To be completely accurate, one has still to check that these four numbers are algebraic conjugate to each other,—by √ checking that the polynomial x4 − 2 is irreducible. This step completes the proof 4 that the number 2 is not totally real. Duplication of a cube The duplication of a cube leads to the equation x3 − 2 = 0. √3 Proposition 34.15. The root 2 is not in the constructible field. Hence a cube cannot be doubled in volume by a ruler and compass construction.
34.8.
Proof. The cubic polynomial x3 − 2 is irreducible over the rational numbers, as shown below in Lemma 34.3. Hence we get from Proposition 34.4 about the extension generated by one root that √3 [Q( 2) : Q] = 3 We know from Corollary 68 that the dimension [Q(k) : Q] is√a power of two for every constructible 3 number k. Since 3 is not a power of two, we conclude that 2 is not a constructible number. Lemma 34.3. The polynomial x3 − 2 has no rational root and hence it is irreducible over the rational numbers. √3 Proof. Suppose towards a contradiction that x = p/q = 2 is the rational root. Here p and q are integers. We can assume that the fraction is cancelled, and hence p and q are not both even. The resulting equation 2q3 = p3 implies that p is even. Since the fraction is cancelled, q is odd. But we can plug p = 2r into 2q3 = p3 = 8r3 and cancel again to get q3 = 4r3 . Hence q is even, which is a contradiction. This contradiction shows that there is no rational root of the polynomial x3 − 2, and hence this polynomial is irreducible over the rational numbers. √n Problem 34.5. Show that the root 2 is constructible if and only if n = 2 s is a power of two. Solution. • For all natural numbers, the polynomial xn − 2 is irreducible over the rational numbers. This follows by applying the Eisenstein criterium with prime p = 2. (Consequently the polynomial has no rational root, neither, but that is not enough to get the conclusions below).
653
•
Because the polynomial xn −2 is irreducible, √n we get from Proposition 34.4 about the extension generated by (an arbitrarily chosen) root 2 its dimension to be √n [Q( 2) : Q] = n
•
We know from Corollary 68 that √n the dimension [Q(k) : Q] is a power of two for every constructible number k. Hence 2 is not √n constructible if n is not a power of two. We see by an easy induction on s that 2 is constructible for n = 2 s any power of two.
•
34.9. Angle trisection Of course the impossibility of angle trisection refers to the general case to trisect an arbitrary angle. There are many special cases of constructible angles that can be trisected, as the examples 90◦ or 45◦ show. We claim that Proposition 34.16. Not every constructible angle can be trisected. Especially an angle of 60◦ can be constructed, but not trisected with ruler and compass. Proof. We use the trigonometric formula 4 cos3 α − 3 cos α = cos 3α and plug in 3α = 60◦ . We get 4 cos3 20◦ − 3 cos 20◦ = cos 60◦ =
1 2
Hence k = 2 cos 20◦ is a root of the cubic equation x3 − 3x − 1 = 0 This cubic polynomial is irreducible over the rational numbers, as shown below in Lemma 34.4. Hence we get from Proposition 34.4 about the extension generated by one element that [Q(cos 20◦ ) : Q] = 3 We know from Corollary 68 that, for every constructible number k, the dimension [Q(k) : Q] is a power of two. Since 3 is not a power of two, we conclude that cos 20◦ is not a constructible number. Lemma 34.4. The polynomial x3 − 3x − 1 has no rational root and hence it is irreducible over the rational numbers. Proof. Suppose towards a contradiction that x = p/q with p and q nonzero relatively prime integers is a rational root of x3 − 3x − 1. p3 3p − = 1 hence both q q3 p3 = 3pq2 + q3 = (3p + q2 )q and q3 = p3 − 3pq2 = p(p2 − 3q2 ) The first relation implies that q divides p3 . Since p and q are relatively prime, this implies q = ±1. The second relation implies that p divides q3 . Since p and q are relatively prime, this implies p = ±1. Hence x = ±1. But neither one of these two numbers is a root of x3 − 3x − 1 = 0. This contradiction shows that there is no rational root of the polynomial x3 − 3x − 1. For a polynomial of degree three, this fact already implies that the polynomial cannot at all be factored over the rational numbers. Hence the polynomial is irreducible.
654
Remark. An even simpler argument can be given, if we use Gauss’ Lemma (Theorem 34.3), or even the rational root test. We need only check for factoring into integer polynomials. Hence in the given case only for a factoring x3 − 3x − 1 = (x + a)(x2 + bx + c) with integer a, b, c. Since ac = −1, the only possibilities are a = 1 or a = −1, which can both be excluded. Corollary 69. Suppose the angle α is constructible. The angle α/3 is constructible if and only if the polynomial x3 − 3x − 2 cos α factors over the constructible field. This happens if and only if the polynomial x3 − 3x − 2 cos α splits in the constructible field. Proof. The polynomial P(x) = x3 − 3x − 2 cos α factors over the constructible field if and only if it has a zero in the constructible field. The three zeros of this polynomial are cos(α/3) and cos(α/3 ± 120◦ ). If either of them is in the constructible field, then the angle α/3 is constructible. Conversely, if the angle α/3 is constructible, then cos(α/3) is a root and the polynomial P factors, and even splits in the constructible field. Example 34.1. The angle α = ∠NAP to be trisected in Plemelj’s construction of the regular 7-gon has √ 27 cos α = √ 28 For the trisected angle, we see that x = 2 cos(α/3) is a root of the polynomial √ 27 3 3 x − 3x − 2 cos α = x − 3x − √ = 0 7 √ We substitute x = (3/7)y and get for the new variable y the equation y3 − 7y − 7 = 0 By the Eisenstein criterium this polynomial is irreducible over the integers. Alternatively, irreducibility follows since there are no rational roots. Hence the field extension [Q(y) : Q] = 3 has dimension three. Hence neither y nor x are constructible numbers. We see that the angle trisection used in Plemelj’s construction of the regular heptagon cannot be done with classical Euclidean tools. √ Example 34.2. Take α = 45◦ . The polynomial x3 − 3x − 2 cos α = x3 − 3x − 2 is not irreducible over the constructible field. Hence we conclude the—already known—fact that the angle α can be √ trisected. Indeed the polynomial already factors in Q( 2): √ √ √ x3 − 3x − 2 = (x + 2)(x2 − 2x − 1) √ √ The polynomial splits in a further two dimensional extension Q( 2, 3) into linear factors √ √ √ √ √ √ 2 + 6 2 − 6 3 x − 3x − 2 = x + 2 x − x − 2 2 corresponding to √ 2 cos(α/3) =
2+ 2
√
√ 6
, 2 cos(α/3 + 120 ) = ◦
√ 2− 2
6
√ and 2 cos(α/3 − 120◦ ) = − 2
655
34.10. Regular polygons We investigate whether a regular polygon with n sides is constructable. We put the polygon into the coordinate plane, with one vertex at point (0, 1) and the origin (0, 0) as center of the circum circle. On giants shoulders. Easier insight is gained from complex numbers. (That is why we are on "giants shoulders".) The complex numbers z = x + iy at the vertices of the polygon are the roots of the polynomial equation zn − 1 = 0 From polar coordinates we get the roots to be zk := ζ k for k = 0, 1, 2, . . . n − 1 with z1 = ζ :=
2πi n
(34.2)
One root is z0 = 1. Division by the factor z − 1 shows that the remaining roots are the zeros of the integer polynomial n−1 zn − 1 X k Kn (z) = = z z−1 k=0 For the proof of impossibility of constructions, the factoring of Kn (z) is needed. 34.11. The 17-gon construction The regular heptadecagon is a constructible polygon (that is, one that can be constructed using a compass and unmarked straightedge), as was shown by Carl Friedrich Gauss in 1796 at the age of nineteen. This proof represented the first progress in regular polygon construction in over 2000 years. It can be found in his Disquisitiones Arithmeticae and is reprinted in his Werke (1870-77), vol. I. Gauss’s proof of the constructibility of the relies on the fact that constructibility is equivalent to expressibility of the trigonometric functions of the angle 360◦ /17 in terms of arithmetic operations and square root extractions. We use radian measure for angles in this section. According to Theorem 34.2, the number 2 cos(2π/17) is constructible if and only if there exists a tower Q = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn 3 2 cos
2π 17
Since 17 is a prime, Proposition 34.19 shows that the polynomial Φ17 (z) =
z17 − 1 z−1
is irreducible over the integers. We use Proposition 34.4 about an extension generated by one element. Since deg Φ17 = 16, we conclude that " ! # 2πi Q exp : Q = 16 17 We give an explicit construction of the tower with three two dimensional extensions ! 2π Q = F0 ⊂ F1 ⊂ F2 ⊂ Q cos 17
(34.3)
One further quadratic extension is needed to adjoin the imaginary parts i sin 2π 17 , the calculation of √ 2 which uses simply sin α = ± 1 − cos α.
656
The following semi empirical approach avoids the primitive roots from number theory. Define 2π j 17 Clearly c j as a function of the integer j is symmetric and has period 17. We get the roots in question from j = 1 . . . 8. Moreover the addition theorem for cosine implies c j = 2 cos
2 cos α cos β = cos(α + β) + cos(α − β) ci · c j = ci+ j + ci− j The tower (34.3) in now reconstructed backwards. The third step in the tower (34.3) constructs the ci from four sums of pairs of the ci . How can one find these four pairs? To get two roots ci and c j by means of a quadratic equation, we need to know both the sum ci + c j and the product ci c j = ci+ j + ci− j . To have the product available, the four pairs have to be chosen according to the following rules: If ci and c j are paired, then ci+ j and ci− j are paired, too. If ci and c j are paired, then c2i and c2 j are paired, too. We see that c1 and c2 cannot be paired. With a bid of experimentation, we find the four pairs to be {c1 , c4 } {c2 , c8 } {c3 , c5 } {c6 , c7 } and guess that F2 = Q(c1 + c4 ) = Q(c2 + c8 ) = Q(c3 + c5 ) = Q(c6 + c7 ) To get the previous extension F2 /F1 , we need to find out which two pairs have to be added. A bid of calculation shows 8 8 X X (c1 + c4 )(c2 + c8 ) = ci and (c3 + c5 )(c6 + c7 ) = ci i=1
i=1
whereas in other such products some ci occur more than once. We guess that F1 = Q(c1 + c4 + c2 + c8 ) = Q(c3 + c5 + c6 + c7 ) For the first extension we need still the product (c1 + c4 + c2 + c8 )(c3 + c5 + c6 + c7 ) = 4
8 X
ci
i=1
and the total sum
P8
i=1 ci .
Before going on, recapitulate what we have obtained so far:
Lemma 34.5. Let c j be any real valued function of the integer j with the properties: c j = c− j = c17− j and ci · c j = ci+ j + ci− j for all integer i and j. Then c1 c4 = c3 + c5 and c3 c5 = c2 + c8 and c2 c8 = c6 + c7 and c6 c7 = c1 + c4 (c1 + c4 )(c2 + c8 ) =
8 X
ci and (c3 + c5 )(c6 + c7 ) =
i=1
(c1 + c4 + c2 + c8 )(c3 + c5 + c6 + c7 ) = 4
8 X
ci
i=1 8 X
ci
i=1
From a known
P8
i=1 ci ,
one can calculate all ci with a tree of quadratic equations.
657
The sum of all eight ci can be obtained from a complex geometric series 8 16 8 2πi 17 X X X 2πi k exp 17 − 1 exp 1+ ci = = = 0 hence ci = −1 17 exp 2πi i=1 i=1 17 − 1 k=0 We can now write down the quadratic equations for the extensions. We check numerically which sign of the square root in their solutions corresponds to which partial sum of ci . 1. The polynomial with the roots x1 = c1 + c4 + c2 + c8 and x2 = c3 + c5 + c6 + c7 is
2.
3.
4.
5.
(x − x1 )(x − x2 ) = x2 + x − 4 and the roots are √ √ −1 + 17 −1 − 17 x1 = and x2 = 2 2 √ We have to check numerically that the signs of 17 occur in that arrangement. The polynomial with the roots y1 = c1 + c4 and y2 = c2 + c8 is (y − y1 )(y − y2 ) = y2 − x1 y − 1 and the roots are q x1 ± x12 + 4 y1,2 = hence 2 q q √ √ √ √ −1 + 17 + 34 − 2 17 −1 + 17 − 34 − 2 17 y1 = and y2 = 4 4 We have to check numerically that the signs of the longer root occur in that arrangement. The polynomial with the roots y3 = c3 + c5 and y4 = c6 + c7 is (y − y3 )(y − y4 ) = y2 − x2 y − 1 and the roots are q x2 ± x22 + 4 y3,4 = hence 2 q q √ √ √ √ −1 − 17 + 34 + 2 17 −1 − 17 − 34 + 2 17 y3 = and y4 = 4 4 We have to check numerically that the signs of the longer root occur in that arrangement. The polynomial with the roots z1 = c1 and z2 = c4 is (z − z1 )(z − z2 ) = z2 − y1 z + y3 and the roots are q y1 ± y21 − 4y3 z1,2 = 2 We have to check numerically that the signs occur in that arrangement. The polynomial with the roots z3 = c2 and z4 = c8 is (z − z3 )(z − z4 ) = z2 − y2 z + y4 and the roots are q y2 ± y22 − 4y4 z3,4 = 2
658
6.
We have to check numerically that the signs occur in that arrangement. The polynomial with the roots z5 = c3 and z6 = c5
is
(z − z5 )(z − z6 ) = z − y3 z + y2 and the roots are q y3 ± y23 − 4y2 z1,2 = 2 2
7.
We have to check numerically that the signs occur in that arrangement. The polynomial with the roots z7 = c6 and z8 = c7
is
(z − z7 )(z − z8 ) = z − y4 z + y1 and the roots are q y4 ± y24 − 4y1 z3,4 = 2 2
We have to check numerically that the signs occur in that arrangement. For the root z1 , I have finished the calculation and have obtained q √ √ 2π = −1 + 17 + 34 − 2 17 16 cos 17 s q q ! √ √ 2 √ √ −1 + 17 + 34 − 2 17 + 16 + 16 17 − 16 34 + 2 17 + Gauss’ book Disquisitiones Arithmeticae gives—in modern notation: r q q q √ √ √ √ √ 2π = −1 + 17 + 34 − 2 17 + 2 17 + 3 17 − 34 − 2 17 − 2 34 + 2 17 16 cos 17 One can check the result numerically, and check algebraically that the two radicals under the last root are equal. Proposition 34.17. With l = 1 . . . 8 and s1 , s2 , s3 = ±1, we obtain all eight values q √ √ 2πl 16 cos = −1 + s1 17 + s2 34 − 2s1 17 17 r q q √ √ √ + 2s2 s3 17 + 3s1 17 − s2 34 − 2s1 17 − 2s1 s2 34 + 2s1 17
in the following order
l 1 2 3 4 5 6 7 8
s1 +1 +1 −1 +1 −1 −1 −1 +1
s2 +1 −1 +1 +1 +1 −1 −1 −1
s3 +1 −1 +1 −1 −1 −1 +1 +1
659
Of course, all these empirical checks of signs are still unsatisfactory. More important, one wants to set up a more systematic procedure at the very beginning. It is the merit of Gauss to have found satisfactory answers to these problems. Thus he opens up the possibility to construct the polygons for other Fermat primes. The list of the seventeen remainders of 3k modulo 17 for k = 0, 1, . . . , 16 is [ 1, 3, −8, −7 , −4, 5, −2, −6 , −1, −3, 8, 7 , 4, −5, 2, 6 ] We see they are all different. Hence one calls 3 a primitive root modulo 17. Consequently the seventeen numbers ! 2πi k dk = exp 3 17 are a permutation of the roots of z17 − 1 = 0. After pairing conjugate complex roots dk and dk+8 , we get the ordered list [ c1 , c3 , c8 , c7 , c4 , c5 , c2 , c6 ] = dk + dk+8 for k = 0, 1, . . . , 7 The pairs formed at the beginning of the empirical procedure, and sums of four in the next step are c1 + c4 c2 + c8 c3 + c5 c6 + c7 c1 + c4 + c2 + c8 c3 + c5 + c6 + c7
= d0 + d4 + d8 + d12 = d2 + d6 + d10 + d14 = d1 + d5 + d9 + d13 = d3 + d7 + d11 + d15 = d0 + d2 + d4 + d6 + d8 + d10 + d12 + d14 = d1 + d3 + d5 + d7 + d9 + d11 + d13 + d15
One can see from sums of dk much more easily how to set up the tree of quadratic equations. Indeed in Proposition 34.17 l ≡ (3k mod 17)
and k =
1 (1 − s1 ) + (1 − s2 ) + 2(1 − s3 ) 2
These hints must be enough towards a guess how to generalize the procedure for other Fermat primes, as Gauss did indeed achieve. 34.12.
A short paragraph about Fermat numbers
Definition 34.3 (Fermat number, Fermat prime). The numbers n
Fn = 22 + 1 for any integer n ≥ 0 are called Fermat numbers. A Fermat number which is prime is called a Fermat prime. Lemma 34.6 (Fermat). If p is any odd prime, and p − 1 is a power of two, then p = Fn is a Fermat prime. Proof. Assume p = 1 + 2a and a ≥ 3 odd. Because of the fatorization a = 2n · (2b + 1) one can factor n n n n n p = 1 + 22 ·(2b+1) = (1 + 22 )(1 − 22 + 22 ·2 − · · · + 22 ·2b ) If p is an odd prime, the only possibility is b = 0 and p = Fn .
660
The only known Fermat primes are Fn for n = 0, 1, 2, 3, 4. They are 3, 5, 17, 257, and 65537, see . Fermat numbers and Fermat primes were first studied by Pierre de Fermat, who conjectured 1654 in a letter to Blaise Pascal that all Fermat numbers are prime, but told he had not been able to find a proof. Indeed, the first five Fermat numbers are easily shown to be prime. However, Fermat’s conjecture was refuted by Leonhard Euler in 1732 when he showed, as one of his first number theoretic discoveries: 1
5
F5 = 22 + 1 = 232 + 1 = 4294967297 = 641 · 6700417. Later, Euler proved that every prime factor p of Fn must have the form p = i · 2n+1 + 1. Almost hundred years later, Lucas showed that even p = j · 2n+2 + 1. These results give a vague hope that one could factor Fermat numbers and perhaps settle Fermat’s conjecture to the negative. There are no other known Fermat primes Fn with n > 4. However, little is known about Fermat numbers with large n. In fact, each of the following is an open problem: 1. Is Fn composite for all n > 4? By this anti-Fermat hypothesis 3, 5, 17, 257, and 65537 would be the only Fermat primes. 2. Are there infinitely many Fermat primes? (Eisenstein 1844) 3. Are there infinitely many composite Fermat numbers? 4. Are all Fermat numbers square free? As of 2012, the next twenty-eight Fermat numbers, F5 through F32 , are known to be composite. As of February 2012, only F0 to F11 have been completely factored. For complete information, see Fermat factoring status by Wilfrid Keller, on the internet at http://ww.prothsearch.net/fermat.html 34.13.
Eisenstein’s irreducibility criterium
Proposition 34.18 (The Eisenstein criterium). For a polynomial P to be irreducible in the ring Z[z], it is sufficient that there exists a prime number p such that 1. the leading coefficient of the polynomial is not divisible by p; 2. all other coefficients except the leading one are divisible by p; 3. the constant coefficient is not divisible by p2 . Proof. Assume toward a contradiction that a reducible polynomial would satisfy the criterium. We would get the integer factorization a0 + · · · + ar xr = (b0 + · · · + b s x s ) c0 + · · · + ct xt , (34.4) where r = s + t and s, t, ≥ 1. Multiplying the terms of the product yields a0 a1 a2 ak
=b0 c0 =b0 c1 + b1 c0 =b0 c2 + b1 c1 + b2 c0 =b0 ck + b1 ck−1 + · · · + bk c0
for all k ≥ 0. It is assumed that the prime p, but not p2 divides a0 = b0 c0 . Hence p divides exactly one of the two numbers b0 and c0 . We may assume that p divides c0 , but does not divide b0 . 1
sequence A019434 in OEIS
661
Furthermore, by assumption, p divides a0 , a1 , . . . ar−1 , but not ar . Recursively, we conclude that p divides b0 c1 = a1 − b1 c0 and hence c1 , next b0 c2 = a2 − b1 c1 − b2 c0 and hence c2 , . . . , cr−1 . Hence p divides all coefficients of the factor ct xt + · · · + c0 , since t ≤ r − 1. This implies that p divides all coefficients of the original polynomial ar xr + · · · + a0 , contradicting the assumption that the prime p does not divide the leading coefficient ar . From this contradiction, we see that no factoring of the polynomial into polynomials of lower degree is possible, and hence it is irreducible. Corollary 70. For a polynomial to be irreducible in the ring Z[z], it is sufficient that it does not have any rational zero; at least one of the two leading coefficients is not divisible by p; all its coefficients except possibly the two leading ones are divisible by the same prime number p; the constant coefficient is not divisible by p2 . Problem 34.6. Prove the corollary. Solution. We argue as in the proof of the Eisenstein criterium: Assume toward a contradiction that a reducible polynomial would satisfy the criterium. We would get the integer factorization (34.4), with r = s + t and s, t ≥ 2, because no linear factor can appear. Recursively, we conclude that the prime p divides the coefficients c0 , c1 , c2 , . . . , cr−2 . Since t ≤ r − 2, these are all coefficients of the factor ct xt + · · · + c0 . Again, we conclude that p divides all coefficients of the original polynomial ar xr + · · · + a0 , contradicting the assumption that p does not divide at least one of the two leading coefficients ar−1 or ar . From this contradiction, we see that no factoring of the given polynomial can exist. q √ 5− 5 Problem 34.7. Find the irreducible polynomial with zero z1 = 2 . Find all its algebraic conjugates, and show that the number is a totally real algebraic integer. Solution of Problem 34.7. Simple arithmetic shows that (2z2 − 5)2 − 5 = 0 and hence P(z) := z4 − 5z2 + 5 = 0 is the monic polynomial in the ring Z[z] with zero z1 . Hence z1 is an algebraic integer. The Eisenstein criterium applies with p = 5. Hence its zeros are exactly the algebraic conjugates of z. Obviously, these zeros are s s s s √ √ √ √ 5+ 5 5+ 5 5− 5 5− 5 , − , , − 2 2 2 2 which all four turn out to be real. Lemma 34.7. For any prime p, the binomial coefficients ! p for k = 1 . . . p − 1 k are divisible by p, but not by p2 .
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Proposition 34.19. For any prime p, the polynomial Φ p (z) =
zp − 1 z−1
is irreducible over the integers. Increasing powers. The assertion is true for p = 2. We assume now that p is an odd prime. We substitute z = 1 + x and use the Eisenstein criterium to show that the resulting polynomial P(x) = Φ p (1 + x) is irreducible. The binomial formula implies ! p (1 + x) p − 1 X p k−1 = x P(x) = x k k=1 ! ! ! p p 2 p = p+ x+ x + ··· + x p−3 + px p−2 + x p−1 2 3 p−2 Because of Lemma 34.7, we see that all three assumptions for the Eisenstein criterium (Proposition 34.18) are satisfied. Hence the polynomial P and hence Φ p are irreducible over the integers. Lemma 34.8. For any odd prime p, the binomial coefficients ! p2 for k = 1 . . . p − 1 k are divisible by p. Proposition 34.20. For any prime p, the polynomial 2
Φ p2 (z) =
zp − 1 zp − 1
is irreducible over the integers. Proof. In the case p = 2, we get Φ4 = 1 + z2 . which we can check to be irreducible. We assume now that p is an odd prime. We substitute z = 1 + x and use the Eisenstein criterium to show that the resulting polynomial Φ p2 (1 + x) =
p(p−1) X
bi xi = p + · · · + x p(p−1)
i=0
is irreducible. The definition and the binomial formula imply 2
zp − 1 zp − 1 = Φ p2 (z) · z−1 z−1 p2 (1 + x) − 1 (1 + x) p − 1 = Φ p2 (1 + x) · x x 2 ! ! p p(p−1) p X p2 X X p xl−1 = bi xi · xk−1 l k i=0 l=1 k=1 We have obtained the integer factorization a0 + · · · + ar xr = (b0 + · · · + b s x s ) c0 + · · · + ct xt ,
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which is really possible in this context since it has been constructed above. The degrees are now r = p2 − 1, s = deg Φ p2 = p(p − 1) and t = deg Φ p = p − 1. Multiplying the polynomials and comparing the coefficients yields once more a0 a1 a2 ak
=b0 c0 =b0 c1 + b1 c0 =b0 c2 + b1 c1 + b2 c0 =b0 ck + b1 ck−1 + · · · + bk c0
for all k ≥ 0. We know from Lemma 34.7 and 34.8 that the prime p divides all the coefficients a0 , . . . , a p2 −1 , but a p2 = 1. That is enough to check inductively that p divides all coefficients b0 , . . . , b p2 −p−1 . But the leading coefficient of Φ p2 (1 + x) is b p2 −p = 1. We see that the polynomial Φ p2 (1+x) satisfies all three assumptions for the Eisenstein criterium (Proposition 34.18). Hence Φ p2 (1 + x) and hence Φ p2 (x) are irreducible over the integers. Example p = 3, p2 = 9. We claim that Φ9 (1 + x) =
6 X
bi xi
i=0
is irreducible. The definition and the binomial formula imply z9 − 1 z3 − 1 = Φ9 (z) · z−1 z−1 (1 + x)9 − 1 (1 + x)3 − 1 = Φ9 (1 + x) · x x ! 6 9 X X i h 9 l−1 x = bi xi · 3 + 3x + x2 l i=0 l=1 We have obtained an integer factorization. Multiplying the coefficients yields once more ! ! 9 9 = b0 · 3 , = b0 · 3 + b1 · 3 , 1 2 ! ! 9 9 = b1 · 1 + b2 · 3 + b3 · 3 , = b2 · 1 + b3 · 3 + b4 · 3 , 4 5 ! ! 9 9 = b4 · 1 + b5 · 3 + b6 · 3 , = b5 · 1 + b6 · 3 , 7 8
polynomials and comparing the ! 9 = b0 · 1 + b1 · 3 + b2 · 3 3 ! 9 = b3 · 1 + b4 · 3 + b5 · 3 6 ! 9 = b6 · 1 9
We get the constant coefficients a0 = 9, b0 = 3 and c0 = 3 and the leading coefficient b6 = 1. As in the general case, we to use the inductive argument to conclude that 3 is a divisor of the coefficients b0 , . . . , b5 . That is all we need. All assumptions in the Eisenstein criterium are satisfied and hence Φ9 (1 + x) and Φ9 (x) are irreducible over the integers. Remark. The explicit result Φ9 (1 + x) = 3 + 9x + 18x2 + 21x3 + 15x4 + 6x5 + x6 is not needed. 34.14. The constructible polygons Constructions for the regular triangle, square, pentagon, and polygons with 2h times as many sides had been given by Euclid, but constructions based on the Fermat primes other than 3 and 5 were unknown to the ancients. Carl Friedrich Gauss proved the constructibility of the regular 17-gon in 1796. The first explicit construction of a heptadecagon was given by Johannes Erchinger in 1825. Another method of
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construction uses Carlyle circles and is shown on the internet. Indeed, based on the construction of the regular 17-gon, one can readily construct n-gons with n being the product of 17 with 3 or 5 or both and any power of two. For example, a regular 51-gon, 85-gon or 255-gon and any regular n-gon with 2h times as many sides. Five years after his construction of the 17-gon, Gauss developed the theory of Gaussian periods in his Disquisitiones Arithmeticae (published 1801). This theory allowed him to prove constructibility of an n-gon if n is any Fermat prime. Hence one can construct of a regular 257-gon or 65537-gon. Lemma 34.9. If both a regular n-gon and a regular m-gon are constructible, and n and m are relatively prime, then an 2h · n · m-gon is constructible for any h. Together with this—rather obvious—lemma, Gauss arrives by his theory of Gaussian periods at a sufficient condition for the constructibility of regular polygons: Theorem 34.4 (Gauss 1801). A regular n-gon can be constructed with compass and straightedge if n is the product of a power of 2 and any number of distinct Fermat primes. Gauss stated without proof that the condition given is his theorem is also necessary, but never published his proof. A full proof of necessity was given by Pierre Wantzel in 1837. To set up necessary conditions for constructibility, we need to concentrate on the case that n is the power of an odd prime p and thus assume n = pr . Lemma 34.10. If p is any odd prime, and a polygon with p sides is constructible, then p − 1 is a power of two. Proof. Proposition 34.19 shows that the polynomial zp − 1 z−1 is irreducible over the integers. We use Proposition 34.4 about an extension generated by one element. Since deg Φ p = p − 1, we conclude that " ! # 2π i Q exp :Q = p−1 (34.5) p Φ p (z) =
If the polygon with p sides is constructible, then the number k = cos 2π p lies in the constructible field and hence by Theorem 34.2, the dimension [Q(k) : Q] is a power of two. Hence the dimension p − 1 of the extension (34.5), which is just the double, is a power of two. Lemma 34.11 (Fermat). If p is any odd prime, and p − 1 is a power of two, then p = Fn and Fn is a Fermat prime. Lemma 34.12. If p is any odd prime, a polygon with p2 sides is not constructible. Proof. Proposition 34.20 shows that the polynomial 2
zp − 1 Φ (z) = p z −1 is irreducible over the integers. We use Proposition 34.4 about an extension generated by one element. Since deg Φ p2 = p(p − 1) and Φ p2 is irreducible, we conclude that " ! # 2π i Q exp 2 : Q = p(p − 1) (34.6) p p2
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We argue by contradiction. If the polygon with p2 sides would be constructible, then the number k = cos(2π/p2 ) would in the constructible field and hence by Theorem 34.2, the dimension [Q(k) : Q] would be a power of two. The dimension of the extension (34.6) is just the double, it would be a power of two. But p(p − 1) is not a power of two. This contradiction shows that a polygon with p2 sides, where p is an odd prime, is not constructable. Necessity of Gauss’ condition. Assume that an n-gon is constructible and let p be an odd prime factor of n. Together the Lemma 34.10 and Fermat’s Lemma 34.11 implies that any odd prime factor of n is a Fermat prime. Lemma 34.12 implies that n is not divisible p2 . Hence a regular n-gon can only be constructed if n is the product of a power of 2 and any number of distinct Fermat primes. Gauss stated without proof the necessity of this condition, but never published his proof. A full proof of necessity was given by Pierre Wantzel in 1837. Main Theorem 29 (Gauss-Wantzel Theorem). A regular polygon with n sides is constructible if and only if n = 2h p1 · p2 · · · p s where p2 · · · p s is a product of different Fermat primes. If the anti-Fermat hypothesis is true, there are exactly five Fermat primes, and hence exactly n f actored 31 regular constructible polygons with an odd number of sides. 17 196 611 3 · 65 537 18 327 685 5 · 65 537 n f actored 19 983 055 3 · 5 · 65 537 1 3 F0 20 1 114 129 17 · 65 537 2 5 F1 21 3 342 387 3 · 17 · 65 537 3 15 3·5 22 5 570 645 5 · 17 · 65 537 4 17 F2 23 16 711 935 3 · 5 · 17 · 65 537 5 51 3 · 17 24 16 843 009 257 · 65 537 6 85 5 · 17 25 50 529 027 3 · 257 · 65 537 7 255 3 · 5 · 17 26 84 215 045 5 · 257 · 65 537 8 257 F3 27 252 645 135 3 · 5 · 257 · 65 537 9 771 3 · 257 28 286 331 153 17 · 257 · 65 537 10 1 285 5 · 257 29 858 993 459 3 · 17 · 257 · 65 537 11 3 855 3 · 5 · 257 30 1 431 655 765 5 · 17 · 257 · 65 537 12 4 369 17 · 257 31 4 294 967 295 3 · 5 · 17 · 257 · 65 537 13 13 107 3 · 17 · 257 14 21 845 5 · 17 · 257 15 65 535 3 · 5 · 17 · 257 16 65 537 F4
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35. The Lunes of Hippocrates
Figure 35.1. Just a lune.
35.1.
The historic lunes
Definition 35.1 (Lune). A lune consists of two circular arcs having a common chord and lying on the same side of this chord. The interior of the lune is the the crescent shaped area formed by the difference of the interiors of the corresponding circles. It is bounded by the lune’s two arcs. Hippocrates of Chios (ca. 430 B.C.) posed the problem: (Hippocrates Problem). Find the lunes which are constructible and squarable with straightedge and compass. He gave three examples of constructible lunes. They are obtained beginning with the following two assumptions: (a) The two circular sectors corresponding to the lune’s arcs have the same area. (b) The central angles of the two circular arcs are commensurable. Hippocrates of Chios is credited with discovering three such lunes; two more were discovered in the 18th century. In the 20th century Tschebatorev and Dorodnov (1947) proved these fives are the only ones. Lemma 35.1. Assumption (a) implies the lune is squarable. Proof. Let CD be the common chord of the two arcs of the lune. Let SA be the circular sector corresponding to the lune’s longer arc and have center A, and SB be the circular sector corresponding to the lune’s shorter arc and have center B. Both sectors are delimited by the radiuses from their respective center to the endpoints of common chord CD. The sector SB lies on the other side of the lune’s shorter arc and does not intersect its interior area. We add to the lune the latter sector and then subtract the sector SA , and obtain the kite ACBD. Since by assumption (a) the two sectors have the same area, the lune has the same area as the kite and hence is squarable.
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Lemma 35.2. Assumption (a) and (b) together imply α n b2 = = 2 β m a
(35.1)
with integers n, m ≥ 1. Indeed n < m becomes possible only if one or both or angles α and β are more than 360◦ . Proof. We now use assumption (b). Let the angle ε be the (greatest) common measure of the central angles α and β of the lune’s longer and shorter arcs around centers A and B. Hence α = nε and β = mε with integers n > m ≥ 1. Subdivision of the sector SA yields n congruent sectors with center A, and similarly, subdivision of the sector SB yields m congruent sectors with center B. All sectors we have obtained have the same central angle ε. Hence they are similar. By assumption (a), the two circular sectors SA and SA have equal area, hence the n small sectors of radius a around center A with central angle ε have together the same area as m similar sectors with central angle ε and radius b around center B. Since the areas of similar figures are proportional to the square of their linear dimension, we conclude na2 = mb2 , and finally get equation (35.1). Definition 35.2 (Circular segment). A circular segment is bounded by an arc and a chord. A segment of central angle ε is obtained from the circular sector with the same angle and arc by subtraction or addition of the triangle with vertices at the endpoints of its circular arc and the center of the circle. For a short arc 0 < ε < 180◦ , the triangle is subtracted, for a long arc 180◦ < ε < 360◦ , the triangle is added to the sector. Lemma 35.3. Two circular segments with the congruent central angles and which are both bounded by the long, or both by the short arc are similar. There areas have the same ratio as the squares of their radius. Lemma 35.4. The areas of similar figures are proportional to the square of their linear dimension. Proposition 35.1. Assumption (a) and (b) together imply the lune has the same area as a polygon with vertices on its arcs. Proof. After subdivision of the arcs α and β, we obtain not only similar circular sectors, but also similar segments with the central angle ε. The endpoints of these segments divide the arc α of the lune into n arcs with central angle ε, and the inner arc β of the lune m arcs with central angle ε. The n circular segments around center A have together the same area as the m similar circular segments around center B. This is clear from equation (35.1) and Lemma 35.3. We add to the lune the m segments on the other side of the lune’s inner arc β, and subtract the n segments with vertices on the lune’s external arc α. We obtain a polygon with vertices on the arcs of the lune which has the same area as the lune. For which values of the integers n and m can we construct squarable lunes? Let α = nε and β = mε be the angles of the lune arcs. Hippocrates has found a squarable and constructible lune for the following three cases: (a) n = 2, m = 1 (b) n = 3, m = 1 (c) n = 3, m = 2 The case (a) is easiest to guess: one puts α = 180◦ , β = 90◦ . The longer arc of the lune is the circum-circle of an isosceles right triangle 4CED and hence its center A is the midpoint of the
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hypothenuse CD.√The shorter arc has its center B in the forth vertex of square CEDB. Hence we know that b/a = 2 as required. √ In the case (b), we need to get b/a = 3. On the longer arc α of the lune lie three congruent segments CE, EF, FD. We begin by choosing an arbitrary length l for them. We know that the endpoints of the last and first segment are the endpoints of the common chord. since m = 1, the segment CD is the only √one on the inner arc of the lune. From equation (35.1) and similarity we get the distance |CD| = 3 |CE|. The quadrilateral CEFD is a symmetrical trapezoid, the sides of which are given. We construct the trapezoid and its circum-circle. Thus we get the longer arc of the lune and its center A. Thus we get the central angle ε = ∠CAE. By adding the central angles of the three congruent _
_
_
arcs CAE, EAF and FAD, we see that ∠CAD = α = 3ε is the central angle of the longer arc of the lune. Since m = 1 by assumption, we find the center B of the second circle from ε = ∠CBD. Too, we know that the trapezoid has the same area as the lune. Problem 35.1. Do the construction described above. One begins by drawing a rhombus consisting of two equilateral triangles with a common side, of length one. Then one constructs the trapezoid, its circum-circle, and finally the lune. Answer. A detailed construction is given in the figure on page 668.
Figure 35.2. The constructible and squarable 3 : 1 lune of Hippocrates.
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Figure 35.3. Hippocrates 3 : 1 lune is greater than a semicircle.
Proposition 35.2 (Hippocrates 430 B.C.). The external arc of Hippocrates’ 3 : 1 lune is greater than a semicircle. Proof. We drop the perpendiculars from A and C onto line BD. Since by construction |BD| = √ 3 |AB| > |AC|, the foot points F and G of the perpendiculars lie inside the segment BD. Thus we obtain a rectangle and two right triangles, one of which is 4CDF with the acute angles ∠FDC and ∠FCD. (i) The angle ∠ACD is the sum of the right angle ∠ACF and angle ∠FCD, and hence obtuse. (ii) The triangle 4ACD has the obtuse angle γ. Hence the Pythagorean comparison implies |AC|2 + |CD|2 < |AD|2 . By construction, the three segments BA AC CD are congruent, and hence 2|AB|2 = |AC|2 + |CD|2 < |AD|2 . (iii) By construction and item (ii) |BD|2 = 3 |BA|2 = |BA|2 + 2 |BA|2 < |BA|2 + |AD|2 (iv) Because of item (iii), the Pythagorean comparison in the triangle 4BAD, implies that angle ∠BAD is acute. _
(v) Since the angle ∠BAD is acute and its vertex lies on the arc BAD, Corollary 43 implies that this arc, which is the external arc of Hippocrates’ 3 : 1 lune, is greater than a semicircle. Construction 35.1 (Hippocrates construction of a squarable 3 : 2 lune). Draw a circle C with diameter AB around center K, and the perpendicular bisector p of radius KB. Construct a segment DH such that |DH|2 /|KB|2 = 3/2 and mark a congruent segment EF DH on your ruler. Finally, one places the marked ruler such that the following three requirements are met: 1. The ruler line goes through the point B.
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Figure 35.4. The 3 : 2 lune of Hippocrates.
2. The point marked E on the ruler lies on the circle C. 3. The second point marked F on the ruler lies on the perpendicular bisector p. _
Let G be the reflection image of E across the perpendicular bisector. We draw a circular arc EKBG _
with center L, and a circular arc EFG with center I. Result: We claim that the lune L between the two circular arcs, the pentagon EFGBK, and the kite IGLE have the same area. Problem 35.2. . (a) The circle around L contains three congruent circular segments, which are similar to two congruent circular segments of the circle around I. Which are these segments. (b) Prove the claim in (a) by means of circumference angles and the axial symmetry across the perpendicular bisector p. (c) Prove by means of (b) that the lune L has the same area as the pentagon EFGBK. (d) Similarly to items (a) and (b), we get three congruent circular sectors of the circle around L, which are similar to two congruent circular sectors of the circle around I. Prove that the first three segments have the same area as the latter two. (e) Prove by means of (d) that the lune L between the two circular arcs has the same area as the kite IGLE.
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Answer. (a) The circle around L contains three congruent circular segments since |EK| = |KB| by construction and |EK| = |BG| by reflection across the bisector p. Let ε be their central angle. (b) We see that ∠ELG = α = 3ε is the central angle of the longer arc of the lune. The circumference angle ∠EGK = ε/2 is half of the central angle ∠ELK = ε . The neusis used in the construction puts points E, F and B on a line. By reflection across the bisector p, we see that the points G, F and K are on a line, too. Hence ∠EGK = ∠EGF is a circumference angle of the inner arc of the lune, too. We get the corresponding central angle ∠EIF = ε. By reflection across the bisector p, we get ∠GIF = ε. We see that ∠EIG = β = 2ε is the central angle of the inner arc of the lune. √ (c) We choose a scale such√that |EK| = |KB| = 1 is a unit segment. Hence |CD| = |CH| = ( 3)/2 and |EF| = |HD| = (3/2). From similar triangles 4ELK ∼ 4EIF we get the scaling factor r b |EI| |EF| 3 = = = (35.2) a |EL| |EK| 2 b2 3 α = = (35.3) a2 2 β confirming the requirement (35.1). The n = 3 circular segments around center L have together the same area as the m = 2 similar circular segments around center I. We add to the lune the two latter segments EF and FG onto the other side of the lune’s inner arc β, and cut along the longer arc the three circular segments EK, KB and BG. We obtain the pentagon EFGBK which has the same _
_
area as the nearby lune EFG GBKE. (d) Similarly to items (a) and (b), we get three congruent circular sectors of the circle around L, which are similar to two congruent circular sectors of the circle around I. Because of equation (35.3) the first three segments have the same area as the latter two. (e) This time we add the two circular sectors EIF and FIG to the lune onto the other side of the lune’s inner arc, and cut along the longer arc the three circular sectors ELK, KLB and BLG. We obtain the kite IGLE which has the same area as the lune. Proposition 35.3 (Hippocrates 430 B.C.). The external arc of Hippocrates’ 3 : 2 lune is less than a semicircle. √ Proof. By construction |EF| = 3/2 |EK| > |EK|. In triangle 4EFK the greater angle lies across the longer side, and hence χ = ∠EKF > ∠EFK = η. Since a triangle can have only one obtuse or right angle, we conclude that η is acute. Finally the supplementary angle ∠KFB is obtuse. (i) 2 |KF|2 < |KB|2 . follows from the Pythagorean comparison for the obtuse isosceles triangle 4KFB. (ii) By item (i) and since EK KB, and the length |EF| is given by construction |EK|2 + |KF|2 < |EK|2 +
1 3 |KB|2 = |KB|2 = |EF|2 2 2
(iii) The angle χ = ∠EKF is obtuse. by Pythagorean comparison for triangle 4EKF.
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Figure 35.5. Hippocrates 3 : 2 lune is smaller than a semicircle.
Figure 35.6. Hippocrates squared a suitable union of a circle and a lune.
(iv) Since the three points K, F, G lie on a line η = ∠EKG is obtuse by item (iii). Since the vertex of _
angle ∠EKG lies on the arc EKG, Corollary 43 implies that this arc is less than a semicircle.
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According to the account of van der Waerden, Hippocrates squared a union of a circle and a lune in a special case, constructed in the following way. A√hexagon is inscribed into a circle of radius 1, and into a concentric smaller circle C of radius 1/ 6 with the same center O. A circular arc which is tangent at the endpoints A and C of two adjacent sides AB and BC of the larger hexagon _
become the inner arc, and the arc ABC of the unit circle the external arc of the lune L. The two arcs of the lune intersect at 30◦ . Proposition 35.4 (Hippocrates). The disjoint union L ∪ C has the same area as the union of the triangle 4ABC and the hexagon inscribed into the circle C. Proof. Let S, S0 and S00 be three congruent circular segments between the unit circle and the inscribed hexagon. The first two of them, segments S, S0 , have the vertices A, B and B, C, respectively. Let T be the large circular segment with vertices A and C. The circular segments S and T are similar, since both have arcs√intersecting at an angle of 30◦ . Since twice the altitude of the equilateral triangle 4ABO is |AC| = 3 |AB|, we conclude that area(T ) =3 area(S) area(S ∪ S0 ∪ S00 ) = area(T ) Only two circular segments S and S0 are cut away from the external arc of the lune L. The third segment S00 has the same area as the six circular segments chopped away from the small circle C. The big circular segment T is added into the inner arc of the lune, to obtain the triangle 4ABC. Meanwhile, the smaller hexagon is left from the smaller circle. The disjoint union L ∪ C has the same area as the union of the triangle 4ABC and the hexagon inscribed into the smaller circle C. 35.2. Historic remark B.L. van der Waerden’s book Science Awakening [38], contains a detailed account of Hippocrates’ work, as far as it has been reconstructed. The book [25] "The Ancient Tradition of Geometric Problems" by Wilbur Richard Knorr contains interesting historic information about Hippocrates’ work, too. Hippocrates of Chios (ca. 430 B.C.) is credited with discovering three squarable lunes; two more were discovered in the 18th century. The manner in which Hippocrates squared his lunes can be learned from a famous fragment by Simplicius (ca. 530 A.D.). According to his own statement, Simplicius had copied word by word from the History of Mathematics of Eudemus (ca. 335 B.C.). Hippocrates’ proofs were preserved through the History of Geometry compiled by Eudemus of Rhodes, which has also not survived, but which was excerpted by Simplicius of Cilicia in his commentary on Aristotle’s Physics. Many scholars of history have attempted to reconstruct this lost work of Eudemus. (Obviously, no such attempt of any historic reconstruction is intended in the present notes.) Another line of information about Hippocrates of Chios comes from Alexander of Aphrodisias. Alexander was the teacher of Simplicius, the most learned and reliable among the commentators of Aristotle. Heath concludes that, in proving his result, Hippocrates was also the first to prove that the area of a circle is proportional to the square of its diameter. Hippocrates’ book on geometry with the title "Elements" in which this result appears, has been lost, but may have formed the model for Euclid’s Elements. Van der Waerden given the following judgement of Hippocrates:
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Looking at all these developments as a whole, we see in the first place that Hippocrates mastered a considerable number of propositions from elementary geometry. The "Elements of Geometry", which he has written according to the Catalogue of Proclus, must have contained a large part of Books III and IV of Euclid, as well as the contents of Books I and II, which were "baby food" for the Pythagoreans. Hippocrates knows the relation between inscribed angles and arcs, the construction of the regular hexagon; he knows how to circumscribe a circle about a triangle and he knows that a circle can be circumscribed about an isosceles trapezoid. He is familiar with the concept of similarity and he knows that the areas of similar figures are proportional to the squares of homologous sides. He knows not only the Theorem of Pythagoras for the right triangle (Euclid I.47) but also its generalization for obtuse- and acute-angled triangles (Euclid II.12 and II.13). Furthermore, he is able to square an arbitrary rectilinear figure, i.e. to construct a square with the same area. By means of this, he knows how to construct lines whose squares have the ratio 3 : 2 or 6 : 1 to the square on a given line. Still more important for the evaluation of the mathematical level, reached in Athens during the second half of the fifth century and of Hippocrates in particular, is the excellent demonstrative technique and the high requirements of rigor demanded in the proofs. Hippocrates is not satisfied merely to construct the lunules and to conclude from the drawings that the external boundary is greater than or less than a semicircle; he wants and succeeds to prove this rigorously. One has to remember that the operation with inequalities is a very late achievement of modern mathematics, about which even Euler did not worry much. 35.3.
Some historic and less historic exercises
Lemma 35.5. The areas of similar figures are proportional to the square of their homologous sides. Proposition 35.5. Three similar figures are attached to the sides of a right triangle, with scales proportional to the sides of the triangle. The sum of the areas of the figures put onto the legs of the triangle equals the area of the figure put onto the hypothenuse. Problem 35.3. Provide drawings with four different examples for this proposition. Convince yourself that the Proposition follows from the Lemma above and the Pythagorean Theorem. Answer. Biography of Alhazen, the polymath Abu Ali al-Hasan ibn al-Haytham Born: 965 in (possibly) Basra, Persia (now Iraq) Died: 1040 in (possibly) Cairo, Egypt Alhazen was born in Basra, in the Iraq province of the Buyid Empire. He probably died in Cairo, Egypt. During the Islamic Golden Age, Basra was a "key beginning of learning", and he was educated there and in Baghdad, the capital of the Abbasid Caliphate, and the focus of the "high point of Islamic civilization". During his time in Buyid Iran, he worked as a civil servant and read many theological and scientific books.
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Figure 35.7. On a generalization of the Pythagorean Theorem.
One account of his career has him called to Egypt by Al-Hakim bi-Amr Allah, ruler of the Fatimid Caliphate, to regulate the flooding of the Nile, a task requiring an early attempt at building a dam at the present site of the Aswan Dam. After his field work made him aware of the impracticality of this scheme, and fearing the caliph’s anger, he feigned madness. He was kept under house arrest from 1011 until al-Hakim’s death in 1021. During this time, he wrote his influential Book of Optics. After his house arrest ended, he wrote scores of other treatises on physics, astronomy and mathematics. He later traveled to Islamic Spain. During this period, he had ample time for his scientific pursuits, which included optics, mathematics, physics, medicine, and practical experiments. http://www-history.mcs.st-andrews.ac.uk/Biographies/Al-Haytham.html http://en.wikipedia.org/wiki/Alhazen%27s_problem In elementary geometry, Alhazen attempted to solve the problem of squaring the circle using the area of lunes (crescent shapes), but later gave up on the impossible task. The two lunes formed from a right triangle by erecting a semicircle on each of the triangle’s sides, inward for the hypotenuse and outward for the other two sides, are known as the lunes of Alhazen; they have the same total area as the triangle itself. Problem 35.4. Provide a drawing of the two lunes of Alhazen. Explain and prove how the lunes can be squared.
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Figure 35.8. The lunes of Alhazen can be squared.
Answer. Because of the generalization of the Pythagorean Theorem stated as Proposition 35.5, the union S of the semicircles erected onto the legs of the triangle has the same area as the semicircle T erected onto the hypothenuse. The intersection S ∩ T of these figures is the sum of the two circular segments over the legs. We subtract the intersection and conclude that the figures S \ T , which is the union of the two lunes, and T \ S = 4ABC, which is a triangle, have the same area. Problem 35.5. In the figure on page 677 is shown another lune construction. This time three circular arcs have been used, but two of them fit together with a common tangent. Explain how this lune is squared, and why squaring the lune is possible. Construction 35.2 (Another constructible and squarable lune). Into the semicircle with diam_
eter AB, we inscribe the right triangle 4ABC. Let point R be the midpoint of circular arc ARB _
opposite to the semicircle BCA. We draw the segment RC. The perpendiculars dropped onto the segment RC from vertices A and B have the foot-points Q and P, respectively. _
_
Onto the legs of triangle 4ABC are put quartercircles BC around P, and CA around Q. Onto _
the hypothenuse is put the quartercircle BA around R. These three quartercircles form a lune L. Squaring the lune. The circular segments erected onto the three sides of triangle 4ABC are similar since all three are bounded by a quartercircle. According to Proposition 35.5, the union S of the circular segments erected onto the legs of the triangle has the same area as the circular segment T erected onto the hypothenuse. We can obtain the lune L from the triangle 4ABC by adding the union S and subtracting circular segment T . Hence the the lune L has the same area as the triangle 4ABC. Another line of information about Hippocrates of Chios comes from Alexander of Aphrodisias. According to Alexander, Hippocrates began with an isosceles right triangle. Two congruent lunes
677
Figure 35.9. Can you square this lune?
are formed by the semicircle on the legs as external arcs, and the semi-circle circumscribed about the triangle as the inner arc. He proved that the sum of the areas of these two lunes is equal to the area of the triangle. Problem 35.6. Provide a drawing with named points, and explain the reasoning. Answer. The reasoning how to square this lune is the same as for the lunes of Alhazen. Here is how Hippocrates, again according to Alexander, tried to find a second squarable and constructible lune. Take an isosceles trapezoid formed by the diameter of a circle and three consecutive sides of an inscribed regular hexagon. Problem 35.7. Look at the graph on page 679. Prove that the sum of the areas of a semicircle on a side of the hexagon and the three lunes formed by the semicircles on three sides of the hexagon and by the semicircle circumscribing the trapezoid, is equal to the area of the trapezoid. Remark. Now there is a tempting speculation: if it were possible to "square" the three lunes, it would be possible to "square" the semicircle and hence the circle!
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Figure 35.10. Can you square this lune?
Answer. Since the side of an inscribed hexagon equals the radius of the circle, and the diameter is double the radius, the union S of four semicircles over sides of the hexagon has the same area as the big semicircle T . We place one of the four semicircles aside. Thus the intersection S ∩ T of the figures consists of the circular segments over three side of the hexagon, or trapezoid ABCD. We subtract the intersection and conclude that the figures S \ T , which is the union of three lunes together with the semicircle put aside, and T \ S = ABCD, which is the trapzoid, have the same area. 35.4.
The lune equation
Proposition 35.6. Any lune satisfying assumption (a) and (b) satisfies the lune equation r sin(nε/2) n =± sin(mε/2) m
(35.4)
with integers n, m ≥ 1. Proof. The common chord of the two arcs of the lune has the length |CD| = 2a sin(α/2) = 2b sin(β/2) Since α = nε and β = mε and na2 = mb2 , the equation is easy to confirm.
679
Figure 35.11. One can square this simple lune.
Figure 35.12. Can one square the circle?
Problem 35.8. The figure on page 680 is a numerical production of Hippocrates’ 3 : 1 lune, obtained from the lune equation sin(3ε/2) √ = 3 sin(ε/2) Convince yourself that the area of the lune equals the area of the kite ACBD. In the figure on page 681 you see the graph for the second solution of the same lune equation. Convince yourself that in this case, the area of the lune equals the area of the kite ACBD plus twice the area of the smaller circle around A.
680
Figure 35.13. A numerical production of Hippocrates’ 3 : 1 lune.
Answer. The figure on page ?? is a numerical production of Hippocrates’ 3 : 1 lune. The union S of the three congruent circular sectors ADD0 , AD0 D00 and AD00C has the same area as the circular sector BDC = T . The lune L = ADD0 D00C is obtained from the kite BDAC by adding the union S and subtracting the circular sector T . Hence the the lune has the same area as the kite BDAC.
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Figure 35.14. The area of the lune equals twice the area of smaller circle plus the area of the kite.
Answer. The figure on page 681 is the graph for the second solution of the same lune equation. In order to translate the explanation above to this case, one has to count area positive or negative according to the orientation of their boundary. The union S of the three congruent circular sectors ADD0 , AD0 D00 and AD00C has the area of two small circles CA around A plus the left sector smallADC. The area of S equals the area of the circular sector BDC = T . The area of the kite BDAC is negative because of the clockwise orientation. The big lune L = bigADC − smallADC is obtained from the negative kite BDAC by adding the union S and subtracting the circular sector T . −area( CADB) = −area( CADB) + area(S) − area(T ) = − (area( CADB) + area(T )) + area(S) = −area(big ADC) + 2 × area(CA ) + area(small ADC) = −area(L) + 2 × area(CA ) area(L) = area( CADB) + 2 × area(CA ) Hence the big lune has the same area as the kite CADB plus twice the area of the smaller circle . Remark. If one or both or angles α = nε and β = mε are more than 360◦ , we have really not squared the lune as drawn. Instead we have squared combined figure consisting of the lune plus q = bnε/360◦ c circles of radius a minus p = bmε/360◦ c circles of radius b.
682
We get |CD| = 2a sin(α0 /2) = (−1)q 2a sin(α/2) = (−1)q 2a sin(nε/2) |CD| = 2b sin(β0 /2) = (−1) p 2b sin(β/2) = (−1) p 2b sin(mε/2) where the primed angles are in the range (0◦ , 360◦ ). The sign in the lune equation turns out to be r n sin(nε/2) p+q = (−1) sin(mε/2) m Indeed n < m or the minus sign in the lune equation both become possible, but only in such a situation. Problem 35.9. Solve the lune equation sin(nε/2) =± sin(mε/2)
r
n m
for n = 3, m = 2 and 0 < ε < 180◦ . (a) Set up the quadratic equation and find the exact root expressions for cos ε. (b) Find the numerical values for ε in degrees. (c) Which values correspond to the historic case of Problem 2, which to Problem 3. Answer. sin(3ε/2) sin(ε/2) cos ε + cos(ε/2) sin ε = sin ε 2 sin(ε/2) cos(ε/2) 2 cos ε + 1 = √ 2(cos ε + 1)
=
cos ε cos ε + 2 cos2 (ε/2) + cos(ε/2) = 2 cos(ε/2) 2 cos(ε/2)
From the lune equation we see that x = cos ε satisfies the quadratic equation (2x + 1)2 3 = 2(x + 1) 2 2 4x + x − 2 = 0 The solutions are:
√ −1 ± 33 cos ε = 8 There are two real root, and both correspond to real angles. The numerical values are cos ε ≈ .5930703308 or − .8430703308 and ε ≈ 53.6◦ or 147.5◦ in degrees. The lunes are drawn in the figures on page 670 and page 683. Problem 35.10. The figure on page 683 shows another 3 : 2 lune construction. (a) The circle around L contains three congruent circular segments, which together turn more than a full circle, and which are similar to two congruent circular segments of the circle around I. Which are these segments.
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Figure 35.15. Another 3 : 2 lune.
(b) Prove the claim in (a) by means of circumference angles and the axial symmetry across the perpendicular bisector p. (c) Similarly to items (a) and (b), we get three congruent circular sectors of the circle around L. Together they are a full circle plus a sector. There are two congruent circular sectors of the circle around I which are similar to former ones. Prove that the first three segments have the same area as the latter two. (d) Prove by means of (c) that lune minus circle have the same difference of areas as the kite IGLE. Problem 35.11. Find a lune and a circle in the figure on page 684 with squarable difference of areas. Convince yourself that they have the same difference of areas as the self-crossing pentagon EFGBK, were the area of the upper quadrilateral is counted positive, and the area of the lower triangle is counted negative. Convince yourself directly that the signed area of the self-crossing pentagon EFGBK, which is really EFGNBKN, and the area of the kite IGLE are equal. _
_
Answer. The lune L is bounded by the long arc GFE and the short arc EMG. The lune minus the lower circle K around L is squarable.
684
Figure 35.16. In this case, the 3 : 2 lune minus the lower circle is squarable. _
_
Indeed, the two circular segments FG and FE inside the upper circle have the same area as _
_
_
the three circular segments EK, KB, and BG inside the lower circle. All these five segments have congruent central angles ε ≈ 147◦ . We subtract the two first segments from the lune L, and the _
_
three latter segments from the circle K. The intersection EMG N of the circular segments EK and _
BG appears twice. Thus we get the differences _ _ _ L − FG + FE = EMG F _ _ _ _ K − EK + KB + BG = 4BNK − EMG N _ _ _ _ _ L − K = L − FG + FE − K + EK + KB + BG _
_
= GMEF − 4BNK + EMG N = GNEF − 4BNK The last difference has the signed area of the self-crossing pentagon EFGBK. We can also see directly that this difference equal in area to the kite IGLE. Indeed, we subtract the two congruent triangles 4GFI and 4FEI inside the upper circle and add the three congruent triangles 4EKL, 4KBL, and 4BGL inside the lower circle. The sum of the areas of the former two is equal the sum of the areas of the latter three triangles.
685
Subtraction of the former two and addition of the latter three figures from the difference GNEF − 4BNK yields the kite IGLE, which has hence the same area. 35.5.
Vieta’s and Euler’s lunes
Problem 35.12. Solve the lune equation sin(nε/2) = sin(mε/2)
r
n m
for n = 4, m = 1 and 0 < ε < 180◦ . (a) Set up the cubic equation and find from Cardano’s formula the exact root expression for 2 cos(ε/2). (b) Find the numerical value for ε in degrees. (c) Draw the lune including its segments and sectors. Answer.
Figure 35.17. The 4 : 1 lune of Vieta.
h i sin(2ε) 2 sin ε cos ε 4 sin(ε/2) cos(ε/2) cos ε = = = 4 cos(ε/2) 2 cos2 (ε/2) − 1 sin(ε/2) sin(ε/2) sin(ε/2) Hence x = 2 cos(ε/2) satisfies the cubic equation x3 − 2x − 2 = 0. From Cardano’s formula we get with b = −2, c = −2: s s s s r r r r 2 3 3 3 3 3 c c b c c2 b3 19 19 2 cos(ε/2) = − + + + − − + = 1+ + 1− 2 4 27 2 4 27 27 27
686
There is only one real root. The numerical value is 2 cos(ε/2) ≈ 1.769292354 and ε ≈ 55.58◦ in degrees. The lune is draw in the figure on page 685. Problem 35.13. Solve the lune equation sin(nε/2) =± sin(mε/2)
r
n m
for n = 5, m = 1 and 0 < ε < 180◦ . (a) Set up the quadratic equation and find the exact root expressions for cos ε. (b) Find the numerical values for ε in degrees. (c) How many real angles occur as solutions.
Figure 35.18. The constructible 5 : 1 lune of Euler.
Answer. sin(5ε/2) = 4 cos2 ε + 2 cos ε − 1 sin(ε/2) From the lune equation we see that x = cos ε satisfies the quadratic equation √ 4x2 + 2x − 1 = ± 5 The solutions are:
q cos ε =
−1 ±
√ 5±4 5
4 There are two real roots and two complex roots. Only the root with both plus signs corresponds to a real angle. The numerical value is cos ε ≈ .6835507455 and ε ≈ 46.9◦ in degrees.
687
Problem 35.14. Solve the lune equation sin(nε/2) =± sin(mε/2)
r
n m
for n = 5, m = 3 and 0 < ε < 180◦ . (a) Set up the quadratic equation and find the exact root expressions for cos ε. (b) Find the numerical values for ε in degrees. (c) How many real angles occur as solutions.
Figure 35.19. The constructible 5 : 3 lune of Euler.
Answer. 4 cos2 ε + 2 cos ε − 1 sin(5ε/2) = sin 3(ε/2) 2 cos ε + 1 From the lune equation we see that t = cos ε satisfies the quadratic equation √ √ 3(4t2 + 2t − 1) = σ 5(2t + 1) The solutions are with σ = ±1: q √ √ √ σ 5 − 3 ± 20 + 2σ 15 cos ε = √ 4 3
688
For σ = +1, there are two real roots, both correspond to a real angle. The numerical values are cos ε ≈ .8330386705 or − .6875414461 and ε ≈ 33.6◦ or 133.4◦ in degrees. For σ = −1, there are two real roots, only one corresponds to a real angle. The numerical values are cos ε ≈ −.0674839681 or − 1.078013256 and ε ≈ 93.9◦ in degrees. Remark. Viëta found around 1593 the 4 : 1 lune leading to a cubic equation. The constructible and squarable 5 : 1 and 5 : 3 lunes were found in 1766 by Martin Johan Wallenius. Too, Leonard Euler made the same discovery around 1771, published in "Solutio problematis geometrici circa lunules a circulis formatas". The two lunes from Euler and the two not so obvious ones by Hippocrates are once more obtained algebraically from the lune equation in an article by Th. Clausen [10] published in 1840. The results of Hippocrates were apparently not known at that time. Except for Hippocrates, none of these authors consider a combination of lune and circle to be squared. Remark. A popular account of some of the material from this section is contained in William Dunham’s book "Journey through Genius". A short introduction and some biographic information is given by http://en.wikipedia.org/wiki/Quadrature_of_the_lune
Figure 35.20. A constructible 5 : 3 lune minus circle around A.
36. Transcendental Numbers and the Constructibility of the Lunes 36.1. About transcendental numbers A number α is called algebraic if there exists a nonzero integer polynomial p ∈ Z[x] such that p(α) = 0. In that case α is called a root of the polynomial p. The set of all real or complex algebraic numbers is denoted by A. As stated in Theorem 34.1, the set of algebraic numbers is both a countable and√a field. Obviously, all rational numbers are √ algebraic, but the converse is not true. We know that 2 and i = −1 are two important examples of irrational but algebraic numbers. The real or complex numbers, which are not algebraic are called transcendental.
689
Figure 35.21. The overlapping polygon has the same signed area as the kite ADBC.
Figure 35.22. A constructible 5 : 3 lune plus circle around B minus circle around A.
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Definition 36.1. The numbers α1 , α2 , . . . αn are called Z-linearly independent if the dependence relation k1 α1 + k2 α2 + · · · + kn αn = 0 with integers k1 , k2 , . . . kn holds only for all k1 = k2 = · · · = kn = 0. The numbers γ1 , γ2 , . . . γn are called linearly independent over the algebraic numbers if the dependence relation β1 γ1 + β2 γ2 + · · · + βn γn = 0 with algebraic numbers β1 , β2 , . . . βn holds only for all β1 = β2 = · · · = βn = 0. Obviously, independence over the integers is equivalent to linear independence of the rationals Q. Theorem 36.1 (Lindemann-Weierstrass Theorem). Let the algebraic numbers α1 , α2 , . . . αn be distinct. Then the exponentials eα1 , eα2 , . . . eαn are linearly independent over the algebraic numbers. Hence n X βi eαi , 0 i=1
for any algebraic numbers β1 , β2 , . . . βn , unless β1 = β2 = · · · = βn = 0. The Lindemann-Weierstrass Theorem immediately yields the classical results that e, π, ln 2 are transcendental. • Assume towards a contradiction that Euler’s number e is algebraic. We put α1 = 0, α2 = 1, β1 = −e, β2 = 1 and get the contradiction −e · e0 + 1 · e1 , 0. (Hermite 1873) • Assume towards a contradiction that π is algebraic. We put α1 = 0, α2 = iπ, β1 = β2 = 1 and get the contradiction 1 · e0 + 1 · eßπ , 0. (Lindemann 1882) • Assume towards a contradiction that ln 2 is algebraic. We put α1 = 0, α2 = ln 2, β1 = −2, β2 = 1 and get the contradiction −2 · e0 + 1 · eln 2 , 0. Corollary 71. Let the distinct algebraic numbers α1 , 0, α2 , 0, . . . αn , 0 be all nonzero and the algebraic numbers β1 , 0, β2 , 0, . . . βn , 0 be nonzero. Then the sum n X
βi eαi
is transcendental.
i=1
Theorem 36.2 (Baker 1967). Let the numbers eα1 , eα2 , . . . eαn be algebraic and the numbers α1 , α2 , . . . αn be Z-linearly independent. Then n X
βi αi
is transcendental
i=1
for any algebraic numbers β1 , β2 . . . βn , unless β1 = β2 = · · · = βn = 0. √
• •
Baker’s Theorem immediately yields the classical results that eπ and 2 2 are transcendental. Assume towards a contradiction that eπ is algebraic. We put α1 = π, α2 = iπ, β1 = −i, β2 = 1 and get the β1 α1 + β2 α2 = (−i) · π + 1 ·√iπ = 0 is transcendental. √ Assume towards a contradiction that 2 2 is algebraic. We put α1 = 2 ln 2, α2 = ln 2, β1 = √ 2, β2 = −2 and get the contradiction that β1 α1 + β2 α2 = 0 is transcendental.
Theorem 36.3 (Gelfond-Schneider Theorem). Let t be irrational and a , 0, 1 be any two algebraic numbers. Then at is transcendental. Indeed all values of the power at are transcendental.
691
Reason. The numbers α1 := ln a and α2 = t ln a are by assumption Z-linearly independent. Assume towards a contradiction that at is algebraic. We put β1 := −t, β2 := 1 and get the contradiction that β1 α1 + β2 α2 = 0 is transcendental. Theorem 36.4 (M. Waldschmidt). Given are two complex numbers t and z. Suppose that t is irrational and z , 0. Then at least one of the numbers t, ez , etz is transcendental. The exponentials ez and etz are defined by the power series of the exponential function. Reason. The numbers α1 := z and α2 = tz are by assumption Z-linearly independent. Assume towards a contradiction that all three numbers t, ez , etz are algebraic. We put β1 := −t, β2 := 1 and get the contradiction that β1 α1 + β2 α2 = (−t) · z + 1 · tz = 0 is transcendental. Remark. The Theorem 36.4 of M. Waldschmidt is just a clarified version of the famous Theorem 36.3 of Gelfond-Schneider. Indeed, for any two algebraic numbers a and t, where t is irrational and a , 0, 1, the Theorem of M. Waldschmidt implies that all values of the power at are transcendental: To see this, we put z := ln a + 2πik with arbitrarily chosen values of ln a and the integer k. One gets a = ez , and etz is one of many values of at ,—it still depends on the integer k since z , 1. From Waldschmidt’s Theorem one concludes that at least one of the three numbers t, ez , etz is transcendental. Since the first two numbers t and ez are algebraic by assumption, we see that the third number etz ∈ at is transcendental. Hence all values of at are transcendental. Corollary 72 (Alan Baker). Under the assumption that the numbers eα1 , eα2 , . . . eαn and β1 , . . . βn are algebraic, the sum n X βi αi i=1
is either algebraic or the sum equals zero. (The assumption that the numbers α1 , α2 , . . . αn are Z-linearly independent has been dropped). Remark. A thorough account of the results just reviewed is contained in Alan Baker’s book [3] on "Transcendental Number Theory". The book "Making Transcendence Transparent" of Burger and Tubbs [8] is readable and informative, too. Finally one can recommend the internet source http://en.wikipedia.org/wiki/Baker%27s_theorem 36.2. Which lunes are algebraic, which constructible? We look once more at the generic figure 35.1. I call a lune algebraic for which the radius a and b, the angle functions sin α, cos α, sin β, cos β of the central angles, and the area of the lune are algebraic numbers. Theorem 36.5. All algebraic lunes satisfy Hippocrates’ two basic assumptions (a) The two circular sectors corresponding to the lune’s arcs have the same area. (b) The central angles of the two circular arcs are commensurable. Proof. Since the area of the kite ACDB is algebraic, and the area of the lune L is assumed to be algebraic, their difference is algebraic. Since area(L) = area(SA ) + area(ACDB) − area(SB )
692
we conclude that the difference of the areas of the circular sectors SA and SB : area(SA ) − area(SB ) = αa2 − βb2 is algebraic, too. The angles α and β at the centers A and B, corresponding to the two arcs of the lune have to be measured in radian measure. The assumptions imply that eiα , eiβ and a, b and the area αa2 − βb2 are all algebraic. We use Alan Baker’s Corollary 72 with α1 := iα, α2 := iβ and β1 := a2 , β2 := −b2 . We conclude that the the sum β1 α1 + β2 α2 = αa2 − βb2 = 0 Hence Hippocrates’ assumption (a) holds. We now use the Theorem 36.4 of Gelfond Schneider and M. Waldschmidt with x = β/α and z = iα. Assume towards a contradiction that x is irrational. We would conclude that at least one of the numbers β/α, eiα , eiβ is transcendental. The latter two are algebraic by assumption and β a2 = α b2 is algebraic, too, by the first part of the proof. Hence β/α is rational, confirming Hippocrates’ assumption (b). Main Theorem 30 (N.G. Tschebatorev and A.W. Dorodnov 1947). There exist only five squarable and constructible lunes, corresponding to the cases in which n : m is 2 : 1 , 3 : 1 , 3 : 2 , 5 : 1 , 5 : 3. Remark. The final result was obtained by A. W. Dorodnow, after many earlier partial results. • E. Landau showed in 1903 that the p : 1 lune is not constructible if p is a prime but not a Fermat prime. • L. Tschakaloff showed in 1926 that the 17 : 1 lune is not constructible, neither any p : m lune where p is a prime and p > m. • N. Tschebotarov [36] showed in 1934 that no n : m lune with n, m both odd and m > 5 is constructible. 36.3. Irreducibility of the lune equation The lune equation (35.4) can be written with the complex variable z = eiε . Since sin(nε/2) zn − 1 = z−(n−1)/2 sin(ε/2) z−1 we get in the new variable √
√ m sin(nε/2)) − (± n) sin(mε/2)) = 0 √ zn − 1 √ zm − 1 m − (± n)z(n−m)/2 =0 z−1 z−1
After taking the square, we get an integer polynomial equation: !2 !2 zn − 1 zm − 1 n−m Pn,m (z) := m − nz =0 z−1 z−1
(36.1)
Proposition 36.1. We assume that p = n be an odd prime and p > m. Then the polynomial Pn,m (z) in the squared lune polynomial equation (36.1) is irreducible over the integers.
693
Theorem 36.6 (L. Tschakaloff 1926). If p is a prime, but not a Fermat prime, and m < p, the p : m lune is not constructible. Proof. We proceed similarly as in the proof of Proposition 34.19. We use the substitution z = x+1. Recall that the binomial formula implies that is the new variable x, all coefficients of ! p (1 + x) p − 1 X p k−1 = x R p (x) := x k k=1 ! ! ! p p 2 p = p+ x+ x + ··· + x p−3 + px p−2 + x p−1 2 3 p−2 except the leading ones are divisible by p. Similarly, we substitute z = 1 + x into the squared lune polynomial Pn,m (z) = mR2p (x) − p(1 + x) p−m R2m (x) = Qn,m (x) and use the Eisenstein criterium, given as Proposition 34.18, to show that the resulting polynomial is irreducible. Indeed, all coefficients of Qn,m (x) except the leading one are divisible by p. The leading coefficient is m − p which is not divisible by the prime p. The constant coefficient is mp2 − pm2 = mp(p − m) which is divisible by p, but not by p2 . Hence all assumptions of the Eisenstein criterium are satisfied and the polynomials Q and P p,m are irreducible. Remark. Similar results are contained in the article [29] which is translated from Postnikov’s 1963 Russian book on Galois theory. Note that we have only dealt with the most easy special case of the main theorem of N.G. Tschebatorev and A.W. Dorodnov. Biographic information is given in http://en.wikipedia.org/wiki/N._G._Chebotarev 36.4.
Tschebychev polynomials
Definition 36.2 (Tschebychev polynomials). The Tschebychev polynomials of first and second kind are defined by sin(n + 1)t T n (cos t) = cos nt and Un (cos t) = sin t for integers n = 0, 1, 2, . . . . Proposition 36.2. Both polynomials T n and Un satisfy the same recursion formula T n+1 = 2xT n − T n−1 Un+1 = 2xUn − Un−1 with the initial data T 0 = U0 = 1 and T 1 = x but U1 = 2x, respectively. Both T n and Un are integer polynomials of degree n. They are even for even n, odd for odd n. Moreover T n (1) = 1 and Un (1) = n + 1 for all n. The T n satisfy the composition formula T n ◦ T m = T nm . The polynomials 2T n (v/2) and Un (v/2) are integer polynomials of the variable v.
694
n 0 1 2 3 4 5 6 7 8 9
Tn 1 x −1 + 2x2 −3x + 4x3 1 − 8x2 + 8x4 5x − 20x3 + 16x5 −1 + 18x2 − 48x4 + 32x6 −7x + 56x3 − 112x5 + 64x7 1 − 32x2 + 160x4 − 256x6 + 128x8 9x − 120x3 + 432x5 − 576x7 + 256x9
Un 1 2x −1 + 4x2 −4x + 8x3 1 − 12x2 + 16x4 6x − 32x3 + 32x5 −1 + 24x2 − 80x4 + 64x6 −8x + 80x3 − 192x5 + 128x7 1 − 40x2 + 240x4 − 448x6 + 256x8 10x − 160x3 + 672x5 − 1024x7 + 512x9
Lemma 36.1 (binomial expansion). bn/2c X
! n n−2m 2 T n (x) = x (x − 1)m 2m m=0 bn/2c X n+1 ! Un (x) = xn−2m (x2 − 1)m 2m + 1 m=0
(36.2)
(36.3)
Proof. Let x = cos t. Use eint = cos nt+i sin nt = (cos t+i sin t)n and separate the binomial formula into real and imaginary parts: ! n X n cos nt + i sin nt = (cos t)n−k (i sin t)k k k=0 bn/2c bn/2c X n! X n! m n−2m 2m T n (x) = cos nt = (−1) (cos t) (sin t) = xn−2m (x2 − 1)m 2m 2m m=0 m=0 ! b(n−1)/2c X n sin nt = (−1)m (cos t)n−2m−1 (sin t)2m+1 2m + 1 m=0 ! b(n−1)/2c X n Un−1 (x) = xn−1−2m (x2 − 1)m 2m + 1 m=0 Lemma 36.2 (generating function). X
T n (x) rn =
1 − xr 1 + r2 − 2xr
(36.4)
Un (x) rn =
1 1 + r2 − 2xr
(36.5)
n≥0
X n≥0
We see once more that both polynomials T n and Un satisfy the same recursion formula.
695
Proof. Let x = cos t. One separates a complex geometric series into real and imaginary parts: X 1 1 − e−it r eint rn = = 1 − eit r |1 − eit r|2 n≥0 X 1 − r cos t + ir sin t rn cos nt + irn sin nt = (1 − r cos t)2 + r2 sin2 t n≥0 X 1 − r cos t rn T n (cos t) = 1 + r2 − 2r cos t n≥0 X r sin t r sin t rn−1 Un−1 (cos t) = 2 1 + r − 2r cos t n≥1 The Tschebychev polynomials of second kind are for even and odd index are even and odd, respectively. U2n (x) = Pn (4x2 − 4) and U2n+1 (x) = 2x Qn (4x2 − 4) Lemma 36.3. The polynomials P and Q with the independent variable s = 4x2 − 4 have the generating functions X 1+s Pk (s) rn = (36.6) 1 − (2 + s)r + r2 k≥0 X 1 (36.7) Ql (s) rn = 1 − (2 + s)r + r2 l≥0 Hence both satisfy the same the recursion formula: Pn+2 = (2 + s)Pn+1 − Pn
and
Qn+2 = (2 + s)Qn+1 − Qn
n 0 1 2 3 4 5 6 7 8
Pn 1 3+s 5 + 5s + s2 7 + 14s + 7s2 + s3 9 + 30s + 27s2 + 9s3 + s4 11 + 55s + 77s2 + 44s3 + 11s4 + s5 13 + 91s + 182s2 + 156s3 + 65s4 + 13s5 + s6 15 + 140s + 378s2 + 450s3 + 275s4 + 90s5 + 15s6 + s7 17 + 204s + 714s2 + 1122s3 + 935s4 + 442s5 + 119s6 + 17s7 + s8
n 0 1 2 3 4
Qn 1 2+s 3 + 4s + s2 4 + 10s + 6s2 + s3 5 + 21s + 21s2 + 8s3 + s4
We put n = 2k or n = 2l + 1 into the binomial formula for U2k (x) = Pk (s) or U2l+1 (x) = 2x Ql (s) and get
696
Lemma 36.4. Pk (s) = 4−k 2Ql (s) = 4−l
! k X 2k + 1 (4 + s)k−m sm 2m + 1 m=0 ! l X 2l + 2 (4 + s)l−m sm 2m + 1 m=0
(36.8)
(36.9)
Lemma 36.5. With the new independent variable s = 4x2 − 4, the polynomials P and Q are monic integer polynomials with positive coefficients. The constant coefficients are Pk (0) = 2k + 1 and Ql (0) = l + 1. Assume now that p is an odd prime. In this case, all coefficients of the polynomial P(p−1)/2 (s) except the leading one are divisible by p. The constant coefficient equals p. The polynomial Pk is related to the construction of a regular 2k + 1-gon. Indeed, its zeros are zi = −4 sin2
i 180◦ 2k + 1
for i = 1, 2, . . . , k
Problem 36.1. Confirm that P7 is divisible by P1 · P2 . Calculate the quotient P7 /(P1 · P2 ). Find its zeros zi and the corresponding central angles i · 24◦ for the 15-gon. Which four integers i do you obtain, and why? Answer. The quotient P7 = 1 + 8s + 14s2 + 7s3 + s4 P1 · P2 have zeros −4 sin2
i 180◦ 15
for i = 1, 2, 4, 7. These are the integers less than 8 relatively prime to 15.
Proposition 36.3. Let p an odd prime. Then the polynomial P(p−1)/2 (s), the Tschebychev polynomial U p−1 (x), and the cyclotomic polynomial Φ p (z) are irreducible. Proof. We know by Lemma 36.5 that all coefficients of the polynomial P(p−1)/2 (s), except the leading one, are divisible by p. The constant coefficient equals p. By the Eisenstein criterium, given as Proposition 34.18, we conclude that the polynomial P(p−1)/2 (s) is irreducible. We put the substitution z = eit into the cyclotomic polynomial and get z p − 1 z(p−1)/2 − z(1−p)/2 = z−1 z1/2 − z−1/2 sin((p − 1)t/2) = = U p−1 (cos(t/2)) = P(p−1)/2 (−4 sin2 (t/2)) sin(t/2)
Φ p (z) =
We see that any factoring of Φ p (z) into integer polynomials entails an integer factoring of the Tschebychev polynomial U p−1 (s) with s = cos(t/2), and an integer factoring of the polynomial P(p−1)/2 (s) with independent variable s = −4 sin2 (t/2). We have ruled out such a factoring. Hence the Tschebychev polynomial U p−1 (s) and finally the cyclotomic polynomial Φ p (z) are irreducible, too. The lune equation sin(nε/2) =± sin(mε/2)
r
n m
697
can be written with the Tschebychev polynomials of second kind Un (cos t) =
sin(n + 1)t sin t
We use instead of x = cos(ε/2) the new independent variable s = 4x2 − 4 = −4 sin2 (ε/2) = 2 cos ε − 2 The Tschebychev polynomials of second kind are expressed by the new polynomials P and Q using the independent variable s. Lemma 36.6 (Some reduction of the lune equation). For n and m both odd, we get the lune polynomial equation √ √ m P(n−1)/2 (2 cos ε − 2) − (± n) P(m−1)/2 (2 cos ε − 2) = 0 (36.10) which is of half the degree, and easier to solve. For n odd and m even, we get the original lune polynomial equation √ √ m P(n−1)/2 (−4 + 4 cos2 (ε/2)) − (± n)2 cos(ε/2)Q(m−2)/2 (−4 + 4 cos2 (ε/2)) = 0 (36.11) The half angle needs to be used and there is no reduction of the degree. An interesting peculiarity occurring in the case n = 9, m = 1 is mentioned by Postnikov in the survey article [29]. We get the lune polynomial equations s4 + 9s3 + 27s2 + 30s + 9 + 3σ0 = 0 again for the variable s = 2 cos ε − 2 and σ0 = ±1. The case with the minus sign σ0 = −1 can indeed be solved by square roots! The factoring s4 + 9s3 + 27s2 + 30s + 12 = ((s + 2)2 + ω(s + 2) − 2) · ((s + 2)2 − ω(s + 2) − 2) √ with ω = (1 + −3)/2 leads to the four roots q √ √ −9 + σ1 −3 + σ2 30 − 2σ1 −3 si = 4 with σ1 = ±1 and σ2 = ±1. Only because these turn out to be complex roots does one not obtain another constructible lune. The polynomial s4 + 9s3 + 27s2 + 30s + 3 for the case with the plus sign σ0 = 1 can be factored into quadratic polynomials by Descartes method, too. One obtains two complex solutions, and two real solutions, one of which corresponding to a lune. But the root cannot be solved by square roots, and hence the lune is not constructible.
698
37. Alhazen’s problem 37.1.
Elementary cases of Alhazen’s problem
(Alhazen’s Problem). Given is a circular mirror, and two points inside its disk. A light ray is emanated from one of these points, and has to be directed to the mirror in such a manner that the second point is reached after one reflection. Find all possible locations for points of reflection.
Figure 37.1. Alhazen’s problem.
The problem is illustrated with the figure on page 698. Let the circle for the mirror have center O and radius 1. Let the light source be at point A and the receiver at point B. Let a = |OA|, b = |OB| and γ = ∠AOB. After one reflection at point R, the light ray has to be directed to point B. By the law of light reflection θ = ∠ORA ∠ORB are congruent angles. We have to determine the possible points R of reflection. Too, we want to know the angles ρ = ∠AOR and θ. Remark. The direct straight ray from A to B should not be counted as a solution, except in the case that A or B lie directly on the mirror. In these special cases, the direct ray is a limiting position of the ray with a reflection, and hence has to be counted as a solution. Only for special cases can the solution be obtained by compass and straightedge. Nevertheless, these are useful to get a feeling and control. Here are such examples, in increasing degree of difficulty. Problem 37.1. The mirror is supposed to be the unit circle with center O. We assume the point A of light source lies on the mirror, take angle ∠AOB = 180◦ , and assume that the receiver B has distance |OB| = 1/2 from the center of the mirror. Give a geometric construction for all four solutions of Alhazen’s problem.
699
Figure 37.2. An easy special case of Alhazen’s problem.
−−→ Answer. The direct ray AB has to be counted as a solution, and the light ray on the same ray going beyond point B and reflected back to this point, is counted as a second one. There two further easy solutions with ρ = ±120◦ , one of which is constructed in the figure on page 699. Problem 37.2. The mirror is supposed to be the unit circle with center O. We assume the point ◦ A of light source lies √ on the mirror, take angle ∠AOB = 135 , and assume that the receiver B has distance |OB| = 2/2 from the center of the mirror. Give a geometric construction for all four solutions of Alhazen’s problem. Answer. One rather easy solution with ρ = 90◦ is constructed in the figure on page 700. The direct −−→ ray AB has to be counted as a second solution. Two further solutions are obtained with ρ = ±120◦ , following the light rays from a previous example. Problem 37.3. Assume the light source A and receiver B lie on the same ray emanating from the center O of the mirror. Convince yourself that there exist only the two obvious solutions of Alhazen’s problem. Answer. Obviously there exist the two solutions, for which the reflection points are the endpoints of the diameter on the line AB. Suppose now that the point R of reflection does not lie on the line −−→ AB. By the las of reflection, the incoming ray AR and the reflected ray r0 lie on different sides of −−→ −−→ line OR. Hence the reflected ray r0 cannot intersect the ray OA = OB. Hence one does not obtain any further solution. Problem 37.4. Assume light source A and receiver B have the same distance from the center O of the mirror M. Give a geometric construction to solve Alhazen’s problem. There exist cases with two solutions, and cases with four solutions. They can be distinguished by drawing the circle C through the three points A, B and O. Explain, and give a reason.
700
Figure 37.3. An second easy special case of Alhazen’s problem.
Answer. The intersection points of the perpendicular bisector p of AB with the mirror M are the two obvious points of reflection. Thus one obtains two symmetric solutions. But in some cases, there exist further solutions, for which the point of reflection does not lie on the perpendicular bisector p, and which thus are nonsymmetric. Indeed the reflection points for the nonsymmetric solutions of Alhazen’s problem are exactly the intersection points of the mirror circle M with the circle C through points A, B and O. Reason. Suppose there exists a solution of Alhazen’s problem with reflection point R not on p. This situation is illustrated in the figure on page 701 with R := R4 . The first part (a) confirms that any additional reflection point necessary is an intersection point of the mirror circle M with the circle C through points A, B and O. The second part (b) shows that it is sufficient to take an intersection point of the mirror circle M with the circle C, in order to get a additional reflection point. (a) necessity The two triangles 4ARO and 4BRO, have congruent sides AO BO, the common side OR OR and by the reflection law congruent angles ∠ARO ∠BRO. But nevertheless, they are not congruent. In other words, we have a SSA situation with two non-congruent solutions. Using the sin-theorem for triangles 4OAR and 4OBR, one checks that the angles at points A and B have the same sinus sin ∠OAR sin ∠ORA sin ∠ORB sin ∠OBR = = = |OR| |OA| |OB| |OR| Since ∠OAR , ∠OBR in the non-symmetry case, we conclude that ∠OAR + ∠OBR = 180◦ . The same result can also be obtained by proposition 7.60 about the restricted SSA congruence in the case with non-uniqueness.
701
Figure 37.4. Alhazen’s problem for a = b.
By Euclid III.22, the opposite angles of a circular quadrilateral add up to two right angles (See theorem 15.7). Moreover, a convex quadrilateral where the opposite angles add up to two right angles is circular. (See proposition 15.2 to get this converse statement.) Hence we conclude that the quadrilateral AOBR is circular. Thus point R is indeed an intersection point of the mirror circle M with the circle C through points A, B and O. (b) sufficiency Conversely, suppose point R is an intersection of circles M and C. Since OA OB, and point R does not lie on the arcs OA nor OB, the angles ∠ORA and ∠ORB are circumference angles of congruent arcs. Hence by corollary 44, they are congruent angles ∠ORA ∠ORB. Thus we have indeed obtained a solution of Alhazen’s problem. Problem 37.5. Assume that the center O of the mirror lie between the light source point A , O and the receiver B. Find a compass and straightedge construction for this special case of Alhazen’s problem. Actually there exists cases for which four points of reflections exist: two of them are obvious, the two other ones are constructed via Apollonius’ Theorem 26.4. How can the cases with four solutions be distinguished from those with only two solutions. Answer. The two obvious points of reflection are the endpoints of the diameter on the line AB. To construct the other two not so obvious solutions, one uses Apollonius’ Theorem 26.4. Recall that the Apollonius circle is the set of all points from which two consecutive segments lying on the same line appear under congruent angles. Because of the law of reflection, the consecutive segments AO and OB appear under congruent angles ∠ARO ∠ORB. We need the Apollonius circle with the
702
Figure 37.5. Alhazen’s problem for center O between A and B.
diameter OP, where A, O, B, P are four harmonic points. According to definition 26.1, this means that the cross ratio AO · BP = −1 (AB, OP) = BO · AP for the directed segments. In the figure on page 702, we have provided an easy construction of the fourth harmonic point P. Moreover, we have drawn Apollonius’ circle with the diameter OP. For the case that the Apollonius circle and the mirror circle intersect, we get two two additional solutions of Alhazen’s problem. Thus there exist four solutions. For the case that the Apollonius circle and the mirror circle do not intersect, we may conclude that Alhazen’s problem has only the two obvious solutions with points of reflection at the endpoints of the diameter on the line AB. Lemma 37.1. Let the receiver B, the center O and the light source A lie on one diameter, and assume that the light source lies directly on the mirror. For such a situation, Alhazen’s problem has two obvious solutions: the direct light ray, and a solution with the reflection point R2 at the opposite end of the diameter on the line AB. Two further solutions may exist or not, depending on the exact location of the receiver. Let S be the point on this diameter such that the center O lies between S and the light source A and |OS | = 13 |OA|. (i) For the receiver lying in the segment S A, Alhazen’s problem has only the two obvious solutions; (ii) for the receiver lying inside the segment R2 S , Alhazen’s problem has four solutions. Besides the two obvious solution, there are two further ones. These are constructed by means of four harmonic points P, B, O, A and Apollonius’ circle with diameter OP. Proof. The reader may easily check that the four points R2 ∗ S ∗ O ∗ A are harmonic points. The case that light source and receiver lie on the same radius has already been handled by problem 37.3. For other position B of the receiver, let P ∗ B ∗ O ∗ A be four harmonic points, too.
703
(i) For the receiver lying in the segment S A, the fourth harmonic point P lies inside or on the mirror disk. Hence the Apollonius circle and the mirror circle do not intersect. We conclude that Alhazen’s problem has only the two obvious solutions with light rays on the line AB. (ii) For the receiver lying inside the segment R2 S , the fourth harmonic point P lies outside the mirror disk. Hence the Apollonius circle and the mirror circle intersect in two points. We get two additional solutions of Alhazen’s problem. Thus there exist four solutions. 37.2. Trigonometric solution We now address the main question of this section, how to solve Alhazen’s problem by means of trigonometry. Referring to the figure on page 698, given are the segments a, b and the angle γ. We set up equations for the unknown angle of reflection θ and angle ρ, which determines the point R of reflection. The sin-theorem for triangles 4OAR and 4OBR yields sin θ sin(θ + ρ) sin θ sin(θ + γ − ρ) = and = a 1 b 1 We want to eliminate θ in order to obtain a trigonometric equation with the only unknown ρ. We use the addition theorem for the sin-function to isolate θ in the sums θ + ρ and θ + (γ − ρ) and get sin θ = a sin θ cos ρ + a cos θ sin ρ
and
sin θ = b sin θ cos(γ − ρ) + b cos θ sin(γ − ρ)
and finally two expressions for tan θ that need to be equal. Multiplying with the denominators yields a polynomial equation. tan θ =
a sin ρ b sin(γ − ρ) = 1 − a cos ρ 1 − b cos(γ − ρ)
(37.1)
Alternatively, one can argue as follows: we have obtained a system of two linear equations with the unknowns sin θ and cos θ. The system has a nontrivial solution (sin θ, cos θ) , (0, 0). Hence the determinant has to be zero. In both ways we obtain the same equation a sin ρ[1 − b cos(γ − ρ)] − b sin(γ − ρ)(1 − a cos ρ) = 0 We can use the addition theorem to simplify, and get the equation a sin ρ + b sin(ρ − γ) − ab sin(2ρ − γ) = 0
(37.2)
This is the main equation from which to calculate the solutions of Alhazen’s problem. Provided one allows for angle ρ all values on the entire circle −180◦ < ρ ≤ 180◦ , we get indeed all solutions of Alhazen’s problem. Afterwards, for each solution, we may obtain the respective values for angle θ of reflection from equation (37.1). Problem 37.6. The polynomial P(ρ) = a sin ρ + b sin(ρ − γ) − ab sin(2ρ − γ)
(37.3)
the zeros of which determine the solutions of Alhazen’s problem can be factored in some special cases, among them are those from problems 37.5, 37.4, and ??. Find these factorizations, and use them to solve Alhazen’s problem completely for these special cases. (This problem needs fluency with trig calculations.)
704
Answer to the case from problem 37.5 . Here γ = 180◦ , and the polynomial (37.3) is P(ρ) = a sin ρ + b sin(ρ − 180◦ ) − ab sin(2ρ − 180◦ ) = (a − b) sin ρ + 2ab sin ρ cos ρ = sin ρ · a − b + 2ab cos ρ The first factor gives the zeros ρ1 = 0◦ and ρ2 = 180◦ . The second factor gives the zeros ρ3,4 = ± arccos
b−a 2ab
Answer to the case of problem 37.4 . We need the identities sin 2x = 2 sin x cos x and sin(x + y) + sin(x − y) = 2 sin x cos y A−B A+B cos sin A + sin B = 2 sin 2 2 Under the assumption a = b, the polynomial (37.3) is P(ρ) = a[sin ρ + sin(ρ − γ)] − a2 sin(2ρ − γ) 2ρ − γ γ 2ρ − γ 2ρ − γ = 2 sin cos − 2a2 sin cos 2 2 2 2 " # 2ρ − γ γ 2ρ − γ = 2a sin · cos − a cos 2 2 2 The first factor gives the zeros ρ1 = γ/2 and ρ2 = 180◦ + (γ/2). The second factor gives the zeros γ γ ρ3,4 = ± arccos a−1 cos 2 2 √ Answer to the case from problem ?? . Here a = 2/2, b = 1, γ = 135◦ , and the polynomial (37.3) is P(ρ) = a sin ρ + b sin(ρ − 135◦ ) − ab sin(2ρ − 135◦ ) = sin ρ cos 45◦ − sin(ρ + 45◦ ) + ab sin(2ρ + 45◦ ) = sin 45◦ [− cos ρ + sin(2ρ + 45◦ )] = sin 45◦ [sin(ρ − 90◦ ) + sin(2ρ + 45◦ )] √ 3ρ − 45◦ −ρ − 135◦ = 2 sin · cos 2 2 The first factor gives the zero ρ2 = 135◦ , ρ3 = 15◦ , and ρ4 = −105◦ . The second factor gives only ρ1 = 45◦ . Problem 37.7. Find the complete set of all solution for Alhazen’s problem with the values a = 1/2, b = 1, γ = 135◦ , already considered in problem ??. Provide a drawings containing all four cases.
705
38. The Quadrix of Hippias 38.1. The quadratrix The quadratrix of Hippias is generated in the following way. In Cartesian (x, y) coordinates, we take two objects: • a horizontal line y = πθ 2 ; • a turning ray with x = r cos θ, y = r sin θ and r ≥ 0. We may imagine the parameter t = πθ 2 as time. With increasing time, the horizontal line y = t is moving up while the ray is rotating counterclockwise. Their intersection points (x, y) yield the quadratrix. The motion of the two objects is coordinated in such a way that for time t = 0 the process is started with the x-axis and a ray from the origin pointing to the right. At time t = 1, we obtain the horizontal line y = 1 and a ray along the positive y-axis. Hence their intersection point (0, 1) lies on the quadratrix . With the real parameter t, the parametric equations for the quadratrix are x = t cot
πt 2
y=t
(38.1)
The angle θ = πt2 has to be measured in radians. Obviously, we can extend the parametric equations to all real parameter t. We see that the x-axis turns out to be the axis of symmetry. Theorem 38.1 (Hippias and Dinostratus). We assume that besides straightedge and compass, the quadratrix is available as a tool for constructions. Under these assumptions • any angle can be partitioned into any given number of congruent parts; • the number π can be constructed, and hence the circle can be squared. Remark. The invention of the quadratrix is credited to Hippias of Elis (born around 460 B.C.). He considered the curve only for 0 ≤ t ≤ 1. Hippias used his curve for trisection of an angle. Dinostratus (ca. 350 B.C.) is usually credited with applying it for squaring of the circle. These arguments are assuming that the quadratrix is available as an entire curve at the beginning of the constructions. With this assumption, the arguments are mathematically rigorous. 38.2. Constructible points on the quadratrix We need to make clear that Hippias’ constructions cannot be achieved with straightedge and compass alone. Neither the trisection of an angle of 60◦ , nor squaring of the circle can be achieved by compass and straightedge. Indeed, we shown in Proposition 38.2 that the only points on the quadratrix (38.1) which are constructible with compass and straightedge are those obtained via the constructible regular polygons. We now ask, which points of the quadratrix are constructible with compass and straightedge. Question. Is the point were the quadratrix intersect the x- axis constructible? Answer. For parameter t = 0, we have to find the intersection of the x-axis with the ray from the origin pointing to the right. In this case, the above construction fails, because we do not obtain a unique intersection point. Nevertheless we can obtain the point (x0 , 0) of the quadratrix on the x-axis as a limit. With a bid of calculus, we get x0 = lim t cot t→0
πt t cos(πt/2) t 2 = lim = lim = t→0 sin(πt/2) t→0 sin(πt/2) 2 π
(38.2)
Since Lindemann’s work, it is known that π is transcendental, hence we conclude that the point of the quadratrix on the x-axis exists as a limit, but is not constructible.
706
Nevertheless, we can construct infinitely many points on the quadratrix. Suppose that the regular polygon with N sides is constructible. Hence the polygon with q = 4N sides is constructible, too. For the integers q ≥ 3 and p , 0, the vertices of the regular q-gon lying in the first quadrant are those on the rays with polar angles θ = πp 2q for p = 1, 2, . . . , q. By dividing the vertical segment from (0, 0) to (1, 0) into q congruent parts, we construct the corresponding horizontal lines. Thus for the time parameters t = qp with p = 1, 2, . . . , q, the horizontal line and the ray have both been constructed. As intersection points, we get on the quadratrix the constructible points
Figure 38.1. Construction of 6 points on the quadratrix by means of a regular 24-gon.
x=
p cos πp 2q
q sin πp 2q p y= q
with p = 1, 2, . . . , q.
(38.3)
Problem 38.1. In the figure on page 706, several points of the quadratrix have been constructed by means of a regular 24-gon. In the first quadrant, one obtains on the quadratrix the points C1 through C6 = B.
707
•
•
Find exact root expressions of the Cartesian coordinates of C2 , C3 , C4 , C6 and mark the points,—in the figure drawn in red,—by their coordinates. (This part can be done with high school knowledge.) Find exact root expression for Cartesian coordinates of C1 , C5 and mark the two points,—in the figure drawn in red,—by their coordinates.
Answer.
k 15◦ 15◦ 30◦ 45◦ 60◦ 75◦ 90◦
◦ cos√k 15 √
◦ sin√k 15 √
2+ 6 4√ 3 √2 2 2 1 √ √2 6− 2 4
6− 2 4 1 √2 2 √2 3 √ √2 6+ 2 4
0
1
cot k 15◦ √ 2+ 3 √ 3 1 √
x=
3
√3 2− 3 0
k 6
y=
2+ 3 6√ 3 3 1 √2 3 √2 10−5 3 6
k 6 1 6 1 3 1 2 2 3 5 6
0
1
◦ cot k 15 √
We do the construction of points of the quadratrix successively using the regular polygons with 8, 16, 32, . . . sides, and obtain a dense set of constructible points on the quadratrix. Proposition 38.1. All points (x, y) of the quadratrix, for which both coordinates are algebraic have rational coordinate y , 0. Indeed, if y , 0 is rational, then x is always algebraic. Proof of Proposition 38.1. We suppose that (x, y) is a point on the quadratrix (38.1) of which both coordinates are algebraic. Since the point ( π2 , 0) is the only point on the quadratrix with y = 0, but its first coordinate is not algebraic, we conclude y , 0 . Hence yx is algebraic. A small calculation yields πt iπt exp iπt πt cos 2 exp iπt + 1 x 2 + exp − 2 = cot = = i =i πt iπt iπt y 2 exp iπt − 1 sin 2 exp 2 − exp − 2
Solving for the exponential yields exp iπt =
x + iy x − iy
(38.4)
which hence turns to be algebraic, too. Hence y = t , 0, and t and eiπt are both algebraic. Distinguish the cases that t is irrational or rational. For irrational parameter t, we invoke the Theorem 36.4 of M. Waldschmidt with z := iπ and t from above. We see that at least one of the numbers t, eiπ , eiπt is transcendental. That is impossible since eiπ = −1 is algebraic, and t and eiπt are both algebraic. Alternatively, we can put a := −1 into the Gelfond-Schneider Theorem 36.3. Since the irrational t and a := −1 , 0, 1 are two algebraic numbers, one concludes that at = eiπt is transcendental. After this contradiction, there is only left the possibility that t , 0 is rational. For these points, w := eiπt is always algebraic since it is a root of the polynomial equation wq − 1 = 0. 1 Proposition 38.2. The only points on the quadratrix (38.1) which are constructible with compass and straightedge have coordinates p cos πp 2q x= q sin πp 2q (38.5) p y= q Gauss has even shown that the polynomial equation wq − 1 = 0 can always be solved by radicals. The proof is really not easy, but contained in a good algebra textbook. 1
708
with either q = 1, 2 and p odd, or q ≥ 3, p , 0. In the latter case the fraction terms and a regular polygon with q sides has to be constructible.
p q
has to be in lowest
Proof of Proposition 38.2. For q = 1 and p odd, one gets x = 0, y = p and for q = 2 and p odd, one gets ±x = y = 2p , which are rational and constructible coordinates. For y = qp in lowest terms with p , 0 and q ≥ 3, the discussion above confirms that the points enumerated are indeed constructible. Conversely, let (x, y) be the coordinates of constructible point of the quadratrix (38.1). Hence both coordinates are a fortiori algebraic. From the Proposition 38.1 above, we know that y = t = p iπt is algebraic. The assumptions imply equation (38.4) above. Hence we q , 0 is rational, and e iπt see that e = (x + iy)/(x − iy) is even constructible. Let the fraction qp be put into lowest terms. Thus the greatest common divisor of p and q equals one. By the extended Euclidean algorithm, there exist integers n and m such that np − mq = 1 . Since eiπt is constructible, we conclude that eiπnt = exp
iπnp iπ(1 + mq) iπ = exp = (−1)m exp q q q
is constructible, too. Hence w = eiπ/q , w2 , w3 , . . . , w2q are all constructible. These are just the coordinates of a regular 2q-gon in complex notation. We conclude that a regular polygon with q sides is constructible. We have obtained by the proposition all constructible points (x, y) on the quadratrix. Remark. To work in a coordinate system, it is more convenient to put the quadratrix into a position different from the the historical one, with x and y from above switched. This has been done inn the figure on page 709 and page 710. Take the points A = (−1, 0), O = (0, 0), B = (1, 0). Beginning −−→ −−→ with the ray OA and the vertical line through point A, and ending with the ray OB and the vertical line through point B, we imagine that with increasing time the vertical line is evenly shifting to the left and a ray turning clockwise. The points were the line and the ray intersect are the points of the quadratrix. One checks that the quadratrix in this position has the equation y = x cot
πx 2
(38.6)
which can conveniently be graphed. Since the function (38.6) is even, in this more convenient position, the y-axis gets the axis of symmetry. Problem 38.2. In the figure on page 709, the quadratrix is used in a setup with a vertical line moving shifting parallel to the right, and a ray turning clockwise. One obtains the equation y = x cot πx 2 ,—convenient to graph the quadratrix. The figure shows the quadratrix, and how it can be used to trisect the angle of 60◦ . Explain the steps of the construction.
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Figure 38.2. The trisection of an angle of 60◦ by means of the quadratrix.
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Figure 38.3. The partition of an arbitrary angle into five congruent parts, done by means of the quadratrix.
Problem 38.3. In the figure on page 710, the quadratrix with equation y = x cot πx 2 is shown once more. It is now used to partition an arbitrary angle into five congruent parts. We assume that the quadratrix is given at the beginning of the construction. Explain the steps how an arbitrary angle is subdivided into five congruent parts. Answer. The angle to be partitioned is put with vertex O. Its two sides intersect the quadratrix at points Q0 and Q5 . From the intersection points we drop the perpendiculars p and q onto the x-axis AB,—where the quadratrix intersects the unit circle. The segment between the foot points P and Q is divided into five congruent parts PP1 , P1 P2 , P2 P3 , P3 P4 , P4 Q. The standard Euclidean construction for this step is shown in the figure on page 710, too. We draw four further parallels to p and q through the partition points P1 , P2 , P3 , P4 and find their intersection points −−−→ −−−→ −−−→ −−−→ Q1 , Q2 , Q3 , Q4 with quadratrix. The rays OQ1 , OQ2 , OQ3 , OQ4 partition the given angle ∠Q0 OQ5 into five congruent parts.
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39. Advanced Euclidean Triangle Geometry 39.1. Morley’s Theorem One of the most surprising theorems about triangles was discovered about 1905 by Frank Morley. Theorem 39.1 (Morley’s Theorem). The points of intersection of the adjacent trisectors of the angles of any triangle are the vertices of an equilateral triangle.
Figure 39.1. Morley’s Theorem states that the small triangle 4XYZ is always equilateral.
After mentioned this discovery to his friends, Morley published the theorem twenty years later in Japan. Meanwhile it was rediscovered and presented as a problem in the Educational Times. Two solutions were sent in. The one by M.T. Naraniengar 1 is quite elegant, using a circle through the vertex A from the original triangle, and the vertices Y and Z from the small equilateral triangle 4XYZ found in Morley’s theorem. A detailed presentation is contained in the book [11] Geometry Revisited by Coxeter and Greitzer. Below is explained another proof, a bid similar to the one from the book [33] Modern Geometries by James R. Smart. I begin with some remarks. Let any triangle 4ABC be given. −−→ −−→ The adjacent trisectors of the angles α and β are the rays AZ and BZ, intersecting at point Z. Let the two other trisectors of the angles α and β intersect at point W. The triangle 4ABZ has the interior angles α/3, β/3 and 180◦ − z, whereas the triangle 4ABW has the interior angles 2α/3, 2β/3 and 180◦ − 2z with α+β γ z= = 60◦ − 3 3 −−→ −−→ For the triangle 4ABW, the rays AZ and BZ are the interior angle bisectors. Hence they intersect in the center of the in-circle,—this is even a fact from neutral geometry as shown by theorem 11.3. −−→ Hence the ray WZ bisects the angle ∠AW B. 1
Mathematical Questions and Their Solutions from the Educational Times (New Series) 15 (1909), p.47
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Proof of Morley’s Theorem. For any angles α, β, γ such that α + β + γ = 180◦ , we construct a triangle 4ABC having these interior angles, and for which the points of intersection of the adjacent trisectors give rise to an equilateral triangle 4XYZ. By similarity, we see that Morley’s theorem holds for any triangle.
Figure 39.2. The proof of Morley’s Theorem begins with the equilateral small triangle 4XYZ.
We begin by drawing any equilateral triangle 4XYZ. Define the angles α x = 60◦ − 3 β ◦ y = 60 − 3 γ ◦ z = 60 − 3 and erect on the side XY an isosceles triangle 4XYW with base angles z, on the side YZ an isosceles triangle 4YZU with base angles x, and on the side ZX an isosceles triangle 4ZXV with base angles y. All three triangles have to lie in the exterior of triangle 4XYZ. Since x < 60◦ , y < 60◦ , z < 60◦ and x + y + z = 120◦ we see that x + y + z + 60◦ = 180◦ . The lines XV and XW intersect with (vertical) angles x, the lines YU and YW with angles y, and the lines ZU and ZV with angles z. The rays VZ and WY intersect at a point A which lies on the same side of YZ as point U. This statement follows since x + z + x + y = 120◦ + x < 180◦ . Moreover we get α ∠Y AZ = 180◦ − (x + z + x + y) = 60◦ − x = 3 The rays WX and UZ intersect at a point B which lies on the same side of ZX as point V. Moreover we get β ∠ZBX = 180◦ − (x + y + z + y) = 60◦ − y = 3 The rays V X and UY intersect at a point C which lies on the same side of XY as point W. Moreover we get γ ∠XCY = 180◦ − (x + z + y + z) = 60◦ − z = 3 It remains to determine the six angles at A, B, C formed by the other rays. Going on as above, one stubbornly gets more unknown angles than equations obtainable from angle sum of triangles. Instead we use the following key lemma.
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−−→ Lemma 39.1. Let the triangle ABW have the interior angle bisector WZ of angle ∠AW B = 180◦ − 2z and suppose that ∠AZB = 180◦ − z and points W and Z lie on the same side of AB. −−→ −−→ Then point Z is the center of the in-circle of triangle ABW, and rays AZ and BZ bisect the angles at A and B, respectively.
Figure 39.3. Point Z is the center of the in-circle for triangle 4ABW.
Proof. Let ZF, ZG and ZH be the perpendiculars drops from point Z onto the sides BW, AW and AB. Let α/3 = ∠ZAW, α0 /3 = ∠ZAB and β/3 = ∠ZBW, β0 /3 = ∠ZBA. We check that α = α0 by showing that both α < α0 and α > α0 lead to a contradiction. Towards a contradiction, suppose that α < α0 . I leave it to the reader to check that ZG < ZH. Since point Z lies on the bisector from W, we know that ZF ZG. Hence ZF < ZH, which in turn implies β < β0 . From the angle sum in triangle 4ABW, we get 2α0 2β0 α + α0 β + β0 + + 180◦ − 2z < + + 180◦ − 2z 3 3 3 3 α0 + β0 z< 3 α0 + β0 + 180◦ − z > 180◦ 3 contradicting the fact that triangle 4ABZ has the angle sum 180◦ , too. Similarly, the assumption α > α0 leads to a contradiction. Hence α = α0 , as to be shown. −−→ −−→ −−→ From the key lemma, we see that the rays AZ and AW trisect the angle α, and the rays BZ and −−→ BW trisect the angle β. Trisection at vertex C can be shown similarly. Thus the proof of Morley’s Theorem is complete. 180◦ =
39.2.
The Euler line
Theorem 39.2 (The Euler line). The orthocenter, the centroid, and the circum-center of a triangle lie on the Euler line. The centroid trisects the segment joining the orthocenter and the circum center. In the exceptional case of an equilateral triangle, all three centers are equal, but the Euler line is not defined.
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Definition 39.1. Let Ma , Mb and Mc denote the midpoints of the three sides of the triangle. The circum-circle of the midpoint triangle 4Ma Mb Mc is called the Feuerbach circle or nine point circle . These are the famous theorems about the Euler line and the Feuerbach circle: Theorem 39.3 (The nine-point circle). The midpoints Ma , Mb , Mc of the sides of a triangle, the foot points Fa , Fb , Fc of its altitudes, and the midpoints Ha , Hb , Hc of the segments joining the orthocenter H (intersection point of the three altitudes) to the three vertices, all lie on the ninepoint circle, also called Feuerbach circle. The center of the Feuerbach circle is the midpoint of the segment joining the orthocenter and the circum-center. The orthocenter H, the center N of the Feuerbach circle, the centroid S , and the circum-center O have the order H, N, S , O on the Euler line. Furthermore, 6 · |NS | = 3 · |S O| = 2 · |HN| = 2 · |NO| = |HO| and (NO, S H) are harmonic points. Main Theorem 31 (Feuerbach’s Theorem). The Feuerbach circle touches the in-circle and all three ex-circles. Problem 39.1. Find out which of these facts are easy to check for a right triangle. 39.3.
Proof of Euler’s Theorem
Problem 39.2. Prove that S = O if and only if the triangle is equilateral. Remark. For an equilateral triangle, it is easy to see that H = O = S . The Euler line of an equilateral triangle is not defined. Problem 39.3. Prove that the Euler line of an isosceles—but not equilateral—triangle is its axis of symmetry. Problem 39.4. Prove Euler’s theorem for a right triangle. Proof. If the angle at vertex C is a right angle, then C = H and O = M, by Thales’ theorem and its converse. The Euler line is the median CM. By the centroid theorem, the centroid S trisects the segment joining the vertex C and the midpoint M of the opposite side. For a right triangle, these two points are just the orthocenter and the circum center. Thus the centroid trisects the segment joining the orthocenter and the circum center— as stated in Euler’s theorem. Proof of Euler’s theorem. In the special case S = O, the triangle is equilateral, and the orthocenter H = O = S —nothing else is claimed. We now discard this special case and assume that S , O. Let E be the point on line S O such that |ES | = 2|S O|, with centroid S lying between E and O. Let M be the midpoint of side AB and F be the foot point of the altitude dropped from vertex C onto side AB. Euler’s theorem follows easily from the Lemma (**). Point E lies on the altitude FC.
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Figure 39.4. Similar triangles 4S OM ∼ 4S EC are used to prove Euler’s theorem.
Because of Claim (**), the point E lies on all three altitudes of the triangle. Hence it is the orthocenter E = H, and the three points S , O and H lie on a line and |HS | = 2|S O|
(39.1)
|HA| = 2|OMa | , |HB| = 2|OMb | , |HC| = 2|OMc |
(39.2)
as to be shown.
Remark. In this way, Claim (**) yields an independent proof that the three altitudes intersect in one point. Generic case: Assume sides AC and BC are not congruent, and the angle at vertex C is not a right angle. Reason for Claim (**) in the generic case. By the centroid theorem, the centroid S divides the median CM at the ratio 2 : 1. By the construction above, the segment EO is divided by S in the same ratio 2 : 1. Hence the two triangles 4S OM ∼ 4S EC are similar by Euclid VI.6. Hence |EC| = 2|OM|
(39.3)
By Euclid VI.5, the two triangles are equiangular, too. (One can independently get these two statements from Double SAS, Proposition 25.4) Via congruent z-angles, one concludes that the two segments OM and EC are parallel. Because OM is perpendicular to the side AB, the parallel segment EC is perpendicular to AB, too. Hence point E lies on the altitude dropped from C onto side AB, as to be shown.
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Reason for the Claim(**) in the exceptional cases. If the angle at vertex C is a right angle, then C = H and O = M, by Thales theorem and its converse. The Euler line is the median CM. By the centroid theorem, the centroid S trisects the segment joining the vertex C = H and the midpoint M = O of the opposite side. Thus the centroid trisects the segment joining the orthocenter and the circum center— as stated in Euler’s theorem. Now assume that the triangle is isosceles with congruent sides AC BC. Clearly M = F. The Euler line OS is the symmetry axis. But this is one altitude, too. Hence point E lies on the one altitude which is the symmetry axis. Too, one can check that statement (39.3) remains true. 39.4.
Proof of the Nine-Point Theorem
Problem 39.5. Because of symmetry, some of the nine points Ma , Mb , Mc midpoints of the sides Fa , Fb , Fc foot points of altitudes, and Ha , Hb , Hc midpoints of the segments joining the orthocenter H to the vertices can become equal. How many points of them are left in the case of (a) a generic right triangle with γ = R > α > β. (b) an isosceles acute triangle. (c) an isosceles right triangle. (d) an isosceles obtuse triangle. (e) an equilateral triangle.
Proof of the nine-point theorem. The notation and set-up from the proof Euler’s theorem is used, once more. Let N be the midpoint of segment OH on the Euler line. Let M be the midpoint of side AB and F be the foot point of the altitude dropped from vertex C onto side AB. As a new feature, let Hc be the midpoint of segment HC. The nine-point theorem follows easily from the Lemma (*). The three points M, F and Hc lie on a circle N around N. Segment MHc is a diameter of this circle. The radius of circle N is half of the radius of the circum-circle of triangle 4ABC. Because of Claim (*), the circle N turns out to be the same, no matter whether its definition involves vertex A or B or C. Hence it is the nine-point circle. Reason for Claim (*) in the generic case. Again, the main step is the proof for the generic case, which is as follows: Generic case: Assume sides AC and BC are not congruent, and the angle at vertex C is not a right angle.
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Figure 39.5. Congruent triangles 4NOM and 4NHHc are used to get the nine-point circle.
From Euler’s theorem and its proof, we know |HS | = 2|S O| |HC| = 2|OM|
(39.4) (39.5)
The orthocenter H is the intersection point of lines S O and CF. In the generic case, OM k CF are two different parallel lines. They are both perpendicular to AB. Recall that Hc is the midpoint of segment HC. The two triangles 4NOM 4NHHc are congruent by SAS congruence. Indeed ∠NOM ∠NHHc OM HHc NO NH
because these are z-angles for the parallel lines OM k HHc = CF follows by (39.5) and the definition of Hc via 2OM HC 2HHc because N is the midpoint of segment OH
Hence ∠ON M ∠HNHc . Furthermore, these are vertical angles, since N is the midpoint of segment OH, and points C and Hc lie on the opposite side of the Euler line OH than point M. Point N is the midpoint of segment MHc , since N M NHc . The angle ∠MFHc is a right angle. Hence the converse Thales’ theorem shows that the three points M, F and Hc lie on a circle, as claimed. Finally, we check the radius of this circle. Because of result (39.5) and definition of Hc , we get 2OM HC 2CHc . Thus MO and CHc are two parallel congruent segments, and the quadrilateral MOCHc is a parallelogram. Hence OC MHc = 2N M. This confirms that the radius N M of circle N is half of the radius OC of the circum-circle of triangle 4ABC. Reason for Claim (*) in the exceptional cases. If the angle at vertex C is a right angle, then C = Hc = H and O = M. The converse Thales’ theorem shows that the three points M, F and C lie on a circle with diameter MC, which is the radius of the circum-circle.
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Now assume that the triangle is isosceles with congruent sides AC BC. Clearly M = F. We check that N is the midpoint of segment MHc . One checks that 2|OHc | = |O2 H| = 2|HM| and hence OHc HM and NHc N M. 39.5.
Proof of Feuerbach’s Theorem
Proposition 39.1 (Angle between the Feuerbach circle and the sides of the triangle). The Feuerbach circle cuts the triangle side AB at point Mc at angle β − α. Similarly, it cuts the triangle side BC at point Ma at angle γ − β, and the triangle side CA at point Mb at angle α − γ.
Figure 39.6. The angle between the Feuerbach circle triangle side AB is β − α.
Reason. Let t be the the tangent to the Feuerbach circle at point Mc , and let point T be on t on the side of AB opposite to C. Here is the case α < β, as in the drawing. ∠T Mc Fc ∠T Mc Ma − ∠Fc Mc Ma ∠Mc Mb Ma − ∠BMc Ma because, by Euclid III.32, the circumference angle (here of arc Mc Ma ) is congruent to the angle between the chord and the tangent at its endpoint. ∠Ma BMc − ∠BAC
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because opposite angles in a parallelogram are congruent, and because z-angles between parallels are congruent. Finally ∠T Mc Fc ∠CBA − ∠BAC = β − α The case α > β is almost similar, and left to the reader.
Let I be the center of the inscribed circle or in-circle I, and Ic be the center of the ex-circle Ic touching side AB. Let Da , Db and Dc be the points where the in-circle touches triangle sides a, b and c, respectively. Let Ea and Eb be the points on the extensions of triangle sides a and b where the ex-circle Ic touches the extended sides, and let Ec be the point where ex-circle Ic touches the side AB. Considering segment AB the base of the triangle 4ABC, I write D = Dc , E = Ec and M = Mc for simplicity.
Figure 39.7. The touching points of the in-circle and one ex-circle.
Problem 39.6. Calculate the positions of the touching points for the in-circle, and the ex-circle Ic in terms of the triangle sides a, b, c and the half perimeter s := a+b+c 2 . Show that M is the midpoint of segment DE. Answer. The segments on the two tangents to a circle from a point outside to the touching points are congruent. Hence x = |ADb | = |ADc |, y = |BDc | = |BDa |, and z = |CDa | = |CDb |. The three lengths x, y, z satisfy the equations x+y = c y+z = a x +z = b
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which implies x + y + z = s, where
a+b+c 2 is defined to be the half perimeter. Hence x = s − a, y = s − b, z = s − c. Because of |AE| = |AEb |, |BE| = |BEa | and a + |BEa | = b + |AEb | and |BE| + |AE| = c s :=
one concludes |AE| = y = s − b and |BE| = x = s − a. Thus we have calculated the positions of the touching points for the in-circle, and the ex-circle. Together, we got |AD| = |BE| = s − a. Hence M is the midpoint of DE as claimed. Continuing the last problem, we draw the forth common tangent tc = D0 E 0 of in-circle I and the same ex-circle Ic . Note that the other three common tangents are just the sides of 4ABC and their extensions. Let G be the intersection point of the two "inner" common tangents tc and AB between the two circles. Point G is the intersection point of the triangle side AB with the segment IIc . Points D0 and E 0 are the points where tc touches the in-circle and the ex-circle Ic .
Figure 39.8. The angle between the two inner common tangents of in-circle and ex-circle is β − α.
Problem 39.7. Calculate the angle ∠D0GB between the two inner common tangents of in-circle and ex-circle. − → −−→ Answer. The ray CI = CIc is the inner angular bisector at vertex C. Euclid I.32 for an exterior angle of the triangles 4AGC and 4CGB, yields γ ∠CGB = + α 2 γ ∠CGA = + β 2
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Line GC bisects the angles between the two inner tangents, hence ∠CGD0 ∠CGA and ∠CGD0 ∠CGA =
γ +β 2
Now angle subtraction at vertex G implies ∠D0GB = ∠CGB − ∠CGD0 = α − β as to be shown.
Figure 39.9. The common tangent of the in-circle and one ex-circle is parallel to the tangent to the Feuerbach circle.
Proposition 39.2. The fourth tangent tc and the tangent t to the Feuerbach circle at midpoint Mc are parallel.
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Definition 39.2. Four points X, Y, U, V are called harmonic if the two points U and V divide segment XY from inside and outside in the same ratio. In other words, the cross ratio (XY, UV) =
XU · YV YU · XV
= −1
for the directed segments.
Figure 39.10. The points on Apollonius circle have all same ratio of distances from C and D.
On an angular bisector from vertex C, we mark the intersection point G with the opposite side, and the centers I of in-circle and ex-circle Ic . Theorem 39.4. The circle with diameter IIc is indeed an Apollonius circle for segment CG. The points on this circle have all same ratio of distances |CA| |CI| |CIc | = = |GA| |GI| |GIc | from C and G. Especially, vertex C, the intersection point G of the angular bisector with the opposite side, and the centers I of in-circle and ex-circle Ic , are four harmonic points: (CG, IIc ) =
CI · GI GIc · CIc
= −1
Main Theorem 32 (Feuerbach’s Theorem). The Feuerbach circle touches the in-circle and all three ex-circles. Reason. It is enough to show that the Feuerbach circle touches the in-circle I and the ex-circle Ic . We use inversion by the circle δ with center Mc and diameter DE. Question. Calculate the diameter of circle δ, assuming a > b.
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Figure 39.11. Inversion by circle δ maps I 7→ I ,
Ic 7→ Ic ,
AB 7→ AB ,
tc 7→ φ.
Answer. In a problem above, we have shown |AD| = |BE| = s−a. Hence |DE| = |AB|−|AD|−|BE| = c − 2(s − a) = c + 2a − (a + b + c) = a − b. The circular inversion by δ maps the points Mc 7→ ∞ , D 7→ D , E 7→ E , G 7→ F ,
(39.6)
since the points of δ are mapped to themselves, and because (DE, FG) are harmonic points. Question. Why are the four points (FG, DE) harmonic? Answer. The four points F, G, D, E are the foot points of the perpendiculars onto line AB from points C, G, I, Ic . Hence it is enough to show that these are harmonic points. But this is just evident − → −−→ from Apollonius’ theorem, because BI and BIc are two angular bisectors of ∠CBG. The inversions maps the generalized circles I 7→ I ,
Ic 7→ Ic ,
AB 7→ AB ,
tc 7→ φ
Indeed, the in-circle I and the ex-circle Ic are mapped to themselves, because they are orthogonal to the inversion circle δ. The extended triangle side AB is mapped to itself, because the center of inversion Mc lies on the line AB. Indeed any line or circle is mapped to a line or circle. The second inner tangent tc is mapped to a circle which I have defined to be φ. Question. Why does the circle φ touch the in-circle I and the ex-circle Ic ? Answer. Because of conservation of angles, touching generalized circles are mapped to touching generalized circles. The inner tangent tc touches both the in-circle and the ex-circle Ic . Applying circular inversion by δ, we conclude that the corresponding images touch, too. Hence the circle φ touches the images of in-circle and ex-circle, which are again the in-circle I and the ex-circle Ic . Question. The circle φ is defined as the inversion image of tc . Why is φ the Feuerbach circle?
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Answer. I show that circle φ and the Feuerbach circle have points F and Mc , and the tangent at point Mc in common. This implies that the two circles are identical. Because tc is a line, the inversion image φ goes through the center Mc of the inversion circle δ, which is the inversion of the point ∞. Clearly, the two intersection points of tc with δ lie on φ, because they are fixed under inversion. Hence the tangent to φ at point Mc is parallel to tc . By a result above, the two lines tc and t are parallel, too. Hence circle φ and the Feuerbach circle have the same tangent t at point Mc . Furthermore both the Feuerbach circle and circle φ have the further point F in common. Indeed, the Feuerbach circle goes through point F, because it is the foot point of the altitude dropped from vertex C. Circle φ goes through point F, because line tc goes through point G, and incidence is conserved by inversion. 39.6.
Additional questions
Problem 39.8. Discuss the location of the Euler line and the Feuerbach circle with its thirteen remarkable points in the case of (a) a generic acute triangle with γ > α > β. (b) a generic right triangle with γ = R > α > β. (c) a generic obtuse triangle with γ > R > α > β. (d) an isosceles acute triangle. (e) an isosceles right triangle. (f) an isosceles obtuse triangle. (g) an equilateral triangle. Given the Euler line and the Feuerbach circle, one can still not know any of the angles of a triangle. That is clear, because any triangle can be mapped by a composition of translation, rotation and dilation into one having the prescribed Euler line and Feuerbach circle. Problem 39.9 (Construction problem). Construct a triangle from given Euler line, Feuerbach −−→ circle, orthocenter H, and direction of the ray HC. Under which restrictions does the problem has a solution? Is the solution unique? Answer. If the orthocenter lies inside the Feuerbach circle and H , N, there always exists a unique −−→ solution, once the direction of ray HC is specified up to an entire 360◦ turn. The solution is an acute triangle. In the special case H = N, one gets as solutions an entire set of rotated equilateral triangles. If the orthocenter lies outside the Feuerbach circle, a solution exists only under the following restrictions: The distance |NH| needs to be less than the diameter of the Feuerbach circle, and the −−→ ray HC needs to lie between the two tangents from orthocenter H to the Feuerbach circle. Under this restriction, the solution is a unique obtuse triangle. In the special case that H lies on the Feuerbach circle, one gets as solution a unique right triangle. Question. Explain the steps needed for the construction of the triangle 4ABC, and the nine points on the Feuerbach circle.
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−−→ Answer. The intersection of ray HC with the Feuerbach circle yields point Hc , the midpoint of segment HC. One draws the circle with center Hc through H, and gets point C on the given ray. The foot points of the two other altitude Fa and Fb are the intersection points with the Feuerbach circle. Next, one draws the circle with diameter OC, and gets the midpoints of sides Ma and Mb as intersection points with the Feuerbach circle. The second intersection of line HC with the Feuerbach circle yields the foot point Fc , and the line AB as perpendicular to the altitude hc at point Fc . Now one can use the circum-center O and use the circum circle, since the center N of the Feuerbach circle is know to be the midpoint of segment HO. One gets the vertices A and B as intersection points of line AB with the circum-circle. The four remaining triangle sides and altitudes can be drawn with some redundancy, which is good for accuracy. Indeed, on the triangle side a, one has the four points C, Fa , Ma and B. On the triangle side b, one has the four points C, Fb , Mb and A. On the altitude ha , one has three points A, H, Fa , and still gets Ha . On the altitude hb , one has three points B, H, Fb , and still gets Hb . Hence one has constructed the triangle 4ABC and the nine points on the Feuerbach circle. Problem 39.10. Do the construction for several examples (a) the orthocenter inside the Feuerbach circle, no vertex on the Euler line. (b) the orthocenter inside the Feuerbach circle, and vertex C on the Euler line. (c) the orthocenter outside the Feuerbach circle, but no vertex on the Euler line. (d) the orthocenter outside the Feuerbach circle, and vertex C on the Euler line. (e) the orthocenter on the Feuerbach circle. To what type of triangle do these examples lead? Answer. case (a) The orthocenter H inside the Feuerbach circle, but no vertex on the Euler line— leads to an acute triangle. case (c) the orthocenter H inside the Feuerbach circle, and vertex C on the Euler line— leads to an acute isosceles triangle. case (c) the orthocenter H outside the Feuerbach circle, but no vertex on the Euler line— leads to an obtuse triangle. case (d) the orthocenter H outside the Feuerbach circle, and vertex C on the Euler line— leads to an obtuse isosceles triangle. case (e) the orthocenter on the Feuerbach circle—leads to a right triangle. Problem 39.11. Give reasons for the following conjectures: Conjecture 1: The Euler line goes a vertex of a triangle if and only if it is either isosceles or right.
Conjecture 2: The Euler line of an acute triangle cuts the shortest and the longest sides.
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Conjecture 3: The Euler line of an obtuse triangle cuts the two longest sides, but not the shortest side.
Conjecture 4: For an obtuse triangle with γ obtuse, the two triangles 4ABC and 4ABH have the same Feuerbach circle, but different Euler lines.
Conjecture 5: All nine special points on the Feuerbach circle are different, except in the cases of an isosceles, or a right triangle, or an obtuse triangle for which the triangle with orthocenter as one vertex, and two given vertices kept, is isosceles. (With angle γ obtuse in 4ABC, the triangle 4ABH has the same Feuerbach circle, and can be isosceles.) 39.7.
The Simson line
Proposition 39.3 (The Simson Line, first proved by W. Wallace, 1799). The three foot points of the perpendiculars dropped onto the three sides of a triangle lie on a line if and only if the point lies on the circum circle of the triangle. Proof. Let P be any point on the circum circle of 4ABC. Let E, F and G be the foot points of the perpendiculars dropped from point P onto the lines AB, AC, and BC, respectively. We claim that (a) The four points A, P, E and F lie on a circle with diameter AP. (b) The four points B, P, E and G lie on a circle U with diameter BP. (c) The four points C, P, F and G lie on a circle V with diameter CP. Indeed this follows immediately, from the converse Thales theorem 39. To get the situation as draw in figure 39.7, we now assume that point P lies on the arc between B and C on which point A does not lie. From Euclid III.21, we can now conclude (a) The three angles α1 = ∠CAP ∠CBP ∠GEP are congruent. (b) The three angles α2 = ∠BAP ∠BCP ∠GFP are congruent. The angles α = ∠CAB and ∠CPB = 180◦ − α1 − α2 add up to two right angles, by Euclid III.22. The quadrilateral figure EPFG has at its vertices E, P, F the angles α1 , 180◦ − α1 − α2 and α2 —which add up to 180◦ . Hence the quadrilateral is degenerate with the three points E, F and G on one line. If point P lies outside the circum circle as shown in figure 39.7, then ∠CPB < 180◦ − α1 − α2 , and one gets a convex quadrilateral EPFG. If point P lies inside the circum circle, then ∠CPB > 180◦ − α1 − α2 ,and one gets a non convex quadrilateral EPFG. Problem 39.12. Assume that the point P is the intersection of the circum circle with the perpendicular bisector of side BC. Prove the following
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Figure 39.12. By Euclid III.21, one gets congruent angles α1 at vertices A, B, E and congruent angles α2 at vertices A, C, F.
(a) Point P lies on the bisector of the angle α = ∠BAC. (b) One of the two points E and F lies on a side of 4ABC, and the other one of an extension of a second side. (c) The triangle 4EPF is isosceles. (d) The angle between the triangle side a = BGC and its Simson line EGF is x := ∠BGE =
γ−β 2 .
Proof. Point P lies on the circum circle C of 4ABC, and hence outside of 4ABC. The center O of the circum circle is the intersection point of the perpendicular bisectors of all three sides of the triangle. Especially, it lies on the perpendicular bisector m of side BC. Hence all three points O, G and P lie on m. Furthermore G is the midpoint of side BC, and line m is a diameter of the circum circle. Hence the two central angles ∠BOP and ∠COP are congruent. Hence the two corresponding circumference angles ∠BAP and ∠CAP, are congruent, too. This implies that line AP is the angular bisector of ∠BAC: α ∠BAP ∠PAC (*) 2 Thus we have proved item (a). Part (b) and (c) are left to the reader.
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Figure 39.13. The Simson line is broken, if point P does not lie on the circum circle.
Finally, we get the angle between and the triangle side a and its Simson line, claimed in part (d). We use Euclid III.22 for the quadrilateral CFGP. Recall that it has a circum circle V, because of the two right angles ∠CFP ∠CGP = 90◦ . Hence the two angles at the two opposite vertices at C and G add up to two right angles. Hence α2 + β + y + 90◦ = 180◦ . We solve for angle y = ∠CGF and use the angle sum α + β + γ = 180◦ to conclude y = 90◦ − as claimed in part (d).
α γ−β −β= 2 2
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Figure 39.14. The Simson line is broken again.
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Figure 39.15. The Simson line for a special case—one gets five congruent angles
α 2
at vertices A, B, C.F, E.
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Part III
Hyperbolic Geometry "... I bought a little blue cap, such as all the men and boys of the people wear down there, almost like a fez—the boina. I shall wear it in the rest-cure, and other places, perhaps. Monsieur shall judge if it becomes me." "What monsieur?" "Sitting here in this chair." "Not Mynheer Peeperkorn?" "He has already pronounced judgment—he says I look charming in it." "He said that—all of it? Did he really finish the sentence, so it could be understood?" "Ah! It seems Monsieur is out of temper? Monsieur would be spiteful, cutting? He would laugh at people who are much greater and better, and—more hu—man than himself and his—his ami bavard de la Méditerranée, son maître et grand parleur—put together. But I cannot listen—" Thomas Mann The Magic Mountain
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40. Hyperbolic Geometry in the Poincaré Model 40.1.
Inversion
Definition 40.1 (The inverse point). Given is a circle ∂D of radius R. For any given point P, −−→ we define the inverse point P0 with respect to this circle to be the point on the ray OP such that 0 2 | OP| · | OP | = R . A point on ∂D is its own inverse. For the center O itself, the inverse point is the point at infinity. Problem 40.1. Do a example for the construction of the inverse point.
Figure 40.1. Construction of the inverse point
Construction 40.1 (The inverse point). To construct the inverse of a given point P, one uses the the theorems related to the Pythagorean Theorem. One erects the perpendicular h to radius OC at point P. This step is different from the elliptic case! Let C an intersection of the perpendicular with ∂D. Next one erects the perpendicular on PC at point C, and gets a tangent to circle ∂D. The inverse point P0 is the intersection at that tangent with the ray OP. Indeed, by the leg theorem, | OP| · | OP0 | = |OC|2 = R2 . Remark. Alternatively, you can use the Thales’ circle with diameter CP0 . By converse Thales’ theorem, P lies on that circle. Then use the chord theorem Euclid III.36 and get again, same as by leg theorem: | OP| · | OP0 | = |OC|2 = R2 . 40.2. Points and lines in the Poincaré model We explain now the Poincaré disk model of hyperbolic geometry. The reader should recall the basic idea of a model in mathematics, as explained in the passage General remark about models in mathematics. The Poincaré disk model of hyperbolic geometry is built in the Euclidean plane. The Euclidean geometry of the plane is the accepted ambient underlying reality, which one can call the "background ontology". Some cleverly chosen objects and relations from Euclidean geometry are now interpreted as objects and relations of hyperbolic geometry. Hence we have in the Poincaré’s disk model the ambient Euclidean plane as the accepted basic reality. In this plane one builds the hyperbolic geometry as a new secondary level. In order to stress this new situation and distinguish the two levels, I use quotation marks for the notions of hyperbolic geometry.
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Definition 40.2 (Poincaré’s disk model). We denote the open unit disk by D = { (x, y) : x2 + y2 < 1 } The center of D is denoted by O. Its boundary is ∂D = { (x, y) : x2 + y2 = 1 } The circle ∂D is also called the line at infinity. • The points of D are the "points" for Poincaré’s model. • The points of ∂D are called "ideal points" or "endpoints". The ideal points are not points for the hyperbolic geometry. Once the hyperbolic distance is introduced, they turn out to be infinitely far away. • The "lines" for Poincaré’s model are circular arcs perpendicular to ∂D, open at their ideal ends. Problem 40.2. Construct the hyperbolic line through two given points, in the Poincaré disk model. The construction uses the inverse point and the polar elements, as now explained. Lemma 40.1. If a circle passes through a pair of inverse points, it consists entirely of pairs of inverse points, and it intersects ∂D perpendicularly. A circle is a hyperbolic line if and only if it passes through a pair of inverse points P and P0 .
Figure 40.2. A orthogonal circle consists of pairs inverted points
Proof of the Lemma. Assume that a circle C passes through a pair of inverse points, P and P0 . Let −−−−→ −−→ Q be a third point on the circle. We draw the rays OPP0 and OQ. Let Q2 be the second intersection −−→ point of the ray OQ with circle C. Because of the theorem of chords (Euclid III.35), we know that | OP| · | OP0 | = | OQ| · | OQ2 |
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By definition of inverse points | OP| · | OP0 | = 1 and hence | OQ| · | OQ2 | = 1 The point O lies outside the circle C, and hence Q and Q2 lie on the same side of O. This confirms that the second intersection point Q2 = Q0 is the inverse point of Q. Why do the circles C and ∂D intersect perpendicularly? The center O lies outside of C, because circle C contains a pair of inverse points, We construct a tangent from point O to circle C (Just one of the two tangents suffices.) Let T be the point at which the tangent touches circle C. By construction OT ⊥ T l⊥ . We apply the theorem of chords (Euclid III.36) for circle C. Thus we get | OP| · | OP0 | = | OT |2 By definition of inverse points | OP| · | OP0 | = 1 and hence | OT |2 = 1 This shows that the touching point T lies on ∂D, as well as on C. Since OT ⊥ T l⊥ this implies that T l⊥ is a tangent to circle ∂D. By construction, OT is a tangent to circle l. Hence we have got two perpendicular tangents to the two circles l and ∂D at their intersection point T . This reasoning has shown that a circle passing through a pair of inverse points is a hyperbolic line. The converse can be checked, too. We assume l is a hyperbolic line, and show that l contains a pair of inverse points. Let T be an intersection point of the two circles ∂D and l. Hence |OT | 2 = 1 By definition of hyperbolic line, the two circles l and ∂D intersect perpendicularly, Hence the segment OT is a radius of circle ∂D as well as a tangent to circle l. The tangent from center O to the arc l touches circle l at point T . Let P be any point of l, and let P” be the second intersection −−→ point of ray OP with the circle l. By Euclid’s theorem of chords |OT | 2 = |OP| · |OP”| But by the definition of inverse points |OP| · |OP0 | = 1 The three equations imply |OP0 | = |OP”|. Hence P0 = P”, and the inverse point P0 lies on circle l, too. Hence circle l goes through the pair of inverse points P and P0 . Definition 40.3 (The polar of point and line). The perpendicular bisector of P and P0 is called the polar of point P. It is denoted by P⊥ or Pperp . The polar l⊥ of a hyperbolic line l is the (Euclidean) center of the circular arc l. The polar elements—P⊥ for a point P, and l⊥ for a line l—always lie outside the disk D. [Sometimes, we denote the midpoint of the segment PP0 by K 0 . See Klein’s model for the reason.] The definitions of the polar for points and lines are consistent with incidence: Fact. A point P lies on a hyperbolic line l if and only if the polar P⊥ goes through the polar l⊥ .
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Reason. Assume point P lies on the hyperbolic line l. Since l is a hyperbolic line, the arcs l and ∂D intersect perpendicularly. By the Lemma, this implies the arc l goes through both point P and the inverse point P0 . Hence the center l⊥ of the arc l lies on the perpendicular bisector of P and P0 , which we defined to be the polar P⊥ . Conversely, if the polar P⊥ goes through l⊥ , the center l⊥ of the arc l has equal distances from P and P0 . Hence the pair of inverse points P and P0 lie on l. By the Lemma, l is a hyperbolic line. Construction 40.2 (Construct of a hyperbolic line through two given points). Given are two points P and Q. To get the hyperbolic line l through P and Q, one constructs the polar P⊥ and Q⊥ . Their intersection point is the polar l⊥ of the line through P and Q. Question. What happens in the special case that P⊥ and Q⊥ are parallel? How does one get the hyperbolic line PQ in that special case? Answer. The radius OP is perpendicular to P⊥ , and, similarly, the radius OQ is perpendicular to Q⊥ . The lines P⊥ and Q⊥ are parallel if and only if the two radii OP and OQ are parallel. This happens if and only if the three points P, Q and the center of the disk O lie on a Euclidean line . In the special case that P⊥ and Q⊥ are parallel, the Euclidean line through the three points P, Q and the center of the disk O is the hyperbolic line through P and Q, too. Indeed , this line is a diameter of the disk D and hence perpendicular to ∂D. Problem 40.3. Do the construction of a hyperbolic line through two given points, for an example.
Figure 40.3. Construction of a hyperbolic line
Problem 40.4. Complete the following sentences, referring to the drawing below.
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Figure 40.4. Recognize inverse and polar elements.
(a) The point D has the inverse image point A. (b) The Euclidean line l is the polar of point A. (c) The hyperbolic line α has the polar C. (d) Point O is mapped to B via a reflection by the hyperbolic line α. (e) Point B has as polar the Euclidean line B⊥ , that is not drawn. (f) Since B lies on β, the polar B⊥ goes through the point β⊥ . (g) Since A lies on α, the polar A⊥ goes through the point α⊥ . (h) The polar of point A is line l. (g) The polar of line α is point C. (h) Indeed, line l goes through the point C. 40.3.
Introduction of metric properties
Definition 40.4. The "angles" for Poincaré’s model are the usual Euclidean angles between tangents to the circular arcs. For the definition of a hyperbolic distance, one needs the cross ratio. The cross ratio of four point A, B, P, Q is defined as |AP| · |BQ| (AB, PQ) = |BP| · |AQ| Remark. Remember A B P→Q ↓ ↑ B A P→Q
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Let A, B be any two points. We denote the hyperbolic line through A and B by l and the ideal endpoints of this line by P and Q. We name those endpoints such that P ∗ B ∗ A ∗ Q. Definition 40.5. The hyperbolic "distance" is defined by s(A, B) = ln(AB, PQ)
(40.1)
Two segments AB and XY are called "congruent" iff s(A, B) = s(X, Y). Remark. I denote the usual Euclidean distance of A and B by |AB|, or AB in case a plus or minus value can be meaningfully assigned. But the hyperbolic distance is denoted by s(A, B). Proposition 40.1 (Additivity of the distance). Let A, B, C be three points on a hyperbolic line, and assume that B lies between A and C. Then s(A, B) + s(B, C) = s(A, C) Reason. Again let P and Q be the ideal endpoints of the line through A, B, C. By definition of the hyperbolic distance |AP| · |BQ| s(A, B) = ln(AB, PQ) = ln |BP| · |AQ| |BP| · |CQ| s(B, C) = ln(BC, PQ) = ln (40.2) |CP| · |BQ| |AP| · |CQ| s(A, C) = ln(AC, PQ) = ln |CP| · |AQ| Hence s(A, B) + s(B, C) = ln(AB, PQ) + ln(BC, PQ) = ln [(AB, PQ) · (BC, PQ)] |AP| · |BQ| |BP| · |CQ| |AP| · |CQ| = ln · = ln (40.3) |BP| · |AQ| |CP| · |BQ| |CP| · |AQ| = ln(AC, PQ) = s(A, C)
as to be shown
Proposition 40.2 (Distance from the center). The hyperbolic distance of a point A form the center O is s(O, A) = 2 tanh−1 |OA| (40.4) Proof. One can take for A a point with coordinates (a, 0), and B has coordinates (0, 0). The ideal endpoints of the horizontal diameter of D are P = (−1, 0) and Q = (1, 0). Hence the cross ratio is (AO, PQ) =
|AP| · |OQ| (1 + a) · 1 = |OP| · |AQ| 1 · (1 − a)
and the hyperbolic distance is s(A, O) = ln(AO, PQ) = ln
(1 + a) · 1 = 2 tanh−1 a = 2 tanh−1 |OA| 1 · (1 − a)
Question. What happens if point A is very near to O? What happens if point A approaches the boundary ∂D? Answer. If point A is near to the center O, the tangent approximation gives s(O, A) ' 2 |OA|. If point A approaches the boundary ∂D, we get |OA| → 1 and hence s(O, A) → ∞. This confirms the Poincaré disk is a model for the unbounded hyperbolic plane.
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40.4. The angle of parallelism This is indeed a remarkable feature of hyperbolic geometry! Here are the basic definitions and facts about the angle of parallelism. Given a line l and a point P not on l, one wants to know what are the parallels to a through P. The asymptotic (or limiting) parallel rays r+ , r− from vertex P are the two rays that do not intersect line l, but all rays in the interior of the angle ∠(r+ , r− ) do intersect the line l. We drop the perpendicular from P onto line l. Let F be the foot point of that perpendicular and s = s(P, F) be its hyperbolic length. Definition 40.6 (Angle of parallelism). The angle of parallelism is the angle between either of the asymptotic parallel rays and the perpendicular from P onto line l. The angle of parallelism depends only on the hyperbolic distance s. Following Lobachevskij, one defines a special function, called π(s), giving the angle of parallelism π for a segment of hyperbolic length s. The function π(s) is explicitly given by a remarkable formula. Proposition 40.3 (Lobachevskij’s formula for the angle of parallelism). For any point P and line l, the angle of parallelism π(s) relates the hyperbolic distance s = s(P, F) from P to the foot point F of the perpendicular dropped on the line l. Indeed, by the formula tan
π(s) = e−s 2
(40.5)
Figure 40.5. How to calculate the angle of parallelism.
Remark. In case one does not assume the axiom of completeness, the quantities on both sides of claim (40.5) are still well defined in the ordered field for segment lengths. Only solving for s requires the logarithmic function ln to be defined—to which end one needs the axiom of completeness.
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Proof. Let X+ = (1, 0), X− = (−1, 0) and Y+ = (0, 1), Y− = (0, −1) be the ideal endpoints of the horizontal and vertical diameter of D. The hyperbolic distance s = s(O, P) is defined to be s = s(O, P) = ln(OP, Y+ Y− )
(40.6)
in terms of the cross ratio (OP, Y+ Y− ). By definition, this cross ratio is (OP, Y+ Y− ) =
|OY+ | · |PY− | |PY− | 1 + |OP| = = |PY+ | · |OY− | |PY+ | 1 − |OP|
(40.7)
Formulas (40.6) and (40.7) allow to calculate the hyperbolic distance s = s(O, P) from the Euclidean distance y = |OP|. Indeed 1+y es = (40.8) 1−y Let R be the intersection point of the horizontal radius X− O with the Euclidean tangent at point P to the asymptotic ray r− . The two tangents from point R to the circle r− form the isosceles 4X− RP. Let its congruent base angles be β = ∠RX− P ∠RPX− . The angle sum in the right 4X− OP implies that β + (β + π(s)) + 90◦ = 180◦ and hence π(s) = 45◦ − β 2 From the definition of the tangent function from the right 4X− PO, one gets tan β = |OP| = y
(40.9)
(40.10)
One now deducts the final claim from (4)(5) and (6). There are several variants to do that, using more trigonometry, or more geometry. Here is a version relying on trigonometry. We use the addition theorem of tangent and get tan
π(s) tan 45◦ − tan β 1 − tan β = tan(45◦ − β) = = ◦ 2 1 + tan 45 · tan β 1 + tan β
Because of (40.10) and (40.8) we conclude tan as to be shown.
π(s) 1 − y = = e−s 2 1+y
Problem 40.5. Given an angle α, state a construction, in the Poincaré model, (and using the underlying Euclidean geometry where needed), to find point P on the segment OY for which π(OP) = α is the angle α as given. Actually do an example for this construction! Again, this can be done by putting point P and line l in a suitable special position. We choose the line to be the horizontal diameter of D. Its ideals endpoints are denoted by X+ and X− . We choose point P to lie on the positive y-axis. (Indeed, any point and line can be mapped to that special position by a composition of two or three hyperbolic reflections.) In this arrangement, the foot point of the perpendicular from P onto l is the center O. Let Y+ and Y− be the ideal endpoints of the perpendicular. The asymptotic parallel rays for point P and line X+ X− are the hyperbolic −−−→ −−−→ rays r+ = PX+ and r− = PX− . Construction 40.3 (Construction of a segment from given angle of parallelism). Let Q be the ideal endpoint in the positive quadrant such that ∠Y+ OQ = α. Next draw chord X− Q and let point P its intersection point with with OY+ . We claim that the segment OP has the angle of parallelism α.
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Figure 40.6. For given line l and angle of parallelism α = 40◦ , point P is constructed.
Reason for validity of the construction. The angle sum in the isosceles 4X 0 OQ implies that 2β + (90◦ + α) = 180◦ . By comparison with formula (5) above this implies α = π(OP), as claimed. Remark. Too, we get a more geometric proof of Lobachevskij’s formula (40.5), avoiding the addition theorem for the tangent function. Repeatedly, we shall use Euclid III.20, and Apollonius’ Lemma 40.2 about the angular bisector. Lemma 40.2. The angular bisector of any triangle cuts the opposite side in the ratio of the lengths of the two adjacent sides. Proposition 40.4 (Euclid III.20). The angle at the the center of a circle is twice the angle with its vertex at a point of the circumference, and subtending the same arc. −−→ Proof. By Thales’ theorem, 4Y− Y+ Q is a right triangle. Ray QP is an angular bisector of the right ◦ angle. Indeed, by Euclid III.20, ∠X− QY− = 45 , because arc X− Y− has the central angle of 90◦ . To calculate the ratio in (40.5), we now use the Lemma 40.2 about the angular bisector. Hence |PY+ | |QY+ | = |PY− | |QY− |
(40.11)
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Figure 40.7. A more geometric proof of Lobachevskij’s formula for angle of parallelism.
By Euclid III.20, ∠Y+ Y− Q = π(s) 2 , because arc Y+ Q has the central angle of π(s) by construction. Hence the definition of the tangent function implies tan
π(s) |QY+ | |PY+ | 1 − y = = = 2 |QY− | |PY− | 1 + y
Now we use (40.8) from item 3, and (40.11) above to get Lobachevskij’s formula (40.5) once again. 40.5.
Hyperbolic reflection
Proposition 40.5. The reflection by a hyperbolic line is the same mapping as the inversion by that circle—for the underlying Euclidean plane. Proof. Let l be any hyperbolic line and A a point not lying on l. By Il we denote the inversion by circle l, and the inverted images are denote by subscript l. Note that we study a different inversion here, not the inversion P 7→ P0 by circle ∂D! By definition of inversion, the inverted image Al of A satisfies |l⊥ A| · |l⊥ Al | = rl2 (40.12) where rl is the radius of the circular arc l. To need to show that Al can be obtained independently by hyperbolic reflection. To this end, we draw the hyperbolic line p through points A and Al and let S be the intersection of l and p. By definition of reflection, we need to check that
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Figure 40.8. The inversion by an orthogonal circle is a hyperbolic reflection.
(i) Lines l and p are perpendicular. (ii) The distances s(A, S ) = s(S , Al ) are equal. The first step is to show that lines l and p intersect perpendicularly. We draw the tangent from the polar l⊥ to circle p, and let T be the touching point of that tangent which lies inside D. By definition, point T lies on circle p. By the theorem of chords (Euclid III.36) |l⊥ A| · |l⊥ Al | = |l⊥ T |2
(40.13)
Comparison of (40.12) and (40.13) implies that |l⊥ T | = rl . Hence point T lies on circle l, too, and thus T = S is the intersection of circles l and p. The segment l⊥ T is a tangent to circle p as well as a radius of circle l. By Euclid III.16, radius and tangent of circle l are perpendicular to each other. Hence, at the intersection point T , the tangent to circle p is perpendicular to the tangent to circle l. Next we check item (ii). Any circle perpendicular to l is mapped to itself by the inversion Il . This follows because inversion maps (generalized) circles to itself and preserves angles. Note that those circles are only mapped to themselves as a set of points, not point by point. Especially, the inversion Il maps both circles ∂D as well as p to themselves. Hence Il maps the intersection p ∩ ∂D to itself. But p ∩ ∂D = {P, Q} consists just of the two ideal endpoints of the hyperbolic line p. Because Il maps the interior of circle l to the exterior, and vice versa, and one of the points P and Q lies in the interior and the other in the exterior of circle l, we conclude that the inversion Il maps P 7→ Q , Q 7→ P , A 7→ Al , Al 7→ A , S 7→ S Especially, this implies that the three points l⊥ , P, Q lie on a Euclidean line.
(40.14)
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Proposition 40.6 (Characterization of perpendicular lines). If two hyperbolic lines l and p intersect each other perpendicularly, then the polar l⊥ of one line l and the ideal endpoints P and Q of the other line p lie on a Euclidean line. The converse of this statement holds, too. Now we can finally confirm claim (ii): s(A, S ) = s(S , Al ). By the definition of hyperbolic distance s(A, S ) = ln(AS , QP) , s(S , Al ) = ln(S Al , QP)
(40.15)
The inversion Il maps points according to (3), and the cross ratio is preserved by inversion. Hence (AS , QP) = (Al S , PQ)
(40.16)
As a last step, we use the elementary fact about cross ratios that (AB, CD) = (BA, DC). Hence (40.15) and (40.16) imply s(A, S ) = s(S , Al ), as to be shown.
Figure 40.9. The inversion by line l maps the given point A to the center O. Furthermore, B and C are mapped to Br and Cr.
Proposition 40.7 (An especially useful reflection). For a given point A, there exists a unique the hyperbolic line l, the inversion by which maps A to the center O. Problem 40.6. Explain a construction for this hyperbolic line l. −−→ Construction. Draw the radial ray OA. Erect on it the perpendicular at point A. Let P be one of the two points where the perpendicular intersects the circle ∂D. Erect the perpendicular on the −−→ radius OP. This is a tangent to the circle ∂D. It intersects the ray OA in the inverse point A0 . The polar of the reflection line is l⊥ = A0 . The hyperbolic line l is given, in the Poincaré model, by the circle with center l⊥ through point P. 40.6.
Proof of the SAS axiom via the Poincaré model
Proposition 40.8 (SAS congruence for the Poincaré Model). Given are two triangles 4ABC and 4XYZ, with the angles at vertices A and X, and adjacent sides pairwise congruent: ∠CAB ∠ZXY , s(A, B) = s(X, Y) , s(A, C) = s(X, Z) Then the two triangles are congruent.
(40.17)
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Figure 40.10. Sorry, the triangles 4ABC and 4XYZ are not congruent.
Reason. By construction (a) above, there exists a hyperbolic line α, the inversion Iα by which maps A to O. In the same way, there exists a hyperbolic line φ such that the inversion Iφ maps point X to the center O. We map 4ABC by the inversion Iα to get the congruent 4OBαCα 4ABC. Similarly, we map 4XYZ by the inversion Iφ to get the congruent 4OYφ Zφ 4XYZ. Moreover, assumption (1) and transitivity imply ∠Cα OBα ∠Zφ OYφ , s(O, Bα ) = s(O, Yφ ) , s(O, Cα ) = s(O, Zφ )
(40.18)
Now it is quite easy to show that the two triangles 4OBαCα and 4OYφ Zφ are congruent (still in the hyperbolic sense!). Indeed, lines through the center O are lines both in the Euclidean and hyperbolic sense. The usual Euclidean reflection across such a line is a hyperbolic reflection, too. −−−→ −−−→ One such reflection will map ray OBα to ray OYφ , and 4OBαCα to 4OB0C 0 . Hence one gets a hyperbolic congruence 4OB0C 0 4OBαCα 4ABC (40.19) −−→0 −−−→ and the rays OB = OYφ are equal. In case that points C 0 and Zφ lie on different sides of ray OBα , one maps 4XZY by a reflection across that ray to get a hyperbolic congruence 4OY 0 Z 0 4OYφ Zφ 4XZY
(40.20)
In case that points C 0 and Zφ lie on the same side of ray OBα , one needs no second reflection, but puts 4OY 0 Z 0 := 4OYφ Zφ , and one gets formula (40.20) once more.
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Figure 40.11. SAS congruence.
Now, after those further mappings by reflections, the triangles 4OB0C 0 and 4OY 0 Z 0 are very easy to compare. Indeed (40.18) implies ∠C 0 OB0 ∠Z 0 OY 0 , s(O, B0 ) = s(O, Y 0 ) , s(O, C 0 ) = s(O, Z 0 )
(40.21)
−−→ −−→ and the rays OB0 = OY 0 are equal, and points C 0 and Z 0 lie on the same side of that ray. everybody can check that this simply implies that B0 = Y 0 and C 0 = Z 0 . (Actually, one uses the uniqueness for the lay off of angles and segments in Euclidean geometry.) Now (40.19) and (40.20) and transitivity imply that the hyperbolic congruence 4ABC 4XYZ as to be shown. 40.7.
Their are enough rigid motions
Theorem 40.1 (There are enough rigid motions). Given are two points P and Q, and rays rP and rQ with these vertices. There exist exactly two rigid motions, which map P to Q and rP to rQ . Proof of the Theorem about enough rigid motions. As explained in proposition 40.7, there exist an inversion Il which maps the given point P to the center O, and, similarly, an inversion Ik which maps the given point Q to the center O. There exists a reflection Im by a line m through O which maps the ray Il (rP ) to the ray Ik (rQ ). The composition Ik ◦ Im ◦ Il of these three mappings is a rigid motion which maps point P to point Q and ray rP to ray rQ .
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The second rigid motion which maps point P to point Q and ray rP to ray rQ , too, is given by the composition Ik ◦ Im ◦ Il ◦ Ir where Ir is the reflection across the line of rP . From the lemma below, we see that these two mappings are the only rigid motions with the required properties. Lemma 40.3. Given any point P and ray rP . There exist exactly two rigid motions, which leave the point P and the ray rP fixed. One of them is the identity, the other one is the reflection Ir across the line of rP . Proof. Let φ be a rigid motion leaving the point P and the ray rP fixed. As easily checked, φ leaves all points on the line of rP fixed. Furthermore, the mapping either exchanges the halfplanes of this line or leaves them invariant. In the first case, the mapping φ is the identity. This is an easy consequence of the unique transfer of segments and angles. Consider the second case that the mapping φ exchanges the halfplanes. Let Ir be the reflection across the line of rP , and define the new mapping ψ := φ ◦ Ir . The mapping ψ leaves the point P and the ray rP fixed, and leaves the halfplanes invariant. Hence, by the reasoning above, ψ is the identity and hence φ = Ir . The following explanation is almost unnecessary, as long as we set up the hyperbolic geometry only in the Poincaré model: Given is any point A, and ideal point (endpoint) E. Let AE denote the line through A with endpoint E. For any line l, one ray of which is a limiting parallel to AE, we simply say that l "goes through the ideal point E". Corollary 73. Let any two points P and Q, and two ends E and F be given. There exist exactly two rigid motions, which map P to Q and the ray PE to the ray QF. Corollary 74. Every hyperbolic rigid motion of the Poincaré disk can be realized by a composition of 0, 1, 2, 3 or 4 hyperbolic reflections. Corollary 75. Realizing these hyperbolic reflections with inversion by circles or lines produces a unique extension of any rigid motion as a bijective mapping of the underlying Euclidean plane. The extended rigid motion conserves the pairs of inverse points. Proof. The inversion Il which realizes the hyperbolic reflection by the line l has the following property: If the inversion Il maps point A to point B, then it maps the inverse point A0 to the inverse point B0 —or written in one formula: [Il (A)]0 = Il (A0 ) To check this conjecture, take any two hyperbolic lines c and d intersecting in point A. The corresponding circles in the Poincaré disk model have the inverse image A0 as their second intersection point. The inversion Il maps c and d into two circles Il (c) and Il (d), with the intersection points Il (A) and Il (A0 ). Since both Il (c) and Il (d) are orthogonal to the line of infinity δD, they consist of pairs of inverse points. Hence [Il (A)]0 ∈ Il (c) ∩ Il (d) = {Il (A), Il (A0 )} and [Il (A)]0 = Il (A0 ) as claimed. Every hyperbolic rigid motion of the Poincaré disk can be realized by a composition φ of up to four inversions by hyperbolic lines. Since [φ(A)]0 = φ(A0 ) holds for any such composition, too, we see that the extended rigid motion conserve pairs of inverse points, too.
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Proposition 40.9. Given a hyperbolic line l and two points A and B symmetric to this line, any rigid motion φ maps these objects to an image line and two image points symmetric with respect to the image line. Hence φ(Il (A)) = Iφ(l) (φ(A))
and
Iφ(l) = φ ◦ Il ◦ φ−1
Theorem 40.2. Given any three different ends E, F and G, as well as their images three different ends E 0 , F 0 and G0 . exactly one rigid motion, which maps E to E 0 , F to F 0 , and G to G0 , 40.8.
Horocycle
Problem 40.7. For the following propositions and theorem, provide drawings in the Poincaré disk model, using compass and straightedge. Copy, use and complete the drawings to prove the statements. Definition 40.7 (horocycle). A horocycle H around the endpoint E through point A consists of all points Al obtained from A by a reflection across any line l through the endpoint E. Problem 40.8. Explain why a rigid motion maps a horocycle bijectively onto a horocycle. Lemma 40.4. In the Poincaré disk, the horocycle around the ideal point E through the center O is depicted by the circle with diameter OE.
Figure 40.12. The set of the reflective images of center O across all lines with endpoint E yields the horocycle around E through O.
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Proof. In the figure on page 747, the center O is reflected across the lines l and k with the common endpoint E. As shown in proposition 40.5, the hyperbolic reflection is an inversion by an orthogonal circle. We now argue using the underlying Euclidean plane. and use the construction 40.1 to obtain the inverse point Ol . 1 The inverted point Ol is midpoint of the chord between the end E and F of arc l. Too, it is the intersection of the ray Ol⊥ with the chord between the end E and F of arc l, and we get a right angle ∠OOl E. By the converse Thales’ theorem, we see that Ol lies on a circle with diameter OE. Hence points of the horocycle around the ideal point E through the center O, as given by definition 40.7, lie on the circle with diameter OE. As easily seen, the converse holds, too. Lemma 40.5. In the Poincaré disk, the horocycle around the ideal point E through any point A is depicted by the circle through A touching the line of infinity δD at the ideal point E. Proof. By theorem 40.1, there exists a rigid motion φ which maps point A to the center O, but leaves the end E fixed. The horocycle H through O is depicted as a circle through O touching the line of infinity ∂D from inside at the ideal point E. The inverse image φ−1 (H) is depicted as a circle through A touching the line of infinity ∂D from inside at the ideal point E. Clearly the inverse image is a horocycle, too. Definition 40.8 (tangent to a horocycle). Given is the horocycle H around the endpoint E through point A. The line t through A perpendicular to AE is called the tangent to the horocycle at point A. Proposition 40.10. The tangent to a horocycle at any point B meets the horocycle only at this touching point. Any other line through the touching point cuts the horocycle in two points B , C. Proof. There exists a rigid motion φ which maps point B to the center O. Hence it is enough to prove the claim for a horocycle H through O. Let E be the end around which this horocycle goes. In the Poincaré model H is depicted as a circle with diameter OE. The tangent to H at O is depicted as the diameter of ∂D perpendicular to OE. Any other hyperbolic line through O is depicted as another diameter of ∂D. Hence it intersects H in two points—which clearly are hyperbolic points, as to be shown. Lemma 40.6. For horocycle around the ideal point E through the center O of the Poincaré disk, and any line l with ideal point E, the limit triangle 4OEOl has congruent angles at vertices O and Ol . Proof. By definition of the horocycle, these two points O and Q = Il (O) are mirror images for a hyperbolic reflection across a line l with one end E. There exists a rigid motion ψ which maps line l to a diameter ψ(l) and leaves E fixed. As image of the mapping, we get the horocycle ψ(H) around E through the two points ψ(O) and ψ(Q). As explained above, ψ(Q) = Iψ(l) (ψ(O)) and the two image points are symmetric mirror images by the line ψ(l), Since ψ(l) is a diameter of δD, this holds in the Euclidean sense, too. And since E = ψ(E), and this point is fixed both by the inversion Il and the inversion ψ(l), it lies on the diameter ψ(l). Hence the limit triangle 4ψ(O)ψ(E)ψ(Q) is symmetric to this diameter both in the Euclidean, and the hyperbolic sense. Hence it has congruent base angles, and the limit triangle 4OEQ has congruent base angles, too, as to be shown. 1
This time we go from outside to inside the circle l.
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Lemma 40.7. If point B lies on the horocycle around A through the ideal end E, and point C lies on the horocycle around B through the same ideal end E, then point C lies on the horocycle around A through the ideal end E. Proof. Assume point B lies on the horocycle HA around the ideal end E. Assume point C lies on the horocycle HB around B through the same ideal end E, In the Poincareé model, horocycle HA is depicted as a circle touching the line of infinity δD at the ideal point E from inside, and going through point A. A circle is uniquely determined by two points, and the tangent at one of them. Hence HA = HB . From this, we see that point C lies on this same horocycle, too. Proposition 40.11. For a horocycle around the ideal end E through any two points B and C, the limit triangle 4BCE has congruent angles at vertices B and C. Proof. There exists a rigid motion φ which maps point B to the center O. Let φ(C) = Q. It is enough to prove the claim for the image horocycle φ(H) through O and Q. But this has already been done in the Lemma above. Proposition 40.12. Given any point A and ideal point E. The set of all points B such that the limit triangle 4ABE has congruent angles at vertices A and B is equal to the horocycle around the ideal point E through point A. 40.9.
Circles and hypercircles Recall the easy definitions from neutral triangle geometry.
Definition 40.9 (Circle). Given is a center A and a distance AX. The set of all points with distance from the center O congruent to AX is called a circle. Definition 40.10 (Equidistance line). Given is a baseline l and a distance AX. The set of all points with distance from a baseline l congruent to AX, and lying on one side of this line, are called an equidistance line or hypercycle. Proposition 40.13 (Circles appear as circles). In the Poincaré disk model, a circle appears as a circle inside the disk D. Reason. In the special case that the center of the given circle A = O is the center of the Poincaré disk, the statement is obviously true. Otherwise, we use Proposition 40.7. There exists a unique orthogonal circle l, the inversion by which maps the center A of the given circle A to the center O of the Poincaré disk. By Proposition 40.5, the inversion by l is a hyperbolic reflection. Hence the circle A is mapped to a circle around O, which appears in the Poincaé model as a circle C around O. A second application of the inversion by l maps circle C back to the original given circle A. As we have shown in the section about inversion by circles, inversion maps circles to circles. Hence A appears as a circle inside Poincaré’s model.
Proposition 40.14 (Equidistance lines are circular arcs). In the Poincaré disk model, the set of all points of a given hyperbolic distance d from a given line l lie on two circular arcs.
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Figure 40.13. A circle appears as a circle—evidently for a circle about O, and hence always by means of a useful reflection.
Figure 40.14. An equidistance line is a circular arc with two ideal endpoints.
Reason. It suffices to take for the line l the horizontal diameter EF of ∂D, and choose a point B on the vertical diameter m. The foot point of the perpendicular from B onto l is A = O. By definition, the hyperbolic distance of the two points A and B is s(A, B) = ln(AB, PQ)
(40.22)
where P and Q are the ideal endpoints of the vertical diameter m. Let ε be the circular arc through point B and the two ideal endpoints E, F of line l. Note that the circular ε is neither a line nor a circle in hyperbolic geometry!
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Question. Why is ε not a hyperbolic line? Answer. Circles ∂D and ε do not intersect perpendicularly. We need to show that all points of ε have the same distance from line l. One begins by choosing an arbitrary point Bσ of ε. Let σ⊥ be the intersection of the Euclidean lines l and BBσ . Construct the tangent from σ⊥ to the circular arc ε, and let S be the touching point of the tangent. Let σ be a circular arc through S around σ⊥ . Is σ a hyperbolic line? To answer that question, apply Euclid III.36 to circle ε and chords through σ⊥ . One gets |σ⊥ S |2 = |σ⊥ B| · |σ⊥ Bσ | = |σ⊥ |E · |σ⊥ F| (40.23) This shows that by inversion Iσ across the circular arc σ maps: S 7→ S , B 7→ Bσ , E 7→ F, F 7→ E
(40.24)
The inversion Iσ maps line l to itself. Hence by preservation of angles and generalized circles, the inversion Iσ maps circle ∂D to itself. Is circle ∂D mapped to itself point by point? Answer. No, formula (40.24) shows that the two points E, F of circle ∂D are interchanged by the mapping Iσ . Because inversion by σ maps the entire circle ∂D to itself, circles ∂D and σ intersect perpendicular, and hence σ is a hyperbolic line. Hence, as shown in the lecture, inversion by circle σ is a hyperbolic reflection. Next we map the points A, P and Q of m by this reflection, and get the reflected points Aσ , Pσ and Qσ . The two points Pσ and Qσ are the ideal endpoints of line of the reflected line mσ . Because line mσ and l are perpendicular to each other, the perpendicular from Bσ onto line l has the foot point Aσ , and hence d(Aσ , Bσ ) is the distance form point Bσ to line l. Because the cross ratio is conserved by circular inversion, we get s(Aσ , Bσ ) = ln(Aσ Bσ , Pσ Qσ ) = ln(AB, PQ) = s(A, B) This shows that points B and Bσ have the same distance from line l. Because point Bσ was chosen arbitrarily, we conclude that all points of the arc ε have the same distance from line l. The second arc the points of which have the same distance from l is produced by reflection across line l. 40.10. We have obtained all circle-like curves It can happen in hyperbolic geometry that three points lie neither on a line nor a circle. We have seen three types of circle-type curves: circle, ´ equidistance line, and horocycle. We now obtain—via the Poincaremodel—a satisfactory theorem confirming that we have found all types of circle-type curves. Theorem 40.3. In the hyperbolic plane, any three different points lie either on a line, a circle, a ´ model horocycle, or an equidistant line. In the Poincaredisk a circle is depicted as a circle lying inside the disk D. a hyperbolic line is a circular or straight arc, which has two intersection points with the line of infinity ∂D, and intersects it perpendicularly. an equidistance line is a circular or straight arc, which has two intersection points with the line of infinity ∂D, but not a perpendicular angle of intersection. a horocycle is a circle touching the line of infinity δD from inside This is simply the remaining possible case!
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40.11. Towards the Klein model The relation of hyperbolic lines and their chords is useful for the translation from Poincaré’s to Klein’s model. Proposition 40.15 (Hyperbolic lines and their chords). Given are two or more hyperbolic lines l, l1 , l2 , . . . all through one point P. (a) The chords between the ideal endpoints of lines l, l1 , l2 . . . all intersect in one common point K, too. (b) The inverse point K 0 is the midpoint of the segment PP0 between P and its inverse point P0 . −−→ (c) The location of the point K on the ray OP is given by |OK| =
2 |OP| 1 + |OP| 2
(40.25)
For the proof, we need to recall, from Euclidean geometry of circles II, the definition of power p(O, C) of a point O with respect to a circle C: For any circle C and point O, let p(O, C) = |OA| · |OB| where A and B are the two intersection points of a line l through O with the circle C. If a second line k intersects the circle C in the points P and Q, by Euclid III.35 and III.36, |OA| · |OB| = |OP| · |OQ| Hence the power p does not depend on the choice of the line l, and thus is well defined. The power is negative for points inside the circle and positive for points outside the circle.
Figure 40.15. The chords of a bundle of hyperbolic lines through a common point P intersect in a common point K.
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Proof of (a). Pick one line l, let E, F be the ideal endpoints of l, and define point K as the intersection of chord EF with chord PP0 , where P0 is the inverse image of P. Actually, EF is the common chord of ∂D and l, and hence p(K, ∂D) = |KE| · |KF| = p(K, l) We now calculate the power of K relative to l. Since PP0 is the common chord of all hyperbolic lines through point P. p(K, l) = |KE| · |KF| = |KP| · |KP0 | Clearly the last two formulas imply p(K, ∂D) = |KP| · |KP0 | which confirms that p(K, ∂D) is independent of the choice of the line l. Since point K lies on the −−→ ray OP, we conclude point K is independent of the choice of the line l. (We may repeat the same process for another line l1 . It would be possible that we get another intersection point K1 of chord E1 F1 with chord PP0 . But, as we have shown, K1 = K does hold.) Proof of (b). We draw the circle t with diameter PP0 . Let K 0 be its center, and RS be the common chord of circles ∂D and t. Since t passed through the pair of inverse points P and P0 , the circles t and ∂D are orthogonal to each other. Hence ∠ORK 0 = 90◦ and RK is the altitude of the right 4ORK 0 . Now |OK| · |OK 0 | = |OR| 2 = 1 follows from the leg theorem in this 4OS K 0 . Hence the center K 0 of circle t is the inverted point of the foot point K. By construction, K 0 is the midpoint of diameter PP0 . Proof of (c). Since the inverse point K 0 is the midpoint of the segment PP0 between P and its inverse point P0 , we use the definition of inverse points twice and get |OK| =
1 2 2 |OP| 2 |OP| = = = 0 0 0 2 |OK | |OP | + |OP| |OP | · |OP| + |OP| 1 + |OP| 2
which confirms claim (40.25).
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41. Geometric Constructions in the Poincaré Disk Before we start. These constructions are done using compass and straightedge of the ambient Euclidean plane and are to be described in terms of the Euclidean geometry of that ambient plane. Use the conventions from the section about Poincaré’s model to keep your explanations short! For each problem, provide drawings and set up a step by step construction process.
Figure 41.1. How to construct a hyperbolic line through two given points A and B.
41.1.
Basic constructions from neutral geometry
Problem 41.1. For two given points A and B, construct the hyperbolic line l through A and B. Construction 41.1. One needs the inverse point of either A or B, say A0 . The hyperbolic line l is modelled by a circle through A, A0 and B. Its center l⊥ is found as intersection point of the perpendicular bisectors of the sides of 4AA0 B. One can, for example construct the perpendicular bisector A⊥ of segment AA0 , and the perpendicular bisector of AB. Finally, one draws a circular arc around l⊥ through point A. This arc passes through points B, A0 and B0 , too. Problem 41.2. For two given points A and B, construct the perpendicular bisector µ. Hint: Let A0 , B0 be the inverse points of A, B by ∂D. Both pairs A, B and A0 , B0 are inverse points by circle µ. ←−→ ← → Construction 41.2. The Euclidean lines AB and A0 B0 intersect in the point µ⊥ . The perpendicular bisector in modelled by a circle around µ⊥ perpendicular to ∂D. One needs still to get the correct radius. The ideal endpoints µ1 , µ2 of µ are the touching points of the tangents from µ⊥ to ∂D. These points are constructed via Thales’ theorem. Indeed µ1,2 lie on a circle with diameter Oµ⊥ . Finally µ is the circle about µ⊥ through µ1,2 .
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Figure 41.2. Construct the line bisector for two given points A and B.
Figure 41.3. Erect a perpendicular on line δ at point P, using Construction 41.3.
← → Problem 41.3. Given is a hyperbolic line δ = EF, with ideal endpoints E and F, and a point P on the line δ. Use the given drawing, with P, P⊥ , δ⊥ , E, F already available to erect the perpendicular σ on the line δ at point P. Use the ideal endpoints S and T of σ to check the accuracy of your drawing. In this first variant, make use of the ideal endpoints. ← → Construction 41.3. The Euclidean lines EF and P⊥ intersect at point σ⊥ . Finally, to get σ is easy, one simply draws a circular arc with center σ⊥ through point P. Remark 1. One need not even construct the tangent from σ⊥ to ∂D. Since δ ⊥ σ, it is known that the ideal endpoints S and T of σ and δ⊥ lie on a Euclidean line. This fact can be used to check the accuracy of the construction.
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Figure 41.4. Erect a perpendicular on line δ at point P, using the right angle as explained in Remark 2. Finally, combining the two constructions yields better accuracy.
Remark 2. As an alternative construction, one can also erect the perpendicular on the radial ray −−→ Pδ⊥ at vertex P. Again, that perpendicular intersects line p⊥ at the polar point σ⊥ . Using both constructions gives another possibility for better accuracy.
Figure 41.5. Drop a perpendicular onto line δ from the given point P.
← → Problem 41.4. Given is a hyperbolic line δ = EF, with ideal endpoints E and F, and a point P not on the line δ. Use the given drawing, with P, P⊥ , δ⊥ , E, F already available to erect the perpendicular σ on the line δ at point P. Use the ideal endpoints S and T of σ to check the accuracy of your drawing.
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← → Construction 41.4. The Euclidean lines EF and P⊥ intersect at point σ⊥ . Finally, to get σ is easy, one simply draws a circular arc with center σ⊥ through point P. ← → Problem 41.5. Given is a hyperbolic line δ = EF, and a point P on the line δ, and an angle α. Use the given drawing, with P, P⊥ , δ⊥ already available. Construct two lines ε1 , ε2 through point P that form the given angle α with the given line δ.
Figure 41.6. Transfer a given angle to a given ray. The example takes α = 10◦ .
−−→ Construction 41.5. Transfer the given angle α onto both sides of the radial ray Pδ⊥ at vertex P. The two new sides produced by angles α intersect line p⊥ in the polar points ε⊥1 , ε⊥2 . Now it is straightforward to get ε⊥1 , ε⊥2 , because one has already the point P on these lines. One simply draws two circular arcs with centers ε⊥1 and ε⊥2 through point P. Problem 41.6. For two given intersecting lines α and β, construct an angle bisector δ. Explain how you can get the second bisector. Construction 41.6. Let A1 , A2 and B1 , B2 be the ideal endpoints of lines α and β. One angular ←−−→ ←−−→ bisector δ has polar δ⊥ at the intersection of the Euclidean lines A1 B1 and A2 B2 . The second ←−−→ ←−−→ angular bisector δ2 has polar its δ⊥2 at the intersection of the Euclidean lines A1 B2 and A2 B1 . Finally one has to draw the hyperbolic lines δ and δ2 . They are modelled by circular arcs with centers δ⊥ and δ⊥2 through the intersection point P of lines α and β.
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Figure 41.7. Construction of the angular bisectors.
Alternative Construction. Let P be the intersection point of lines α and β. Let ε be the bisector of ∠α⊥ Pβ⊥ , and let ε2 be the outer bisector of that angle. (Indeed ε2 is perpendicular to ε.) One angular bisector δ of the given lines α and β has its polar δ⊥ at the intersection of the Euclidean ←−−→ lines α⊥ β⊥ and ε. The second angular bisector δ of lines α and β has its polar δ⊥2 at the intersection ←−−→ of the Euclidean lines α⊥ β⊥ and ε2 .
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Figure 41.8. Construction of the middle line of two parallel lines.
41.2.
Typically hyperbolic constructions
Problem 41.7. For two divergently parallel lines α and β, construct the middle line δ. The reflection by this line maps α to β. Construction 41.7. Let A1,2 and B1,2 be the ideal endpoints of lines α and β. The numbering is such that, as one moves around the boundary circle δD, point A1 lies adjacent to B1 , and point A2 ←−−→ adjacent to B2 . The middle line δ has polar δ⊥ at the intersection of the Euclidean lines A1 B1 and ←−−→ A2 B2 . (This intersection point lies outside the disk D, whereas the intersection of the Euclidean ←−−→ ←−−→ lines A1 B2 and A2 B1 lies inside D.)
Figure 41.9. Construction of the common perpendicular of two parallel lines.
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Problem 41.8. For two divergently parallel lines α and β, construct the common perpendicular µ. Construction 41.8 (Variant 1). The common perpendicular µ has polar µ⊥ at the intersection of ←−−→ ←−−→ the Euclidean lines A1 A2 and B1 B2 .
Figure 41.10. Variant 2 for the construction of the common perpendicular of two parallel lines— and a figure showing both variants.
Problem 41.9. For two divergently parallel lines α and β, find a second way to construct the common perpendicular µ. ←−−→ Construction 41.9 (Variant 2). Draw the Euclidean line c = α⊥ β⊥ connecting the polars of the two given lines. The M be the foot point of the perpendicular dropped from O onto this line c. The inverse point M 0 = µ⊥ . It is constructed by putting the tangents to circle ∂D at the intersection points c with ∂D. Those two tangents and the perpendicular OM intersect all three in the point µ⊥ . Question. Describe and explain the simultaneous construction of the middle line and the common perpendicular of two parallel lines done in the figure on page 761. 41.3.
Circle constructions
Problem 41.10. For two points A and B, construct a circle ε with center A through point B. (a) Assume the prescribed center A , O and point B do not lie on the diameter OA. Moreover, assume the hyperbolic line r = AB has already been constructed.
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Figure 41.11. Given two parallel lines α and β, the middle line δ and the common perpendicular µ are constructed simultaneously.
Figure 41.12. Construction of a circle with given center A through point B.
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Figure 41.13. Construction of a circle with given center A through point B, which is on or close to line OA.
Figure 41.14. How to construct of a circle around O with given hyperbolic radius.
(b) Assume the prescribed center A , O and point B lie on the diameter OA. Construction 41.10. One needs to find the Euclidean quasi-center M of circle ε. It lays both on
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the segment OA, and on the tangent to the arc r at point B. (a) In this case, these are two different lines which intersect in the quasi-center M. (b) If center A, disk-center O and point B lie on a line—or even for a point B lying too close to the diameter OA—we construct at first a second point B” on the circle. To this end, we erect the hyperbolic perpendicular a to OA at point A, and reflect point B by line a. Since B and the reflection image B” have the same hyperbolic distance from A, both lie on the circle to be constructed. The circle is perpendicular to line OA, and hence its center M is the intersection of the perpendicular bisector of BB” with segment OA. Problem 41.11. For two points A and B, construct a circle ε with center O and radius equal to the hyperbolic distance s(A, B). Construction 41.11. Proceed as in the example "A useful reflection" to get the hyperbolic line α, the reflection by which maps point A to the center O. This reflection maps point B to image point −−→ Bα . One has to construct Bα via inversion by α. To this end, one draws the radial ray A0 B, and a perpendicular on it through point B, which intersect circle α in point Q. Then one erects the −−→ perpendicular on A0 Q at point Q. It intersect the ray A0 B in the inversion point Bα . The circle around O through point Bα has the hyperbolic radius s(A, B) = s(Aα , Bα ) = s(O, Bα ) as required.
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41.4.
Triangle constructions
Proposition 41.1 (Construction of a triangle from its three angles). In hyperbolic geometry, its three angles determine a triangle up to congruence. Indeed, for any given angles α, β, γ with sum α + β + γ < 180◦ , there exists a triangle, unique up to congruence. To achieve a construction in the Poincaré disk, it is convenient to chose the vertices first, and determine the disk D in a second step. Such a procedure does not introduce a real restriction, because, in an extra step, a linear delation can be used to map the entire figure such that the disk falls into any disk given in advance. Furthermore, we shall choose vertex C at the center of disk D. Construction 41.12. Let σ = α + β + γ be the angle sum and δ = 180◦ − α − β − γ be the defect. One begins by drawing ∠Ac⊥ B = δ as given. Then draw a circle c with center c⊥ of any radius. We can assume that A and B lie on that circle, and construct the tangents to c at A and B. Next, we lay off the angles α at vertex A, and β at vertex B, with intersecting tangent rays as one of their legs, and the second legs outside of circle c. Those second legs intersect at point C = O, and form an angle γ. Finally, one needs the boundary ∂D of the Poincaré disk. To this end, one constructs the tangents from point C to circle c. The touching points S , T of these tangents can be constructed via Thales’ theorem. Indeed S and T lie on a circle with diameter c⊥C. Finally the boundary ∂D is the circle about C through S and T . (Indeed, the points S , T are the ideal endpoints of the triangle side AB, too.) Problem 41.12. Do the construction with given angles α = 40◦ , β = 50◦ , γ = 60◦ . Proposition 41.2. In hyperbolic geometry, its three angles determine a triangle up to congruence. Indeed, for any given angles α, β, γ with sum α + β + γ < 180◦ , there exists a triangle, unique up to congruence. Too, there exist asymptotic triangles for which one, two, or even all three of its vertices are ideal endpoints. In the disk model, these ideal endpoints simply lie on ∂D. The angle at an ideal vertex is zero. Construction 41.13 (Construction of an asymptotic triangle from its two angles). We assume that vertex C is ideal, and hence γ = 0. For simplicity, we take for vertex A the center of disk −−→ D. Hence side AC and the ray AB are hyperbolic as well as Euclidean lines. They form the given angle α at vertex A, and can be constructed immediately. Next we need to get a⊥ , the polar to the side a = BC. Since point C lies on the side a, the polar ⊥ C goes through the polar a⊥ . But the polar C ⊥ is simply the tangent to ∂D at the ideal point C. Thus we get already one coordinate for point a⊥ . We need to use the given angle β at vertex B to get a second coordinate. Let σ = α + β + γ be the angle sum and δ = 180◦ − α − β − γ be the defect. The Euclidean quadrilateral ABa⊥C has the angles α at vertex A, 90◦ at vertex C, 90◦ + β at vertex B, and hence δ at vertex a⊥ .
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Figure 41.15. Two examples for the construction of a triangle from three given angles.
−−→ Hence the isosceles 4BCa⊥ has two base angles σ2 . Thus the ray CB form an angle σ2 with the tangent Ca⊥ , and an angle 2δ with the radius CA. Hence one transfers angle 2δ to vertex C with one side CA. Now point B lies on the other side of this angle. Problem 41.13. Do the construction with given angles α = 60◦ , β = 45◦ , γ = 0◦ . Proposition 41.3 (SAA Construction of a triangle). In hyperbolic geometry, a triangle is determined, up to congruence, by giving one side, one adjacent angle, and the angle opposite
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Figure 41.16. Construction of an asymptotic triangle with angles α = 60◦ , β = 45◦ , γ = 0◦ .
to that side. Indeed, for any given angles α, γ with sum α + γ < 180◦ , and segment of length c, there exists such a triangle, unique up to congruence. I give two variants for a construction. Explanation of construction variant 1. We choose to put the vertex B, where no angle is specified, at the center of the Poincaré disk. Let A be any point such that segment AB has the length c as required. The construction uses the inverse point A0 and the polar A⊥ , too. Next we construct line −−→ b through point A, which produces angle α with AB as required. To this end, one transfers angle − − → 90◦ − α at vertex A with one side AA0 . The intersection of the other leg of this angle with A⊥ yields point b⊥ . Now line b is simply a circular arc with center b⊥ through point A. −−→ The harder part is to get vertex C. Let C ∗ be the ideal endpoint of ray AC such that we get angle ∠BAC ∗ = α as required. −−−−→ Transfer angle 90◦ + γ to vertex C ∗ with one side C ∗ b⊥ . Let a∗ be the other side of that angle. We have thus produced the given angle γ at vertex C ∗ , with one side a ray tangent to b, and ray a∗ the other side. Let σ be the circle around b⊥ through O and let O∗ be the intersection point of that circle with the ray a∗ . The remaining part of the construction uses a Euclidean reflection by the bisector of angle ∠Ob⊥ O∗ . Too, the reflection maps angle γ to vertex C, with one side a ray tangent to b, and the ray a = CO as other side. Answer. Remark. Here is a slightly different way to get vertex C. One transfers angle γ to vertex A, with one −−→ leg being the tangent to side b, the other leg in the same half plane as AB. (Alternatively, transfer −−→ angle γ − α with one leg AB.) Let D be the intersection point of the second leg with σ, the circle around b⊥ through point O. Finally, one has to draw a circle around B of Euclidean radius AD. It intersects the arc b in two points C1 and C2 . One of the 4ABCi for i = 1, 2 has angle γ at vertex Ci . (The other one has the supplementary angle 180◦ − γ, and has not to be taken into account.) Explanation of construction variant 2. We choose to put the vertex A and the given angle α at the center O = A of the Poincaré disk. Let B be any point such that segment AB has the length −−→ −−→ c as required. Let E be the ideal endpoint of ray AC = AE such that we get angle ∠BAE = α as required, and F be the ideal end of the opposite ray.
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Figure 41.17. Example for the SAA construction of a triangle.
The harder part is to get vertex B, to which end we use an equidistance line. By definition, an equidistance line is the set of all points with same distance from a given line. Let ε be the equidistance line through vertex B consisting of all points with same distance from line OE. The line ε is modelled by a circular arc though point B and the ideal ends E and F. as explained in the section on the Poincaré model. −−→ We transfer angle γ to vertex O = A with one side OE. Let c∗ be the other side of that angle, ∗ and B be its intersection with the equidistance line ε. The remaining part of the construction uses a hyperbolic reflection by line r, which is constructed to map B to B∗ , and the line FOE to itself. The polar r⊥ is the intersection of the Euclidean lines BB∗ and OE. Finally vertex C is the reflective image of O. This is the intersection −−→ of the common chord of δD and r with ray OE. Too, the reflection maps angle γ to vertex C, with the sides CO and CB. Problem 41.14. We choose α = 40◦ and γ = 29◦ . For disk D and point A already given, do the construction as described above. Report the angle β from your construction. Remark. The actual value of β does depend to the choice of the hyperbolic distance s(A, B). 41.5.
The altitudes and the orthocenter
Problem 41.15. Read the proof and of the proposition about the altitudes. Provide a drawing with all relevant entities named consistently.
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Figure 41.18. Variant 2 for the SAA construction. This variant uses an equidistant line and a hyperbolic reflection by r.
Proposition 41.4. In hyperbolic geometry, if two altitudes of a triangle intersect, then all three altitudes intersect in one point. Proof. We use the Poincaré disk model of hyperbolic geometry. In the given 4ABC, let H be the intersection point of the two altitudes dropped from vertices A and C. By means of a hyperbolic reflection, we can put H = O into the center of the Poincaré disk. As usual, let a, b, c denote the sides of the triangle opposite to its vertices A, B, C. We use the polar triangle with vertices a⊥ , b⊥ , c⊥ polar to sides a, b, c. This is a triangle in the Euclidean sense. For an acute triangle, it has the Poincaré disk in its interior. ("swallowing that disk") The altitude hA of 4ABC dropped from A passes through the center of the Poincaré disk. Hence it is a Euclidean straight line. Since hA is perpendicular to the opposite side a, it passes through the ←−−→ ← → polar point a⊥ . Furthermore, line hA = OA is perpendicular to A⊥ = b⊥ c⊥ . Similarly, altitude hC passes through c⊥ and is perpendicular to C ⊥ . Hence lines hA and hC are the altitudes of the polar triangle 4 a⊥ , b⊥ , c⊥ , too. We can use the fact, known from Euclidean geometry, that its altitudes intersect in one point, which is indeed H. Hence line b⊥ H is an altitude of the polar triangle and hence it is perpendicular to line B⊥ . Thus point B lies on line b⊥ H, which means that the three points B, H, b⊥ lie on one straight line. This implies that side b of perpendicular to the altitude hb , which is the altitude hB of the original triangle, too, and passes through point H. Remark. The three altitudes of an acute triangle always intersect. For an obtuse triangle, the altitudes may or may not intersect. Problem 41.16. Use the Poincaré model to construct a triangle with an orthocenter, but no circumcenter— and another triangle without an orthocenter, but a circum-center.
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Figure 41.19. The three altitudes.
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Figure 41.20. A triangle with an orthocenter, but no circum-center— another triangle without an orthocenter, but a circum-center.
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42. Hyperbolic Geometry in Klein’s Model 42.1. Setup of Klein’s model The second important model for hyperbolic geometry goes back to Felix Klein. The reader should recall the basic idea of a model in mathematics, as explained in the passage General remark about models in mathematics. Again, one uses the Euclidean plane as ambient underlying reality ("background ontology"). We put into the Euclidean plane the open unit disk D = {(x, y) : x2 + y2 < 1} with the boundary ∂D = {(x, y) : x2 + y2 = 1} The center of D is denoted by O. Definition 42.1 (Basic elements of Klein’s model). The points of D are the "points" for Klein’s model. The points of ∂D are called "ideal points" or "endpoints". The ideal points are not points of the hyperbolic plane. Once the hyperbolic distance is introduced, the points of ∂D turn out to be infinitely far away. Hence we call ∂D the "circle of infinity". The "lines" for Klein’s model are straight chords. Poincaré’s and Klein’s model differ, because lines are represented differently, and—even more importantly—the hyperbolic isometries are given by different types of mappings. In Poincaré’s model, the hyperbolic reflections are realized as inversions by circles. In Klein’s model, the hyperbolic reflections are realized quite differently. Indeed, hyperbolic reflections are projective mappings, which leave the circle of infinity ∂D invariant. The developing Klein’s model based on projective geometry is postponed to the subsection about the projective nature of Klein’s model. I shall now use a rather simple-minded different approach: there exists an isomorphism which is a translation from Poincaré’s to Klein’s model. Because we already know that Poincaré’s model is a consistent model for hyperbolic geometry, the translation implies that Klein’s model is a consistent model for hyperbolic geometry, too. Proposition 42.1 (The mapping from Poincaré’s to Klein’s model). The point P in Poincaré’s −−→ −−→ model is mapped to a point K in Klein’s model by requiring that the rays OP = OK are identical and 2 |OP| |OK| = (42.1) 1 + |OP| 2 The mapping (42.1) keeps the ideal endpoints fixed, and it takes a circular arc l ⊥ ∂D to the corresponding chord with the same ideal endpoints. Indeed, the mapping (42.1) is a translation of Poincaré’s to Klein’s model , since the points and lines of Poincaré’s model, are mapped to points and lines of Klein’s model, preserving incidence. Reason. As shown in the last proposition of the section on Poincaré’s model, point K is the −−→ intersection of ray OP with the chord between the ideal endpoints of any arc l ⊥ ∂D through point P. This chord k is a hyperbolic line in Klein’s model. Clearly all points of arc l are mapped to points of chord k by the same construction, and hence are all given by mapping (42.1). For the Poincaré disk model, it has been very useful to define polar elements outside the closed disk D. We shall use polar elements for Klein’s model, too. As a first step, we define and construct the inverse point P0 of any given point P, in the way explained in the section about the Euclidean geometry of circles II.
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Definition 42.2 (The polar elements for Klein’s model). The polar l⊥ of a line l is the intersection point of the tangents to ∂D at its ideal endpoints. −−→ The Klein polar or projective polar K ⊥ of a point K is perpendicular to the ray OK at the inverse point K 0 . A few clarifying remarks are in place: For both the Poincaré and the Klein model, the polar of a line is the intersection point of the tangents to the circle of infinity at the ideal ends. But the mapping from the points to their polar elements are different for the two models. For clarification, I use the terms Poincaré polar and projective polar. 1 The difference occurs because the points are mapped from Poincaré’s to Klein’s via the isomorphism (42.1), but the polar elements are the same for both models, hence the mappings from points to their polar are different for the two models. Proposition 42.2. The Poincaré polar of a point P is the perpendicular bisector of the segment PP0 between the given point and its inverse. The projective polar of a point K is the perpendicular −−→ to the ray OK at the inverse point K 0 . The following diagram for the mapping (42.1), the Poincaré polar and the projective polar is commutative: isomorphism (42.1)
point P of Poincaré’s model −−−−−−−−−−−−−→ point K of Klein’s model perpendicular bisector of PP0 y yperpendicular to OK at K 0 Poincaré’s polar P⊥
projective polar K proj⊥
Proof. By item (b) from the last proposition from the section on the Poincaré model, the inverse point K 0 is the midpoint of the segment PP0 between P and its inverse point P0 . Hence the Poincaré polar P⊥ is identical to the Klein polar K proj⊥ . In figure 42.1, Poincaré’s elements are drawn in blue, Klein’s elements in brown, and the polar elements are green. This should make clear the meaning of the commutative diagram in proposition 42.2. Proposition 42.3. The definitions of the projective polar of points and lines are consistent with incidence: A point K lies on a hyperbolic line k if and only if the projective polar K proj⊥ goes through the polar k⊥ . Proof using the development above. As shown in the last proposition in the section about Poincaré’s model, the point K lies on a chord k if and only if the Poincaré point P lies on the arc l ⊥ ∂D with the same endpoints. This happens if and only if the polar l⊥ lies on P⊥ . But by proposition 42.2, these polar elements are identical with those of Klein’s model: l⊥ = k⊥ and P⊥ = K proj⊥ . Hence, expressing everything in Klein’s model, we conclude that a point K lies on a hyperbolic line k if and only if the polar K proj⊥ goes through the polar k⊥ . 1
The simple term polar is common usage for the projective polar. It refers to the pole and polar relation studied in projective geometry. The term is also used in common software packages for geometry. On the other hand, GoodmanStrauss uses the term polar for the Poincaré polar, as I do in the sections on Poincaré’s model.
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Figure 42.1. A bundle of hyperbolic lines l, l1 through a common point P in Poincaré’s model. They intersect in a common point K in Klein’s model, too. Their polar elements are identical for both models.
Figure 42.2. If line l goes through point K, then the polar l⊥ lies on the polar K proj⊥ .
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Direct independent proof. Assume that point K lies on line l. We need to check whether the polar K proj⊥ goes through l⊥ . As drawn in figure 42.1, let L be the foot point of the perpendicular dropped from center O onto line l. The definitions of the polar and the inverse point can easily be seen to imply l⊥ = L0 . −−→ Let L00 be the intersection of ray OL and the polar K proj⊥ . This construction ensures that the polar K proj⊥ goes point L00 . The triangles 4OKL and 4OL00 K 0 are equiangular. Hence, by Euclid VI.4, their sides are proportional to each other: |OK| |OL00 | = |OL| |OK 0 | By definition of the inverse point |OK| · |OK 0 | = 1 and hence |OL| · |OL00 | = 1 which shows that L00 is the inverse point of L. Now L00 = L0 and L0 = l⊥ imply L00 = l⊥ . Hence the polar K proj⊥ goes through l⊥ , as to be shown. The converse follows as easily. Before discussing the metric properties and congruence, we need to clarify some terms about the use of any mathematical models, as Klein’s or Poincaré’s: Definition 42.3. A theorem or a feature of a figure is part of neutral geometry if and only if it can be deducted assuming only the axioms of incidence, order, congruence. The facts of neutral geometry are valid in both Euclidean and hyperbolic geometry—as well as the more exotic non-Archimedean geometries. Definition 42.4. A feature of a figure drawn inside Klein’s model (as for example an angle, midpoint, altitude or bisector) is called absolute if it is valid both for the underlying Euclidean plane, on which the model is based, and the hyperbolic geometry inside the model. Remark. Here are some features that are absolute, valid both as features of hyperbolic geometry and in the underlying Euclidean plane: An angle with the center of Klein’s disk appears as an absolute angle. A right angle of which one side is a diameter appears absolute. A perpendicular bisector or an angle bisector which is a diameter appears absolute. Reason. We know that angles are depicted undistorted in Poincaré’s model. For the cases mentioned above, the angles are left undistorted by the mapping from Poincaré’s to Klein’s model. Hence they are appear absolute in Klein’s model. Next, we can translate orthogonality. We have shown that in Poincaré’s model two hyperbolic lines l and p intersect each other perpendicularly, if and only if the polar l⊥ of one line l and the ideal endpoints P and Q of the other line p lie on a Euclidean line. Since the polar of a line is easily translated, we get the following criterium for perpendicular lines in the Klein model: Proposition 42.4 (Perpendicular lines). In Klein’s model, two hyperbolic lines l and p intersect each other perpendicularly, if and only if the polar l⊥ of one line l lie on the (ultra ideal extension) of the other line p. Remark. Since being perpendicular is a symmetric relation, the polar l⊥ of line l lies on the (ultra ideal extension) of line p if and only if the polar p⊥ of the second line p lies on the (ultra ideal extension) of the first line l.
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Figure 42.3. Two perpendicular lines l and p.
Remark. All other angles are distorted in Klein’s model, and can only be defined via the isometries—which are projective mappings. For the definition of a hyperbolic distance, one needs the cross ratio. The cross ratio of four point K, L, E, F is defined as KE · LF (KL, EF) = LE · KF The way the endpoints E and F of the segment KL appear in this fraction can be remembered by means of the diagram: K→L L←K
E→ F E→ F
Definition 42.5 (Hyperbolic distance and congruence of segments). Let K, L be any two points. Let the hyperbolic line through K and L denoted by l, and the ideal endpoints of this line by E and F. We name those endpoints such that E ∗L∗K ∗F. The hyperbolic distance or simply "distance" of points K and L is defined by s(K, L) =
1 KE · LF 1 ln(KL, EF) = ln 2 2 LE · KF
(42.2)
As usual, the length of a segment is the distance of its endpoints. Two segments are called "congruent" if they have the same length.
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Figure 42.4. The polar of the first perpendicular line lies on the extension of the second one, and vice versa.
42.2. Angle of parallelism To check that formula (42.2) is the correct translation of the distance function of Poincaré’s model, we use this definition of distance to derive the same formula for the angle of parallelism as is valid in Poincaré’s model. Proposition 42.5 (The angle of parallelism in Klein’s model). For any point P and line l, the angle of parallelism π(s) relates the hyperbolic distance s = s(P, Q) from P to the foot point Q of the perpendicular dropped on the line l—by the remarkable Bolyai-Lobachevsky formula π(s) tan = e−s (42.3) 2 Proof. Similar to the proof for Poincaré’s model, we use a special position to get easy calculations. But for Klein’s model, we need to choose a different special arrangement than for Poincaré’s! Because angles can be easily measured only at the center, we have to choose P = O. Moreover, we can choose the foot point Q on the positive y-axis. Thus we get at vertex Q an absolutely correct right angle between the line l and the perpendicular PQ. Let U and V be the ideal endpoints of line l, and in terms of coordinates put V = (x, y). Let Y+ = (0, 1) and Y− = (0, −1) be the ideal ends of the perpendicular OQ. From the definition (42.2) of the hyperbolic distance, we easily get s = s(O, Q) =
1 1 OY+ · QY− 1 1 + y ln(OQ, Y+ Y− ) = ln = ln 2 2 QY+ · OY− 2 1 − y
in terms of the cross ratio (OQ, Y+ Y− ) and the Euclidean distance y = OQ. Applying the exponential function yields s 1+y s e = (42.4) 1−y
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Figure 42.5. The angle of parallelism in Klein’s model.
The angle of parallelism is defined to be the angle between the perpendicular and the asymptotic parallel. For point O and line QV, the distance point to line is s = |OQ|, and the corresponding angle of parallelism is π(s) = ∠VOQ. Next, we need to get tan π(s) 2 . By Euclid III.21, the angle at π(OQ) the circumference is half the angle at the center. Hence 2 = ∠VY− Q. Now the definition of the tangent function, used for the right triangle 4V QY− , is tan
VQ x π(s) = = 2 1 + y Y− Q
Because V = (x, y) is an ideal endpoint, it lies on the unit circle ∂D. Hence Pythagoras’ theorem yields x2 + y2 = 1. One can eliminate x and get s s p 1 − y2 x (1 + y)(1 − y) 1−y = = = 1+y 1+y 1+y (1 + y)2 and hence π(s) tan = 2
s
1−y 1+y
(42.5)
Because of the factor one half in the definition of the hyperbolic distance, everything fits well! Formulas (42.4) and (42.5) yield s π(s) 1−y tan = = e−s 2 1+y which is just Bolyai’s formula (42.3) for the angle of parallelism.
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As a further benefit of Klein’s model, I shall confirmed Bolyai’s construction of the asymptotic parallel ray. Construction 42.1 (Bolyai’s Construction of the Asymptotic Parallel Ray). Given is a line l and a point P not on that line. Drop the perpendicular from P onto line l and let Q be the foot point. Erect the perpendicular onto PQ at point P. One gets a line m parallel to l. Choose a second point R on line l, and drop the perpendicular from that point onto m. Let S be the foot point. So far, we have got a Lambert quadrilateral PQRS . Now one draws a circle of radius QR around the −−→ center P. Let B be the intersection point of that circle with segment RS . Thus one gets a ray PB asymptotically parallel to the given line l.
Figure 42.6. Bolyai’s construction of the asymptotic parallel ray.
Reason. Similar as in proposition 42.5, we use Klein’s model with disk D and put the point P = O at the center of the disk. We can put the foot point Q on a vertical diameter of D. With this layout, all three right angles of the Lambert quadrilateral PQRS appear as right angles in Klein’s model. Indeed, PQ is a vertical radius of D and PS is a horizontal radius of D, and hence the right angles at vertices P, Q and S are absolute right angles. Let the line l = QR have ideal ends U and V. Next, we draw the line c = OV. Let V 0 be its second ideal end. The lines RS and c intersect (why?). 1 We call the intersection point B. Thus line c = OB has the ideal endpoints V and V 0 . The following argument refers to the underlying Euclidean geometry (not to the hyperbolic geometry!). In the drawings, I indicate an hyperbolic right angle by a square, but a right angle for the underlying Euclidean geometry gets a doubled arc. 1
Points R and S lie on different sides of line OV.
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By Thales’ theorem, ∠V 0 UV is a right angle for the underlying Euclidean geometry. Hence, again in the underlying Euclidean geometry, the three lines UV 0 , QP and RS are parallel. Now one uses similar triangles and gets the following proportions: QV QU RU RV QV · RU RV · QU
= = =
OV OV 0 BV 0
=1
BV OV · BV 0 BV · OV 0
Hence the cross ratios and hyperbolic distances are equal, and segments QR and OB are congruent: (QR, VU) = (OB, VV 0 ) s(Q, R) = s(O, B) QR OB
(42.6)
We have checked that this construction produces a line c, which has a common ideal end V with −−→ line l. Hence Bolyai’s construction yields an asymptotic parallel ray OB to line l through point O. Because of the congruence (42.6), Bolyai’s construction works inside the hyperbolic plane without using the ideal endpoint V. Indeed, the construction works without need of a model like Klein’s or Poincaré’s—and was discovered by Bolyai long before either model was known. The essential step is to draw a circle of radius QR around the center P. One get is the intersection point B of that circle with segment RS , and finally the limiting parallel OB.
Figure 42.7. By Martin’s theorem, ∠BCF = π(CA) if and only if ∠ACF = π(CB).
Theorem 42.1 (George Martin’s Theorem). If the angle between an altitude and a side of a triangle is the angle of parallelism of the other side adjacent to the vertex, then the angle between the same altitude and this second side is the angle of parallelism of the first side. To put it into definite terms: Let the triangle 4ABC have acute angles at vertices A and B, and let F be the foot point of the altitude dropped from vertex C onto the side AB. In that situation, ∠BCF = π(CA) if and only if ∠ACF = π(CB)
(42.7)
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Figure 42.8. Proving Martin’s theorem in Klein’s model with vertex C = O in the center.
Proof. We can use Klein’s model with center C = O. In that case, the angles in assertion (42.7) appear absolute, since both have vertex C. Let D be the ideal end of the perpendicular to the ray −−→ CA erected at point A, and lying on the same side as point B. We put the Euclidean circum circle C around the right triangle 4DAC. We now assume that ∠BCF = π(CA). In the drawing on page 780, this angle is named π(b), and the corresponding segment length is s(C, A) = b. From the definition of the angle of parallelism, we conclude ∠DCA = π(CA) = ∠BCF Now the angle sum in the two right triangles 4BCF and 4DCA implies ∠FBC ∠ADC Hence ∠ABC = ∠FBC ∠ADC, and the congruence of circumference angle implies that all four points A, B, C and D lie on the circle C— all this is true only in the Euclidean sense. Hence by Thales’ theorem the angle ∠DBC is right, and indeed absolutely right. Hence the definition of the angle of parallelism implies ∠BCD = π(CB) −−→ −−→ The sum angle ∠BCA can now be partitioned in two ways, using either the interior rays CD or CF. By angle additions and substraction at vertex C one arrives at ∠BCA = ∠BCD + ∠DCA − ∠BCF = π(CB) + β − β = π(CB) In the drawing on page 780, this angle is named π(a), corresponding to the length s(C, B) = a. We are now able to solve the problem converse Bolyai’s construction.
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−r with Problem 42.1 (Find the segment from its angle of parallelism). Let a point C, a ray → vertex C, and the acute angle α be given. Construct a segment CB on the given ray which has the given angle α as its angle of parallelism π(CB) = α.
Figure 42.9. Since ∠BCF = π(CA) by construction, Martin’s theorem implies that ∠ACF = π(CB) as requested.
Construction 42.2 (Construction of a segment with given angle of parallelism). Let CA = b be any other segment, for which we have already constructed the angle of parallelism π(b), using Bolyai’s construction. We transfer this angle π(b) = β −r , and secondly the given angle α onto the newly produced ray, turning in the onto the any ray → −−→ −r the sum angle β + α. same direction. We thus obtain a third ray CA which forms with the first ray → This sum is still less than two right angles, since it is the sum of two acute angles. Onto the third ray, the segment CA of hyperbolic length S (C, A) = b is transferred. We now drop the perpendicular c from point A onto the middle ray. We obtain a foot point F − − −r . Let B be the on the middle ray → m. Indeed the perpendicular to → m intersects the original ray → intersection point. We claim that the segment CB has the angle of parallelism as required: π(CB) = α The main idea. We can now use Martin’s theorem: since ∠BCF = π(CA) by construction, Martin’s theorem implies that ∠ACF = π(CB) as requested.
Detailed reason for validity. The argument above does not explain, why the perpendicular c and −r intersect in the hyperbolic sense, as claimed. the ray → We need to repeat the details for a proper justification. We can use Klein’s model with center C = O. In that position, the angles α and β constructed above appear absolute, because they all
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−−→ have vertex C. Let D ∈ ∂D be the ideal end of the perpendicular to the ray CA at point A, lying on the same side as foot point F. By the definition of the angle of parallelism, ∠DCA = π(CA) = β In the sense of Euclidean geometry, both ∠ABC = ∠FBC = R − π(CA) and ∠ADC = R − π(CA). The congruence of these angles ∠ADC ∠ABC implies that the four points A, B, C and D lie on a circle C in the Euclidean sense. This circle is the Euclidean circum circle of the right triangle 4DAC. By the converse Thales theorem, this circle has diameter CD, which is a radius of Klein’s disk. Hence circle C touches the line of infinity ∂D from inside at the ideal endpoint D. We can conclude that point B lies inside −r intersect in the hyperbolic sense in point B, Klein’s disk. Thus the perpendicular c and the ray → as claimed. The remaining details are easy by now: By Thales’ theorem, the angle ∠DBC is a right angle. This is indeed an absolute right angle, because its side BC goes through the center of the Klein disk. Hence, by the definition of the angle of parallelism, π(CB) = ∠DCB = ∠ACB − ∠DCA = α + β − β = α
as to be shown.
Problem 42.2. For a given angle α, construct a segment with angle of parallelism pi(a) = α. Explain Bolya’s construction in elementary steps, and minimize the number of transfers needed. Answer. Let α = ∠ACQ be the given angle, with AC QC = b conveniently chosen. We erect the perpendiculars l at Q and p at C onto the segment QC. On the perpendicular l, we transfer a segment QR QC, and drop the perpendicular from point R onto p. The foot-point is called S . We draw the circle about C through points A and Q. It intersects the segment RS in point −−→ −−−→ B1 . Bolya’s construction tells that the rays QR and CB1 are asymptotically parallel, and hence ∠QCB1 = π(CQ). We drop the perpendicular from point A onto the segment CQ. By Martin’s theorem, it −−−→ intersects the asymptotic ray CB1 in a point B, and the segment CB has the angle of parallelism α as requested. Problem 42.3. Give an illustration of this construction in Klein’s model with vertex C at the center of the disk. 42.3.
Projective nature of Klein’s model
Definition 42.6 (The projective plane). On the set R3 \{(0, 0, 0)}, an equivalence relation is defined by setting any two points (x, y, z) ∼ (λx, λy, λz) with λ , 0 equivalent. The quotient space by this equivalence R3 /∼ is called the projective plane and denoted by PR2 . Too, the points of R3 \(0, 0, 0) are called the homogeneous coordinates of the projective plane. The projective plane has as an subset the improper line consisting of the points with the homogeneous coordinates (x, y, 0). The projective plane can be viewed as the union of the usual Euclidean plane with the improper line. 1 For the points of usual Euclidean plane one can use for example as homogeneous coordinates (x, y, 1). For convenience, we define the z-slice plane P1 := {(x, y, 1) : x, y ∈ R} 1
Because of its improper points and line, the projective plane has a different topological structure that the Euclidean plane. Especially, the Restricted Jordan Curve Theorem 5.13 from the section on incidence geometry, is not valid for the projective plane.
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Figure 42.10. Construction of the segment a = OB from given angle of parallelism α.
For any nonsingular 3 × 3 matrix A, the linear mapping x ∈ R3 7→ Ax ∈ R3 , induces a mapping φA : x ∈ PR2 7→ x0 = Ax ∈ PR2 , by means of the homogeneous coordinates. For the points of the Euclidean plane, this mapping is given by the fractional linear transformations a11 x1 + a12 x2 + a13 a21 x1 + a22 x2 + a23 x10 = and x20 = (42.8) a31 x1 + a32 x2 + a33 a31 x1 + a32 x2 + a33 The mapping of the points on the improper line can be rather easily deduced from these formulas. The details can be left to the reader. Definition 42.7 (Projective mapping). The extensions of the fractional linear mappings (42.8) with det A , 0, to the projective plane are called projective mappings. Main Theorem 33. Given are four points x1 , x2 , x3 , x4 ∈ PR2 with no three of them lying on a line. Similarly, there are given any four image points x10 , x20 , x30 , x40 ∈ PR2 . Once more, it is assumed that no three of them lie on a line. Then there exists exactly one projective mapping which takes xi to xi0 for i = 1, 2, 3, 4. Proof. Let x1 , x2 , x3 , x4 ∈ R3 \ {0} be any homogenous coordinates of the four given points xi ∈ PR2 . Since any four vectors in R3 are linearly dependent, there exists λ1 , λ2 , λ3 , λ4 ∈ R such that λ1 x1 + λ2 x2 + λ3 x3 + λ4 x4 = 0
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Since by assumption no three of the four points xi lie on a line, all four λi , 0 are nonzero. The matrix A = [λ1 x1 , λ2 x2 , λ3 x3 ] maps p1 = (1, 0, 0) 7→ λ1 x1 , p2 = (0, 1, 0) 7→ λ2 x2 , p3 = (0, 0, 1) 7→ λ3 x3 , p4 = (1, 1, 1) 7→ −λ4 x4 , and is nonsingular. The induced projective mapping φA : x ∈ PR2 7→ x0 = Ax ∈ PR2 , takes the points fi 7→ xi for i = 1, 2, 3, 4. By means of composition A0 ◦ A−1 , we find a projective mapping such that xi ∈ PR2 7→ xi0 ∈ PR2 where the four preimages xi and images xi0 can be arbitrarily prescribed.
Main Theorem 34 (Projective invariance of the cross ratio). Let P1 , P2 , P3 , P4 be any four points on a line and Q1 , Q2 , Q3 , Q4 their images by a projective mapping. Then the four points Qi lie on a line, too. The cross ratios (P1 P2 , P3 P4 ) = (Q1 Q2 , Q3 Q4 ) are equal. Definition 42.8. The projective mappings which leave the circle of infinity ∂D invariant are called automorphic collineations. With composition of mappings as group operation, the automorphic collineations form a group. The following fact shows that it is a rather large group. Proposition 42.6. Given is any point A inside the disk D and its image A0 inside the disk. Too, there is given any ideal point U on the circle of infinity and its image U 0 , again on the circle of infinity. Then there exist exactly two automorphic collineations mapping A to A0 and U to U 0 . One of these two mappings preserves the orientation, the other one reverses the orientation of the hyperbolic plane. Proof. Let x1 := U and x2 be the second ideal end of the chord UA. Let x3 be the intersection point of the tangent to the circle of infinity ∂D at points x1 and x2 . Let x4 be one of the intersection points of the line of ∂D with the line Ax3 . No two of the four points x1 , x2 , x3 , x4 lie on a line. The same construction has to be done with primed image points. By the Main Theorem 33, there exists a projective mapping taking xi to xi0 for i = 1, 2, 3, 4. This mapping takes the tangents to ∂D at points xi to the tangents at xi0 for i = 1 and i = 2. Counting multiplicity, five points of the circle of infinity are mapped to five other points of this line. Since a projective mapping takes a conic section to a conic section, and a conic section is determined by five of its points, 1 the entire circle of infinity is mapped to itself. Hence the projective mapping constructed above is an automorphic collineation. Because of the arbitrary choice of x40 among the two intersections of line A0 x30 and ∂D, one gets two solutions. These are the only solutions, because lines are mapped to lines and tangents to ∂D are mapped to tangents to ∂D. Proposition 42.7. Given any three different ideal points U, V, W on the circle of infinity and three different images U 0 , V 0 , W 0 , again on the circle of infinity. Then there exists exactly one automorphic collineation mapping U 7→ U 0 , V 7→ V 0 , W 7→ W 0 . It depends on the order of the given points, whether the mapping preserves or reverses the orientation of the hyperbolic plane. 1
This fact can be shown using Pascal’s magic hexagon.
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Figure 42.11. There exist two automorphic collineations mapping A to A0 and U to U 0 .
Answer (Proof of Proposition 42.7). Let x1 be the intersection point of the tangents at U and V to ∂D, and x10 be the intersection point of the tangents at U 0 and V 0 to ∂D. Of the four points x1 , U, V, W no three lie on a line, and similarly for the points x10 , U 0 , V 0 , W 0 By the Main Theorem 33, there exists exactly one projective mapping taking x1 7→ x10 and U 7→ U 0 , V 7→ V 0 , W 7→ W 0 . This mapping takes the tangents to ∂D at points U and V to the tangents at U 0 and V 0 . Counting multiplicity, five points of the circle of infinity are mapped to five other points of this line. Since a projective mapping takes a conic section to a conic section, and a conic section is determined by five of its points, the entire circle of infinity is mapped to itself. Hence the projective mapping constructed above is an automorphic collineation. On the other hand, the obtained mapping is unique, since any automorphic collineation with the required property necessarily maps x1 to x10 . Proposition 42.8. Given any ideal point U on the circle of infinity and any two different points A and A0 on the tangent to the circle ∂D at point U. Then there exists exactly one automorphic collineation which keeps U fixed, map A to A0 and has no other fixed point. This mapping preserves the orientation of the hyperbolic plane. Proof of Proposition 42.8. Let X and X 0 be the touching points of the tangents from points A and A0 to ∂D. We let Z 0 be the intersection point of these tangents and draw the line b = UZ 0 . Let Y 0 be the second end of line b, let Y be the second end of line AY 0 . Of the four points A, X, Y, U nor of the four points A0 , X 0 , Y 0 , U no three lie on a line. By the Main Theorem 33, there exists exactly one projective mapping taking A 7→ A0 , X 7→ X 0 , Y 7→ Y 0 , and leaving point U 7→ U fixed. This mapping takes the tangents to ∂D at points U and X to the tangents at U and X 0 since A is mapped to A0 . Furthermore point Y ∈ ∂D is mapped to Y 0 ∈ ∂D. Counting multiplicity, five points of the circle of infinity are mapped to five other points of this circle. Hence the prescribed mapping is an automorphic collineation.
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Figure 42.12. An automorphic collineation that transports points along horocycles.
On the other hand, the obtained mapping is unique, since any automorphic collineation with the required property necessarily maps A 7→ A0 , X 7→ X 0 , Y 7→ Y 0 and leaving point U 7→ U fixed. It is left to the reader to check that X, Y, U and X 0 , Y 0 , U define the same orientation of the circle ∂D. Problem 42.4. Convince yourself that the automorphic collineation constructed above is the composition Rb ◦ Ra of the two reflections across lines a and b. Solution. The reflection Ra takes X 7→ X, Y 7→ Y 0 , leaving point U 7→ U fixed. The reflection Rb takes X 7→ X 0 , Y 0 7→ Y 0 , leaving point U 7→ U fixed. Hence the composition Rb ◦ Ra takes X 7→ X 0 , Y 7→ Y 0 , leaving point U 7→ U fixed. Since it is an automorphic collineation and preserves the orientation, it is uniquely specified by this property. Since the given mapping S is an orientation preserving automorphic collineation with the same property, we conclude S = Rb ◦ Ra .
The development of Klein’s model is based on the fact: Main Theorem 35. The automorphic collineations are the isometries of the hyperbolic plane. They leave the hyperbolic distances and angles invariant. Proof. The invariance of the distances is an easy consequence of the projective invariance of the cross ratio. The hyperbolic distance any two points K and L is defined by s(K, L) =
1 1 KE · LF ln(KL, EF) = ln 2 2 LE · KF
where E and F are the ideal endpoints of the line KL, ordered such that E ∗ L ∗ K ∗ F.
(42.9)
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Any automorphic collineation maps these four points to points E 0 , L0 , K 0 , F 0 , lying again on one line, and E 0 and F 0 lying on the circle of infinity ∂D. Hence the latter two points are the ideal ends of the image line K 0 L0 and have the hyperbolic distance s(K 0 , L0 ) =
1 1 K 0 E 0 · L0 F 0 ln(K 0 L0 , E 0 F 0 ) = ln 2 2 L0 E 0 · K 0 F 0
(42.10)
Hence the invariance of the cross ratio by projective mappings implies KE · LF LE · KF
=
K 0 E 0 · L0 F 0 L0 E 0 · K 0 F 0
and hence s(K, L) = s(K 0 , L0 ), as to be shown.
The invariance of the angles allows for measurement of angles. To this end, one maps the given angle by an automorphic collineation, taking its vertex A to the center O. Because the image angle has its vertex at the center, it appears as an absolute angle and can be measured by the geometry of the Euclidean plane. Secondly the construction used in the proof of proposition 42.6 confirms once more the criterium for right angles given in proposition 42.4. Our next goal is the construction of a hyperbolic reflection. It turns out that the key figure is the ideal quadrilateral. Proposition 42.9 (An ideal quadrilateral produces five right angles). Let the ideal quadrilateral ABCD have diagonals intersecting at P, and drop perpendicular l from point P onto side AB. This perpendicular l is perpendicular to both opposite sides AB and CD. Next we erect the perpendicular p onto l at point P. This second perpendicular p is perpendicular to the other two opposite sides BC and DA. Furthermore, the three lines AB, CD and p meet in one ultra ideal point, and the three lines BC, DA and l meet in another one.
Proof. There exists an automorphic collineation that maps the point P to the center P0 = O. Hence the special position with the intersection of the diagonals at the center of the disk can be achieved by means of an automorphic collineation, which is an isometry of the hyperbolic plane. But in that special position, the diagonals and the lines l and p all intersect at the center O. Thales’ theorem now implies that the ideal quadrilateral ABCD appears as a rectangle. The five right angles appear as absolute right angles, valid both in the underlying Euclidean plane and in hyperbolic geometry. Reason via the Poincar´e model. Put the figure into the Poincaré model, and use the special position with the intersection of the diagonals at the center of the disk. As shown in the section on the Poincaré model, this special position with P = O can be achieved by means of hyperbolic reflections. But in that special position, the diagonals and the lines l and p are absolute straight lines, and the ideal quadrilateral ABCD appears as a rectangle. The five right angles are immediate to confirm. We now use the ideal quadrilateral for the construction of a hyperbolic reflection.
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Figure 42.13. An ideal quadrilateral produces five right angles.
Figure 42.14. Construction of the reflective image K 0 for a given point K and reflection line l.
Construction 42.3 (Hyperbolic reflection of a point by a given line). Given is a reflection line l and a point K not on line l. We want to construct the reflective image of K 0 of K by the line l. Choose any parallel to l with ideal ends B and C, and get the polar BC ⊥ as intersection point of the tangents at B and C to the circle of infinity ∂D. The line p through the points K and BC ⊥ is the perpendicular dropped from K onto line l. The foot point P is the intersection of lines p and l.
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Finally, to get the reflective image, we draw the line CP with the second end A, and BP with the second end D, thus producing an ideal quadrilateral ABCD. The reflection point K 0 is the intersection of side AD with the perpendicular p. Remark. The four points K, P, the polar BC ⊥ , and the intersection of lines AB and CD lie all on the perpendicular p. This property should be used to achieved better accuracy.
Figure 42.15. Construction of the true hyperbolic angle ∠OKF.
Construction 42.4 (True hyperbolic angle). Given is a hyperbolic angle ∠OKF. We draw the ends E and F of the line EF and its polar C = EF ⊥ . The circle around C through the ends E and F intersects the segment OK in point P. We draw segments EB and FA through point P and obtain the ends A and B. These points are actually the endpoints of a diameter. The true hyperbolic angle is ω = ∠KOA. Proposition 42.10 (Distortion of angles). The measure of the hyperbolic angle ∠OKF can be obtained by the construction from the figure on page 789 by a reflection across the point P. One obtains the true hyperbolic angle ω = ∠KOA with vertex at the center. By the formula √ tan ω = tan α 1 − r2 (42.11) the true angle ω is calculated from the apparent angle α. Justification of the construction. The mapping from Poincaré’s to Klein’s model as explained in proposition 42.1 gives 2 |OP| |OK| = (42.1) 1 + |OP| 2
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With the short-hands r = |OK| and p = |OP| we get 2p 1 − r (1 − p)2 (1 − r)(1 + p) (1 − p) = = , and 1 + r (1 + p)2 (1 + r)(1 − p) (1 + p) 1 + p2 Hence point P is the hyperbolic midpoint of segment OK. The hyperbolic point reflection across P maps K 7→ O, E 7→ B, F 7→ A. Hence points A, B and O lie on a line, and the angle ∠PKF is reflected to the congruent angle ∠PKA, as claimed. r=
Justification of the formula (46.21). With the cos theorem for triangles 4PAO and 4PBO we get |PA|2 = 1 + p2 − 2p cos ω , |PB|2 = 1 + p2 + 2p cos ω |PA|2 |PB|2 = (1 + p2 )2 − 4p2 cos2 ω = (1 − p2 )2 + 4p2 sin2 ω We apply the sin theorem for all four triangles 4PAO, 4PBO, 4PEK, 4PFK and get sin2 ω |PE||PF| sin x sin y |KE||KF| = |PA||PB| sin2 α |OA||OB| sin x sin y sin2 ω |PA||PB| |KE||KF| = |KE||KF| = |PA|2 |PB|2 2 |PA||PB||PE||PF| sin α |PE||PF| h i (1 − r)(1 + r) = (1 − p2 )2 + 4p2 sin2 ω (1 − p)2 (1 + p)2 Reminding r=
2p 4p2 (1 − p2 )2 r2 4p2 2 2 , r = , 1 − r = , = 1 + p2 (1 + p2 )2 (1 + p2 )2 1 − r2 (1 − p2 )2
we get !2 2p sin2 ω 2 (1 − r2 ) = 1 − r2 + r2 sin2 ω sin ω 1 + = 1 − p2 sin2 α (1 − r2 ) sin2 α 1 − r2 sin2 α 1 − sin2 α cos2 ω = 1 − r2 sin2 α 2 tan ω = (1 − r2 ) tan2 α sin2 ω =
Moreover, we see that the angle ω is acute if and only if the angle α is acute, ω is right if and only if α is right, and ω is obtuse if and only if α is obtuse. Hence we get formula (46.21) with the positive square root. 42.4. Engel’s Theorem In this paragraph, we state and prove Engel’s theorem. This theorem constructs a bijective correspondence between right triangles and Lambert quadrilaterals. Klein’s model is convenient for the proof, which uses a refinement of Bolyai’s construction. Recall that quadrilaterals with three right angles are called Lambert quadrilaterals. Right triangles and Lambert quadrilaterals both have five basic pieces (angles or sides), and can be constructed once any two of them are given. Proposition 42.11 (Transfer of a segment on its line). Let E and F be the ideal endpoints of line c. We put any two ideal points A, B on one of the two arcs from E to F, and two more points V, W on the other arc on ∂D from E to F. Let P2 P3 and Q2 Q3 be the segments on line l cut out by the angles ∠V AW and ∠V BW, respectively. The two segments P2 P3 and Q2 Q3 are hyperbolically congruent.
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Figure 42.16. Transfer of a segment along its line.
Proof. We need these two basic facts: (Invariance of the cross ratio for central projections). Given are any two lines across a bundle of four rays with common vertex. Let P1 , P2 , P3 , P4 and Q1 , Q2 , Q3 , Q4 be the intersection points of the rays with the two lines across the bundle. Then the cross ratios (P1 P2 , P3 P4 ) = (Q1 Q2 , Q3 Q4 ) are equal. The same statement holds for two lines intersecting two congruent ray bundles. (Euclid’s III.21). Two angles from points of a circle subtending the same arc are congruent. −−−−−−−−−−−→ −−−−−−−−−−−→ By Euclid III.21, the two ray bundles A(E, V, W, F) and B(E, V, W, F) are congruent. The two ray bundles intersect line EF in the four points E, P2 , P3 , F and E, Q2 , Q3 , F, respectively. Now we use fact 1 with E = P1 = Q1 and F = P4 = Q4 and get (P2 P3 , EF) = (Q2 Q3 , EF) as to be shown.
and
s(P2 , P3 ) = s(Q2 , Q3 )
For s given, one defines s∗ to be the hyperbolic length for which π(s) + π(s∗ ) = 90◦ . The pieces of 4ABC are denoted in Euler’s standard fashion. Proposition 42.12 (Engel’s Theorem). There is a bijective (one-to-one and onto) correspondence between right triangles and Lambert quadrilaterals PQRS . The Lambert quadrilateral PQRS and the right triangle 4ABC are matched by putting one leg of the Lambert quadrilateral onto one leg of the right triangle, and one of the outer right angles of the Lambert quadrilateral onto the right angle of the triangle. Thus A = P and C = S . The correspondence is established by requiring (42.13) and any one of the other four statements (42.12),(42.14),(42.15) or (42.16).
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In that way, the five pieces of the triangle and of the Lambert quadrilateral are matching as follows: a := BC has angle of parallelism π(a) = ∠QRS b := AC = PS c := AB = QR ∗ α := ∠BAC = π(l ) with l = PQ β := ∠CBA = π(m) with m = RS
(42.12) (42.13) (42.14) (42.15) (42.16)
Figure 42.17. The correspondence of a Lambert quadrilateral and a right triangle.
Proof. We use Klein’s disk model with disk D and choose as center O = A = P. The bijective correspondence of 4ABC and PQRS is given by requiring that (i) A = P and C = S . −−→ (ii) vertex B lies on the ray S R. (iii) Hypothenuse AB and quadrilateral side QR have a common ideal endpoint V. In the figure on page 778 about Bolyai’s construction of the asymptotic parallel ray, one can see at once that requirements (i)(ii)(iii) define a bijection between right triangles and Lambert quadrilaterals. For this correspondence, claim (42.13) is obvious. Claim (42.14) and (42.15) both follow from Bolyai’s construction. Reason of claim (42.15). Indeed point P and line UV have the angle of parallelism ∠V PQ, as −−→ angle between the perpendicular PQ and the asymptotic parallel ray PV. Because ∠V AQ is the complementary angle of α = ∠CAB, one gets π(PQ) = 90◦ − α and hence α = π(s(P, Q)∗ ). This confirms claim (42.15). Reason of claim (42.12). As shown in figure 42.4, we draw line V 0 S and let W be its second ideal endpoint. Draw line US and let X be its second ideal endpoint. The lines RS and PS are perpendicular, and intersect in S . The ideal quadrilateral V 0 UWX has its diagonals intersecting in S . Its side UV 0 is perpendicular to PS . Hence by proposition 42.9,
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the opposite side WX is perpendicular to PS , too. Furthermore, the other two opposite sides UW and V 0 X are perpendicular to RS . Thus the quadrilateral V 0 UWX, its diagonals, and the two perpendicular lines RS and PS produce the English flag and five right angles, as shown in figure 42.4. Now we complete the proof of claim (42.12): Let H be the intersection point of the two perpendicular lines UW and RS . It can happen that H lies inside the segment S B, or inside the segment BR, or H = B. The figure 42.4 shows the case that H lies inside the segment BR. Let E and F be ideal endpoints of line RS = EF. As explained in proposition 42.11 about the −−−−−−−−−−−−→ −−−−−−−−−−−→ transfer of a segment on its line, the two ray bundles U(E, V, W, F) and V 0 (E, V, W, F) are congruent. Hence the two segments cut out on line EF are congruent: (HR, EF) = (S B, EF)
and
s(H, R) = s(S , B) = a
(42.17)
The angle of parallelism of segment RH is ∠URH, because it is the angle between the asymptotic −−→ parallel ray RU and the perpendicular RH, for point R and line UW. Hence π(RH) = ∠URH. Combined with (42.17), we conclude Hence π(a) = π(RH) = ∠URH = ∠QRS as claimed in (42.12).
Figure 42.18. Find the English flag.
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Figure 42.19. Proof of claim (42.12): a has angle of parallelism π(a) = ∠URH.
Reason of claim (42.16). This is done in the same manner we just have proved (42.12). This time, we use line US and let X be its second ideal endpoint. As shown above, the two segments UW and V 0 X are both perpendicular to RS . Let G be the intersection point of the perpendicular lines V 0 X and S R. −−−−−−−−−−−−→ −−−−−−−−−−−→ Again by Euclid III.21, the two ray bundles U(E, V, X, F) and V 0 (E, V, X, F) are congruent. The two ray bundles intersect line EF in the four points E, R, S , F and E, B, G, F, respectively. Hence proposition 42.11 implies (RS , EF) = (BG, EF)
and
s(R, S ) = s(B, G) = m
(42.18)
Referring to point B and line V 0 X, we see that the angle of parallelism of segment BG is ∠V 0 BG, −−→ because it is the angle between the asymptotic parallel ray BV 0 and the perpendicular BG. Hence π(BG) = ∠V 0 BG. Combining with (42.18), we conclude that π(m) = π(RS ) = π(BG) = ∠V 0 BG = ∠ABC = β, as claimed in (42.16). 42.5.
The Hjelmslev quadrilateral
Theorem 42.2. Hjelmslev’s theorem 27.2 about the quadrilateral with right angles at two opposite vertices holds in hyperbolic and neutral geometry, too.
Problem 42.5. Prove the angle congruence of Hjelmslev’s Theorem in hyperbolic geometry. Assume that the Euclidean theorem has already been shown. Use Klein’s model, and put into the center of the disk the vertex of the pairs of angles that you want to compare.
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Figure 42.20. Proof of claim (42.16): m = RS has angle of parallelism π(m) = ∠V 0 BG.
Figure 42.21. The angle congruence of Hjelmslev holds in hyperbolic geometry, too.
Proof of the angle congruence in hyperbolic geometry. I use Klein’s model with the vertex B put into the center of the disk. Because angles at the center of the disk are absolute (not distorted), it suffices to compare the Euclidean angles at B. The right angles of the Hjelmslev quadrilateral are absolute right angles, too, since one side of them is a diameter of the Klein disk. Hence the hyperbolic Hjelmslev quadrilateral appears in Klein’s model as a Euclidean Hjelmslev
796
quadrilateral, too. Now we can use the Euclidean version of the theorem, and conclude Euclidean congruence of the angles β := ∠ABD and β0 = ∠GBC. Because angles at the center of the disk are not distorted, the Euclidean congruence implies the hyperbolic congruence. Problem 42.6 (Open problem). Prove the segment congruence of Hjelmslev’s Theorem in hyperbolic geometry. Assume that the Euclidean theorem has already been shown and use Klein’s model.
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43. Hyperbolic Triangle Geometry
Figure 43.1. The trigonometry of the right triangle.
43.1.
Hyperbolic trigonometry
Theorem 43.1 (The hyperbolic right triangle). The sides and angles of any hyperbolic right triangle with γ the right angle satisfy sinh a and sinh c tanh b cos α = and tanh c cosh c = cosh a cosh b cos α cosh a = and sin β cosh c = cot α cot β sin α =
sinh b sinh c sinh a cos β = sinh c
sin β =
(43.1) (43.2) (43.3)
cosh b =
cos β sin α
(43.4) (43.5)
Proof of the formula (43.2) for the cos of an angle. The formula has to be derived from the geometry in the Poincaré and Klein model, which are best used together. We depict the given right triangle 4ABC in the Poincaré model with vertex A = O at the center. The opposite side is depicted as a circle around a⊥ . Let the transformation to the Klein model produce triangle 4ALK. Still, the right angle at vertex C is transformed to a right angle at K, since AK is a diameter.
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We use the underlying Euclidean geometry for the 4ALK and obtain cos α =
|OK| tanh b = |OL| tanh c
We have used that the hyperbolic distance to the center in the Klein model is given by b = s(O, K) =
1 1 + |OK| ln = tanh−1 |OK| 2 1 − |OK|
Hence |OK| = tanh b, and similarly |OL| = tanh c.
Proof of the formula (43.1) for the sin of an angle. We use the same setting as in the proof given above. Once more, the given right triangle 4ABC is depicted with vertex A = O at the center. In the Poincaré disk model, the opposite side is depicted as a circle around a⊥ = K 0 . The translation to Klein’s model produces the points C 7→ K and B 7→ L, as well as for their images C 0 7→ K 0 and B0 7→ L0 by circular inversion. The underlying Euclidean geometry for the 4K 0 L0 B yields sin β =
|BL0 | |BL0 | |BB0 | = = 0 0 |BK | |CK | |CC 0 |
We use now the hyperbolic distance to the center in the Poincaré model: b = s(O, C) = ln
1 + |OC| = 2 tanh−1 |OC| 1 − |OC|
Hence |OC| = tanh b2 and |CC 0 | = |OC 0 | − |OC| = Similarly we get |BB0 | =
2 sinh c .
2 sinh(b/2) cosh(b/2) − = cosh(b/2) sinh(b/2) sinh b
Thus the required sin is sin β =
|BB0 | sinh b = |CC 0 | sinh c
All remaining formulas can be derived using only trigonometric identities. Proof of the Pythagorean formula (43.3). Because of the Pythagorean theorem in its Euclidean form, we know that tanh2 b sinh2 a + tanh2 c sinh2 c sinh2 c · cosh2 b = sinh2 b · cosh2 c + sinh2 a · cosh2 b 1 = cos2 α + sin2 α =
(1 + sinh2 c) · cosh2 b = sinh2 b · cosh2 c + (1 + sinh2 a) · cosh2 b cosh2 c · cosh2 b − sinh2 b · cosh2 c = cosh2 a · cosh2 b cosh2 c = cosh2 a · cosh2 b cosh c = cosh a · cosh b
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A side in terms of two angles (43.4) . We get from the hyperbolic Pythagorean theorem cosh a =
cosh c tanh b sinh b cos α = : = cosh b tanh c sinh c sin β
Similarly, we get cos β sin α
cosh b =
Formula (43.5) now follows immediately from the hyperbolic Pythagorean theorem (43.3).
Problem 43.1 (The hyperbolic area of circle, calculated in Archimedes fashion). We use a regular n-gon with radius of circum-circle R to get the area of the circle from the of the defect, taking the limit n → ∞. (a) We calculate cot α for a right triangle with hypothenuse R and β = πn . (b) From the defect δ = n-gon to be
π 2
− α − β of one of 2n right triangles, we add up the total defect for the ∆n = 2n
π 2
− α − 2π
(c) We need the first term in the expansions a0 + an1 + na22 + . . . for the quantities π tan − α = cot α = 2 π −α= 2 ∆n =
(43.6) (43.7) (43.8)
From the last expansion, we calculate limn→∞ ∆n , which is the area of the circle with radius R. Answer. (a) The regular n-gon consists of n nonoverlapping isosceles triangles. We drop the perpendiculars from the center O and get 2n right triangles. Each one of them has the angle β = πn at the center and the hypothenuse R. From the trigonometry of the right triangle we get cot α cot β = cosh R π cot α = (cosh R) tan n (c) We need the first term in the expansions a0 + an1 + an22 + . . . for the quantities π π π − α = cot α = (cosh R) tan = (cosh R) + O(n−3 ) tan 2 n n π π π −3 − α = arctan (cosh R) + O(n ) = (cosh R) + O(n−3 ) 2 n n π −2 ∆n = 2n − α − 2π = 2π(cosh R − 1) + O(n ) 2 R lim ∆n = 2π(cosh R − 1) = 4π sinh2 n→∞ 2 which is the area of the circle with radius R.
(43.9) (43.10) (43.11) (43.12)
800
43.2.
The orthocenter
Theorem 43.2 (The conditional orthocenter—hyperbolic version). The three altitudes of an acute or right triangle always intersect in one point. For an obtuse triangle, the three altitudes may intersect or not. There are three possible cases: (i) The three altitudes of an obtuse triangle intersect in one point. (ii) The three altitudes of an obtuse triangle are all divergent parallel to each other. There exists a line p perpendicular to all three altitudes. (iii) The three altitudes of an obtuse triangle are all asymptotically parallel to each other. Corollary 76. 1 If any two altitudes intersect, then all three altitudes intersect in one point. If the altitudes of two sides of a triangle have a common perpendicular, then the altitudes of all three sides have a common perpendicular. If the altitudes of two sides of a triangle are asymptotically parallel, then the altitudes of all three sides are asymptotically parallel.
Figure 43.2. If any two altitudes intersect, one can put the intersection point at the center of Klein’s disk. Three drawings are given: for an acute, right, or obtuse triangle.
We need to clarify some terms about the use of any mathematical models, as Klein’s or Poincaré’s: Definition 43.1. A theorem or a feature of a figure is part of neutral geometry if and only if it can be deducted assuming only the axioms of incidence, order, congruence. The facts of neutral geometry are valid in both Euclidean and hyperbolic geometry—as well as the more exotic non-Archimedean geometries. Definition 43.2. A feature of a figure drawn inside Klein’s model (as for example an angle, midpoint, altitude or bisector) is called absolute if it appears in the same way both for the underlying Euclidean plane, on which the model is based, and the hyperbolic geometry inside the model. 1
The first two sentences of the Corollary are stated in neutral geometry. A proof of that part in neutral geometry is clearly valid a real—but not Klein-bottle—of wine.
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Remark. Here are some features that appear absolute, both as features of hyperbolic geometry and in the underlying Euclidean plane: An angle with the center of Klein’s disk appears as an absolute angle. A right angle of which one side is a diameter appears absolute. A perpendicular bisector or an angle bisector which is a diameter appears absolute. Proof, using Klein’s model of hyperbolic geometry. As stated in Proposition 11.6, any two altitudes of an acute or right triangle do intersect. It can, but does not need to happen that the altitudes of an obtuse triangle intersect. These are simple facts in neutral geometry. If any two altitudes of the triangle intersect, we obtain an easily understandable picture in Klein’s model: just put the intersection point H into the center of Klein’s disk. In the figure on page 800, I have given illustrations for an acute, right and an obtuse triangle. Why does that picture come out that simple? Here are the logical steps for the reason: The right angles at the foot points of the two intersecting altitudes appear undistorted as absolute right angles. We can now apply Theorem 25.3 from Euclidean geometry to the Euclidean triangle in the underlying Euclidean plane of Klein’s model. Hence the third Euclidean altitude passes through the intersection point of the two other ones, which we have put at the center O. Because the right angle at the foot point of the third altitude is depicted undistorted, too, all three altitudes are both Euclidean as well hyperbolic altitudes. Hence the three hyperbolic altitudes intersect at the center H. For an obtuse triangle, two new cases (ii) and (iii) do occur in hyperbolic geometry. They are illustrated in the figure on page 801. Of course, one can no longer put the orthocenter in the center of Klein’s disk, because it does not exist! Instead, we put the vertex C with the obtuse angle into
Figure 43.3. The altitudes of an obtuse triangle may or may not intersect. A convenient drawing puts the vertex with the obtuse angle at the center of Klein’s disk.
the center of the disk. In that way, still the three right angles at the foot points of the altitudes are depicted as absolute right angles. All three altitudes are both Euclidean as well as hyperbolic altitudes. Depending where the altitudes intersect, we get the three cases: (i) The intersection point H lies inside Klein’s disk. The three altitudes intersect in one point. (ii) The intersection point H lies outside Klein’s disk. The three altitudes are all divergent parallel to each other. There exists a line l perpendicular to all three altitudes. (iii) The intersection point H lies on the boundary of Klein’s disk. The three altitudes are all asymptotically parallel to each other.
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In the "genuine hyperbolic case" (ii), point H lies outside Klein’s disk. Hence H is not a point of the hyperbolic plane, but a so-called ultra-ideal point. The polar of point H yields a common perpendicular l to the three altitudes. Recall that the boundary circle δD of Klein’s disk is called the circle of infinity. Finally, we get the "borderline case" (iii) if point H lies on the circle of infinity. Neither two of the three altitudes intersect, nor do any two of them have a common perpendicular. 43.3. About the circum-circle We begin by recalling Propositions 11.2 and 11.4 from neutral triangle geometry. Theorem 43.3 (The perpendicular bisectors and their meaning). Given any triangle, three different cases can occur. "As seen in Euclidean geometry" The three perpendicular bisectors intersect in one point. The triangle has a circum-circle, and the intersection point is the center of the circum-circle. This case always occurs for an acute or right midpoint-triangle. Too, it can—but does not need to occur—if the midpoint-triangle is obtuse. The orthocenter of the midpoint triangle 4Ma Mb Mc , is the circum-center of the larger original triangle 4ABC, too. H2 = O (H2O) "The genuine hyperbolic case" The three bisectors are all divergently parallel to each other. There exists a line l perpendicular to all three bisectors. All three vertices have the same distance from line l. Hence there exists an equidistance line through the three vertices of the triangle. "The borderline case" The three perpendicular bisectors are all asymptotically parallel to each other. Neither two of the three bisectors intersect, nor do any two of them have a common perpendicular. The three vertices lie neither on a circle nor an equidistance line. There exists a horocycle through the three vertices of the triangle. In hyperbolic geometry, all three cases do occur, whereas Euclidean geometry leads always to the first case. Corollary 77. Any two perpendicular bisectors of sides of a triangle intersect iff the bisectors of all three sides intersect in one point iff the triangle has a circum-circle. The perpendicular bisectors of any two sides of a triangle have a common perpendicular iff the perpendicular bisectors of all three sides have a common perpendicular iff the three vertices lie on an equidistance line. The perpendicular bisectors of two sides of a triangle are asymptotically parallel iff the perpendicular bisectors of all three sides are asymptotically parallel iff the three vertices lie on a horocycle. Proof for hyperbolic geometry, using Klein’s model. Most statements just repeat Proposition 11.7 from neutral geometry. As a new feature in hyperbolic geometry, we can use Klein’s model. Thus we can confirm that all three mentioned cases actually do occur. Too, the meaning of the "borderline case" is clarified. By Corollary 30, the bisectors of the sides of the original triangle 4ABC are the altitudes of the midpoint-triangle 4Ma Mb Mc . Thus they form six right angles, in neutral geometry.
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Figure 43.4. The bisectors and altitudes of the midpoint triangle inside Klein’s disk model. For a triangle with circum-center, three drawings are given: for an acute, right, or obtuse midpoint triangle.
We start with the case that any two altitudes of the midpoint-triangle do intersect. By Corollary 30, the altitudes of the midpoint-triangle are side bisectors for the original triangle. Now Proposition 11.2 implies that all three side bisectors intersect in one point. This point is the circum-center of the original triangle and the orthocenter of the midpoint-triangle as stated in formula (H2O). We have arrived at the case "As seen in Euclidean geometry", for which the triangle has a circum-circle. The case "As seen in Euclidean geometry" always occurs for an acute or right midpointtriangle, because any two altitudes of an acute or right triangle do intersect by Proposition 11.6. It can, but does not need to happen that the altitudes of an obtuse midpoint-triangle intersect. In Klein’s model an easily understandable picture is obtained by putting point O into the center of Klein’s disk. In the figure on page 803, I have given illustrations for an acute, right and an obtuse midpoint-triangle. Remark. For an acute midpoint-triangle, both vertex C and circum-center O lie on the same side of the longer triangle side AB. For an obtuse midpoint-triangle, vertex C and circum-center O lie opposite sides of the longer triangle side AB. But, indeed, the other two cases already mentioned in Proposition 11.7 do occur in hyperbolic geometry, too. They are illustrated in the figure on page 804. Of course, one can no longer put the circum-center in the center of Klein’s disk, because it does not exist! Instead, we put the midpoint Mc of the longest side in the center of the disk. Of the six right angles between the perpendicular bisectors and the sides of the triangles 4ABC and 4Ma Mb Mc , only four are depicted undistorted. Because the three right angles between the bisectors and the sides of the midpoint-triangle 4Ma Mb Mc are still undistorted, it is straightforward to construct its virtual orthocenter H2 . In the "genuine hyperbolic case", point H2 lies outside Klein’s disk. Hence H2 is not a point of the hyperbolic plane. Instead, the polar of the ultra-ideal point H2 yields a common perpendicular l to the three altitudes of triangle 4Ma Mb Mc . But—since Proposition 30 is a theorem of neutral geometry—it is still true that the altitudes of the midpoint-triangle are the bisectors of the original triangle. Therefore line l is a common perpendicular to the three bisectors of triangle 4ABC. Now still we drop the perpendiculars from the three vertices A, B and C onto line l. We end up with six lines that form right angles with line l! Too, we have reconstructed the situation from Example 11.1. The three points X, Y and Z are the intersections of l with the perpendicular
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Figure 43.5. The altitudes of the midpoint triangle intersect outside Klein’s disk.
bisectors. Once more, we get three Saccheri quadrilateral Y XAB and YZCB and ZXAC. Hence the three vertices A, B and C have congruent distances to line l, as to be shown. Recall that the boundary circle δD of Klein’s disk is called the circle of infinity. Finally, we get the "borderline case" if point H2 lies on the circle of infinity. In that case, the three vertices lie neither on a circle nor an equidistance line. Neither two of the three bisectors intersect, nor do any two of them have a common perpendicular. Problem 43.2. In the case that the three vertices of a triangle lie on an equidistance line, they have congruent distances to the baseline. In the figure of page 804, we get AX BY CY, where X, Y, Z are the foot points of the perpendiculars, dropped from the vertices onto the baseline l. This is hard to believe, especially because segment CZ intersects the opposite side AB of the triangle 4ABC, but the other two segments AX and BY do not intersect any side of the triangle. Redraw the figure in Poincaré’s model. The visible divergence of parallel lines makes existence of a baseline l more intuitive. Problem 43.3. In Klein’s disk model, there is given a triangle 4ABC with midpoint Mc of side AB at the center. Construct the midpoint-triangle, the six right angles and decide which case of Theorem 43.3 did occur. Problem 43.4. Given given is a triangle 4ABC with midpoint Mc of side AB at the center of a Klein disk, and midpoint Mb already specified. Construct the circle of infinity δD of Klein’s disk. Finally get the midpoint-triangle, the six right angles and decide which case of Theorem 43.3 did occur.
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Figure 43.6. The three vertices A, B and C have congruent distances from the baseline l, which is chosen to be the horizontal diameter. The second drawing erases the part of the construction done outside the Poincaré disk.
Figure 43.7. Construction of the perpendicular bisector pb and the point H2 = O.
43.4. Thales’ Theorem in hyperbolic geometry As already mentioned in the introductory section, Thales’ theorem does not hold in hyperbolic geometry. Once it is known that the angle sum of a triangle is less than two right angles, it is easy to see that the angle in a semicircle is acute. Nevertheless, even in neutral geometry, we can get a nice statement about a triangle in a semicircle. Proposition 43.1. Let one side of a triangle be a diameter of a semicircle and the third vertex lie on that semicircle. Then the third vertex and the midpoints of the three sides are a Lambert quadrilateral. Proof. Let AB be the diameter and let C be the third vertex on the semicircle. Let 4Ma Mb Mc be
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Figure 43.8. Construction of the circle of infinity δD and the point H2 = O.
the midpoint-triangle. Because Mc has congruent distances from vertices A and C, it lies on the bisector pb , as follows from Proposition 11.1. Thus angle ∠Mc MbC is a right angle. By the same reasoning, congruent distances Mc B McC imply that Mc lies on the bisector pa , and hence angle ∠Mc MaC is right. The bisector pb bisects the angle ∠AMcC, and the bisector pa bisects the supplementary angle ∠BMcC. As shown in paragraph of in-circle and ex-circles, the bisectors of supplementary angles are perpendicular. Hence pa and pb are perpendicular, and hence angle ∠Ma Mc Mb is right. Thus we have checked that the quadrilateral CMa Mc Mb has three right angles, resulting in a Lambert quadrilateral. Corollary 78. In neutral geometry, the angle in a semicircle is a right angle if and only if a rectangle exists. Reason. If a rectangle exists, we know from the second Legendre theorem that every Lambert quadrilateral is a rectangle. Hence the Lambert quadrilateral constructed in Proposition 43.1 is a rectangle. The right angle at vertex C is now Thales’ right angle in a semicircle. Conversely, if the angle in a semicircle is right, the Lambert quadrilateral constructed in Proposition 43.1 is a rectangle, and hence a rectangle does exist. Remark. The corollary holds without assuming Archimedes’ axiom. Similarly as we did in the introductory section, one can ask what happens for the third vertex inside or outside the semicircle. Thus one is lead to a strengthened version, and a converse to Proposition 43.1.
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Figure 43.9. Construction of the circle of infinity δD and the point H2 = O—in this example it turns out to be outside the disk!
Figure 43.10. A triangle in a semicircle produces a Lambert quadrilateral. For the proof, the extra line we need is—once again—the radius to the third vertex.
Proposition 43.2 (A neutral version of the strengthened Thales’ theorem). Let one side AB of a triangle be a diameter of a circle. If the third vertex C lies inside the circle, the midpoint-triangle has an obtuse angle at the center of the semicircle. The quadrilateral CMa Mc Mb has an obtuse angle at vertex Mc and two acute angles at vertices Ma and Mb . If the third vertex C lies outside the circle, the midpoint-triangle has an acute angle at the center of the semicircle. The quadrilateral CMa Mc Mb has an acute angle at vertex Mc and two obtuse angles at vertices Ma and Mb .
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Corollary 79. Let one side of a triangle be a diameter of a circle. The third vertex C lies on that circle if and only if the midpoint-triangle has a right angle at the center of the semicircle.
Figure 43.11. A triangle with third vertex inside the circle produces an obtuse angle of the midpoint-triangle at center Mc . In case of the third vertex outside the circle, an acute angle is produced.
Proof in Klein’s model of hyperbolic geometry. The drawings on page 808 use Klein’s model with midpoint Mc at the center of the disk. I did not draw the circle of infinity ∂D. The drawings include the semicircle, the original triangle 4ABC, its midpoint-triangle 4Ma Mb Mc . By Corollary 30, the altitudes of the midpoint triangle are the side bisectors of the original triangle. Hence the altitude of the midpoint-triangle dropped from vertex Mb onto the opposite side Ma Mc is equal to the perpendicular bisector pb . It has the foot point F. An additional perpendicular is dropped from Mc onto the side AC, and has the foot point G. Begin by distinguishing the following cases: (i) The midpoint-triangle has an obtuse angle at vertex Mc . (ii) The midpoint-triangle has a right angle at vertex Mc . (iii) The midpoint-triangle has an acute angle at vertex Mc , but an obtuse or right angle at vertex Ma . (iv) The midpoint-triangle has an acute angle at vertex Mc , but an obtuse or right angle at vertex Mb . (v) The midpoint-triangle is acute. To finish the proof of the Corollary, assume that case (ii) occurs. Clearly the "water-point" is H2 = O = F = Mc . The side bisector pb = Mb FH2 and the line GMc are equal, because both are the perpendicular to Mc Ma at point Mc = F. We get a Lambert quadrilateral CMa Mc Mb . Because Mc lies on the bisector pb , both points C and A have congruent distances to Mc . Hence point C lies on the circle with diameter AB.
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Next we consider case (i). By Proposition 11.6, the orthocenter H2 = O of the midpointtriangle lies inside the vertical angle to the obtuse angle ∠Mb Mc Ma . 1 Hence H2 lies on the side of line AB opposite to the points Ma , Mb and C. Answer. Because the angle ∠Ma Mc Mb is obtuse, the foot point F lies on the extension of side Ma Mc of the midpoint-triangle, and hence Mc lies between F and Ma . Sorting again points below and above line AB, we conclude that F and H2 lie below line AB, whereas Ma , Mb and C lie above line AB. −−−−−→ From here, one can check that ray Mb Mc lies inside the right angle ∠CMb H2 . Hence G lies on −−−→ the ray MbC, and GC is shorter than AC. Finally McC is shorter than Mc A, which confirms that C lies inside the circle with diameter AB. It is left to the reader to check that in all remaining cases (iii) through (v) the following can be −−−−−→ deducted: Ray Mb Mc lies inside the right angle ∠AMb H2 . Hence GC is longer than AC. Finally McC is longer than Mc A, which confirms that C lies outside the circle with diameter AB. Problem 43.5. Given is a triangle 4ABC with midpoint Mc of longest side AB at the center of a Klein disk and vertex C in the upper half disk. Use Proposition 43.2, and prove that the hyperbolic midpoints Mb and Ma are below the apparent Euclidean midpoint if and only if the midpointtriangle is obtuse. The hyperbolic midpoints Mb and Ma are above the apparent Euclidean midpoint if and only if the midpoint-triangle is acute. Partial solution. The hyperbolic midpoint Mb is equal to the apparent midpoint of segment AC if and only if Mb = G. By the neutral version of the strengthened Thales’ theorem, given in Proposition 43.2 above, this happens if and only if the angle Ma Mc Mb is right. The assertion follows by continuity, once is hold in at least one example for obtuse and acute triangles. This can be established for some extreme isosceles triangles. 43.5.
The centroid
Theorem 43.4 (The Centroid). The three medians of a triangle intersect in one point. Remark. The ratio in which the intersection point divides the medians can be different from 2 : 1. Proof. I use Klein’s model of hyperbolic geometry, to given a proof for the hyperbolic case. Given is a triangle 4ABC. A convenient position is achieved by putting triangle side AB on the horizontal diameter with its midpoint Mc = O in the center of the disk. The midpoints of the two other sides of the triangle do not appear as absolute midpoints, because of the distortion of distances in Klein’s model. Nevertheless, we can take advantage of Theorem 11.1. The perpendicular bisector pc of one side AB is perpendicular to the line l through the midpoints Ma and Mb of the other two sides BC and CA. Because this perpendicular bisector pc is a vertical diameter of the disk D, the right angle between pc and l appears as an absolute right angle. Hence, in the sense of underlying Euclidean geometry, the lines AB and Ma Mb get parallel . We are just back to the figure studied in the construction 31.3 of a Euclidean parallel— in the section about constructions by restricted means. By the Theorem of the harmonic quadrilateral (see Theorem 31.1), the three lines CMc , AMa and BMb intersect in one point, as to be shown. 1
Clearly, the orthocenter can be an ideal or ultra-ideal point of Klein’s model.
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Figure 43.12. The three medians intersect in one point because the harmonic quadrilateral ABMb Ma has Euclidean parallel sides.
43.6.
The in-circle, ex-circles, and the orthic triangle
Problem 43.6 (Conditional ex-circle—hyperbolic version). Provide the details of the proof for the following proposition. Part is easy to verify in Klein’s model—once one knows how to interpretate Klein’s model. If any two of the exterior bisectors at vertices A and B, and the interior bisector of the third angle ∠BCA intersect, then all these three bisectors intersect in one point. In that case, the triangle has an ex-circle touching side AB from outside, and the extensions of the two other sides. If any two of these three bisectors have a common perpendicular, then three have a common perpendicular. Furthermore, there exists a unique equidistance line touching side AB from outside, and the extensions of the two other sides. If any two of these three bisectors are asymptotically parallel, then all three are asymptotically parallel. Furthermore, there exists a unique horocycle touching side AB from outside, and the extensions of the two other sides. Theorem 43.5. For an acute triangle, the altitudes are the inner angular bisectors of the orthic triangle. For an obtuse triangle, only the altitude dropped from the obtuse angle is an inner angular bisector. The other two altitudes are exterior angular bisectors of the orthic triangle. Proof. This is easy to verify in Klein’s model! One needs to choose the center of the disk differently for an acute and an obtuse triangle. For an acute triangle, I put the orthocenter H at the center of the disk. Again the altitudes as well as the angular bisectors and the in-circle of the orthic triangle become absolute. For an obtuse triangle, I put vertex C at the center of the disk. Still the right angles at the foot points of the three altitudes appear as absolute right angles. But only the altitude dropped from the obtuse angle is an inner angular bisector of the orthic triangle. The other two inner bisectors are just the extensions of the two sides of the triangle at the obtuse angle. The other two altitudes get exterior angular bisectors, since it is a fact of neutral geometry that interior an exterior angular bisectors are perpendicular to each other.
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Figure 43.13. The altitudes are angular bisectors of the orthic triangle.
Figure 43.14. Centers for any ex-circle of the orthic triangle can only be the three vertices or the orthocenter of the original triangle.
Problem 43.7. Prove that the orthic triangle of an acute triangle has an in-circle and three excircles. The orthic triangle of an obtuse triangle has an in-circle and either two or three ex-circles. The orthic triangle has three ex-circles if and only if the original triangle has an orthocenter. Proof. For an acute triangle, one can put any vertex at the center of Klein’s disk, too. One sees that the center becomes the intersection of two exterior and one interior angular bisector, and one gets an ex-circle around that center, all depicted in Klein’s model absolutely. (Of course, one can
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only depict one of the four circles in question absolutely at a time!). For an obtuse triangle, one can put the two vertices with acute angles at the center, and get two ex-circles of the orthic triangle. To get the third ex-circle, one needs to put the orthocenter at the center of Klein’s disk, which is possible if and only if the orthocenter exists.
Figure 43.15. The triangle has no circum-circle—drawn in Klein’s model!
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44. The Poincaré Half-Plane Model In the first subsection, we construct the half-plane model via an isometric mapping of the disk to the half plane. We obtain the hyperbolic lines and distances as expected. Both the distance of two points and the Riemann metric of the Poincaré’s half-plane are calculated via the isometry. The next section explains the Euler-Lagrange equation from the calculus of variations. In the following sections, we reconstruct all features of the Poincaré’s half-plane model, taking the Riemann metric as starting point. At first, the curve of minimal distance between any two given points is calculated from the Euler-Lagrange equation. It turns out to be a circular arc with center on the boundary line of the half-plane, or—in a special exceptional case—a Euclidean line perpendicular to the boundary line. These minimizing lines specify the hyperbolic line connecting the two given points. The minimum of the hyperbolic length of any connecting curve determines the hyperbolic distances between these two points. Using the Riemann metric, the length of the hyperbolic segment is obtained by integration. The hyperbolic distance turns out to be the logarithm of the cross ratio of the two endpoints of the given segment, and the ideal endpoints of the hyperbolic line through these two points. 44.1. Poincaré half-plane and Poincaré disk The open unit disk is denoted by D = {(x, y) : x2 + y2 < 1}, and its boundary is ∂D = {(x, y) : x2 + y2 = 1}. We denote the upper open half-plane by H = {(u, v) : v > 0}. Its boundary is just the real axis ∂H = {(u, v) : v = 0}. We shall construct the half-plane model via an isometric mapping of the disk to the half plane. It is convenient to use the complex variables z = x + iy and w = u + iv. We use the notation w = u − iv for the conjugate complex. Proposition 44.1 (Isometric mapping of the half-plane to the disk). The linear fractional function 1−z (44.1) w=i 1+z is a conformal mapping, and a bijection from C ∪ {∞} to C ∪ {∞}. The inverse mapping is z=
i−w i+w
(44.2)
These bijective mappings preserves angles, the cross ratio, the orientation, and map generalized circles to generalized circles. The unit disk D = {z = x + iy : x2 + y2 < 1} is mapped bijectively to the upper half-plane H = {w = u + iv : v > 0}. Especially, one easily checks that z = −1 7→ w = ∞ ,
z = i 7→ w = 1 ,
z = 1 7→ w = 0 ,
z = 0 7→ w = i
Problem 44.1.. (a) We check whether the mapping (44.1) maps indeed the boundaries ∂D → 7 ∂H and find the restriction of the mapping to the boundary. Confirm that a point z = eiθ is mapped to w = tan 2θ . Do not separate real and imaginary parts. (b) Only now separate now real and imaginary parts. Use the mapping (44.1) to confirm the identities θ sin θ 1 − cos θ tan = = 2 1 + cos θ sin θ
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(c) Use the inverse mapping (44.2) to confirm the identities cos θ =
θ 2 tan2 2θ
1 − tan2 1+
and
sin θ =
2 tan 2θ 1 + tan2
θ 2
Solution of part (a). We plug z = eiθ into formula (44.1) and get w=i
1−z 1 − eiθ e−iθ/2 − eiθ/2 2 sin(θ/2) θ =i = i = = tan iθ −iθ/2 iθ/2 1+z 1+e 2 cos(θ/2) 2 e +e
The Poincaré half-plane model of hyperbolic geometry is constructed from the disk model via this isometry. One translates the definitions from the section on the Poincaré disk model to the half-plane model and arrives at the following conventions: The points of H are the "points" for Poincaré’s model. The points of ∂H are called "ideal points" or "endpoints". Those are not points for the hyperbolic geometry. The "lines" for Poincaré’s model are circular arcs, or—in a special case—Euclidean lines perpendicular to ∂H. The "angles" for Poincaré’s model are the usual Euclidean angles between tangents to the circular arcs. Remark. The inversion image of any point P = (u, v) obtained by reflection across the real axis is P0 = (u, −v). In complex notation, reflection by the real axis is complex conjugation: point P = w = u + iv is reflected to P0 = w = u − iv. Remark. The one-point compactification C ∗ = C ∪ {∞} of the complex plane is well-known and useful in complex analysis, especially it is possible to define regularity and power series of functions in a neighborhood of ∞. Only a linear fractional mapping as for example mapping (44.1) and its inverse are naturally extended to bijective continuous mappings 1 C ∪ {∞} 7→ C ∪ {∞} Hence the half-plane of Poincaré’s model gets just one point ∞. This point can occur as ideal end of a line, circle, equidistance lines or horocyle. Especially for the half-plane, this state of affairs is a bid contrary to the common imagination. Indeed, we have a totally different definition and usage of improper points in the projective completion from projective geometry. Given are any two points A and B. Let P and Q be the ideal endpoints of the hyperbolic line through A and B. These points are named in a way that A, B, P, Q occur in this order during an entire turn around the circle. For the definition of a hyperbolic "distance" and of "congruence of segments", the Definition 40.5 and the preservation of the cross ratio are used as starting points. Thus we arrive at the following Definition 44.1 (Distance and Congruence). The hyperbolic distance of points A and B is given by |AP| · |BQ| s(A, B) := ln(AB, PQ) = ln (44.3) |AQ| · |BP| Two segments AB and XY are called "congruent" iff s(A, B) = s(X, Y). 1
They are indeed the only analytic mappings with such an extension
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Since the mapping (44.2) provides an isometry between the half-plane and the disk model, we can directly calculate the Riemann metric for Poincaré’s half-plane: Proposition 44.2 (Riemann Metric for Poincaré’s half-plane). In the Poincaré half-plane, the infinitesimal hyperbolic distance ds of points with coordinates (u, v) and (u + du, v + dv) is (dsH )2 =
du2 + dv2 v2
(44.4)
Proof. The metric is determined by the requirement that i−w i+w provides an isometry from the half-plane to the disk: z=
(44.2)
dsD = dsH
(44.5)
Hence we need to convert the known metric ds2 = 4
dx2 + dy2 (1 − x2 − y2 )2
(40.11)
of the Poinaré disk model to a metric in the half-plane. We calculate the denominator 1 − | z|2 =
4v |i + w|2 − |i − w|2 (w + i)(w − i) − (i − w)(−i − w) = = 2 2 |i + w| |w + i| |w + i|2
and the derivative of the mapping (44.2): dz 2 =− dw (w + i)2 Putting these two results into formula (40.11) yields 4(dx2 + dy2 ) 4| dz|2 = (1 − x2 − y2 )2 (1 − |z|2 )2 !2 !2 2 2 2 |w + i|2 dz |w + i|2 2 2 | dw| = 4 = 4 | dw| dw 4v 4v (w + i)2 | dw|2 du2 + dv2 = 2 = v v2 Thus formula (44.4) arises from the isometry (44.5) of the half-plane and the disk. ds2 =
44.2. The Euler-Lagrange equation The basic problem of the calculus of variations is to determine the curve y = y(x) between two given points (a, y(a)) and (b, y(b)) for which the prescribed functional Z b L[y] := F(x, y, y0 ) dx a
assumes an extremum (minima or maxima), or simply becomes stationary. 1 It turns out that the stationary curves for the functional L[y] satisfy the Euler-Lagrange equation d ∂F ∂F − = 0 0 dx ∂y ∂y 1
In physical applications, the functional is obtained from first principles of physics.
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To derive the Euler-Lagrange equation, we take a pencil of connecting curves y = y(x, p) depending smoothly on a parameter p, and differentiate the functional L[y(., p)] by the parameter p. It is customary to denote the derivative of any quantity by p with the symbol δ and call it the variation of this quantity. One obtains Z b d d F(x, y, y0 ) δ L[y] = L[y] = dx dp dp a " # Z b ∂ F(x, y, y0 ) ∂ y ∂ F(x, y, y0 ) ∂ y0 = · + · dx ∂y ∂p ∂y0 ∂p a ! " #b Z b " # ∂ F(x, y, y0 ) ∂ y d ∂ F(x, y, y0 ) ∂ y ∂ F(x, y, y0 ) ∂ y · · = · − dx + ∂y0 ∂p a ∂y ∂p dx ∂y0 ∂p a The boundary terms vanish for a problem with prescribed endpoints (a, y(a)) and (b, y(b)) of the curve. Hence we obtain !# Z b" d ∂F ∂F δ L[y] = − δy dx ∂y dx ∂y0 a Since the variation δy of the curve can be chosen to be any smooth function of x, the Lemma of the calculus of variations shows that the expression in the bracket has to vanish identically. Thus we obtain the Euler-Lagrange differential equation. Lemma 44.1 (Lemma of the calculus of variations). Let g(x) be a piecewise continuous function and suppose that Z 1
η(x)g(x)dx = 0
0
for all functions η ∈ C ∞ . Then the function g is identically zero. Proof. We show that for a continuous function g , 0 the assertion Z 1 η(x)g(x)dx = 0 0
does not hold for all functions η ∈ C ∞ . Assume g , 0 and g continuous. Hence there exists 0 < a < 1 such that g(a) > 2ε > 0. There are cases where you need to go with the negative −g and get the following reasoning for the negative function −g. Because of the continuity of g there exists δ > 0 such that |x − a| < 2δ implies |g(x) − g(a)| < ε and hence g(x) > g(a) − ε > ε. There exists a continuous, and even C ∞ function η ≥ 0 such that η(x) = 0 for |x − a| > 2δ and η(x) = 1 for |x − a| < δ. Hence Z 1 Z a+2δ Z a+δ η(x)g(x)dx = η(x)g(x)dx ≥ η(x)g(x)dx = 2δε > 0 0
a−2δ
a−δ
Hence the assumption that 1
Z
η(x)g(x)dx = 0
0
for all functions η ∈ C ∞ does not hold. As a contrapositive, the assumptions that a continuous function g , 0 satisfies Z 1 η(x)g(x)dx = 0 0
for all functions η ∈ C ∞ imply g(x) = 0 for all x.
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44.3. The curve of minimal hyperbolic length We want to find the curve of minimal hyperbolic length connecting two given points. In this problem, it turns out to be more convenient to use the right half plane {(x, y) : x > 0} as model of hyperbolic geometry. The corresponding Riemann metric is p dx2 + dy2 ds = (44.6) x The hyperbolic length of any curve y = y(x) between two given points (a, y(a)) and (b, y(b)) is given by the functional Z b p 2 dx + dy2 L[y] := x a Choosing the variable x as independent, we get Z b p 2 Z b p dx + dy2 1 + y02 = dx x x a a In the variational problem occurs the function p 1 + y02 F(x, y, y ) = x The Euler-Lagrange equation becomes particularly simple. Since the variable y does not occur in the functional F, we can immediately perform one integration and get p d ∂ 1 + y02 = 0 dx ∂y0 x p ∂ 1 + y02 = c ∂y0 x y0 = c p x 1 + y02 0
y02 = (1 + y02 )c2 x2 y02 (1 − c2 x2 ) = c2 x2 y0 = √
cx
1 − c2 x 2 Here c denotes a constant independent on x. Of course the value of c can still depend on the coordinates of the endpoints. The last line is a first order differential equation. If c = 0, we get the solution y = const. The minimizing curve is a Euclidean line perpendicular to the boundary. If c , 0, we do a further integration and obtain Z cx dx y = y0 + √ 1 − c2 x 2 We substitute v = 1 − c2 x2 and dv = −2c2 xdx to obtain Z 1 dv y = y0 − √ 2c v √ v = y0 − c √ = y0 ∓ c−2 − x2 This is the equation of an circular arc with center (0, y0 ) and radius |c|−1 .
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44.4. The minimum of hyperbolic length I go now back to the more commonly used upper half-plane. For the convenience of the reader, I use variables x and y. The upper half plane is {(x, y) : y > 0} and has the metric p dx2 + dy2 ds = (44.7) y The minimum of the hyperbolic length of any connecting curve determines the hyperbolic distances between two points. Given are two points A with Euclidean coordinates (xA , yA ) and B with coordinates (xB , yB ). In the case xA , xB , the minimizing curve of connection is a circular arc with center on the x-axis. 1 The equation of such an arc is p y = + r2 − (x − x0 )2 (44.8) The radius r > 0 and the center (x0 , 0) are to be determined from the coordinates of the two points A and B. Problem 44.2. We can check directly that the function (44.8) is a solution of the Euler-Lagrange equation for the functional p 1 + y02 0 F(x, y, y ) = y (a) Confirm that
∂F 1 + y02 + y00 y d ∂F − = p dx ∂y0 ∂y y2 (1 + y02 ) 1 + y02
(b) Check that the derivatives of function (44.8) for the upper half-circle satisfy y02 + yy00 + 1 = 0. The hyperbolic length of the arc is Z xB p 2 Z xB p dx + dy2 1 + y02 s(A, B) = = dx y y xA xA We need to differentiate the square root composite function (44.8) occurring inside the integral and get x − x0 y0 = p 2 r − (x − x0 )2 r2 r2 1 + y02 = 2 = 2 2 r − (x − x0 ) y We plug into the distance functional and obtain Z xB p 1 + y02 s(A, B) = dx y xA Z xB r dx = y2 x Z AxB r dx = 2 − (x − x )2 r 0 xA 1
I leave the special case xA = xB as an exercise.
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This integral can be calculated by means of the partial fraction decomposition r 1 1 = + 2 2 2(r − x + x0 ) 2(r + x − x0 ) r − (x − x0 ) Z xB Z xB dx dx + s(A, B) = 2(r + x − x ) 2(r − x + x0 ) 0 xA x # xB " A 1 1 ln(r + x − x0 ) − ln(r − x + x0 ) = 2 2 xA It remains to check that this result agrees with formula (44.3). We calculate the logarithm of the cross ratio of the two endpoints A and B of the given segment, and the ideal endpoints P and Q of the hyperbolic line through these two points. We assume xA < xB . The points A, B, P, Q occur during a clockwise turn around the circle. The Euclidean coordinates of the endpoints are (xP , yP ) = (x0 + r, 0) and (xQ , yQ ) = (x0 − r, 0). Hence the cross ratio, and its logarithm are |AP| · |BQ| (AB, PQ) = |AQ| · |BP| 1 |AP|2 1 |BP|2 ln(AB, PQ) = ln − ln 2 |AQ|2 2 |BQ|2 #x " 1 (r − x + x0 )2 + y2 A = ln 2 (r + x − x0 )2 + y2 xB #x " 1 −2(x − x0 )r + 2r2 A = ln 2 2(x − x0 )r + 2r2 xB " #x 1 −x + x0 + r A = ln 2 x − x0 + r xB in agreement with the result (??). In the special case xA = xB , the minimizing curve is a Euclidean line perpendicular to the x-axis. We leave the calculation of the distance in the special case as an exercise. 44.5.
Some useful reflections in the half-plane
Problem 44.3. Check how the mapping 1−z 1+z maps the boundaries ∂D 7→ ∂H. Confirm that a point z = eiθ is mapped to w = tan 2θ . Use the inverse mapping i−w z= i+w to confirm the identities w=i
cos θ =
θ 2 tan2 2θ
1 − tan2 1+
and
sin θ =
(44.1)
(44.2)
2 tan 2θ 1 + tan2
θ 2
Problem 44.4. Let S α denote the reflection across the line with ends α ∈ R and ∞. Confirm that S α (w) = 2α − w
(44.9)
for any α ∈ R. Use this result for an easy check that S α+β = S α S 0 S β holds for any α, β ∈ R.
(44.10)
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Problem 44.5. Let Mγ denote the reflection across the line with ends γ > 0 and −γ. Confirm that Mγ (w) =
γ2 w |w|2
(44.11)
for any γ > 0. Use this result for an easy check of Mγδ = Mγ M1 Mδ
(44.12)
Problem 44.6. Confirm that S 0 Mγ = Mγ S 0 and Mγ M1 S α M1 Mγ = S αγ2 for any α, γ ∈ R.
(44.13)
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45. Hilbert’s Axiomatization of Hyperbolic Geometry The Uniformity Theorem from the section about the natural axiomatization of geometry classifies the Hilbert planes into three basic types. In the definition 12.1 from the section "Towards a Natural Axiomatization of Geometry", we have agreed to call a Hilbert plane, in which the angle sum of every triangle is less than two right angles, a semi-hyperbolic plane. Problem 45.1. Convince yourself that in a semi-hyperbolic plane, there exist at least two parallels to a given point to a given line. Is the converse true? 45.1. Some basic facts about semi-hyperbolic planes Recall that by definition 5.16, two rays are called coterminal if one of them can be obtained from the other one by extension or deletion of a segment. Being coterminal defines an equivalence relation among rays. Definition 45.1 (Limiting parallel ray). A ray k with vertex A is called a limiting parallel to ray h −−→ with vertex B if these two rays do not intersect, but each ray from vertex A inside the angle ∠(AB, k) intersects the ray h. It is easy to see: if a ray k is a limiting parallel to ray h, then the same ray k is a limiting parallel to any ray h0 coterminal with h, too. We drop the perpendicular from vertex A onto the line of h and obtain the foot point C. Definition 45.2 (Angle of parallelism). The angle of parallelism is the angle between the limiting parallel ray and the perpendicular from A onto line l. The angle of parallelism depends only on the hyperbolic length (congruence class) of the segment AC. Problem 45.2. Convince yourself of the last assertion. Convince yourself that in a semi-hyperbolic plane, the angle of parallelism is always acute, if it exists. Be aware that this statement does not imply the existence of the limiting parallel ray. Definition 45.3. Following Lobachevskij, one defines a special function, called π(s), giving the angle of parallelism π for a segment of hyperbolic length s. We shall use the notation π∗ (s) = 90◦ − π(s) For a semi-hyperbolic plane, these are is indeed partial functions, since existence of the limiting parallel ray is not postulated. Problem 45.3. Prove in neutral geometry: Two lines which intersect a transversal with congruent z-angles have a common perpendicular. Remark. The illustration wants to make clear that one does not know from the beginning that the three points E, F and M lie on a line. After one has proved that they do, one has found the common perpendicular. Proof. Let a transversal line t intersect the two lines m and l at points A and B, producing zangles α β which are assumed to be congruent. Too, we may assume these angles to be acute. Indeed, in the special case that they are right angles, the given transversal already is the common perpendicular. In the case that they are obtuse, by proposition 7.15, their supplements are a pair of congruent acute z-angles.
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Figure 45.1. Two lines traversed congruent z-angles have a common perpendicular.
From the midpoint M of segment AB, we drop the perpendiculars onto the given lines m and l. The foot-points E and F produce the triangles 4MEA and 4MFB. By Proposition 7.46(i), any right triangle has two acute angles opposite to its legs. Hence the triangles 4MEA and 4MFB have acute angles at their respective vertices A and B. Since the z-angles α and β are acute, too, their sides lying on the lines m and l are the rays −−→ −−→ AE ⊆ m and BF ⊆ l. Hence α = ∠MAE and β = ∠MBF. By definition 7.20 of z-angles, they −−→ −−→ lie on opposite sides of the transversal. In our case α and β have the sides AE and BF lying on opposite sides of the transversal. We conclude that the foot points E and F lie on opposite sides of the transversal t. The two triangles 4MEA 4MFB are congruent by SAA congruence. Hence they have congruent angles ∠AME ∠BMF. Since A ∗ M ∗ B and points E and F lie on opposite sides of the transversal t, these are vertical angles and E ∗ M ∗ F. The three points E, F and M lie on a line, which is the common perpendicular to be found. 45.1.1. The equivalence relation of limiting parallelism Theorem 45.1. Being limiting parallel or coterminal defines an equivalence relation among rays. The corresponding equivalence classes are called ends or ideal points. Let AC be a perpendicular onto line l with foot-point C. Somewhat anticipating such a result, −−→ we denote the end of the limiting ray AP by α. Furthermore, we denote the limiting ray with vertex −→ A and end α by Aα. −−→ −→ −→ Lemma 45.1. If AP = Aα is limiting parallel ray to line l, the "cutted" coterminal ray Pα is a limiting parallel to line l, too.
823
Figure 45.2. Cutting the limiting parallel ray.
−−→ Proof. Drop the perpendicular from point P onto line l, let F be the foot-point. Take any ray PQ −−→ inside the angle ∠FPα. The ray AQ, from the original vertex A, lies inside the angle ∠CAα and hence intersects line l in a point S . We use Pasch’e axiom with triangle 4AS C and line PQ. We −−→ conclude that ray PQ intersects segment CS , and hence line l in a point T . −→ Since Q is an arbitrarily chosen point inside the angle ∠FPα, we see that the ray Pα is a limiting parallel to line l. −→ −−→ −→ Lemma 45.2. If ray Aα is limiting parallel to line l, the coterminally extended ray RA = Rα is a limiting parallel to line l, too.
Figure 45.3. Extending a limiting parallel ray.
−−→ Proof. Drop the perpendicular from point R onto line l, let F be the foot-point. Take any ray RQ inside the angle ∠FRα. If this ray intersects segment FC, we are ready. Otherwise it intersects segment AC, say at point X. Because of the exterior angle theorem for triangle 4RXA, we get for the angle of parallelism ∠CAα > ∠AXR = ρ −−→ We transfer the angle ∠AXR = ρ onto the perpendicular AC to produce a third ray s. Because of the estimate above, the ray s lies inside the angle ∠CAα. Hence it intersects line AC, say at point S . −−→ We use Pasch’e axiom with triangle 4AS C and line RQ. We conclude that ray RQ intersects segment CS , and hence line l in a point T . −→ Since point Q is an arbitrarily chosen point inside the angle ∠FRα, we see that the ray Rα is a limiting parallel to line l.
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Lemma 45.3. If a ray k is a limiting parallel to ray h, any ray k0 coterminal with k is a limiting parallel to any ray h0 coterminal to h. Lemma 45.4. Assume the two lines l = CB and m = AD have the common perpendicular AC. −−→ −−→ Then the ray AD is not limiting parallel to the ray CB. This statement holds in any semi-elliptic Hilbert plane, too.
Figure 45.4. Lines with a common perpendicular diverge.
Proof. We consider the case of a semi-hyperbolic plane. Let AC be the common perpendicular. We choose a second point B on line l. Because the geometry is semi-hyperbolic, the triangle 4ABC has angle sum less two right angles and hence α + β < R. We transfer the angle β to vertex A to −−→ produce congruent z-angles, and get a new ray AE. Since α + β < R, this ray lies in the interior of −−→ the right angle ∠CAD (with side AD lying on m). Ray AE does not intersect line l, because there exists a common perpendicular EF of lines l and AE, as explained in Problem 45.3. −−→ −−→ Because ray AE lies in the interior of angle ∠CAD and does not intersect line l, the ray AD is not a limiting parallel to line l. The modification necessary for the semi-elliptic case is shown in
Figure 45.5. Lines with a common perpendicular diverge in the semi-elliptic case, too.
the figure on page 824.
Lemma 45.5 (Symmetry). If ray l is limiting parallel to ray k, then ray k is limiting parallel to ray l.
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Figure 45.6. If l is limiting parallel to k, then k is limiting parallel to l.
−−→ Proof. Let ray k0 be a limiting parallel to ray l = CB. We may assume that the vertex C of ray l is the foot point of the perpendicular dropped from the vertex A of k0 onto l. We want to check that ray l is a limiting parallel to ray k0 . To this end, we draw the an arbitrary ray r0 inside the angle ∠BCA, and check whether it intersects the ray k0 . Let F 0 be the foot-point of the perpendicular from A onto the ray r0 . Since the hypothenuse AC is the longest side of the right triangle 4ACF 0 , we can construct a point F between A and C such that AF AF 0 . We produce the congruent angles ∠FAF 0 ∠(k, k0 ), with the new ray k inside the −−→ angle ∠(AC, k0 ). Since k0 is assumed to be a limiting parallel to ray l, the new ray k does intersect l, say in point L. The perpendicular r erected on AC at point F is (divergent) parallel to the ray l. From Pasch’s axiom, applied to triangle 4ACL and the perpendicular r, we conclude that the ray r does intersect segment AL, say at a point S . The triangles 4AFS and 4AF 0 S 0 are congruent, actually by ASA congruence. They are obtained from each other by a rotation around point A. The point S 0 which we have obtained, is the intersection of rays r0 and k0 . Since the ray r0 can be chosen arbitrarily inside the angle ∠BCA, we see that each ray inside this angle intersects the ray k0 . Hence the ray l is a limiting parallel to ray k0 . −→ −−→ Lemma 45.6. Let ray k = Aα be a limiting parallel to ray l = CB and the vertex C be the foot point of the perpendicular dropped from the vertex A onto l. The perpendicular r on segment AC, erected at any point F between A and C does intersect the limiting ray k. −−→ Proof. By Lemma 45.4, there exists a ray s = CD that remains inside the strip between the divergent parallel rays l and r. By Lemma 45.5, the rays s and k intersect, say at point S . We can now apply Pasch’s axiom to triangle ACS and ray r. We conclude that this ray intersects either segment CS or segment AS . The first case is excluded since ray s lies in the half plane of r as its vertex C. Hence we get an intersection point T of segment AS ⊂ k and ray r.
826
Figure 45.7. Limiting parallel ray k to l comes closer than the ray r.
Proposition 45.1 (Strict monotonicity of the Lobachevskij function). If AB AC, and a corresponding limiting ray exists, then π∗ (AB) = π∗ (AC). If AB < AC, and the corresponding limiting rays exist, then π∗ (AB) < π∗ (AC). Lemma 45.7. If two rays are both limiting parallel to a third ray, they are limiting parallel to each other.
Figure 45.8. Transitivity, case where rays k and h lie in different half-planes of l.
Proof. We assume that the rays k and h are both limiting parallel to a third ray l. There are two cases to be considered, as shown in the figures on page 826 and 827. Consider the case with two rays k and h lying in different half-planes of a third ray l. By plane separation,
827
Figure 45.9. Transitivity, case where rays k and h lie in the same half-plane of l.
we can choose the vertices K, H and L of the three rays lying on a line, with vertex L between K and H. −−→ Take any ray r with vertex K inside the angle ∠(KH, k), and check whether r intersects h. Since ray k is limiting parallel to ray l, the rays r and l intersect in a point S . We can now cut rays r and l to the common vertex S . Because of Lemma 45.5 and Lemma 45.1, the (cutted) ray l is limiting parallel to ray h. Hence the (cutted) ray r intersects ray h. Thus we have checked that rays k and h are limiting parallel. Secondly, we consider the case assuming the two rays k and h lie in the same half-plane of a third ray l. We can choose the vertices K, H and L of the three rays lying on a line, with vertex H between K and L. It is enough to check whether ray k is a limiting parallel to ray h. −−→ Consider any ray r with vertex K inside the angle ∠(KH, k), and check whether r intersects h. Since ray k is limiting parallel to ray l, the rays r and l intersect in a point R. We use Pasch’s axiom for the triangle 4KLR and the line of h. We conclude that the ray h intersects segment KR ⊂ r. Thus we have checked that rays k and h are limiting parallel. −→ −→ 45.1.2. Limiting triangles From two ray Aγ which is a limiting parallel to ray Bγ, we get a limiting triangle with two proper vertices A and B and the end γ as third vertex. By definition, it is assumed that the vertices A, B and γ of a limiting triangle do not line on a line. Proposition 45.2 (SAL-Congruence Theorem). Assume two limiting triangles have a pair of congruent sides, one pair of congruent angles adjacent to these sides. Then the two triangles are congruent. Hint . This fact is a direct consequence of the extended ASA-Theorem 7.13, already explained in the section on neutral triangle geometry, together with Lemma 45.5. Proposition 45.3 (ASAL-Congruence Theorem). Given is a limiting triangle and a segment congruent to its proper side. The two angles at the vertices of this side are transferred to the endpoints of the segment, and reproduced in the same half plane. Then the newly constructed rays are limiting parallel. Hence one gets a second congruent limiting triangle.
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Hint . This fact is a consequence of the extended ASA-Theorem 7.13, already explained in the section on neutral triangle geometry. From Lemma 45.4 and Lemma 45.3, and finally problem 45.3, we conclude Proposition 45.4 (Hilbert III.1). On two lines with a common perpendicular do not lie on limiting parallel rays. On two lines which transversed by a third one with congruent z-angles do not lie limiting parallel rays. Proposition 45.5 (Limiting exterior angle theorem). For a limiting triangle, an exterior angle is greater than the nonadjacent interior angle. Hence the sum of the two interior angles of a limiting triangle is less than two right angles.
Figure 45.10. The exterior angle theorem for a limiting triangle; with the two congruent angles β, the case drawn is impossible.
Proof. Let S (α) denote the exterior angle supplementary to angle α. The last proposition 45.4 excludes the case that S (α) = β. −→ We assume S (α) < β towards a contradiction. Choose a point D on the ray opposite to Aγ and −−→ produce the angle ∠DAE β. If β > S (α), the newly produced ray AE lies inside the angle ∠BAγ. −→ −→ −−→ −→ Since ray Aγ is a limiting parallel to ray Bγ, the newly produced ray AE intersects ray Bγ, say at point C. Now the exterior angle theorem is violated for triangle 4ABC. Thus the case S (α) < β is ruled out. Hence S (α) > β is the only possibility left. Remark. By proposition 7.29, taking supplements reverses the order. We conclude that α < S (β), which is no contradiction. Proposition 45.6 (AAL-Congruence Theorem). Assume two limiting triangles have two pairs of congruent angles adjacent to their proper side. Then the two triangles are congruent. Hint . This is consequence of Lemma 45.3 about z-angles and strict monotonicity of the Lobachevskij function ( Proposition 45.1).
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45.2.
The hyperbolic parallel postulate
Definition 45.4 (Hilbert’s Hyperbolic Parallel Postulate). Given a line l and a point P not on l, there exist two rays r+ , r− from vertex P which do not lie on the same line and do not intersect the line l, but nevertheless any ray from the vertex P in the interior of the angle ∠(r+ , r− ) intersects the line l. Proposition 45.7. In a semi-elliptic plane do not exist any limiting parallel rays. Hence a plane satisfying Hilbert’s Hyperbolic Parallel Postulate is semi-hyperbolic.
Figure 45.11. There exist no limiting parallel rays in the semi-elliptic case.
Proof. The arguments leading to the exterior angle theorem 45.5 for the limiting triangle even hold in the semi-elliptic case. 1 −→ Assume towards a contradiction that a semi-elliptic plane contains a ray Aγ which is limiting parallel to a line l. We drop the perpendicular from point A onto line l to get the foot-point F. Let π(AF) = ∠FAγ be the angle of parallelism. It is immediately possible to construct a Saccheri quadrilateral ABFG, with point B on the −→ same side of perpendicular AF as the ray Aγ. Let δ = ∠FAB ∠GBA be the top angle of this Saccheri quadrilateral. Too, we can construct a limiting triangle 4ABγ. It has the angles ∠γAB = δ − π(AF)
and ∠ABγ = δ + π(AF)
Hence the exterior angle theorem 45.5 implies 2R − δ + π(AF) > δ + π(AF) We conclude that δ < R and hence the plane is semi-hyperbolic.
Definition 45.5 (Hyperbolic Hilbert plane). A hyperbolic plane is a semi-hyperbolic Hilbert plane for which Hilbert’s Hyperbolic Parallel Postulate holds. Remark. I repeat the assumption that the plane is semi-hyperbolic, since it is not obvious at all that no limiting rays exist in a semi-elliptic Hilbert plane.
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Figure 45.12. Hilbert’s construction of the common perpendicular, case χ > ε.
Figure 45.13. Hilbert’s construction of the common perpendicular, case χ < ε.
45.3.
Construction of the common perpendicular
Proposition 45.8 (Hilbert III.2). For any two lines l and m which neither intersect nor have limiting parallel rays, there exists a common perpendicular.
Construction 45.1 (Construction of the common perpendicular). Given are any two lines l and m which neither intersect nor have limiting parallel rays. We choose two points A and C on the line l and drop the perpendiculars onto line m, with the foot-points B and D. If CD AB, we get the Saccheri quadrilateral ABDC and proceed to the last step below. We may assume that CD > AB. We produce the point E between C and D such that DE BA. −−→ Let λ be the end of ray CA. The exterior angle χ = ∠BAλ of the quadrilateral ABDC at vertex A −−→ −→ is transferred onto the ray ED to produce a new ray n, in the same half-plane of ED as ray Cλ. −→ There exists an intersection point F of the ray Cλ with the ray n. Next the segment EF is −→ transferred onto the ray Aλ to produce congruent segments EF AH. We drop the perpendiculars from points F and H and get on the line m the foot-points G and K. We have obtained a Saccheri quadrilateral FGKH. The perpendicular bisector p of segment GK is the common perpendicular of the two lines l and m. 1
It is not obvious at this point that we refer to a truth about a nonexisting object
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−→ Lemma 45.8. There exists an intersection point F of the ray Cλ with the ray n, and hence ∠DEF ∠BAλ. Let ε = ∠DEA be the top angle of the Saccheri quadrilateral DEAB and χ = ∠BAλ be the transferred angle. There are two cases, and a borderline case: χ > ε: The intersection F lies in the segment CA, as drawn in figure on page 830. χ = ε: The intersection is F = A. −→ χ < ε: The intersection F lies on the extension ray Aλ, as drawn in figure on page 830. In this case EF > AF since the opposite angles in triangle 4AEF are ε + χ > ε − χ. and across the greater angle lies the longer side. It can happen that F = A. −−→ −−→ Proof of Lemma 45.8. λ is the end of ray CA. Let µ denote the end of ray DB and ν the end of ray n. A simple SAL-congruence (see Proposition 45.2) for limiting triangles shows that 4λBA 4νDE
hence
∠λBA ∠νDE
Because of the assumption that the two given lines do not contains asymptotic rays, we know that µ , λ. Hence the limiting triangle 4DBλ exists. The exterior angle theorem for this limiting triangle yields ∠λBµ > ∠λDB For the complementary angle we get the reversed inequality. ∠λBA < ∠λDE Together with the congruence we conclude ∠νDE < ∠λDE −→ The last inequality shows the ray Dν lies in the interior of angle ∠λDE = ∠λDC. Since 4λDC is a −→ −→ limiting triangle, we conclude that ray Dν intersects ray Cλ, say in point T . −→ Now we use Pasch’s axiom for triangle 4CDT and the line of ray n = Eν. The side CD is intersected at point E, but the side DT ⊂ Dν cannot be intersected. We see that ray n intersects the segment CT in some point F. Justification of the construction 45.1. The goal is to produce a Saccheri quadrilateral. As explained in the construction 45.1, we have produced two quadrilaterals DEFG BAHK. It is straightforward to check that they are congruent. Nevertheless, there are several possible cases for the figure obtained. The two congruent quadrilateral may either overlap or not overlap, as seen in the figures on page 830 for case χ < ε; and page 830 for the case χ > ε, respectively. It can happen that F = A. But it cannot happen that F = H. We rule out this coincidence as follows: Indeed, if χ ≤ ε this is impossible since F lies inside the segment AC, but point H not, and hence F ∗ A ∗ H. If χ > ε, then AF < EF HF and hence A ∗ F ∗ H. Hence FGKH is a Saccheri quadrilateral. The perpendicular bisector p of segment GK is the perpendicular bisector of segment FH, too. Further remaining details are given by Proposition 9.3 (Hilbert’s Proposition 36), in the section on Legendre’s theorems. Indeed p is the common perpendicular of the two lines l and m.
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45.4.
The enclosing line
Proposition 45.9. Given any angle, there exists a unique line the opposite rays of which are limiting parallels to the two sides of the angles. A bid more generally, we easily conclude: Proposition 45.10 (Hilbert III.3). For any two rays which are not limiting parallel to each other, there exists a unique line the opposite rays of which are limiting parallels to the two given rays. We call this line the enclosing line of the angle, or the two given rays.
Figure 45.14. Hilbert’s construction of the enclosing line.
Construction 45.2 (Construction of the enclosing line). Given is an angle ∠αOβ with two different ends α , β. We choose points A and B on the two sides of the angle such that OA OB. We draw the limiting rays Aβ and Bα. Next we construct the angular bisectors s and t of the angles ∠αAβ and ∠βBα. The common perpendicular of these two bisectors is the enclosing line with ends α and β.
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Figure 45.15. Justification for the construction of the enclosing line.
Proof of proposition 45.9. Let s be the end of the angular bisector at vertex A and t be the end of −→ the angular bisector at vertex B. The ray Aβ and the angular bisector t intersect in point C, since −→ −→ by the symmetry Lemma 45.5, the ray Bβ is a limiting parallel to ray Aβ. By construction, we get congruent angles ∠CAs ∠CBβ as indicated in the drawing on page 833. Furthermore, at vertex C there are congruent vertical angles ∠ACt ∠BCβ Since across the larger angle lies the longer side, we see that CA > CB. We conclude from the − → − → ASAL-Congruence Theorem 45.3 that the rays As and Ct cannot be limiting parallel. Neither can they intersect—that would produce a triangle 4ACQ congruent to the existing limiting triangle 4BCβ what is impossible. Actually we would have the situation considered in −−→ −→ Problem ??, and hence again the rays AC and Bβ would intersect, what is impossible. We see that the lines s and t neither intersect nor contain limiting parallels. Hence, by Proposition 45.8 (Hilbert III.2), the lines of s and t have a common perpendicular.
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We have still to check that the common perpendicular has the ends α and β. Denote the ends of the common perpendicular by p and p0 . Let F and G be the points on s and t were the common −−→ −→ −→ perpendicular intersects them. We draw the three limiting rays Fα, Fβ, and Gβ. We get the congruence ∠αF p ∠βG p0 because of the axial symmetry across the bisector of ∠AOB. We get the congruence ∠αF p ∠βF p0 because of the axial symmetry across the axis As. Hence ∠βG p0 ∠βF p0 . If p0 would not be limiting parallel to β, a limiting triangle 4FGβ would exist. Hence the exterior angle theorem 45.5 would imply ∠βGp0 > ∠βFG = ∠βF p0 . This would contradict the congruence obtained above. Hence the common perpendicular’s end p0 is β, and similarly the other end p is α, as to be shown. Problem 45.4. Use the enclosing line to prove that Lobachevskij’s function π(s) assumes all acute angles, for a hyperbolic plan. Solution. Given any acute angle α, let ∠AOB = 2α and construct the enclosing line αβ. Let point F be the intersection of the enclosing line with angle bisector of ∠AOB. For point O and the enclosing line, the angle of parallelism is the given angle ∠AOF = α. Proposition 45.11. Lobachevskij’s function π(s) is a strictly decreasing bijection from the lengths s of segments to the acute angles. 45.5. The three reflections Theorem In this subsection, we refer repeatedly to the section on neutral triangle geometry. Let S a , S b , S c denote the reflections across the side bisectors of the sides a, b, c, respectively, of triangle 4ABC. Proposition 45.12 (The three reflections theorem for a triangle with a circum-circle). Assume that triangle 4ABC has a circum-circle and let O be its center. Let S A , S B , S C denote the reflections across the lines OA, OB, OC, respectively. The compositions of these reflections satisfy Sa ◦ Sb ◦ Sc = SB Similarly, we get the relations SA SB SC Sa Sb Sc
= Sc ◦ Sa ◦ Sb = Sb ◦ Sa ◦ Sc = Sa ◦ Sb ◦ Sc = Sc ◦ Sb ◦ Sa = Sb ◦ Sc ◦ Sa = Sa ◦ Sc ◦ Sb = Sc ◦ SA ◦ Sb = Sb ◦ SA ◦ Sc = Sa ◦ SB ◦ Sc = Sc ◦ SB ◦ Sa = Sb ◦ SC ◦ Sa = Sa ◦ SC ◦ Sb
(45.1)
Proof. As already stated in Proposition 11.2 from the section about neutral triangle geometry, the three side bisectors intersect at the center O of the circum-circle—the existence of which has been assumed. The composite mapping M := S a ◦ S b ◦ S c ◦ S B maps both points O and B to themselves, mapping the latter point via B 7→ B 7→ A 7→ C 7→ B. Since the mapping M is an isometry, it maps all points of line OB to themselves. Because of preservation of orientation, the mapping cannot exchange the two half plane of line OB. Hence the mapping M is the identity. Any reflection S is an involution, which means S ◦ S is the identity. The remaining claims follow easily from this fact.
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Corollary 80. The product S a ◦ S b ◦ S c of any three reflections across axes which intersect in a common point is a reflection across a forth axis through this point. Problem 45.5. In a semi-Euclidean plane, three lines a, b and c intersecting in one point are given. Construct a triangle for which a, b and c are the side bisectors. Prove that there exist exactly two such triangles with a given circum-circle.
Figure 45.16. Euclidean construction of a triangle with three given side bisectors.
Solution of Problem 45.5. Take any point P on the side bisector b and construct its reflection images Pa and Pc across lines a and c, respectively. The line OB is the perpendicular bisector of Pa Pc . We can choose any point B on this line as one of the vertices of the triangle, and get the other vertices A and C by reflections across the given side bisectors c and a, respectively. Question. Explain the reason for this construction. Answer. The mapping S B = S a ◦ S b ◦ S c maps point Pc via Pc 7→ P 7→ P 7→ Pa . Hence points Pc and Pa are reflection images across the axis of S B . Vertex B lies on this axis. We have explained in Theorem 11.5 that not every triangle has a circum-circle. Indeed this Theorem recasts the almost tragical errors of Farkas Bolyai in the present day context of neutral geometry. Question. Why do I consider the errors of Farkas Bolyai almost tragical. Find an impressive historic source. Farkas Bolyai to his son János:
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You must not attempt this approach to parallels. I know this way to its very end. I have traversed this bottomless night, which extinguished all light and joy of my life. I entreat you, leave the science of parallels alone . . . . I thought I would sacrifice myself for the sake of the truth. I was ready to become a martyr who would remove the flaw from geometry and return it purified to mankind. I accomplished monstrous, enormous labors; my creations are far better than those of others and yet I have not achieved complete satisfaction. . . . I turned back when I saw that no man can reach the bottom of the night. I turned back unconsoled, pitying myself and all mankind. I admit that I expect little from the deviation of your lines. It seems to me that I have been in this regions; that I have traveled past all reefs of this infernal Dead Sea and have always come back with broken mast and torn sail. The ruin of my disposition and my fall date back to this time. I thoughtlessly risked my life and happiness—aut Caesar aut nihil.
But the young Bolyai was not deterred by his father’s warnings. We exemplify now a second important case occurring in hyperbolic geometry. Proposition 45.13 (The three reflections theorem for a triangle with vertices on an equidistance line). Similar to Proposition 11.4 from the section on neutral triangle geometry, we assume that the three side bisectors have a common perpendicular p. Let S A , S B , S C denote the reflections across the lines through the vertices A, B, C, respectively, which are perpendicular to p. The compositions of these reflections satisfy the same relations (45.1). Proof. We consider the same composite mapping M := S a ◦S b ◦S c ◦S B . This mapping takes point B to itself, and the line p as a set to itself. Hence the foot point of the perpendicular q dropped from B onto p, and indeed all points of perpendicular q, are mapped to themselves. The perpendicular q ⊥ p is the symmetry axis of S B . Since mapping M preserves the orientation, it cannot exchange the two half plane of the axis. Hence the mapping M is the identity. In the hyperbolic plane, it turns out to be especially interesting to deal with the case left open by the last two propositions. Proposition 45.14 (Hilbert III.4). In a hyperbolic plane is given a triangle 4ABC, which has two side bisectors a and c with a common end ∞, and vertex B lying in the strip between the two limiting parallel rays. Then all three side bisectors have a common end. Proof. The common end can only occur for the two rays on a and c running from the midpoints of the respective side through the interior of the triangle 4ABC. As a consequence of Pasch’s axiom, these rays intersect the third side CA at points P and Q, respectively. Each point of a side bisector has equal distance to the two endpoints of this side. We get together with the triangle inequality |PC| = |PB| < |PQ| + |QB| = |PQ| + |QA| = |PA| |QA| = |QB| < |QP| + |PB| = |QP| + |PC| = |QC| Hence the midpoint M of triangle side AC lies between points P and Q, and therefore in the strip between the other two side bisectors a and c. As stated in Proposition 11.2, either all three side bisectors intersect at one point, or neither any two of them intersect. We conclude that the side bisector b cannot intersect neither a or c, since it is assumed these two side bisectors do not intersect. Hence the entire line b lies in the strip between bisectors a and c. −−−→ Any ray with vertex M different from M∞ lying on the same side of CA as ∞ does intersect −−→ −−→ either ray P∞ or ray Q∞—otherwise the latter two rays would not be limiting parallels. Hence all −−−→ −−→ −−→ three rays M∞, P∞ and Q∞ are limiting parallels.
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Figure 45.17. Three limiting parallel side bisectors.
Question. Convince yourself that none of the three opposite rays can be limiting parallels. Proposition 45.15. For any triangle in the hyperbolic plane exactly one of the three alternatives occurs: (a) The bisectors of all three sides intersect in one point. The triangle has a circum-circle. (b) The bisectors of all three sides have a common perpendicular. The three vertices lie on an equidistance line. (c) The bisectors of all three sides have a common end. Proof. Assume that neither case (a) or (b) occurs. By Proposition 11.2 and Proposition 11.4 from the section about neutral triangle geometry and Hilbert’s Proposition 45.8, we conclude that any two of the three side bisectors have a common end. −−−−→ −−−−→ We may assume, by choosing the names, the rays Ma ∞ and Mb ∞ from the midpoints of the sides BC and BA, respectively, have the common end ∞. The common end can only occur for the two rays on a and c running from the midpoints of the respective side through the interior of the triangle 4ABC. The vertex B can only lie in the strip between the two limiting parallel rays. −−−−→ As explained in Proposition 45.14, the third side bisector b contains the ray Mb ∞ with the same common end.
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Proposition 45.16. In a hyperbolic plane is given a triangle 4ABC, for which all three side bisectors have a common end ∞. Let S A , S B , S C denote the reflections across the lines ∞A, ∞B, ∞C, respectively. The compositions of these reflections satisfy the same relations (45.1). Conversely, we can now solve the construction problem 45.5 for any triangle in a hyperbolic plane. Proposition 45.17 (Generalization of Hilbert III.5 ). We expect three cases:
1
In any Hilbert plane, for any three rays with a common intersection point, there exists a triangle for which these are the side bisectors. In any Hilbert plane, for any three rays with a common perpendicular, there exists a triangle for which these are the side bisectors. In a hyperbolic plane, for any three rays with a common end, there exists a triangle for which these are the side bisectors. In all three cases, the reflections S a , S b , S b across the side bisectors, and appropriate reflections S A , S B , S C across axes through the vertices A, B, C satisfy the relations (45.1). If the axes of S a , S b , S c have an intersection point O, the axes of reflection S A , S B , S C pass through point O, too. If the axes of S a , S b , S c have a common perpendicular p, the axes of reflection S A , S B , S C are perpendicular to p, too. If the axes of S a , S b , S c have a common end ∞, the axes of reflection S A , S B , S C have the end ∞, too. Proof. We proceed as in the solution of Problem 45.5, which yields the first statement and construct the axis S B using relation S B = sa ◦ S b ◦ S c from (45.1). Take any point P on the side bisector b and construct its reflection images Pa and Pc across lines a and c, respectively. The perpendicular bisector of Pa Pc is the axis of the reflection S B . We can choose any point B on this line as one of the vertices of the triangle, and get the other vertices A and C by reflections across the given side bisectors c and a, respectively. In all cases, the axis of the reflection S B passes through vertex B. The meaning of the axis of the reflection S B differs in the three cases, in the way as explained. Proposition 45.18. For any triangle, the following three statements are equivalent: (a) The perpendicular bisectors of two sides have a common end. (b) The three vertices lie on an horocycle. (c) The bisectors of all three sides have a common end. 1
In a semi-elliptic plane, or a semi-hyperbolic plane that is not hyperbolic, there are further awkward cases possible.
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45.6. Construction of Hilbert’s field of ends In any hyperbolic plane, our results allow us to extend the domain of the mapping of reflection across any line and include all ends (which are also called ideal or improper points). We agree that proper segments are never congruent to rays or lines, and proper angles are never congruent to improper angles with an end as vertex. We can extend the obvious and well-known facts about preservation of incidence and congruence by the mapping of reflection by allowing to replace points by ends. Question. Write down as many different instances of this idea as you find. Given is any hyperbolic plane. Three different ends are chosen arbitrarily and named 0, 1 and ∞. Question. Why do there exist three different ends? We drop the perpendicular from end 1 onto the line 0 ∞ and call the foot point i. This foot point is a point of the hyperbolic plane, but not an end. 1 Question. Explain the construction to drop the perpendicular from a given end onto a given line. Remark. As an alternative, we may as well begin by choosing only two ends, and name them 0 and ∞. Then we choose a point on the line 0 ∞, naming it i. At this point, we erect the perpendicular, and call one end of it 1. We construct a field F, the elements of which are the ends different from ∞. We now use reflections—and their composition as mappings— to define the operations of addition and multiplication of ends. 45.6.1. Addition of ends For any end α , ∞, let S α denote the reflection across the line α ∞. The sum of two ends α, β , ∞ is defined by the requirement S α+β = S α ◦ S 0 ◦ S β
(45.2)
Because of Proposition 45.17 (Hilbert’s proposition III.5), there exists an end γ such that S γ equals the left-hand side of formula (45.2). Question. Give the explicit construction of the sum α + β. Answer. Take any point P on the line 0 ∞ and construct its reflection images Pα and Pβ across lines α ∞ and β ∞, respectively. The line (α + β) ∞ is the perpendicular bisector of the segment Pα Pβ . Question. Check that the operation of addition is commutative and associative. We easily see that 0 is indeed the neutral element of addition. Given is any end α , ∞. The additive inverse (−α) can be obtained from S −α = S 0 ◦ S α ◦ S 0
(45.3)
Question. Give the explicit construction of the additive inverse −α. Answer. The line (−α) ∞ is the mirror image of the line α ∞ across the symmetry axis 0 ∞. Hence we can take any point P on the line α ∞ and construct its reflection image across line 0 ∞. We get a point P0 of the line (−α) ∞, and hence its end −α. The ends in the same half plane of line 0 ∞ as 1 are called positive, the ends in the opposite half plane are negative. I call the hyperbolic half plane containing rays of end 1 the positive semi-region. 2
Question. Check that the sum of two positive ends is positive. 1 2
I use names in agreement with the Poincaré half plane model from the previous section. In the Poincaré half-plane model, this is the quadrant of u + iv with u > 0 and v > 0.
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45.6.2. Multiplication of ends For any positive end µ , ∞, let Mµ denote the reflection across the line (−µ) µ. Lemma 45.9. The lines 0 ∞ and (−µ) µ are perpendicular. Hence S 0 ◦ Mµ = Mµ ◦ S 0
(45.4)
commutes for all µ > 0.
Figure 45.18. The lines 0 ∞ and (−µ) µ are perpendicular.
Proof. Let p be any point of line µ ∞ and P0 its mirror image across line 0 ∞. Let F be the midpoint of PP0 . This is the point were the line 0 ∞ is intersected perpendicularly. We draw the −−−−−→ −→ rays Fµ and F (−µ). Let G the intersection point of the enclosing line (−µ) µ with line 0 ∞. By ASAL and SAL congruence, we get two pairs of congruent limiting triangles 4µFP 4(−µ)FP
and 4µFG 4(−µ)FG
The congruent supplementary angles ∠µGF ∠(−µ)GF at vertex G are right angles.
The product of two positive ends β, γ > 0 is defined by the requirement Mβγ = Mβ ◦ M1 ◦ Mγ
(45.5)
Because of the second part of Proposition 45.17, there exists an end µ such that Mµ equals the left-hand side of formula (45.5). Question. Give the explicit construction of the product βγ. Answer. Take any point Q on the line (−1) 1, for example i. Construct its reflection images Qβ and Qγ across lines (−β) β and (−γ) γ, respectively. The line (−βγ) (βγ) is the perpendicular bisector of the segment Qβ Qγ. Its end βγ > 0 lies in the positive semi-region.
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Question. Check that the operation of multiplication is commutative and associative. We easily see that 1 is indeed the neutral element of multiplication. Given is any positive end α > 0. The multiplicative inverse α−1 can be obtained from Mα−1 = M1 ◦ Mα ◦ M1
(45.6)
Question. Give the explicit construction of the multiplicative inverse α−1 . Answer. The line (−α−1 ) α−1 is the mirror image of the line (−α) α across the symmetry axis (−1) 1. Hence we can take any point Q on the line (−α) α and construct its reflection image across line (−1) 1. We get a point Q0 of the line (−α−1 ) α−1 . We drop the perpendicular onto line 0 ∞. Because of Lemma 45.9, this is the required line (−α−1 ) α−1 . 45.6.3.
The distributive law
Lemma 45.10. Mγ ◦ Mα = Mγ/α ◦ M1 Mγ ◦ Mα ◦ Mγ = Mγ2 α−1 Mγ (α) = γ2 α−1
(45.7) (45.8) (45.9)
The ends α and γ2 α−1 are mirror images across the line (−γ) γ. Proof. The first formula (45.7) can be checked using the definition 45.5 of the product and the fact that Mα is an involution. To check the second formula (45.8) is left to the reader, too. Let β = Mγ (α) be the mirror image of end α across the line (−γ) γ. The mapping Mγ ◦ Mα ◦ Mγ on the left-hand side of the second formula (45.8) takes β 7→ α 7→ α 7→ β and exchanges the ends 0 and ∞. Hence it is the reflection across the perpendicular dropped form β onto line 0 ∞, and hence Mγ ◦ Mα ◦ Mγ = Mβ Now the second formula (45.8) implies that β = γ2 α−1 , and hence formula (45.9). In other words, the mirror image of end α across the line (−γ) γ is γ2 α−1 . Problem 45.6. Prove by a similar argument that S γ (α) = 2γ − α
(45.10)
S γ2 α−1 = Mγ ◦ M1 ◦ S α−1 ◦ M1 ◦ Mγ S γ2 α = Mγ ◦ M1 ◦ S α ◦ M1 ◦ Mγ
(45.11) (45.12)
Lemma 45.11.
Proof. The right-hand side of formula (45.11) maps ∞ 7→ ∞
and
β 7→ α 7→ α−1 7→ α−1 7→ α 7→ β
Hence this mapping equals the reflection S β , again with β = γ2 α−1 . It can be left to the reader to check the last formula (45.12). Lemma 45.12. γ2 α + γ2 β = γ2 (α + β)
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Proof. Use the definition 45.2 of the sum on the left-hand side, formula (45.12) twice, and the commutativity (45.4): S γ2 α+γ2 β = S γ2 α ◦ S 0 ◦ S γ2 β = Mγ ◦ M1 ◦ S α ◦ M1 ◦ Mγ ◦ S 0 ◦ Mγ ◦ M1 ◦ S β ◦ M1 ◦ Mγ = Mγ ◦ M1 ◦ S α ◦ S 0 ◦ M1 ◦ Mγ ◦ Mγ ◦ M1 ◦ S β ◦ M1 ◦ Mγ = Mγ ◦ M1 ◦ S α ◦ S 0 ◦ S β ◦ M1 ◦ Mγ = Mγ ◦ M1 ◦ S α+β ◦ M1 ◦ Mγ = S γ2 (α+β) Lemma 45.13. For every µ > 0, there exists a square root such that γ2 = µ and γ > 0. Proof. Let i and K be the intersection points of the lines (−1) 1 and (−µ) µ intersect line 0 ∞ at right angle, respectively. We erect the perpendicular on the midpoint C of segment i K. Let −γ and γ be the ends of the perpendicular. We easily check that this construction implies Mµ = Mγ ◦ M1 ◦ Mγ Indeed, both sides map point K to itself, the right-hand side via K 7→ i 7→ i 7→ K. Both side leave the line 0 ∞ invariant and reverse the orientation. Hence these mappings are equal. By definition of the field operations this means that µ = γ2 . Thus we have constructed the required square root. Problem 45.7. For every α, there exists a half such that 2η = α. Use Sα = Sη ◦ S0 ◦ Sη and get this fact by an analogous argument. Proposition 45.19. Hilbert’s field F of ends is an ordered Euclidean field. Proof. We have checked that the sum of two positive ends is positive. Our definition makes the product of two positive ends positive, too. One defines a total order by requiring that α < β if and only if β − α > 0. Thus one gets an ordered field. To check whether F is Euclidean, we need to find a square root for any given µ > 0, as explained in Lemma 45.13 above. The distributive law is now easily deducted from Lemma 45.12. 45.7. Reconstruction of the complex field and the half-plane model From Hilbert’s field of ends F, we can construct the half-plane H = {(u, v) : u, v ∈ F and v > 0}. We endow the plane with the Euclidean geometry and get a Euclidean plane, precisely with the properties stated by Definition 24.3 in the section about the natural axiomatization of geometry. In a second step, we use this Euclidean geometry as the underlying theory, upon which to reconstruct the Poincaré’s half-plane model of hyperbolic geometry. Proposition 45.20. Given are ends u, v ∈ F with v > 0. All lines with ends u1 , u2 such that u1 u2 − u (u1 + u2 ) + u2 + v2 = 0 √ √ go through one point P. Especially, this point is the intersection of the lines u ∞ and − u2 + v2 u2 + v2 . Proof. All lines through point i, except 0 ∞, have ends u1 , u2 ∈ F satisfying u1 u2 + 1 = 0
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Problem 45.8. Convince yourself of this statement. Equations (45.10) and (45.9) imply u + β = S u/2 ◦ S 0 (β)
(45.13)
vα = M √v ◦ M1 (α)
(45.14)
for any ends u, v, α and β. The mapping M √v ◦ M1 maps ends u1 7→ vu1 =: u01 and u2 7→ vu2 =: u02 and point i 7→ Q, hence Q = M √v ◦ M1 (i) Hence all lines through point Q, except 0 ∞, have ends u01 , u02 satisfying u01 u02 + v2 = 0 The mapping S u/2 ◦ S 0 takes ends u01 7→ u + u01 = u + vu1 =: u001 and u02 7→ u + u02 = u + vu2 =: u002 and point Q 7→ P, hence P = S u/2 ◦ S 0 (Q) Hence all lines through point P, except u ∞, have ends u001 , u002 satisfying u001 u002 − u (u001 + u002 ) + u2 + v2 = 0 Point P is obtained from the point i by the mapping M √v ◦ M1 , which is called a translation along 0 ∞, The mapping S u/2 ◦ S 0 , used in a second step, is called a rotation around ∞. Every point P of the hyperbolic plane can be obtained in this way. Indeed, given any point P, there exists a ray P ∞. Let u ∈ F be the second end of line P ∞. Let Q the mirror image of P across the line (u/2) ∞. Let −γ γ be the perpendicular bisector of segment i Q, and finally put v := γ2 . According to this construction, we have obtained ends u, v ∈ F, v > 0 and got mappings such that P = S u/2 ◦ S 0 (Q) Q = M √v ◦ M1 (i) P = S u/2 ◦ S 0 ◦ M √v ◦ M1 (i) We give the point P the coordinates u + iv. We see by Proposition 45.20 that the Euclidean equation of a hyperbolic line through point P with ends u1 and u2 is ! ! u001 + u002 2 u001 − u002 2 2 u− +v = 2 2 except the additional case of "vertical" line with arbitrary v and u as given by point P. In the interpretation of the Euclidean coordinate geometry, we see that the hyperbolic lines through an arbitrary point P are depicted as circular arcs and lines perpendicular to the u-axis. The u-axis becomes the line at infinity for the half-plane model. The construction above shows that enough rigid motions exist in order map any point P and −→ ray with vertex P to the point i and the ray i∞. Hence one can define hyperbolic congruence of segments and angles. In a second step, we use the Euclidean geometry in the plane {(u, v) : u, v ∈ F} as the underlying theory to reconstruct Poincaré’s half-plane model of hyperbolic geometry. Theorem 45.2 (Reconstruction of the half-plane model). Any hyperbolic plane is isomorphic to the Poincaré model with the half-plane H = {(u, v) : u, v ∈ F and v > 0} over the Hilbert field F of ends.
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45.8.
The angle unboundedness postulate
Proposition 45.21. In a hyperbolic Hilbert plane, Aristole’s Angle Unboundedness Axiom 12.2 holds.
Figure 45.19. Existence of asymptotically parallel rays implies Aristole’s Axiom.
Proof. Let the acute angle θ = ∠(m, n) be given. Reflection across the side n yields the double angle 2θ = ∠(m, m0 ). As shown in the proof of Hilbert’s foundation of hyperbolic geometry, the two rays m and m0 have an inclosing line l. One checks that l and n intersect perpendicularly, say at point P. Let p be the ray on l with vertex P which is asymptotically parallel to ray m. Let µ denote their common end. We transfer any given segment onto this ray to produce the (arbitrarily long) segment PQ. We drop the perpendicular from Q onto ray m. Since we are staying on one side of n, we obtain the foot point F. The angle ∠FQµ is an acute angle of parallelism. Hence the angle ∠PQF is obtuse. We erect onto line l the perpendicular ray g at Q inside this angle. This ray is inside the angle ∠PQF, and by the crossbar theorem, intersects segment PF, say at point G. We apply Pasch’s axiom to triangle 4APF and line g, where A is the vertex of angle θ. Since g and n are both perpendicular to PQ, they are parallel. But ray g and side PF intersect in point G. Hence by Pasch’s axiom, line g intersects the third side AF. Call X the intersection point. We drop the perpendicular from point H onto line n and obtain the foot point Y. Finally, we have the Lambert quadrilateral QXY P, with an acute angle χ = ∠QXY at vertex X. By Lemma 12.1 either side adjacent to the acute angle is longer than the respectively opposite side. Hence we conclude XY > PQ as required for Aristole’s angle unboundedness axiom to hold.
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Proposition 45.22. Given is a semi-hyperbolic Hilbert plane, for which Aristole’s Axiom holds. It two lines have a common perpendicular, then the distance between two lines is arbitrarily large. In other words, for any given segment, there exists a point on one of the two lines from which the distance to the other line is longer than the given segment. Corollary 81. In a hyperbolic plane, the distance between two lines with a common perpendicular is arbitrarily large.
Figure 45.20. In a hyperbolic plane, the distance between two lines with a common perpendicular is arbitrarily large.
Proof. Let PQ the segment on the common perpendicular from line m onto line l. From any second point A of line m, we drop the perpendicular onto line l and get the foot point B. Now QBAP is a Lambert quadrilateral, with an acute angle ∠PAB. We drop the perpendicular from point P onto line AB and get the foot point C. We have −−→ produced a second Lambert quadrilateral QBCP, which has an acute angle ∠QPC. The ray PC lies inside the right angle ∠QPA, and hence point C lies between A and B. −−→ We use Aristole’s axiom for the angle θ = ∠APC. Hence there exists a point X on the side PA for which the segment XY dropped onto the other side is longer than the given segment S T : XY > S T We drop from point X the perpendicular onto line l and get the foot point D. The points X and D lie on different sides of line PC, since points A and X lie on one side, and the points Q, B and D lie on the other side. Hence the segment XD intersects line PC in a point E, and XD > XE The segment XE is the hypothenuse of the right triangle 4XY E, and hence its longest side XE > XY Hence for any given segment S T , there exists a point X on line m such that the distance XD > XE > XY > S T from point X the foot point D on line l is longer than the given segment S T .
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46. Gauss’ Differential Geometry and the Pseudo-Sphere 46.1. Introduction Through the work of Gauss on differential geometry, it became clear— after a painfully slow historic process—that there is a model of hyperbolic geometry on surfaces of constant negative Gaussian curvature. One particularly simple such surface is the pseudo-sphere. According to Morris Kline, it is not clear whether Gauss himself already saw this nonEuclidean interpretation of his geometry of surfaces. Continuing Gauss work, Riemann and Minding have thought about surfaces of constant negative curvature. Neither Riemann nor Minding did relate curved surfaces to hyperbolic geometry (Morris Kline III, p.888 etc). But, independently of Riemann, Beltrami finally recognized that surfaces of constant curvature are non-Euclidean spaces. Due to the ideas forwarded by Gauss, mathematicians have in the end advanced to the concept of a curved surface as a space of its own interest. Gauss’ work implies that there are nonEuclidean geometries on surfaces regarded as spaces in themselves. An obvious and important idea is finally spelled out! As we explain in detail below, Beltrami shows that one can realize a piece of the hyperbolic plane on a rotation surface of negative constant curvature. This surface is called a pseudo-sphere. But this new discovery comes with a disappointment: by a result of Hilbert, there is no regular analytic surface of constant negative curvature on which the geometry of the entire hyperbolic plane is valid (see Hilbert’s Foundations of Geometry, appendix V). Concerning models of hyperbolic geometry, the final outcome turns out to be a trade off between the pseudo-sphere and the Poincaré disk. Both have their strengths and weaknesses. The pseudo-sphere is a model for a limited portion of the hyperbolic plane. Both angle and length are represented correctly. The arc length of a geodesic is the correct hyperbolic distance. Furthermore, because of the constant Gaussian curvature, on the pseudo-sphere a figure may be shifted about and just bending will make it conform to the surface. The situation is similar to the more familiar case of Euclidean geometry on a circular cylinder or cone. As everybody knows, on a circular cylinder, a plane figure can be fitted by simply bending it, without stretching and shrinking. On the other hand, only the Poincaré disk is a model for the entire hyperbolic plane. Here only angles are still represented correctly, but the price one finally has to pay is that hyperbolic distances are distorted. The hyperbolic lines become circular arcs, perpendicular to the ideal boundary. One can see the distortion easily in Esher’s superb artwork, based on tiling of the hyperbolic plane with congruent figures. The trade off just explained makes the isometry between the pseudo-sphere into the Poincaré disk especially interesting. One such isometric mapping is explicitly constructed below. Hilbert’s result gets rather natural, too. As explained below, in the sense of hyperbolic geometry, the boundary of the pseudo-sphere turns out to be an arc of a horocyle. 46.2. About Gauss’ differential geometry Karl Friedrich Gauss had devoted an immense amount of work to geodesy and map making, starting 1816. This stimulus leads to his definitive paper in differential geometry of 1827: "Disquisitiones Generales circa Surperficies Curvas" . In this work, Gauss introduces the basics of curved surfaces, and goes far beyond. The real benefit is that, due to the ideas forwarded by Gauss, mathematicians have in the end advanced to the concept of a curved surface as a space of its own interest. To begin with, one imagines a curved surface to be embedded into three dimensional space R3 , and given by some parametric equations x = x(u, v) , y = y(u, v) , z = z(u, v)
(46.1)
The distance ds of neighboring points on the surface with parameters (u, v) and (u + du, v + dv) is given by the first fundamental form ds2 = Edu2 + 2Fdudv + Gdv2
(46.2)
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The first fundamental form is straightforward to calculate from the parametric equations (46.1) since E = xu2 + y2u + z2u F = xu xv + yu yv + zu zv (46.3) G = xv2 + y2v + z2v follows from elementary vector calculus. The geodesics on curved surfaces are defined to be the shortest curves lying on the given surface, connecting any two given points. Gauss’ work sets up the differential equation for the geodesics. Gauss introduces the two main curvatures, called κ1 , κ2 . They turn out to be simply the extremal curvatures of normal sections of the surface. A new important feature is the 2 Gaussian curvature, called K. Gauss shows that K = LN−M , the quotient of the determinants of the EG−F 2 second and first fundamental form. But, even simpler, the Gaussian curvature turns out to be the product of the two principle curvatures: K = κ1 κ2 (46.4) Gauss shows the remarkable fact that this curvature is preserved during the process of bending the curved surface inside a higher dimensional space, without stretching, contracting or tearing it. On the contrary, the two main curvatures are changed by flexing the surface. There are actually at least two different proofs for this fact contained in Gauss’ work. The first one depends on Gauss’ characteristic equation " # " # 1 ∂ F ∂E 1 ∂G 1 ∂ 2 ∂F 1 ∂E F ∂E K= − + − − (46.5) 2H ∂u EH ∂v H ∂u 2H ∂v H ∂u H ∂v EH ∂u √ where H = EG − F 2 . Obviously, any such equation implies that the Gaussian curvature depends only on the first fundamental form. The first fundamental form is preserved, if one bends the curved surface in three space, without stretching, contracting or tearing it. Therefore the functions E, F, G, H which determine the first fundamental form depend only on the parameters (u, v), but do not depend at all on how—or even whether at all—the surface lies in a three dimensional space. Because of the Gauss’ characteristic equation (46.5), the same is true for the Gaussian curvature K. Because of all that, one says that the Gaussian curvature is an intrinsic property of the curved surface. 46.3.
Riemann metric of the Poincaré disk
Proposition 46.1 (Riemann Metric for Poincaré’s Model). In the Poincaré model, the infinitesimal hyperbolic distance ds of points with coordinates (x, y) and (x + dx, y + dy) is (dsD )2 =
4(dx2 + dy2 ) (1 − x2 − y2 )2
(46.6)
Reason. The fact that angles are measured in the usual Euclidean way implies that ds2 = 2 2 E(x, py)(dx + dy ). The rotational symmetry around the center O implies that E(x, y) = E( x2 + y2 , 0). Hence q ds2 = E( x2 + y2 , 0)(dx2 + dy2 ) (46.7) Now it is enough to calculate the distance of the points (x, 0) and (x + dx, 0). The hyperbolic distance of a point (x, 0) from the center (0, 0) is 2 tanh−1 x, as we have derived in Proposition 40.2
848
in the section on the Poincaré disk model. See formula (40.11) there, which is of course the primary origin of the hyperbolic distance! Taking the derivative by the variable x yields ds d d 1+x 1 1 2 = (2 tanh−1 x) = ln = + = dx dx dx 1 − x 1 + x 1 − x 1 − x2 4 ds2 = dx2 (1 − x2 )2 Hence E(x, 0) =
4 (1−x2 )2
(46.8)
and q E(
x2 + y2 , 0) =
4 (1 − x2 − y2 )2
(46.9)
Now formulas (46.7) and (46.9) imply the claim (46.14).
Problem 46.1 (Hyperbolic circumference of a circle). Calculate the circumference of a circle of hyperbolic radius R. We use the Poincaré disk, put the center of the circle at the center O of the disk. In polar coordinates, the Riemann metric is ds2 = 4
dx2 + dy2 dr2 + r2 dθ2 = 4 (1 − x2 − y2 )2 (1 − r2 )2
(a) Calculate the hyperbolic length R of a segment OA with Euclidean length |OA| = r < 1. R (b) Get the circumference C = ds of the circle around O, at first in terms of the Euclidean radius |OA| = r < 1. (c) Get the circumference C of this circle in terms of the hyperbolic radius R. Solution. We take O as center of the circle, and point A on the circumference. Let r = |OA| denote the Euclidian radius, and R = s(O, A) be the hyperbolic radius. The hyperbolic radius R can be found directly for the Riemann metric (46.14). One needs partial fractions to do the integral. # Z r Z r Z r" 2dr 1 1 dr = [− ln(1 − r) + ln(1 + r)]r0 R= ds = = + 2 1 − r 1 + r 1 − r 0 0 0 = 2 tanh−1 r Remark. Of course, we can go back once more to Proposition 40.2, formula (40.11) from the section on the Poincaré disk model and get R = s(O, A) = 2 tanh−1 |OA| = 2 tanh−1 r. We solve R = 2 tanh−1 r for the Euclidean radius and get r = tanh
R eR/2 − e−R/2 eR − 1 = = 2 eR/2 + e−R/2 eR + 1
For the usual Euclidean polar coordinates (r, θ) we get the Euclidean arc length: Z 2π q Z 2π √ Z 2π 2 2 2 2 2 LEucl = dx + dy = dr + r dθ = r dθ = 2πr 0
0
(46.10)
0
The first line holds for any smooth curve. In the second line, we go to the special case of a circle . For a circle, the coordinate r is constant and hence dr = 0, and the factor r can be pulled out of the integral.
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Now the distance along the circumference is measure in the hyperbolic metric (46.14) from Proposition 46.1. Hence the calculation above is modified to Z 2π p 2 Z 2π √ 2 dx + dy2 dr + r2 dθ2 Lhyp = 2 = 2 1 − x2 − y2 1 − r2 0 0 (46.11) Z 2π 2r 4πr = dθ = 1 − r2 0 1 − r2 We have found the correct hyperbolic arc length. But still, one needs to use the formula r = express r in terms of the hyperbolic distance R. Lhyp =
eR −1 eR +1
to
4π(eR − 1) 4πr = 1 − r2 [1 − ( eRR −1 )2 ](eR + 1) e +1
4π(eR − 1)(eR + 1) 4π(e2R − 1) = 4eR (eR + 1)2 − (eR − 1)2 R −R = π(e − e ) = 2π sinh R =
(46.12)
Proposition 46.2 (The circumference of a circle). In hyperbolic geometry, the circumference of a circle of hyperbolic radius R is 2π sinh R. Problem 46.2. The hyperbolic circumference of a circle is much larger than the Euclidean circumference. Let R = 1, 2, 5, 10 and estimate how many times the radius fits around the circumference of a circle of that radius. Answer. A simple calculation yields R 1 2 5 10
(2πsinh R)/R 7.38 11.39 93.25 6919.82
Problem 46.3 (Hyperbolic area of a circle). For a circle of hyperbolic radius R, calculate the area A. Again, we use the Poincaré disk, put the center of the circle at the center O of the disk. For the area, we use the formula from differential geometry Z 2π Z r √ EG − F 2 dr dθ A= 0
0
The first fundamental form is the Riemann metric. It has been already given by formula (46.14), and transformed to polar coordinates in the previous problem. ds2 = 4
dx2 + dy2 dr2 + r2 dθ2 = 4 = Edr2 + 2Fdrdθ + Gdθ2 (1 − x2 − y2 )2 (1 − r2 )2
(a) Get the area of the circle, at first in terms of the Euclidean radius |OA| = r < 1. (b) Get the area A of this circle in terms of the hyperbolic radius R.
(46.13)
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(c) Check that dA =C dR (a) The first fundament form (46.13) yields √
H=
EG − F 2 =
4r (1 − r2 )2
and hence the hyperbolic area of a circle of Euclidean radius r is A=
2π
Z 0
r
Z
√
EG − F 2 dr dθ = 2π
r
Z 0
0
4r dr (1 − r2 )2
This integral is solved with the substitution u = r and du = 2rdr. " #u Z u 4π 4π 4πr2 2 du A = 2π = = − 4π = 2 (1 − u) 0 (1 − r2 ) (1 − r2 ) 0 (1 − u) 2
This is the area in terms of the Euclidean radius |OA| = r < 1. (b) The hyperbolic radius R has already been calculated in the previous problem. We solve R = 2 tanh−1 r for the Euclidean radius and get r = tanh r2 =
(eR − 1)2 (eR + 1)2
and
R eR/2 − e−R/2 eR − 1 = = 2 eR/2 + e−R/2 eR + 1 (eR + 1)2 − (eR − 1)2 4eR 1 − r2 = = (eR + 1)2 (eR + 1)2
Now plug this formula into the result from part (a) and get A=
4πr2 π(eR − 1)2 = π(eR − 1 + e−R ) = 2π(cosh R − 1) = 2 eR (1 − r )
An alternative formula is A=
π(eR − 1)2 R = π(eR/2 − e−R/2 )2 = 4π sinh2 R e 2
(c) We have obtained in Proposition 46.2 from the section on the Poincaré disk model, that the hyperbolic circle of radius R has the circumference C = 2π sinh R. On the other hand, differentiating the result of (b) gives dA d cosh R = 2π = 2π sinh R = C dR dR as to be shown. Problem 46.4. Use the fundamental form for the Poincaré disk model (dsD )2 = to calculate its Gaussian curvature.
4(dx2 + dy2 ) (1 − x2 − y2 )2
(46.14)
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Answer. Formula (46.14) implies that the functions in the first fundamental form are E = G = H = 4(1 − x2 − y2 )−2 and F = 0. Hence, with x = u and v = y, we get from formula (46.5) " # " # ! 1 ∂E 1 ∂ 1 ∂E 1 ∂2 1 ∂ ∂2 − + − =− K= + ln E 2E ∂x E ∂x 2E ∂y E ∂y 2E ∂x2 ∂y2 ! ! 1 ∂2 ∂2 1 ∂ −2x ∂ −2y 2 2 =+ + ln(1 − x − y ) = + + E ∂x2 ∂y2 E ∂x (1 − x2 − y2 ) ∂y (1 − x2 − y2 ) ! −2 1 − x2 − y2 + 2x2 1 − x2 − y2 + 2y2 (−2) · 2 = = −1 + = 2 2 2 2 2 2 E 4 (1 − x − y ) (1 − x − y ) By the way, the result K = −1 motivates the annoying factor 4 in formula (46.14). 46.4.
Riemann metric of Klein’s model
Proposition 46.3 (Hilbert-Klein Metric). In the Klein model, the infinitesimal hyperbolic distance ds of points with coordinates (X, Y) and (X + dX, Y + dY) is ds2 =
dX 2 + dY 2 − (XdY − YdX)2 (1 − X 2 − Y 2 )2
(46.15)
Proof. We shall derive this metric using the transformation from the Poincaré to the Klein model. As stated in Proposition 42.1, the mapping from a point P in Poincaré’s model to a point K in Klein’s model is 2 |OP| |OK| = (42.1) 1 + |OP| 2 −−→ −−→ requiring that the rays OP = OK are identical. We use Cartesian coordinates and put P = (x, y) for Poincaré’s model and K = (X, Y) for the points in Klein’s model. Finally we put r2 = x2 + y2 and R2 = X 2 + Y 2 . From the mapping (42.1), we get 2x 2y and Y = 2 1+r 1 + r2 The Riemann metric for Poincaré’s model has been derived in Proposition 46.1 to be X=
ds2 = 4
(46.16)
dx2 + dy2 = E dx2 + 2F dxdy + G dy2 (1 − x2 − y2 )2
Here E, F, G denotes the fundamental form for the Poincaré model in terms of (x, y). In the following we shall use the matrix " # " # 4 1 0 E F (46.17) = F G (1 − x2 − y2 )2 0 1 From the fact that the transformation from Poincaré’s to Klein’s model is a passive coordinate transformation, we know that the infinitesimal hyperbolic distance ds of points is left invariant. Because of the invariance, the fundamental form E, F, G for the Klein model has to satisfy ds2 = E dX 2 + 2F dXdY + G dY 2 = E dx2 + 2F dxdy + G dy2 We take for now (x, y) as independent variables. From chain rule ∂ X ∂ Y " # ∂ X E F ∂x ∂x ∂ X ∂∂xY F G ∂ Y ∂y ∂y ∂x
calculus, we know that by means of the ∂X ∂y ∂Y ∂y
" # E F = F G
(46.18)
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It now remains to carry out the arithmetic. The superscript T denotes transposition of matrices and the superscript −1 denote inversion of matrices. As usual, we use ∂ X ∂ X D X ∂x ∂y = ∂ Y ∂ Y Dx ∂x ∂y as shorthand for the Jacobi matrix of the transformation (46.16). We need to solve the equation (46.18) for the new fundamental form E, F, G to obtain " # # D X T " E F E F DX = F G F G Dx Dx " # # D X T,−1 " E F E F D X −1 = F G Dx F G Dx The Jacobi matrix of the transformation (46.16) is obtained explicitly from equations (46.16) to be # " DX 2 1 − x2 + y2 −2xy = −2xy 1 + x 2 − y2 Dx (1 + x2 + y2 )2 The determinant is Det
h i h i 4(1 − r2 ) DX 4 4 = 1 − (x2 − y2 )2 − 4x2 y2 = 1 − r4 = 2 4 2 4 Dx (1 + r ) (1 + r ) (1 + r2 )3
Hence the inverse turned out to be " # D X −1 (1 + r2 )3 2 1 + x 2 − y2 2xy = 2xy 1 − x 2 + y2 Dx 4(1 − r2 ) (1 + r2 )2 # " 1 + r 2 1 + x 2 − y2 2xy = 2xy 1 − x 2 + y2 2(1 − r2 ) With the fundamental form from formula (46.17) and the inverse Jacobi matrix just obtained plugged into equation (??), we calculate " " #2 # 4 (1 + r2 )2 E F 1 + x 2 − y2 2xy = 2xy 1 − x2 + y2 F G 4(1 − r2 )2 (1 − r2 )2 " # (1 + r2 )2 1 + 2x2 − 2y2 + (x2 − y2 )2 + 4x2 y2 4xy = 4xy 1 − 2x2 + 2y2 + r4 (1 − r2 )4 " # (1 + r2 )2 (1 + r2 )2 − 4y2 4xy = 4xy (1 + r2 )2 − 4x2 (1 − r2 )4 This is the new fundamental form. We need still to introduce the new coordinates (X, Y). We use the short hands r2 = x2 + y2 and R2 = X 2 + Y 2 . By means of equation (46.16) we get 1 − X2 =
(1 + r2 )2 − 4x2 (1 + r2 )2 − 4y2 2 , 1 − Y = (1 + r2 )2 (1 + r2 )2 4xy (1 − r2 )2 2 XY = , 1 − R = (1 + r2 )2 (1 + r2 )2
Thus the new fundamental form miracously simplifies to be " # " # 1 E F 1 − Y2 XY = XY 1 − X2 F G (1 − R2 )2
,
(46.19)
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For the line element we get from this fundamental form ds2 = E dX 2 + 2F dXdY + G dY 2 (1 − Y 2 )dX 2 + 2XYdXdY + (1 − X 2 )dY 2 (1 − R2 )2 dX 2 + dY 2 − (XdY − YdX)2 = (1 − X 2 − Y 2 )2 =
Problem 46.5 (Gaussian curvature of the Hilbert-Klein metric). Use Gauss’ characteristic equation (46.5) and check directly that the Hilbert-Klein metric (46.15) from proposition 46.3 has constant Gaussian curvature K = −1. We use polar coordinates X = r cos θ, Y = r sin θ and convert formula to dr2 + r2 (1 − r2 )dθ2 ds2 = (46.20) (1 − r2 )2 since this simplifies the calculation considerably. Answer. We have to put u = r and v = θ. The first fundamental form and its coefficients become ds2 = Edr2 + 2Fdrdθ + Gdθ2 E = (1 − r2 )−2 , F = 0 , G = r2 (1 − r2 )−1 √ H = EG − F 2 = r(1 − r2 )−3/2 Hence we get for the Gaussian curvature K from the characteristic equation " # " # ∂ ∂G ∂ −1 ∂ r2 2HK = − H −1 =− r (1 − r2 )3/2 ∂r ∂u ∂r ∂r 1 − r2 " # 2 2 i ∂ h ∂ −1 2 3/2 2r(1 − r ) − r (−2r) 2 −1/2 r (1 − r ) 2(1 − r ) = − =− ∂r ∂r (1 − r2 )2 = (−2)(−1/2)(1 − r2 )−3/2 (−2r) = −2r(1 − r2 )−3/2 = −2H We get the constant Gaussian curvature K = −1, as expected. Problem 46.6 (Distortion of angles by the Hilbert-Klein metric). We use polar coordinates to simplify the calculation, and the corresponding contravariant components for the tangent vectors. At a point K with polar coordinates X = r cos θ, Y = r sin θ are attached the radial tangent vector (ar , aθ ) = (1, 0) and any other tangent vector (br , bθ ). Hence the apparent angle α satisfies cos α = q
br b2r + r2 b2θ
and tan α =
rbθ br
Check with the Hilbert-Klein metric (46.20) that the angle ω between the two vectors in Klein’s model satisfies √ tan ω = tan α 1 − r2 (46.21)
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Answer. The apparent angle α and the hyperbolic angle ω between the two tangent vectors (ar , aθ ) and (br , bθ ) satisfy cos α = q
ar br + r2 aθ bθ q (a2r + r2 a2θ ) · (b2r + r2 b2θ )
cos ω = q
ar br + r2 (1 − r2 )aθ bθ q a2r + r2 (1 − r2 )a2θ · b2r + r2 (1 − r2 )b2θ
In the given example with ar = 1, aθ = 0 the expressions simplify br cos α = q b2r + r2 b2θ
and tan2 α =
r2 b2θ b2r
r2 (1 − r2 )b2θ br cos ω = q and tan2 ω = b2r b2r + r2 (1 − r2 )b2θ Hence
√ tan ω = tan α 1 − r2
46.5. A second proof of Gauss’ remarkable theorem The most enlightened proof that the Gaussian curvature is an intrinsic property of the surface uses Gauss’ notion of integral curvature. For R R any domain G on a given curved surface, the integral curvature is defined as the integral KdA, where dA denotes the area element of the surface. G Take a geodesic triangle 4ABC. Let T denote the region bounded by the geodesics between any three given points A, B, C on the surface. Let α, β, γ be the angles (between tangents) to the geodesics at the three vertices. Gauss proves Z Z KdA = α + β + γ − π (46.22) T
where angles are to be measured in radians. The quantity on the right hand side is the deviation of the angle sum α + β + γ from the Euclidean value 180◦ , respectively π. 1 The quantity α + β + γ − π is called the excess of the triangle 4ABC. For the hyperbolic case, the excess is negative. In that case, one calculates using the excess times −1, which is called defect. In words, Gauss’ theorem tells the following: For a geodesic triangle, the integral curvature equals the excess of its angle sum. This theorem, Gauss says, ought to be counted as a most elegant theorem. I discuss a few immediate, but important consequences of (46.22). First of all, instead of the complicated characteristic equation (46.5), one has a simple property of a geodesic triangle from which to derive the Gaussian curvature in a limiting process. Secondly, as an immediate implications of (46.22), the Gaussian curvature is an intrinsic property of a curved surface. Recall that both geodesics, as well as measurement of area depend only on the first fundamental form. Hence, because of (46.22), the same is true for the Gaussian curvature. Another easy consequence of (46.22) is obtained from the special case of a sphere. For this surface, the Gaussian curvature is constant, and equal to K = R−2 where R is the radius of the sphere. Hence one obtains for the area of a spherical triangle A = (α + β + γ − π)R2 . as was already known before Gauss, e.g. to Lambert. 1
In radian measures, the Euclidean angle sum is π.
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Problem 46.7. Tile a sphere by equilateral triangles. It can be done in three ways: (i) Four triangles with α = β = γ = 120◦ . (ii) Eight triangles with α = β = γ = 90◦ . (iii) N triangles with α = β = γ = 72◦ . Explain and draw these tilings. To which Platonic bodies do the vertices correspond? Determine the surface area of the sphere from (i) and (ii), then get the number N in (iii). Answer. From item (i), the area of the sphere is ! 2π A = 4 · 3 − π R2 = 4πR2 3 The vertices of the four triangles form a tetrahedron. Similarly, item (ii) yields π A = 8 · 3 − π R2 = 4πR2 2 The vertices of the eight triangles form a octahedron. We can now calculate the number N of triangles in the tiling (iii). Because of ! 2π A = N · 3 − π = 4πR2 5 one gets N = 20. The vertices of the twenty triangles form an icosahedron. Here is a further important consequence of equation (46.22). Corollary 82 (A common bound for the area of all triangles). On a surface with negative π constant Gaussian curvature K < 0, the area of any triangle is less than . −K On December 17, 1799, Gauss wrote to his friend, the Hungarian mathematician Wolfgang Farkas Bolyai (1775-1856): As for me, I have already made some progress in my work. However, the path I have chosen does not lead at all to the goal which we seek [deduction of the parallel axiom], and which you assure me you have reached. It seems rather to compel me to doubt the truth of geometry itself. It is true that I have come upon much which by most people would be held to constitute a proof; but in my eyes it proves as good as nothing. For example, if we could show that a rectilinear triangle whose area would be greater than any given area is possible, then I would be ready to prove the whole of Euclidean geometry absolutely rigorously. Most people would certainly let this stand as an axiom; but I, no! It would indeed be possible that the area might always remain below a certain limit, however far apart the three angular points of the triangle were taken. From about 1813 on Gauss developed his new geometry. He became convinced that it was logically consistent and rather sure that it might be applicable. His letter written in 1817 to Olbers says: I am becoming more and more convinced that the physical necessity of our Euclidean geometry cannot be proved, at least not by human reason nor for human reason. Perhaps in another life we will be able to obtain insight into the nature of space, which is now unattainable. Until then we must place geometry not in the same class with arithmetic, which is purely a priori, but with mechanics.
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Problem 46.8. (a) Find two further enlightening statements of Gauss, and comments on all four statements. (i) Are they courageous? (ii) Are they to the benefit of the scientific community? (iii) Are they helpful for the person he addresses? (iv) Are they just against other people? (v) What would you have done in Gauss’ place? (b) Choose two of Gauss’ comments. Write a letter as you imagine you would have written in place of Gauss. Problem 46.9. To test the applicability of Euclidean geometry and his non-Euclidean geometry, Gauss actually measured the sum of the angles of the triangle formed by three mountain peaks in middle Germany: Broken, Hohenhagen, and Inselberg. the sides of the triangle were 69, 85, 197 km. His measurement yielded that the angle sum exceeded 180◦ by 14”.85. (a) Use Herons formula A = good approximation.
√
s(s − a)(s − b)(s − c) to calculate the area of the triangle, in a very
(b) Take R = 6378 km as radius of the earth. Calculate the angle excess for a spherical triangle between the three mountain peaks. You need to convert angular measurements! 1 radian 180◦ 180 · 3600” equals = . π π (c) Is the triangle that Gauss measured actually a spherical triangle. Why or why not? (d) Reflect on the motives why Gauss did his measurement. Find and read some further sources. Think of the following and further motives and possibilities. Did Gauss really just want to (i) check accuracy? (ii) check geometry? (iii) It was just a theoretical thought experiment, not really performed. Answer. (a) Herons formula give the area A =
√
s(s − a)(s − b)(s − c) = 2929.42 km2 .
(b) Take R = 6378 km as radius of the earth. The angle excess for a spherical triangle between the three mountain peaks is A α + β + γ − π = 2 = 7.201 10−5 (46.23) R This is the value in radian measure. Converted to degrees, we get .00413◦ which is 14.9”. (c) Of course the triangle one measures is not a spherical triangle, since light rays do not follow the curvature of the earth.
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46.6. Principal and Gaussian curvature of rotation surfaces Before introducing the pseudo-sphere, we need some facts about the curvature of general rotation surfaces. We take the graph of an arbitrary function y = f (x), and rotate it about the y-axis to produce a rotation surface in three dimensional space. The first principle curvature of a rotation surface in the xy plane is κ1 =
y00
(46.24)
3
(1 + y0 2 ) 2
This is just the curvature of the graph y = f (x). Recall that the perpendicular to the tangent of a curve is called the normal of the curve. The second principal curvature occurs for a section of the surface by a plane P2 , which intersects the xy plane along the normal of the curve y = f (x), and is perpendicular to the xy plane. The second principal curvature is y0 κ2 = (46.25) 1 x(1 + y0 2 ) 2 Proposition 46.4. The Gaussian curvature of the rotation surface produced by rotating the graph of y = f (x) around the y-axis, is the product K=
y0 y00 x(1 + y0 2 )2
(46.26)
The formula (46.24) for a curvature of a plane curve is standard. Finally, since K = κ1 κ2 , formulas (46.24) and (46.25) imply the claim (46.26). Here is an argument to justify (46.25): Let tan β = y0 be tangent of the slope angle for y = f (x), as usual. Calculate sin β. Calculate the hypothenuse AB of the right 4ABC, with vertex A = (x, y) on the curve, leg AC parallel to the x-axis, leg BC on the axis of rotation, and hypothenuse AB perpendicular to the curve. One can show that point B is the center of the best approximating circle 1 . Use this idea to get the main curvature κ2 . in the plane P2 . Hence κ2 = AB Answer. tan β =
AC BC
= y0 Hence sin β =
AC AB
= q
AC 2
2
y0 = p 1 + y0 2
BC + AC 1 sin β y0 κ2 = = = 1 x AB x(1 + y0 2 ) 2
A second proof of Proposition 46.4 . On the surface of rotation, we choose as parameters u = φ the rotation angle, and v = r the distance from the rotation axis. Since the y-axis is the axis of rotation, the surface of rotation gets the parametric representation x = v cos u y = f (v) z = v sin u The derivatives by the parameters are xu = −v sin u yu = 0 zu = v cos u
xv = cos u yv = f 0 (v) zv = sin u
(46.27)
858
Figure 46.1. Curvature of a rotation surface.
From these derivatives, one gets the first fundamental form. We use the general formulas E = xu2 + y2u + z2u F = xu xv + yu yv + zu zv G=
xv2
+
y2v
+
(46.3)
z2v
valid for any surface, and specialize to the surface of rotation given above. Now calculate E, F, G from (46.3) and (46.27), to get the first fundamental form (46.2). Next get the root of the √ determinant H = EG − F 2 . Finally calculate the Gaussian curvature from the characteristic equation (46.5). Problem 46.10. Use the approach as indicated to confirm formula (46.26). Answer. One gets E = v2 , F = 0 and G = 1 + f 0 2 and ds2 = v2 du2 + (1 + f 0 2 )dv2 = r2 dφ2 + (1 + f 0 (r)2 )dr2 q The root of the determinant is H = v 1 + f 0 2 . Because all four quantities E, F, G, H depend only on v, the partial derivatives by u all vanish. Thus Gauss’ characteristic equation (46.5) can be simplified to yield " # " # 1 ∂ F ∂E 1 ∂G 1 ∂ 2 ∂F 1 ∂E F ∂E K= − + − − 2H ∂u EH ∂v H ∂u 2H ∂v H ∂u H ∂v EH ∂u " # " # 1 ∂E 1 d 2v 1 ∂ = − =− 2H ∂v H ∂v 2H dv H
859 1
Now use H = v(1 + f 0 2 ) 2 and go on. We arrive at " # 1 1 d(1 + f 0 2 )− 2 1 d 2v =− q 2H dv H dv v 1 + f 02 h 3i 1 f 0 f 00 = (1 + f 0 2 )− 2 2 f 0 f 00 = q v(1 + f 0 2 )2 2v 1 + f 0 2
K=−
This result is equivalent to formula (46.26), since x = v = r is the distance from the axis of rotation and y = f (x). Problem 46.11. Calculate the Gaussian curvature of a three dimensional sphere of radius a. Answer. The sphere is provided by rotating the graph of x2 + y2 = a2 about the y-axis. Implicit differentiation yields 2x + 2yy0 = 0 and hence x y xy0 − y −x2 /y − y a2 1 · y − xy0 = = = − y00 = − y2 y2 y2 y3 a2 y0 y00 a2 1 = K= = = x(1 + y0 2 )2 y4 1 + x2 /y2 2 (x2 + y2 )2 a2 y0 = −
46.7. The pseudo-sphere The issue is now to find a rotation surface of constant negative Gaussian curvature K = −a−2 . Such a surface is called pseudo-sphere. Problem 46.12. Use the formula (46.26) for the Gaussian curvature, and get a differential equation of first order for the function u := y0 2 . You may begin by getting the derivative du dx . Answer. The derivative of the function u := y0 2 is into the formula (46.26). One gets
du dx
= 2y0 y00 . Next, I put the requirement K = −a−2
y0 y00 1 =− 2 a x(1 + y0 2 )2 2x(1 + y0 2 )2 2y0 y00 = − a2 du 2x(1 + u)2 =− dx a2
(46.28) (46.29) (46.30)
Problem 46.13. Solve the differential equation 2x(1 + u)2 du =− dx a2
(46.30)
by separation of variable. For simplicity, we use the initial data u(a) = 0, and get a curve through the point x = a, u = 0.
860
Answer.
u
Z a
du 2x(1 + u)2 =− dx a2 du 2xdx =− 2 (1 + u)2 a Z x du 2xdx du = − dx 2 (1 + u) a2 0 " #u " 2 # x 1 x − = − 2 1+u a a 0 −
1 x2 +1=− 2 +1 1+u a a2 1+u= 2 x a2 − x 2 u= x2
Problem 46.14. Check that the differential equation √ a2 − x 2 0 y =− x with |x| ≤ a, has the solution √ a + a2 − x2 √ − a2 − x2 + C y = a ln x
(46.31)
Too, find the general solution of the equation √ y =+ 0
Answer.
a2 − x 2 x
√ a − a2 − x2 √ + a2 − x2 + C y = a ln x
Definition 46.1 (Pseudo-sphere). The rotation surface with constant negative Gaussian curvature is called a pseudo-sphere. With curvature K = −a−2 and the y-axis as axis of rotation, its equation is √ a + a2 − x2 − z2 p 2 − a − x2 − z2 y = a ln √ x2 + z2 Problem 46.15. Check, once more, that the Gaussian curvature of the specified surface is K = −a−2 . Answer. Problem 46.16. Check the following fact: The segment on the tangent to the curve (46.31), between the touching point T , and the intersection S of the tangent with the y-axis has always the same length a. For that reason, the curve (46.31) is called tractrix.
861
Figure 46.2. The tractrix has a segment on its tangent of constant length.
Answer. Take the right 4T S C, formed by the segment T S on the tangent at point T , and the perpendicular from T onto the y-axis. 1 We know that y0 = tan α = 2
2
SC TC 2
T S = TC + S C = x2 (1 + y0 2 ) = a2 Problem 46.17. The surface area of of rotation surface, made by rotating y = f (x) about the y-axis, for x1 ≤ x ≤ x2 is Z x2 q S = 2πx 1 + y0 2 dx x1
Calculate the surface of the pseudo-sphere for bounds 0 < x ≤ a. Answer. Because of 1 + y0 2 = 1 + u =
a2 , x2
we get Z a a S = 2πx dx = 2πa2 x 0
We introduce now (φ, r) as two convenient coordinates on the pseudo-sphere. As first coordinate, we choose the angle of rotation φ about the y-axis. The second coordinate is the radius 1
This triangle is different from triangle 4ABC in the figure on page 858.
862
√ r = x2 + z2 measured from the axis of rotation, Up to now it was called x, but now I choose to name it r. The three parameters r, φ, y are cylindrical coordinates of three dimensional space. The first two of them are convenient parameters on the pseudo-sphere. Proposition 46.5 (Riemann Metric for the Pseudo Sphere). The infinitesimal distance ds of points with coordinates (φ, r) and (φ + dφ, r + dr) is ds2 = r2 dφ2 +
a2 2 dr r2
(46.32)
Proof. The distance on the pseudo-sphere is calculated from the usual Euclidean distance for points of the three dimensional space into which the surface is embedded. At first, I convert the distance from Cartesian to cylindrical coordinates. Because the y-axis is the rotation axis, its coordinate stays, but the pair (x, z) is converted to polar coordinates. Hence one gets ds2 = dx2 + dy2 + dz2 = dr2 + r2 dφ2 + dy2
(46.33)
We restrict to points on the pseudo-sphere. Hence the coordinates r and y are related in the same √ a2 −x2 0 way as x and y before. Thus y = − x gets √ dy a2 − r 2 =− (46.34) dr r Now we use (46.34) to eliminate y from (46.33) and get !2 dy dr2 dr a2 2 a2 − r2 2 2 2 dr = r dφ + dr = dr2 + r2 dφ2 + r2 r2
ds2 = dr2 + r2 dφ2 + dy2 = dr2 + r2 dφ2 +
as to be shown. As an alternative, we can use the first fundamental form calculated above. Since 2 2 2 1 + f 0 (r)2 = 1 + a r−r = ar2 , one gets again 2 ds2 = r2 dφ2 + (1 + f 0 (r)2 )dr2 = r2 dφ2 +
a2 2 dr r2
46.8. Poincaré half-plane and Poincaré disk Throughout, we denote the upper open halfplane by H = {(u, v) : v > 0}. Its boundary is just the real axis ∂H = {(u, v) : v = 0}. The open unit disk is denoted by D = {z = x + iy : x2 + y2 < 1}, and its boundary is ∂D = {z = x + iy : x2 + y2 = 1}. The following isometric mapping of the half-plane to the disk is used in this section. It differs from the one used in the previous section by a rotation of the disk by a right angle. We repeat for convenience. Proposition 46.6 (Isometric Mapping of the Half-plane to the Disk). The linear fractional function iw + 1 z= (46.35) w+i is a conformal mapping and a bijection from C ∪ {∞} to C ∪ {∞}. The inverse mapping is w=
1 − iz z−i
(46.36)
863
These mappings preserves angles, the cross ratio, the orientation, and map generalized circles to generalized circles. The upper half-plane H = {w = u + iv : v > 0} is mapped onto the unit disk D = {z = x + iy : x2 + y2 < 1}. Especially w = 0 7→ z = −i ,
w = 1 7→ z = 1 ,
w = ∞ 7→ z = i ,
w = i 7→ z = 0
Proposition 46.7 (Riemann Metric for Poincaré’s half-plane). In the Poincaré half-plane, the infinitesimal hyperbolic distance ds of points with coordinates (u, v) and (u + du, v + dv) is du2 + dv2 v2 The mapping (46.35) provides an isometry between the half-plane and the disk: (dsH )2 =
dsD = dsH
(46.37)
(44.5)
Proof. The metric of the half plane is calculated from the known metric (dsD )2 =
4(dx2 + dy2 ) (1 − x2 − y2 )2
(46.14)
of the Poinaré disk model. The mapping iw + 1 w+i provides an isometry from the half-plane to the disk. The denominator is z=
(46.35)
(w + i)(w − i) − (iw + 1)(−iw + 1) 2iw − 2iw 4v = = 2 2 |w + i| |w + i| |w + i|2 The derivative of the mapping (46.35) is dz 2 =− dw (w + i)2 Putting the last two formulas into (46.14) yields 1 − | z|2 =
4| dz|2 4(dx2 + dy2 ) = (1 − x2 − y2 )2 (1 − |z|2 )2 !2 !2 2 dz |w + i|2 2 2 |w + i|2 2 2 | dw| = 4 | dw| = 4 dw 4v 4v (w + i)2 2 2 2 | dw| du + dv = 2 = v v2 Hence formula (46.37) arises from the isometry (46.35) between half-plane and the disk. ds2 =
46.9.
Embedding the pseudo-sphere into Poincaré’s half-plane
Proposition 46.8. The mapping
a (46.38) r of the pseudo-sphere to the line element dsH of the half-plane w=φ+i
transforms the line element dsPS such that
dsPS = a (dsH )
(46.39)
For a = 1, we get an isometry. This is just the case with Gaussian curvature K = −a−2 = −1. Because an isometry conserves the Gaussian curvature, this shows that the Poincaré half-plane has Gaussian curvature −1.
864
Proof. We separate equation (46.38) into its real- and imaginary part to get u=φ,
a r
v=
(46.40)
Using its derivatives, we plug into dsH 2 =
du2 + dv2 v2
(46.37)
One gets dsH = 2
2 dφ2 + −ar−2 dr2 a2 r−2
=a
−2
a2 r dφ + 2 dr2 r 2
!
2
Now comparing with ds2PS = r2 dφ2 +
a2 2 dr r2
(46.32)
one concludes dsH 2 = a−2 (dsPS )2
and hence equation (46.39) holds.
46.10. Embedding the pseudo-sphere into Poincaré’s disk The next goal is to construct an isometric mapping from the pseudo-sphere to the Poincaré disk. It is convenient to get this mapping as composition of a mapping from the pseudo-sphere to the half-plane, and the conformal mapping from the half-plane to the disk. Proposition 46.9. We take a pseudo-sphere with a = 1. This normalizes the Gaussian curvature to be K = −1. The mapping r − 1 + irφ (46.41) z= rφ + i(r + 1) maps the pseudo-sphere isometrically into the Poincaré disk. Proof. The mapping (46.41) is constructed as a composition of two mappings PS → 7 H 7→ D. Take the mapping PS 7→ H given by equation (46.42), and the mapping H 7→ D given by equation (46.35). The composition of the mapping w=φ+i
1 r
(46.42)
from the pseudo-sphere to the Poincaré half-plane, with the mapping z=
iw + 1 w+i
(46.35)
from the Poincaré half-plane to the Poincaré disk is the required mapping. following mapping i φ + ri + 1 r − 1 + irφ z= = i rφ + i(r + 1) φ+ r +i from the pseudo-sphere to the Poincaré disk. Both mappings (46.42) and (46.35) are isometries, as stated by formulas (46.39) with a = 1 and formula (??). Hence their composition (46.41) conserves the line element: dsPS = dsH = dsD .
865
46.11. About circle-like curves We now go back to the Poincaré disk model. At first, here are a few remarks about circle-like curves. In hyperbolic geometry, there exist three different types of circle-like curves. I define as a circle-like curve a curve which appears to be a circle in the Poincaré model. Recall that ∂D is the boundary circle of the Poincaré disk. Take any second circle C. I call its Euclidean center M the quasi-center. The meaning of C for the hyperbolic geometry of the Poincaré disk depends on the nature of the intersection of the two circles C and ∂D. There are three important cases: (i) The circle C lies totally in the interior D. In that case, it is a circle for hyperbolic geometry. This circle has a center A in hyperbolic geometry. Note that the quasi-center M is different from the center A of C as an object of hyperbolic geometry. (ii) The circle C touches the boundary ∂D from inside, say at endpoint E. In that case, it is a horocycle for hyperbolic geometry. A horocycle has no hyperbolic center, instead it contains an ideal point E. Hence it is unbounded. The hyperbolic circumference of a horocycle is infinite, as follows from part (c) below. (iii) The circle C intersects the boundary ∂D at two endpoints E and F. In that case, the circular arc inside the disk D is either an equidistance line or a geodesic for hyperbolic geometry. A geodesic intersects ∂D perpendicularly. In the case of non perpendicular intersection of ∂D and C, one gets an equidistance line. Actually all points of that equidistance line have the same distance from the hyperbolic straight line with ends E and F. Problem 46.18. Take points Y+ = (1, 0) and O = (0, 0). Find the analytic equation for a horocycle H with apparent diameter OY+ . Answer. In complex notation, point Y+ is i. The quasi-center is M = 2i , and the apparent radius is 1 2 . Hence one gets the equation 2 z − i = 1 2 4 !2 1 1 x2 + y − − =0 2 4 x2 + y(y − 1) = 0 We need another more convenient parametric equation for the horocycle H. Let Z = (x, y) be any point on H and define the circumference angle ρ ∠OY+ Z. Calculate tan ρ in terms of (x, y). Then express x and y in terms of the central angle 2ρ ∠OMZ. Use double angle formulas, and finally express x and y in terms of tan ρ. Answer.
x y = 1−y x sin 2ρ tan ρ x= = sin ρ cos ρ = 2 1 + tan2 ρ 1 − cos 2ρ tan2 ρ y= = sin2 ρ = 2 1 + tan2 ρ
tan ρ =
(46.43)
Problem 46.19. Confirm that the hyperbolic arc length of the arc OZ on the horocycle H is just s = 2 tan ρ.
866
Figure 46.3. Measuring an arc of a horocycle.
Answer. Let t = tan ρ. Differentiation yields t , 1 + t2 t2 y= , 1 + t2
x=
1 − x2 − y2 = 1 − y =
1 1 + t2
Hence the hyperbolic metric (46.14) implies !2 ds = 4(1 − x2 − y2 )−2 dt
dx 1 − t2 = dt (1 + t2 )2 dy 2t = dt (1 + t2 )2 !2 !2 dx dy 1 + = dt dt (1 + t2 )2 ! !2 dx 2 dy = 4 + dt dt
Hence by elementary integration s = 2t = 2 tan ρ. Problem 46.20. Give the representation of the horocycle H with this arc length as parameter. Explain in a drawing, how to measure the arc length on this horocycle. Answer. We get the parametrization 2s 4 + s2 s2 y= 4 + s2
x=
(46.44)
867
Figure 46.4. Isometry of the sliced pseudo-sphere to a half infinity strip.
The hyperbolic arc length of OZ on the horocycle H is the Euclidean length |Y− Z 0 | since s = 2 tan ρ = |Y− Z 0 |. 46.12. Mapping the boundaries There cannot exist an isometry of between the pseudosphere and the half-plane, since they have different topologies. A corresponding problem already arises in Euclidean geometry, for the construction of for an isometry between the cylinder and the plane. At least, there exists a non-invertible homomorphism from the plane onto the cylinder. This homomorphism can be restricted to on isomorphism between a strip of the plane and the sliced cylinder. We return to the hyperbolic case. By slicing the pseudo-sphere, we get an isomorphism of the sliced pseudo-sphere into a strip of the half-plane, and furthermore into part of the disk, too. The pseudo-sphere is sliced along the geodesic in the negative (x, y)-plane, restricting the rotation angle to the half-open interval −π ≤ φ < π. The mapping w=φ+i
1 r
(46.42)
maps the sliced pseudo-sphere onto a half open rectangular domain PS H in the upper half-plane. The boundary of PS H consists of a segment AB with the endpoints A = −π + i, B = π + i, and two −−→ −−→ unbounded rays A∞ and B∞ with vertices A and B parallel to the positive v axis. Furthermore, we map the sliced pseudo-sphere to the Poincaré disk via the isometry (46.41). The image PS D of the pseudo-sphere is a part of the interior of the horocycle H with apparent diameter 0 to i. Problem 46.21. On the pseudo-sphere, we use as parameters the cylindrical coordinates r and φ. The boundary of the sliced pseudo-sphere is given by r = 1, and −π < φ < π. To which curve in the disk D is the boundary mapped by the isometry (46.41)? Answer. The boundary ∂PS D of image PS D consists of three circular arcs. A segment AB of a horocycle H with endpoints iπ iπ A= , B= π + 2i π − 2i
868
Figure 46.5. Isometric image of the sliced pseudo-sphere in the Poincaré disk.
−−−→ −−−→ as well as two geodesic rays AY+ and BY+ with vertices A and B pointing to the ideal endpoint Y+ = i. Problem 46.22. Give a parametric equation for the boundary, with parameter φ, at first in complex notation for z = x + iy. Then separate into real and imaginary parts to get equations for x and y. Answer. Simply put r = 1 into equation (46.41). One gets z=
iφ φ + 2i
To separate real and imaginary parts, one needs a make the denominator real: iφ iφ(φ − 2i) = φ + 2i (φ + 2i)(φ − 2i) iφ2 + 2φ x + iy = 2 φ +4 2φ x= 2 φ +4 φ2 y= 2 φ +4 z=
(46.45) (46.46) (46.47) (46.48)
Problem 46.23. Check that your parametric equation is a circle with center 2i . Answer. This is a parametric equation of a circle with center
i 2
because
i iφ i i(φ − 2i) = − = 2 φ + 2i 2 2(φ + 2i) z − i = 1 2 2 z−
(d) Compare (46.46) with the result tan ρ 1 + tan2 ρ tan2 ρ y= 1 + tan2 ρ
x=
and check that φ = 2 tan ρ. Hence, because of s = 2 tan ρ was shown, one concludes that φ = s = 2 tan ρ. Of course φ = s follows directly because of the isometries PS 7→ H → 7 D.
869
Problem 46.24. Draw sketches of the pseudo-sphere as it appears in the domains PS 7→ H 7→ D. Use different colors for the different vertices and edges of the boundary, but same colors for corresponding objects in all three domains PS 7→ H 7→ D, as they are mapped by our isometries from above. Problem 46.25. The Poincaré disk can be tiled with congruent triangles. Indeed, there exist infinitely many different types of such tilings. I choose a tiling with congruent equilateral triangles, such that at each vertex seven triangles meet. Use Gauss’ remarkable theorem to calculate the hyperbolic area of one such triangle. Answer. I measure angles in radian measure. The angles of one triangle are all α=β=γ=
2π 7
and hence the defect of the angle sum is α+β+γ−π=
6π π −π=− 7 7
Since the Gaussian curvature of Poincaré’s model is K = −1, the area is just the negative of the excess, (this is also called the defect) and is π7 . Problem 46.26. Overlay two drawings, of the tiling by equilateral triangles, and a second drawing of the image PS D of the pseudo-sphere and its boundary ∂PS D . How many of those triangles of the tiling fit entirely into PS D ? How many triangles make up (with bids and pieces!) the total area of PS D ? Answer. Using item 12, one calculated that the total area of the pseudo-sphere is 2π. This equals the area of 14 triangles.
870
List of Figures 2.1 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23
Logical relations of geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . There are two four-point incidence geometries. . . . . . . . . . . . . . . . . . . . . . . There are four five-point incidence geometries. . . . . . . . . . . . . . . . . . . . . . . Too simple, this is not even an incidence geometry. . . . . . . . . . . . . . . . . . . . . A straight fan is an incidence geometry—but still no projective plane. . . . . . . . . . . . Enumerate the pencils for the two models (A) and (B). . . . . . . . . . . . . . . . . . . Two isomorphic six-point incidence geometries . . . . . . . . . . . . . . . . . . . . . . The labelling of the points shows the isomorphism. . . . . . . . . . . . . . . . . . . . . Name the points as to show the isomorphism. . . . . . . . . . . . . . . . . . . . . . . . The names of the points show the two drawings of the Fano-plane are really isomorphic. . . . A nine-point incidence geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . In an affine plane, we find a second point on any given line. . . . . . . . . . . . . . . . . A picture to explain the schedule for five groups of five students each, on six days. . . . . . Fano’s seven-point projective plane . . . . . . . . . . . . . . . . . . . . . . . . . . . The symmetric drawing of the Fano-plane is really isomorphic to the projective completion of the affine plane of order 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fano’s seven-point incidence geometry is selfdual . . . . . . . . . . . . . . . . . . . . The points and lines of the Fano-plane named in a way to confirm self-duality. . . . . . . . The complete quadrilateral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Both illustrations confirm the self-duality . . . . . . . . . . . . . . . . . . . . . . . . n + 1 = 5 pairwise intersecting lines in an affine plane of order n = 4. . . . . . . . . . . . Construction of the plane through a given point parallel to a given plane. . . . . . . . . . . Desargues’ configuration in parallel setting. . . . . . . . . . . . . . . . . . . . . . . . The Theorem of Desargues in parallel setting. . . . . . . . . . . . . . . . . . . . . . . Desargues’ Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Projecting the three-dimensional Desargues to the two-dimensional one. . . . . . . . . . . Desargues implies the converse Desargues. . . . . . . . . . . . . . . . . . . . . . . . Constructing the parallel to two given parallel lines. . . . . . . . . . . . . . . . . . . . . The Theorem of scissors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Little Theorem of scissors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Prove the Theorem of scissors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . How to construct a tiling. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . How to prove the expected incidences;—step 1. . . . . . . . . . . . . . . . . . . . . . How to prove the expected incidences;—step 2. . . . . . . . . . . . . . . . . . . . . . How to prove the expected incidences;—step 3. . . . . . . . . . . . . . . . . . . . . . How to prove the expected incidences;—all steps. . . . . . . . . . . . . . . . . . . . . Lines in the Moulton plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In the Moulton plane, the Desargues Theorem is not valid. . . . . . . . . . . . . . . . . In the Moulton plane, the Little Desargues Theorem is not valid. . . . . . . . . . . . . . . For Pappus’ configuration: If BC 0 k B0 C and AC 0 k A0 C, then AB0 k A0 B. . . . . . . . . . The Theorem of Pappus in the projective setting. . . . . . . . . . . . . . . . . . . . . . Pappus’s Theorem implies Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . . The case with OABC a parallelogram can be handled, too . . . . . . . . . . . . . . . . The Little Pappus Theorem asserts for a hexagon AC 0 BA0 CB0 that has its vertices alternating on two parallel lines: if BC 0 k B0 C and AC 0 k A0 C, then AB0 k A0 B. . . . . . . . . . . . . . . The Little Desargues Theorem asserts for two triangles with vertices on three parallel lines: if AB k A0 B0 and BC k B0 C 0 , then AC k A0 C 0 . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . .
29 40 41 42 42 44 45 45 45 46 47 48 51 55
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
56 57 57 60 62 64 71 72 73 74 75 76 77 79 79 80 81 82 83 84 85 88 89 90 91 92 92 93
.
93
.
94
871
4.24 4.25 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 6.1 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17 7.18 7.19 7.20 7.21
The Little Desargues Theorem implies the Little Pappus Theorem. . . . . . . . . . . . . . In the Moulton plane, the Little Pappus Theorem is not valid. . . . . . . . . . . . . . . . How to get a point inside the given segment AC. . . . . . . . . . . . . . . . . . . . . . Assume that neither A nor C lie between the two other of the three points A, B and C. The construction shows that B does lie between A and C. . . . . . . . . . . . . . . . . . . . Two intersecting segments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bernays’ Lemma would not hold— and line A, E 0 , E, C would be equal to line E 0 , C, D, B—all points collapse onto one line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generic cases in the proof of the plane separation theorem. . . . . . . . . . . . . . . . . Special cases in the proof of the plane separation theorem. . . . . . . . . . . . . . . . . Interior and exterior of an angle are intersection and union of two half planes. . . . . . . . . Interior of the vertical angle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A ray interior of an angle intersects a segment from side to side. . . . . . . . . . . . . . . The Crossbar Theorem is proved using Pasch’s axiom. . . . . . . . . . . . . . . . . . . About any two angles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Any n points lie in the interior or on the boundary of an angle. . . . . . . . . . . . . . . . The segment between two points on different sides of a plane intersects the plane. . . . . . . The interior of a triangle is the intersection of three half planes. . . . . . . . . . . . . . . The three half planes H 0 , K 0 , L0 opposite to the interior ones have empty intersection. . . . . A line through an interior point of a triangle intersects either two sides, or goes through a vertex and intersects the opposite side. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . How to get three rays from given vertex P in a star-shape. . . . . . . . . . . . . . . . . . The three ray star makes a triangle with common vertex in the interior, and conversely. . . . . The nonempty intersection of interiors of two triangles contains a triangle. . . . . . . . . . Three rays pointing from the vertices into the interior of a triangle. . . . . . . . . . . . . . The restricted Jordan Curve Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . PG(3, 2) but not all the lines are drawn. . . . . . . . . . . . . . . . . . . . . . . . . . Uniqueness of segment transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Segment subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Segment comparison for congruence classes . . . . . . . . . . . . . . . . . . . . . . . Transitivity of segment comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . To extend a line—. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . An isosceles triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . How to simply get ASA congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . Transfer of a triangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Extended ASA congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Congruence of supplementary angles, the first pair of congruent triangles . . . . . . . . . . Congruence of supplementary angles, the second pair of congruent triangles . . . . . . . . . Congruence of supplementary angles, the third pair of congruent triangles . . . . . . . . . . Supplementary angles yield points on a line . . . . . . . . . . . . . . . . . . . . . . . Two pairs of supplementary angles yield vertical angles . . . . . . . . . . . . . . . . . . Angle addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . the first pair of congruent triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . the second pair of congruent triangles . . . . . . . . . . . . . . . . . . . . . . . . . . the third pair of congruent triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . Three subcases for angle addition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . The symmetric kite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Which kite is symmetric? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . .
94 95 96
. .
98 99
. . . . . . . . . . . .
100 101 102 110 111 112 113 113 114 117 117 118
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
119 122 123 126 130 131 145 148 149 149 150 150 153 154 155 156 157 158 158 158 159 160 160 161 161 162 164 164
872
7.22 7.23 7.24 7.25 7.26 7.27 7.28 7.29 7.30 7.31 7.32 7.33 7.34 7.35 7.36 7.37 7.38 7.39 7.40 7.41 7.42 7.43 7.44 7.45 7.46 7.47 7.48 7.49 7.50 7.51 7.52 7.53 7.54 7.55 7.56 7.57 7.58 7.59 7.60 7.61 7.62 7.63 7.64 7.65 7.66 7.67
The diagonals of the rhombus bisect each other. . . . . . . . . . . . . . . . . . . . . . Completing Euclid’s proof of (I.7). . . . . . . . . . . . . . . . . . . . . . . . . . . . Completing Euclid’s proof of (I.7). . . . . . . . . . . . . . . . . . . . . . . . . . . . Transitivity of comparison of angles . . . . . . . . . . . . . . . . . . . . . . . . . . . The case (i) where α < β. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The case (ii) where β < α. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . First proof of the hypothenuse leg theorem . . . . . . . . . . . . . . . . . . . . . . . . Drop the perpendicular, the two cases . . . . . . . . . . . . . . . . . . . . . . . . . . Congruences needed in the two cases. . . . . . . . . . . . . . . . . . . . . . . . . . . Construction of an isosceles triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . Erect a perpendicular . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The perpendicular bisector via the kite . . . . . . . . . . . . . . . . . . . . . . . . . . Get the perpendicular bisector via: (a) a convex kite, (b) a nonconvex kite . . . . . . . . . . A point with congruent distances to both endpoints lies on the perpendicular bisector. . . . . Construction of the perpendicular bisector—the natural way . . . . . . . . . . . . . . . . The angular bisector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The essential pair of congruent triangles . . . . . . . . . . . . . . . . . . . . . . . . . Three cases for adding angles: they (i) are supplementary (ii) can be added (iii) or cannot be added . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . If A and G lie on different sides of BC, then C and G lie on different sides of AB. . . . . . . Bisecting a pair of supplementary angles yields perpendicular rays. . . . . . . . . . . . . The impossible situation of a congruent exterior angle . . . . . . . . . . . . . . . . . . . Two ways from B to D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The impossible situation of a smaller exterior angle . . . . . . . . . . . . . . . . . . . . Comparing the other nonadjacent angle . . . . . . . . . . . . . . . . . . . . . . . . . A proof of the exterior angle theorem presupposing bisection of a segment. . . . . . . . . . A pair of z-angles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . No three points have the same distance from a line . . . . . . . . . . . . . . . . . . . . Assumed is ε ≤ β. Why does the ray r intersect the side AC. . . . . . . . . . . . . . . . . Since ε ≤ β < δ the ray r intersects the side AC. . . . . . . . . . . . . . . . . . . . . . . A parallelogram in neutral geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . Addendum to the extended ASA congruence. . . . . . . . . . . . . . . . . . . . . . . . Across the longer side lies the greater angle . . . . . . . . . . . . . . . . . . . . . . . The foot point has the shortest distance . . . . . . . . . . . . . . . . . . . . . . . . . . The triangle inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SAA congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The hypothenuse leg theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Hinge Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Hinge Theorem: turning a long side . . . . . . . . . . . . . . . . . . . . . . . . . The Hinge Theorem: the borderline case . . . . . . . . . . . . . . . . . . . . . . . . . The Hinge Theorem: turning a short side . . . . . . . . . . . . . . . . . . . . . . . . . Hilbert’s construction of the midpoint . . . . . . . . . . . . . . . . . . . . . . . . . . M = A is impossible . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Point A lying between M and B is impossible . . . . . . . . . . . . . . . . . . . . . . . Point A lying between M and B is impossible . . . . . . . . . . . . . . . . . . . . . . . Apply the SAA congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The generic situation for Proposition 7.57, for which we prove: Two segments CX and DY on different sides of XY are congruent if and only if midpoint M of segment CD lies on line l. The third figure shows the case with both conditions true. . . . . . . . . . . . . . . . . . . .
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166 166 167 169 169 170 172 173 174 175 176 177 178 179 180 180 181
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
182 182 184 187 188 189 189 190 191 193 194 195 195 196 197 199 199 201 202 203 203 204 204 205 205 206 207 207
. 208
873
7.68 7.69 7.70 7.71 7.72 7.73 7.74 7.75 7.76 7.77 7.78 7.79 7.80 8.1 8.2 8.3 8.4 8.5 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 10.1 10.2 10.3 10.4 10.5 10.6 10.7
The diagonals of a parallelogram bisect each other. . . . . . . . . . . . . . . . . . . . . A Euclidean example for SSA triangle construction . . . . . . . . . . . . . . . . . . . . Matching two triangles with SSA . . . . . . . . . . . . . . . . . . . . . . . . . . . . Establish two solutions for SSA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reflect a point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reflection by a line l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Where on the boundary can guys A and B meet the quickest? . . . . . . . . . . . . . . . . They meet where the perpendicular bisector of AB intersects the boundary. . . . . . . . . . Where on the boundary can guys A and B meet the quickest? . . . . . . . . . . . . . . . . Where on the boundary can guys A and B meet the quickest? . . . . . . . . . . . . . . . . Reflect a light ray from P to a given point Q. . . . . . . . . . . . . . . . . . . . . . . . Turning the mirror by angle θ turns the reflected ray by the angle 2θ. . . . . . . . . . . . . To get a rhombus without use of reflection. . . . . . . . . . . . . . . . . . . . . . . . The Archimedean axiom. For the case drawn in the figure, it turns out that n = 7. . . . . . . This measurement yields |AB| = 1.1101 . . . as a binary fraction. . . . . . . . . . . . . . . The angular bisector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Many consecutive small angles surpass any angle. . . . . . . . . . . . . . . . . . . . . Congruent angles cut longer and longer segments from a line. . . . . . . . . . . . . . . . Two triangles with same angle sum—and area. . . . . . . . . . . . . . . . . . . . . . . Two triangles with same angle sum, cases (i) and (ii). . . . . . . . . . . . . . . . . . . . A Saccheri quadrilateral is bisected into two Lambert quadrilaterals. . . . . . . . . . . . . The steps to get symmetry of a Saccheri quadrilateral. . . . . . . . . . . . . . . . . . . Getting rectangles of arbitrary width— Is ζ a right angle? . . . . . . . . . . . . . . . . . ε is congruent to both θ and ϕ, hence there are two supplementary right angles θ and ϕ! . . . . If one rectangle exists, why are all Lambert quadrilaterals rectangles? . . . . . . . . . . . To every triangle corresponds a Saccheri quadrilateral. . . . . . . . . . . . . . . . . . . How to get two pairs of congruent triangles, and how to get the top angles. . . . . . . . . . The angle sum 2R implies uniqueness of the parallel. . . . . . . . . . . . . . . . . . . . Halfing an angle with an isosceles triangle. . . . . . . . . . . . . . . . . . . . . . . . For angle sum less 2R, there exist two different parallels. . . . . . . . . . . . . . . . . . The defect and the area of triangles are both additive. . . . . . . . . . . . . . . . . . . . Additivity of the defect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Doubling the defect of a triangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hyperbolic geometry produces an angle at vertex E. . . . . . . . . . . . . . . . . . . . Some statements about a line intersecting a circle in two points. . . . . . . . . . . . . . . The short/long arc lies in the interior/exterior of the central angle. . . . . . . . . . . . . . Construction of a chord through the given point P of given length A0 B0 . . . . . . . . . . . . Can you see why the construction works? . . . . . . . . . . . . . . . . . . . . . . . . How to get the second intersection point of a circle with a line. . . . . . . . . . . . . . . . A neutral construction of the tangents to a circle . . . . . . . . . . . . . . . . . . . . . If any ad hoc construction of the tangents to a circle is available, the circle-line intersection property holds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.8 The tangent is the limiting position of a short secant. Hence the shorter secant T P forms with the tangent the smaller angle ∠RT P. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.9 We use the Hinge theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.10 A circle with points on both sides of a line. . . . . . . . . . . . . . . . . . . . . . . . . 10.11 A segment from inside to outside a circle intersects the circle—proved from the circle-circle intersection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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208 210 211 212 214 215 216 217 217 218 219 219 225 227 228 231 232 233 240 241 243 243 244 244 244 245 246 247 248 249 251 251 252 253 256 257 258 259 260 260
. 261 . 262 . 263 . 267 . 268
874
10.12 The construction of a 60◦ angle at the center O is possible with straightedge compass in neutral geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.13 This construction of an equilateral triangle with Hilbert tools works only in Euclidean geometry 10.14 A segment from inside to outside a circle contains both points inside and outside the circle. . . . 10.15 Dedekind’s axiom implies existence of the intersection point K ∗ of two circles. . . . . . . . . 10.16 A circular arc from inside to outside a circle C contains both points inside and outside that circle. 11.1 A point lies on the perpendicular bisector if and only if its distances from both endpoints are congruent. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Strange perpendiculars to the three sides of a triangle. . . . . . . . . . . . . . . . . . . . . 11.3 In hyperbolic geometry, the perpendicular bisectors can be parallel as shown—but this figure is impossible in Euclidean geometry. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 The perpendicular bisectors of the sides of triangle 4ABC are all parallel. The triangle has no circum-circle—true, but hard to believe! . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 The generic situation for Proposition 11.5, for which the following statement is proved: segments AX and BY are congruent if and only if lines l and pc are perpendicular. . . . . . . . . . . . 11.6 Constructing the Saccheri quadrilateral from its top: if the lines pc and l are perpendicular to each other, then the segments AX and BY are congruent. . . . . . . . . . . . . . . . . . . 11.7 To triangle 4ABC corresponds the Saccheri quadrilateral AFGB. . . . . . . . . . . . . . . 11.8 The altitudes of the midpoint triangle are the side bisectors and form six right angles. . . . . . 11.9 For an obtuse triangle, too, the altitudes of the midpoint triangle are the side bisectors, and form six right angles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.10 The Hjelmsjev line in the generic case. . . . . . . . . . . . . . . . . . . . . . . . . . . 11.11 The ray opposite to the bisecting ray b bisects the vertical angle, because all four red angles are congruent. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.12 Angle addition of two pairs of congruent angles yields congruent supplementary angles—proving that the bisectors of supplementary angles are perpendicular. . . . . . . . . . . . . . . . . 11.13 A point P on the interior angular bisector has congruent distances from both sides the angle ∠(h, k). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.14 Any two angular bisectors intersect at the center of the in-circle. . . . . . . . . . . . . . . . 11.15 The in-circle and one ex-circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.16 The Theorem of Hjelmslev. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.17 The segment congruence in neutral geometry. . . . . . . . . . . . . . . . . . . . . . . . 11.18 Begin defining the lines g0 and h0 by the angle congruences. . . . . . . . . . . . . . . . . . 11.19 Triangle 4ABC has a circum-circle, hence line n intersects line l in its center O—parallels are unique. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.20 The perpendicular bisectors of the sides of triangle 4ABC are all parallel. The triangle has no
270 271 274 275 276 278 281 281 283 284 285 286 286 287 289 290 290 291 293 294 295 296 298 298
circum-circle—true, but hard to believe! There exists two parallels m and n to line l through point P. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.21 If the three midpoints lie on a line, a rectangle exists. . . . . . . . . . . . . . . . . . . . . 11.22 How the hyperbolic case appears in Klein’s model. . . . . . . . . . . . . . . . . . . . . . 12.1 Across the longer side BC is the larger angle δ. . . . . . . . . . . . . . . . . . . . . . . . 12.2 For an acute top angle, the perpendicular dropped from a point P inside the top segment CD is shorter than the opposite sides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 For an acute top angle γ, the perpendicular PQ dropped from a point outside the top segment is longer than the opposite sides BC. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 For an obtuse top angle γ, the perpendicular PQ dropped from a point outside the top segment is shorter than the opposite sides BC and AD. . . . . . . . . . . . . . . . . . . . . . . . .
299 300 300 303 303 304 305
875
12.5 Two Saccheri quadrilaterals with a common middle line MN have either both acute, both obtuse, or both right top angles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Any two Saccheri quadrilaterals can be put into a position such that the middle line of the first one is the base line of the second one, and vice versa. . . . . . . . . . . . . . . . . . . . . 12.7 Does the figure close to a rectangle? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.8 Does the figure close to a rectangle? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.9 Still not always a rectangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.10 Following the ray r. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.11 The points X, Y, Z on the ray r. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.12 What kind of quadrilateral does one get? . . . . . . . . . . . . . . . . . . . . . . . . . . 12.13 About the semi-hyperbolic AAA congruence. . . . . . . . . . . . . . . . . . . . . . . . 12.14 Uniqueness of parallels implies Wallis’ postulate. . . . . . . . . . . . . . . . . . . . . . 12.15 Wallis’ postulate implies uniqueness of parallels. . . . . . . . . . . . . . . . . . . . . . . 12.16 Uniqueness of parallels implies unbounded opening of an angle. . . . . . . . . . . . . . . . 12.17 The second line n , m, different to the double perpendicular m, drawn through point P is along the side of a triangle 4PXR, and hence by Pasch’s axiom intersects line l. . . . . . . . . . . . 12.18 For the angle between two parallels m and n to line l, intersecting at P, the distance from a point on one side to the other side of angle ∠(m, n) is always smaller than the perpendicular PQ. . . . 12.19 Doubling the distance from a point on one side to the other side of angle θ. . . . . . . . . . . 13.1 To every triangle corresponds a Saccheri quadrilateral. . . . . . . . . . . . . . . . . . . . 13.2 Even an obtuse triangle is equicomplementable to its corresponding Saccheri quadrilateral. . . . 13.3 deZolts postulate tells that P and Q are not equicomplementable . . . . . . . . . . . . . . . 14.1 Construction of the perpendicular with just two circles and their intersection. . . . . . . . . . 14.2 Construction of the circle through A and B tangent to line l. . . . . . . . . . . . . . . . . . 14.3 Transfer of a segment with Euclidean tools. . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Alternative construction to transfer of a segment with Euclidean tools. . . . . . . . . . . . . 14.5 Transfer of an angle with Euclidean tools. . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 A segment from inside to outside a circle intersects the circle—proved from the circle-circle intersection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.7 Transfer 1 of an angle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.8 Transfer 2 of an angle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Thales Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Construction of a triangle with angles of 30◦ , 60◦ and 90◦ . . . . . . . . . . . . . . . . . . . 15.3 Three triangles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Construction of a right triangle with hypothenuse c = 5 and leg a = 3. . . . . . . . . . . . . 15.5 Construction of a right triangle with projections p = 3, q = 4. . . . . . . . . . . . . . . . . 15.6 Erect the perpendicular via Thales’ Theorem. The numbers indicate the order of the steps. . . . 15.7 Drop the perpendicular via Thales’ Theorem. The numbers indicate the order of the steps. . . . 15.8 A strengthening of Thales’ Theorem, with vertex C inside the circle . . . . . . . . . . . . . 15.9 A strengthening of Thales’ Theorem, with vertex C outside the circle . . . . . . . . . . . . . 15.10 A strengthening of Thales’ Theorem, still another case with vertex C outside the circle . . . . . 15.11 Where lie the midpoints of all chords through point P? . . . . . . . . . . . . . . . . . . . 15.12 The midpoints of chords lie on Thales’ circle with diameter OP. . . . . . . . . . . . . . . . 15.13 (up-down) 4ABC congruent 4BAD (left-right) 4CAD congruent 4DBC (scissors) 4AMC congruent 4BMD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.14 The congruence of 4BCA to 4CBD implies that the diagonals of a rectangle are congruent . . . 15.15 What steps 1,2 and 3 yield in hyperbolic geometry . . . . . . . . . . . . . . . . . . . . . 15.16 Tangents to a circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
305 306 307 307 308 309 309 310 311 313 314 315 316 317 318 321 321 326 335 336 337 337 338 339 340 342 345 347 347 348 348 349 350 351 351 352 352 353 354 355 356 358
876
15.17 Two circles with no common tangents. . . . . . . . . . . . . . . . . . . . . . . . . . . 15.18 Two circles with one common tangent. . . . . . . . . . . . . . . . . . . . . . . . . . . 15.19 Two circles with two common tangents. . . . . . . . . . . . . . . . . . . . . . . . . . . 15.20 Two circles with three common tangents. . . . . . . . . . . . . . . . . . . . . . . . . . 15.21 Two circles with four common tangents. . . . . . . . . . . . . . . . . . . . . . . . . . . 15.22 Construction of two of the four common tangents for two circles lying outside each other. . . . 15.23 The circumference angle of a short arc is acute, of a long one is obtuse. . . . . . . . . . . . . 15.24 Central and circumference angle of a circular arc . . . . . . . . . . . . . . . . . . . . . . 15.25 Angles in a circle—still two other cases . . . . . . . . . . . . . . . . . . . . . . . . . . 15.26 A special quadrilateral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.27 Opposite angles in a quadrilateral with a circum circle . . . . . . . . . . . . . . . . . . . . 15.28 Use isosceles triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.29 The angle between tangent and chord is congruent to the circumference angle . . . . . . . . . 15.30 The angle between tangent and chord is a limiting case of the circumference angle . . . . . . . 15.31 The sides of an angle cutting two circles at the endpoints of their common chord cut them in two further parallel chords. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.32 Another example for the two-circle lemma. . . . . . . . . . . . . . . . . . . . . . . . . 16.1 The parallel as double perpendicular. . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 The parallel using Thales’ circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 The parallel obtained from a rhombus. . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 The parallel using similar triangles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 The parallel using the harmonic quadrilateral. . . . . . . . . . . . . . . . . . . . . . . . 16.6 Trisection using parallels. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.7 How to trisect a segment but not an angle. . . . . . . . . . . . . . . . . . . . . . . . . . 16.8 Trisection of a segment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.9 Completed trisection with equilateral triangles . . . . . . . . . . . . . . . . . . . . . . . 16.10 What are the angles of 4ABC? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.11 Construction of a triangle with angles of 45◦ , 60◦ and 75◦ . . . . . . . . . . . . . . . . . . . 16.12 A triangle construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.13 A triangle construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.14 A triangle construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.15 Congruent angles produced by the altitudes of a triangle. . . . . . . . . . . . . . . . . . . . 16.16 A triangle construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.17 Six quadrilaterals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.18 A quadrilateral with α + γ = 120◦ and given vertices A, B, C. . . . . . . . . . . . . . . . . . 16.19 A quadrilateral with α + γ = 270◦ and given vertices A, B, C. . . . . . . . . . . . . . . . . . 17.1 Construction of the products ab and ba, with an empirical error. . . . . . . . . . . . . . . . 17.2 Pappus’ theorem in the special case with two isosceles triangles. . . . . . . . . . . . . . . . 17.3 Producing parallel chords . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Three circles help to produce the third pair of parallel segments. . . . . . . . . . . . . . . . 17.5 Circles CA and CC intersect the angle in the parallel chords A0C and AC 0 . . . . . . . . . . . 17.6 Circles CB and CC intersect the angle in the parallel chords B0C and BC 0 . . . . . . . . . . . 17.7 Finally, circles CA and CB help to confirm that AB0 and A0 B are parallel. . . . . . . . . . . . 17.8 Pascal’s circular hexagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.9 Pascal’s circular hexagon tangled . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.10 Hexagrammum Mysticum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.11 Proving Pascal’s circular hexagon configuration . . . . . . . . . . . . . . . . . . . . . . 17.12 Hexagrammum Mysticum works on a hyperbola! . . . . . . . . . . . . . . . . . . . . . .
361 361 361 362 362 363 364 366 368 369 369 370 372 373 374 375 376 377 378 379 380 380 381 382 383 384 384 385 386 387 387 388 389 390 391 392 393 393 394 395 395 396 397 397 398 399 400
877
Construction of the product ab for given segments |OA0 | = a, |OB| = b and |OE| = 1. . . . . . Construction of the products ab and ba, with an empirical error exaggerated. . . . . . . . . Proof of the distributive law ab + ac = a(b + c). . . . . . . . . . . . . . . . . . . . . . The proof of the associative law yields at first c(ab) = a(cb). . . . . . . . . . . . . . . . Prove commutativity ab = ba. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Commutativity ab = ba follows from both lying on the circle. . . . . . . . . . . . . . . . a(bc) = b(ac) both lie on the circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . The legs of equiangular right triangles are proportional. . . . . . . . . . . . . . . . . . . Equiangular triangles are similar. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Addendum to the extended ASA congruence. . . . . . . . . . . . . . . . . . . . . . . . Two isosceles similar triangles with angles of 30◦ and 120◦ . . . . . . . . . . . . . . . . . Inscribing a square into a triangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . For h = a, one can fold the triangle to cover the square twice. . . . . . . . . . . . . . . . For a < h, one can fold the triangle to cover the square twice, and triangle 4A0 B0 C 0 left over. . For a > h, one can fold the triangle to cover the square twice, and part of triangle 4A0 B0 C 0 is covered twice. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.9 Constructing a circle through a point and touching two lines. . . . . . . . . . . . . . . . . 19.10 Constructing a circle through two points touching a line. . . . . . . . . . . . . . . . . . . 19.11 Theorem of chords, case of chords intersecting inside the circle . . . . . . . . . . . . . . 19.12 Theorem of chords, case of chords intersecting outside the circle . . . . . . . . . . . . . . 19.13 Getting the limit C → T, D → T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.14 A trapezoid and the lens equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.15 The trapezoid appearing for an optical lens. . . . . . . . . . . . . . . . . . . . . . . . . 20.1 The common notation for a right triangle . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 The altitude theorem for a right triangle . . . . . . . . . . . . . . . . . . . . . . . . . 20.3 The leg theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4 The Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.5 A pair of triangles 4A0 QB and 4C 0 QD of equal area directly confirms the theorem of chords . 20.6 Two equally good runners start at O and R. Do they better meet at point S or point T . . . . . 20.7 Proof of the parallelogram equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1 Proof of the sin theorem for acute, and for obtuse angles. . . . . . . . . . . . . . . . . . 21.2 Proof of the cos theorem for acute, and for obtuse angles. . . . . . . . . . . . . . . . . . 21.3 For the given data, there are two non-congruent solutions. . . . . . . . . . . . . . . . . . 21.4 Proof of the tangent addition theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1 Archimedes’ approximation of a circular arc . . . . . . . . . . . . . . . . . . . . . . . 22.2 Measuring the area of a regular polygon. . . . . . . . . . . . . . . . . . . . . . . . . . 23.1 A square and a parallelogram that are easily seen to be equidecomposable . . . . . . . . . . 23.2 A square and a parallelogram that are only shown to be equicomplementable . . . . . . . . . 23.3 Euclid’s leg theorem tells the square over the leg is equicomplementable to the rectangle as indicated. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.4 Euclid’s proof of the leg theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.5 Euclid’s proof of the Pythagorean Theorem. . . . . . . . . . . . . . . . . . . . . . . . 23.6 What went wrong? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.7 A dissection proof of the Theorem of Pythagoras. . . . . . . . . . . . . . . . . . . . . . 23.8 Dissection proof of the Pythagorean theorem in a special case . . . . . . . . . . . . . . . 23.9 Dudeney’s dissection of an equilateral triangle and a square of the same area. . . . . . . . . 23.10 Another dissection of an equilateral triangle and a square of the same area. . . . . . . . . . 23.11 A construction of Dudeney’s dissection starting with the square I. . . . . . . . . . . . . .
18.1 18.2 18.3 18.4 18.5 18.6 18.7 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8
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403 404 405 406 407 409 410 412 412 414 416 417 417 418
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418 419 421 422 423 424 425 426 427 427 428 428 429 433 434 436 437 439 440 445 450 453 453
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455 455 456 457 457 458 459 460 461
878
23.12 A construction of Dudeney’s dissection starting with the square II. . . . . . . . . . . . . . . 23.13 A construction of Dudeney’s dissection starting with the square III. . . . . . . . . . . . . . . 23.14 A construction of Dudeney’s dissection starting with the triangle. . . . . . . . . . . . . . . . 23.15 For the calculation of its area, any side of a triangle may be used as its base. . . . . . . . . . . 23.16 Two triangles with same base and different altitudes are not equicomplementable . . . . . . . . 23.17 Herons’ formula for the area of triangle. . . . . . . . . . . . . . . . . . . . . . . . . . . 23.18 An arbitrary triangle is partitioned into six right triangles. . . . . . . . . . . . . . . . . . . 24.1 The intersections C ∩ D = C ∩ l = D ∩ l are equal—point P and line l can be constructed with Hilbert tools. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.1 Do the perpendiculars p and q intersect? . . . . . . . . . . . . . . . . . . . . . . . . . . 25.2 No, the triangle 4OQR cannot exist. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.3 Constructing a triangle of double size. . . . . . . . . . . . . . . . . . . . . . . . . . . . 25.4 Construction and properties of the midpoint triangle. . . . . . . . . . . . . . . . . . . . . 25.5 The three medians intersect in the centroid. . . . . . . . . . . . . . . . . . . . . . . . . 25.6 Every triangle 4ABC is the midpoint triangle of a larger triangle 4A0 B0C 0 . . . . . . . . . . . 25.7 The altitudes are the angular bisectors of the orthic triangle. . . . . . . . . . . . . . . . . . 25.8 The orthic triangle for an obtuse triangle. . . . . . . . . . . . . . . . . . . . . . . . . . 25.9 The Euler line comes from a dilation with center S . . . . . . . . . . . . . . . . . . . . . . 26.1 Menelaus’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.2 The area of the middle triangle is one seventh of the total area. . . . . . . . . . . . . . . . . 26.3 What is the area of the triangle in the middle? . . . . . . . . . . . . . . . . . . . . . . . . 26.4 The Theorem of Cevá . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.5 The harmonic quadrilateral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.6 The Gergonne point. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.7 Three angles of 45◦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.8 How three angles of 45◦ are obtained. . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.9 Find angles α, 90◦ − α and 90◦ − 2α. . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.10 The points on Apollonius circle have all same ratio of distances from points C and G. . . . . . 26.11 The points on Apollonius circle have all same ratio of distances from A and B, and the segments AX and BX appear under equal angles. . . . . . . . . . . . . . . . . . . . . . . . . . . 26.12 The points on this Apollonius circle have all the ratio 2 : 1 of distances from points A and B. . . 26.13 Harmonic points XY, UV from Apollonius circle. . . . . . . . . . . . . . . . . . . . . . . 26.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.15 The attraction by a conducting sphere is equivalent to the attraction by a mirror charge. . . . . 26.16 Albrecht Dürer 1530 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.17 From were does a square look like this? . . . . . . . . . . . . . . . . . . . . . . . . . . 26.18 Reconstruction of the position of the eye. . . . . . . . . . . . . . . . . . . . . . . . . . . 26.19 Reconstruction of the eye position and the object square from its perspective image. . . . . . . 26.20 Tiling of a plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.21 The mapping by perspective in coordinates. . . . . . . . . . . . . . . . . . . . . . . . . 27.1 How the Euclidean egg should look. . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.2 A Euclidean egg—how it is done. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.3 This is a beginning. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.4 Too small. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.5 Too large. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.6 The egg built from inside—just right. . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.7 Triangle construction from given side, sum of the two other sides, and the angle across. . . . . 27.8 In Euclidean geometry, an isosceles solution exists for a + b = 10, c = 6 and γ2 ≤ 36.84◦ . . . .
461 462 464 465 466 473 474 491 492 493 494 495 496 497 499 500 501 502 503 503 504 505 506 506 507 508 509 509 511 512 513 513 515 516 516 517 518 518 520 521 522 523 523 524 525 526
879
27.9 Archimedes’ broken chord . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.10 Triangle construction (27.8) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.11 The second symmetric solution for construction (27.8) . . . . . . . . . . . . . . . . . . . 27.12 The two symmetric solutions of (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.13 Still another variant of triangle construction (27.8) . . . . . . . . . . . . . . . . . . . . . 27.14 The Theorem of Collignon yields two congruent segments PR and QS orthogonal to each other. 27.15 In case of a parallelogram, the two congruent orthogonal segments PR and QS bisect each other at its midpoint. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.16 Two pairs of similar triangles: 4DBC ∼ 4ABE and 4ABD ∼ 4EBC. . . . . . . . . . . . . . 27.17 If E lies between A and C, the point A lies on the arc BD opposite to C. . . . . . . . . . . . 27.18 If point A lies between E and C, the point A lies on the same arc BD as point C. . . . . . . . . 27.19 If C lies between A and E, the point A lies on the arc CD opposite to B. . . . . . . . . . . . 27.20 The Theorem of Hjelmslev. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.21 Euclidean proof of the angle congruence of Hjelmslev. . . . . . . . . . . . . . . . . . . . 27.22 Euclidean proof of the segment congruence of Hjelmslev. . . . . . . . . . . . . . . . . . . 27.23 The theorem of Hjelmslev for a self-intersecting quadrilateral. . . . . . . . . . . . . . . . . 27.24 The proof of Hjelmslev’s theorem for an overlapping quadrilateral. . . . . . . . . . . . . . . 28.1 Two similar triangles appear within the slice of the regular 10-gon. . . . . . . . . . . . . . . 28.2 A simple Euclidean construction of the regular pentagon. . . . . . . . . . . . . . . . . . . 28.3 Euclid’s theorem of secants gives another derivation of the golden mean. . . . . . . . . . . . 28.4 Proving Euclid XIII.10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.5 Proving Euclid XIII.10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.6 A construction of the regular pentagon by Hilbert tools. . . . . . . . . . . . . . . . . . . . 28.7 Still another construction of the pentagon with Hilbert tools. . . . . . . . . . . . . . . . . 28.8 This is a Euclidean construction for the regular pentagon. . . . . . . . . . . . . . . . . . . 28.9 This construction does not yield a regular pentagon. . . . . . . . . . . . . . . . . . . . . 29.1 Construction of the equipower line of two circles. . . . . . . . . . . . . . . . . . . . . . 29.2 Construction of an outer common tangent by similar triangles. . . . . . . . . . . . . . . . . 29.3 Construction of an inner common tangent by similar triangles. . . . . . . . . . . . . . . . . 29.4 Construction of the inverse point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29.5 The gear of Peaucollier. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29.6 The Theorem of chords explains the gear of Peaucollier. . . . . . . . . . . . . . . . . . . 29.7 In case b > a, the gear of Peaucollier constructs the antipodal point. . . . . . . . . . . . . . 29.8 A circle, not going through O is mapped to a circle. . . . . . . . . . . . . . . . . . . . . . 29.9 The common tangent of a circle and its inverse image intersect at O. . . . . . . . . . . . . . 29.10 A circle through O is mapped to a line. . . . . . . . . . . . . . . . . . . . . . . . . . . 29.11 The angle between a circle and the radial ray is conserved. . . . . . . . . . . . . . . . . . 29.12 Relation of the ratio of three points, and ratio of their inverted images. . . . . . . . . . . . . 30.1 A spherical line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 An elliptic line consists of pairs of antipodes . . . . . . . . . . . . . . . . . . . . . . . . 30.3 Construction of the antipode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.4 Construction of an elliptic line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.5 The exterior angle can be congruent or smaller than a nonadjacent interior angle. . . . . . . . 30.6 Two non congruent triangles matching in two angles and the side across one of them. . . . . . 30.7 SAA congruence in neutral geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.8 Another example of two non congruent triangles matching in two angles and the side across one of them. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.9 A spherical triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
527 529 529 530 531 532 534 536 537 537 538 539 540 541 541 542 543 545 545 547 548 550 551 553 554 558 562 562 564 564 565 566 567 568 569 570 570 572 574 574 575 576 577 577 578 579
880
30.10 The Pythagoras lune with a = 45◦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.11 The Pythagoras lune with a = 70◦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.12 The Pythagoras lune with a = 90◦ is highly degenerate. . . . . . . . . . . . . . . . . . . . 30.13 The Pythagoras lune with a = 120◦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.14 The Pythagoras lune with a = 150◦ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.15 The Pythagoras lune with a = 180◦ corresponds to a flat right triangle with sides 3, 4 and 5—it is degenerate, too. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.16 An approximately isosceles lune. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.17 The stereographic projections maps circles to circles. . . . . . . . . . . . . . . . . . . . . 30.18 Antipodes on the sphere are mapped to antipodes in the conformal model. . . . . . . . . . . 30.19 Many congruent angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.20 Points symmetric to the equatorial plane are mapped to inverse points with respect to the equator circle ∂D. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.21 The diameter in the drawing plane is projected symmetrically. . . . . . . . . . . . . . . . . 31.1 Another construction of the Euclidean parallel— the lines are numbered in the order they are constructed. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 The harmonic quadrilateral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Transfer of a given segment AB onto line l at point P. . . . . . . . . . . . . . . . . . . . . 31.4 Constructing a perpendicular line, using only straightedge and unit measure. . . . . . . . . . . 31.5 Rotating a given angle about its vertex. . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Validity of the construction to rotate an angle. . . . . . . . . . . . . . . . . . . . . . . . 31.7 Transferring a given angle to both sides of the given ray. . . . . . . . . . . . . . . . . . . . 31.8 Construction of the midpoint with compass only. . . . . . . . . . . . . . . . . . . . . . . 31.9 Intersection of line and circle with the rusty compass. . . . . . . . . . . . . . . . . . . . . 31.10 What the teacher wanted. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.11 What Hilbert—at school 3—could have done. . . . . . . . . . . . . . . . . . . . . . . . 32.1 Trisection of an acute angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Trisection of an obtuse angle less than 135◦ . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 Trisection of an obtuse angle larger than 135◦ . . . . . . . . . . . . . . . . . . . . . . . . 32.4 Trisection of an acute angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.5 Trisection of an obtuse angle between 90◦ and 135◦ . . . . . . . . . . . . . . . . . . . . . . 32.6 Trisection of an obtuse angle larger than 135◦ . . . . . . . . . . . . . . . . . . . . . . . . 32.7 Trisection with Nicolson’s angle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.8 Trisection by paperfolding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.9 Reflection at the folding line√ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.10 Nicomedes’ construction of 3 2a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.11 The conchoid of Nicomedes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.12 Another gift for Apollo. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.13 Nicomedes’ Solution of the Delian Problem. . . . . . . . . . . . . . . . . . . . . . . . . 32.14 My final gift to Apollo. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.15 The Delian problem, once more . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.16 The Delian problem and the curve of Agnesi. . . . . . . . . . . . . . . . . . . . . . . . . 32.17 Nicomedes’ trisection construction has four solutions . . . . . . . . . . . . . . . . . . . . 32.18 The conchoid gives all four solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.19 Archimedes’ star of Mercedes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.20 Archimedes’ trisection and the conchoid . . . . . . . . . . . . . . . . . . . . . . . . . . 32.21 Archimedes’ trisection and the conchoid . . . . . . . . . . . . . . . . . . . . . . . . . . 33.1 The heptagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
581 582 583 583 584 584 585 586 588 589 589 591 593 593 594 595 595 596 597 598 599 600 601 604 605 605 607 608 608 610 611 612 614 616 618 619 620 621 623 624 626 628 631 631 632
881
33.2 How I begin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.3 Use Archimedes’ trisection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.4 Angle sum in seven parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.5 A false heptagon. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.6 A regular heptagon. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.1 Find the length of segment BY, and AY exactly. . . . . . . . . . . . . . . . . . . . . . . 34.2 Find the length of segment BF exactly. . . . . . . . . . . . . . . . . . . . . . . . . . . 35.1 Just a lune. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.2 The constructible and squarable 3 : 1 lune of Hippocrates. . . . . . . . . . . . . . . . . . . 35.3 Hippocrates 3 : 1 lune is greater than a semicircle. . . . . . . . . . . . . . . . . . . . . . 35.4 The 3 : 2 lune of Hippocrates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.5 Hippocrates 3 : 2 lune is smaller than a semicircle. . . . . . . . . . . . . . . . . . . . . . 35.6 Hippocrates squared a suitable union of a circle and a lune. . . . . . . . . . . . . . . . . . 35.7 On a generalization of the Pythagorean Theorem. . . . . . . . . . . . . . . . . . . . . . . 35.8 The lunes of Alhazen can be squared. . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.9 Can you square this lune? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.10 Can you square this lune? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.11 One can square this simple lune. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.12 Can one square the circle? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.13 A numerical production of Hippocrates’ 3 : 1 lune. . . . . . . . . . . . . . . . . . . . . . 35.14 The area of the lune equals twice the area of smaller circle plus the area of the kite. . . . . . . 35.15 Another 3 : 2 lune. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.16 In this case, the 3 : 2 lune minus the lower circle is squarable. . . . . . . . . . . . . . . . . 35.17 The 4 : 1 lune of Vieta. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.18 The constructible 5 : 1 lune of Euler. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.19 The constructible 5 : 3 lune of Euler. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35.20 A constructible 5 : 3 lune minus circle around A. . . . . . . . . . . . . . . . . . . . . . . 35.21 The overlapping polygon has the same signed area as the kite ADBC. . . . . . . . . . . . . 35.22 A constructible 5 : 3 lune plus circle around B minus circle around A. . . . . . . . . . . . . . 37.1 Alhazen’s problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37.2 An easy special case of Alhazen’s problem. . . . . . . . . . . . . . . . . . . . . . . . . . 37.3 An second easy special case of Alhazen’s problem. . . . . . . . . . . . . . . . . . . . . . 37.4 Alhazen’s problem for a = b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37.5 Alhazen’s problem for center O between A and B. . . . . . . . . . . . . . . . . . . . . . . 38.1 Construction of 6 points on the quadratrix by means of a regular 24-gon. . . . . . . . . . . . 38.2 The trisection of an angle of 60◦ by means of the quadratrix. . . . . . . . . . . . . . . . . . 38.3 The partition of an arbitrary angle into five congruent parts, done by means of the quadratrix. . . 39.1 Morley’s Theorem states that the small triangle 4XYZ is always equilateral. . . . . . . . . . . 39.2 The proof of Morley’s Theorem begins with the equilateral small triangle 4XYZ. . . . . . . . 39.3 Point Z is the center of the in-circle for triangle 4ABW. . . . . . . . . . . . . . . . . . . . 39.4 Similar triangles 4S OM ∼ 4S EC are used to prove Euler’s theorem. . . . . . . . . . . . . . 39.5 Congruent triangles 4NOM and 4NHHc are used to get the nine-point circle. . . . . . . . . . 39.6 The angle between the Feuerbach circle triangle side AB is β − α. . . . . . . . . . . . . . . 39.7 The touching points of the in-circle and one ex-circle. . . . . . . . . . . . . . . . . . . . . 39.8 The angle between the two inner common tangents of in-circle and ex-circle is β − α. . . . . . 39.9 The common tangent of the in-circle and one ex-circle is parallel to the tangent to the Feuerbach circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39.10 The points on Apollonius circle have all same ratio of distances from C and D. . . . . . . . .
633 634 637 638 639 650 651 666 668 669 670 672 672 675 676 677 678 679 679 680 681 683 684 685 686 687 688 689 689 698 699 700 701 702 706 709 710 711 712 713 715 717 718 719 720 721 722
882
39.11 Inversion by circle δ maps I 7→ I , Ic 7→ Ic , AB 7→ AB , tc 7→ φ. . . . . . . . . . . . 39.12 By Euclid III.21, one gets congruent angles α1 at vertices A, B, E and congruent angles α2 at vertices A, C, F. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39.13 The Simson line is broken, if point P does not lie on the circum circle. . . . . . . . . . . . . 39.14 The Simson line is broken again. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39.15 The Simson line for a special case—one gets five congruent angles α2 at vertices A, B, C.F, E. . . 40.1 Construction of the inverse point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.2 A orthogonal circle consists of pairs inverted points . . . . . . . . . . . . . . . . . . . . . 40.3 Construction of a hyperbolic line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.4 Recognize inverse and polar elements. . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.5 How to calculate the angle of parallelism. . . . . . . . . . . . . . . . . . . . . . . . . . 40.6 For given line l and angle of parallelism α = 40◦ , point P is constructed. . . . . . . . . . . . 40.7 A more geometric proof of Lobachevskij’s formula for angle of parallelism. . . . . . . . . . 40.8 The inversion by an orthogonal circle is a hyperbolic reflection. . . . . . . . . . . . . . . . 40.9 The inversion by line l maps the given point A to the center O. Furthermore, B and C are mapped to Br and Cr. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.10 Sorry, the triangles 4ABC and 4XYZ are not congruent. . . . . . . . . . . . . . . . . . . . 40.11 SAS congruence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.12 The set of the reflective images of center O across all lines with endpoint E yields the horocycle around E through O. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.13 A circle appears as a circle—evidently for a circle about O, and hence always by means of a useful reflection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.14 An equidistance line is a circular arc with two ideal endpoints. . . . . . . . . . . . . . . . . 40.15 The chords of a bundle of hyperbolic lines through a common point P intersect in a common point K. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1 How to construct a hyperbolic line through two given points A and B. . . . . . . . . . . . . . 41.2 Construct the line bisector for two given points A and B. . . . . . . . . . . . . . . . . . . 41.3 Erect a perpendicular on line δ at point P, using Construction 41.3. . . . . . . . . . . . . . 41.4 Erect a perpendicular on line δ at point P, using the right angle as explained in Remark 2. Finally, combining the two constructions yields better accuracy. . . . . . . . . . . . . . . . . . . . 41.5 Drop a perpendicular onto line δ from the given point P. . . . . . . . . . . . . . . . . . . . 41.6 Transfer a given angle to a given ray. The example takes α = 10◦ . . . . . . . . . . . . . . . 41.7 Construction of the angular bisectors. . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.8 Construction of the middle line of two parallel lines. . . . . . . . . . . . . . . . . . . . . 41.9 Construction of the common perpendicular of two parallel lines. . . . . . . . . . . . . . . . 41.10 Variant 2 for the construction of the common perpendicular of two parallel lines— and a figure showing both variants. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.11 Given two parallel lines α and β, the middle line δ and the common perpendicular µ are constructed simultaneously. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.12 Construction of a circle with given center A through point B. . . . . . . . . . . . . . . . . 41.13 Construction of a circle with given center A through point B, which is on or close to line OA. . . 41.14 How to construct of a circle around O with given hyperbolic radius. . . . . . . . . . . . . . 41.15 Two examples for the construction of a triangle from three given angles. . . . . . . . . . . . 41.16 Construction of an asymptotic triangle with angles α = 60◦ , β = 45◦ , γ = 0◦ . . . . . . . . . . 41.17 Example for the SAA construction of a triangle. . . . . . . . . . . . . . . . . . . . . . . 41.18 Variant 2 for the SAA construction. This variant uses an equidistant line and a hyperbolic reflection by r. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.19 The three altitudes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
723 727 728 729 730 732 733 735 736 738 740 741 742 743 744 745 747 750 750 752 754 755 755 756 756 757 758 759 759 760 761 761 762 762 765 766 767 768 769
883
41.20 A triangle with an orthocenter, but no circum-center— another triangle without an orthocenter, but a circum-center. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 770 42.1 A bundle of hyperbolic lines l, l1 through a common point P in Poincaré’s model. They intersect in a common point K in Klein’s model, too. Their polar elements are identical for both models. . 773 42.2 If line l goes through point K, then the polar l⊥ lies on the polar K proj⊥ . . . . . . . . . . . . . 773 42.3 Two perpendicular lines l and p. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775 42.4 The polar of the first perpendicular line lies on the extension of the second one, and vice versa. . 776 42.5 The angle of parallelism in Klein’s model. . . . . . . . . . . . . . . . . . . . . . . . . . 777 42.6 Bolyai’s construction of the asymptotic parallel ray. . . . . . . . . . . . . . . . . . . . . . 778 42.7 By Martin’s theorem, ∠BCF = π(CA) if and only if ∠ACF = π(CB). . . . . . . . . . . . . . 779 42.8 Proving Martin’s theorem in Klein’s model with vertex C = O in the center. . . . . . . . . . . 780 42.9 Since ∠BCF = π(CA) by construction, Martin’s theorem implies that ∠ACF = π(CB) as requested. 781 42.10 Construction of the segment a = OB from given angle of parallelism α. . . . . . . . . . . . . 783 42.11 There exist two automorphic collineations mapping A to A0 and U to U 0 . . . . . . . . . . . . 785 42.12 An automorphic collineation that transports points along horocycles. . . . . . . . . . . . . . 786 42.13 An ideal quadrilateral produces five right angles. . . . . . . . . . . . . . . . . . . . . . . 788 42.14 Construction of the reflective image K 0 for a given point K and reflection line l. . . . . . . . . 788 42.15 Construction of the true hyperbolic angle ∠OKF. . . . . . . . . . . . . . . . . . . . . . . 789 42.16 Transfer of a segment along its line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 791 42.17 The correspondence of a Lambert quadrilateral and a right triangle. . . . . . . . . . . . . . . 792 42.18 Find the English flag. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 793 42.19 Proof of claim (42.12): a has angle of parallelism π(a) = ∠URH. . . . . . . . . . . . . . . . 794 42.20 Proof of claim (42.16): m = RS has angle of parallelism π(m) = ∠V 0 BG. . . . . . . . . . . . 795 42.21 The angle congruence of Hjelmslev holds in hyperbolic geometry, too. . . . . . . . . . . . . 795 43.1 The trigonometry of the right triangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . 797 43.2 If any two altitudes intersect, one can put the intersection point at the center of Klein’s disk. Three drawings are given: for an acute, right, or obtuse triangle. . . . . . . . . . . . . . . . 800 43.3 The altitudes of an obtuse triangle may or may not intersect. A convenient drawing puts the vertex with the obtuse angle at the center of Klein’s disk. . . . . . . . . . . . . . . . . . . 801 43.4 The bisectors and altitudes of the midpoint triangle inside Klein’s disk model. For a triangle with circum-center, three drawings are given: for an acute, right, or obtuse midpoint triangle. . . . . 803 43.5 The altitudes of the midpoint triangle intersect outside Klein’s disk. . . . . . . . . . . . . . 804 43.6 The three vertices A, B and C have congruent distances from the baseline l, which is chosen to be the horizontal diameter. The second drawing erases the part of the construction done outside the Poincaré disk. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.7 Construction of the perpendicular bisector pb and the point H2 = O. . . . . . . . . . . . . 43.8 Construction of the circle of infinity δD and the point H2 = O. . . . . . . . . . . . . . . . 43.9 Construction of the circle of infinity δD and the point H2 = O—in this example it turns out to be outside the disk! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.10 A triangle in a semicircle produces a Lambert quadrilateral. For the proof, the extra line we need is—once again—the radius to the third vertex. . . . . . . . . . . . . . . . . . . . . . . 43.11 A triangle with third vertex inside the circle produces an obtuse angle of the midpoint-triangle at center Mc . In case of the third vertex outside the circle, an acute angle is produced. . . . . . 43.12 The three medians intersect in one point because the harmonic quadrilateral ABMb Ma has Euclidean parallel sides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.13 The altitudes are angular bisectors of the orthic triangle. . . . . . . . . . . . . . . . . . 43.14 Centers for any ex-circle of the orthic triangle can only be the three vertices or the orthocenter of the original triangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. 805 . 805 . 806 . 807 . 807 . 808 . 810 . 811 . 811
884
43.15 The triangle has no circum-circle—drawn in Klein’s model! . . . . . . . . . . . . . . . . . 45.1 Two lines traversed congruent z-angles have a common perpendicular. . . . . . . . . . . . . 45.2 Cutting the limiting parallel ray. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.3 Extending a limiting parallel ray. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.4 Lines with a common perpendicular diverge. . . . . . . . . . . . . . . . . . . . . . . . . 45.5 Lines with a common perpendicular diverge in the semi-elliptic case, too. . . . . . . . . . . 45.6 If l is limiting parallel to k, then k is limiting parallel to l. . . . . . . . . . . . . . . . . . . 45.7 Limiting parallel ray k to l comes closer than the ray r. . . . . . . . . . . . . . . . . . . . 45.8 Transitivity, case where rays k and h lie in different half-planes of l. . . . . . . . . . . . . . . 45.9 Transitivity, case where rays k and h lie in the same half-plane of l. . . . . . . . . . . . . . . 45.10 The exterior angle theorem for a limiting triangle; with the two congruent angles β, the case drawn is impossible. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.11 There exist no limiting parallel rays in the semi-elliptic case. . . . . . . . . . . . . . . . . . 45.12 Hilbert’s construction of the common perpendicular, case χ > ε. . . . . . . . . . . . . . . . 45.13 Hilbert’s construction of the common perpendicular, case χ < ε. . . . . . . . . . . . . . . . 45.14 Hilbert’s construction of the enclosing line. . . . . . . . . . . . . . . . . . . . . . . . . . 45.15 Justification for the construction of the enclosing line. . . . . . . . . . . . . . . . . . . . . 45.16 Euclidean construction of a triangle with three given side bisectors. . . . . . . . . . . . . . . 45.17 Three limiting parallel side bisectors. . . . . . . . . . . . . . . . . . . . . . . . . . . . 45.18 The lines 0 ∞ and (−µ) µ are perpendicular. . . . . . . . . . . . . . . . . . . . . . . . . . 45.19 Existence of asymptotically parallel rays implies Aristole’s Axiom. . . . . . . . . . . . . . 45.20 In a hyperbolic plane, the distance between two lines with a common perpendicular is arbitrarily large. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46.1 Curvature of a rotation surface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46.2 The tractrix has a segment on its tangent of constant length. . . . . . . . . . . . . . . . . . 46.3 Measuring an arc of a horocycle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46.4 Isometry of the sliced pseudo-sphere to a half infinity strip. . . . . . . . . . . . . . . . . . 46.5 Isometric image of the sliced pseudo-sphere in the Poincaré disk. . . . . . . . . . . . . . . . 46.6 Archimedes at work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46.7 Feuerbach at work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46.8 Feuerbach at work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
812 822 823 823 824 824 825 826 826 827 828 829 830 830 832 833 835 837 840 844 845 858 861 866 867 868 885 886 887
885
Figure 46.6. Archimedes at work
886
Figure 46.7. Feuerbach at work
887
Figure 46.8. Feuerbach at work
