1. MEMO, Eisenstadt, Austria Individual competition, September 22, 2007
1. Let a, b, c, d be positive real numbers with a + b + c + d = 4. Prove that a2 bc + b2 cd + c2 da + d2 ab ≤ 4. 2. A set of balls contains n balls which are labeled with numbers 1, 2, 3, . . ., n. Suppose we are given k > 1 such sets. We want to colour the balls with two colors, black and white, in such a way that (a) the balls labeled with the same number are of the same colour, (b) any subset of k + 1 balls with (not necessarily all different) labels a1 , a2 , . . ., ak+1 satisfying the condition a1 + a2 + . . . + ak = ak+1 , contains at least one ball of each colour. Find, depending on k, the greatest possible number n which admits such a colouring. 3. Let k be a circle and k1 , k2 , k3 and k4 four smaller circles with their centres O1 , O2 , O3 and O4 respectively on k. For i = 1, 2, 3, 4 and k5 = k1 the circles ki and ki+1 meet at Ai and Bi such that Ai lies on k. The points O1 , A1 , O2 , A2 , O3 , A3 , O4 , A4 , lie in that order on k and are pairwise different. Prove that B1 B2 B3 B4 is a rectangle. 4. Determine all pairs (x, y) of positive integers satisfying the equation x! + y! = xy .
Each problem is worth 8 points. The order of the problems does not depend on their difficulty. Time: 5 hours Time for questions: 45 min
1. MEMO, Eisenstadt, Austria Team competition, September 23, 2007
5. Let a, b, c, d be arbitrary real numbers from the closed interval [ 12 , 2] satisfying abcd = 1. Find the maximal value of µ ¶µ ¶µ ¶µ ¶ 1 1 1 1 a+ b+ c+ d+ . b c d a
6. For a set P of five points in the plane in general position, we denote the number of acute-angled triangles with vertices in P by a(P ) (a set of points is said to be in general position if no three points lie on a line). Determine the maximal value of a(P ) over all possible sets P . 7. Let s(T ) denote the sum of the lengths of the edges of a tetrahedron T . We consider tetrahedra with the property that the six lengths of their edges are pairwise different positive integers, where one of them is 2 and and another one of them is 3. 1. Find all positive integers n for which there exists a tetrahedron T with s(T ) = n. 2. How many such pairwise non congruent tetrahedra T with s(T ) = 2007 exist? Two tetrahedra are said to be non congruent, if one cannot be transformed by reflections with respect to planes, translations and/or rotations into the other. (It is not necessary to prove that the tetrahedra are not degenerated i.e. have positive volume.) 8. Determine all positive integers k with the following property: there exists an integer a such that (a + k)3 − a3 is a multiple of 2007.
Each problem is worth 8 points. The order of the problems does not depend on their difficulty. Time: 5 hours Time for questions: 45 min
2nd MEMO, Olomouc, Czech Republic Individual competition, September 6, 2008
I–1: Let (an )∞ n=1 be any increasing sequence of positive integers with the following property: for each quadruple of indices (i, j, k, m), where 1 ≤ i < j ≤ k < m and i + m = j + k, the inequality ai + am > aj + ak holds. Determine the least possible value of a2008 . I–2: Consider a n × n chessboard, where n > 1 is an integer. In how many ways can we put 2n − 2 identical stones on the chessboard (each on another square) such that no two stones lie on the same diagonal? (By a diagonal we mean a row of squares whose diagonals of one direction lie on the same line). I–3: Let ABC be an isosceles triangle with |AC| = |BC|. Its incircle touches AB and BC at D and E, respectively. A line (different from AE) passes through A and intersects the incircle at F and G. The lines EF and EG intersect the line AB at K and L, respectively. Prove that |DK| = |DL|. I–4: Find all integers k such that for every integer n, the numbers 4n + 1 and kn + 1 are relatively prime.
Each problem is worth 8 points. The order of the problems does not depend on their difficulty. Time: 5 hours Time for questions: 45 min
2nd MEMO, Olomouc, Czech Republic Team competition, September 7, 2008
T–1: Find all functions f : R → R (so, f is a function from the real numbers to the real numbers) such that xf (x + xy) = xf (x) + f (x2 )f (y) for all real numbers x and y. T–2: In a given n-tuple of positive integers with n ≥ 2, we choose in each step a pair of numbers and replace each of them by their sum, i.e. we make the transformation (. . . , a, . . . , b, . . .) → (. . . , a + b, . . . , a + b, . . .). Determine all values of n for which, after a finite number of steps, we can get an n-tuple of identical numbers from any initial n-tuple. T–3: Given an acute-angled triangle ABC, let E be a point situated on the different side of the line AC than B, and let D be an interior point of the line segment AE. Suppose that ∠ADB = ∠CDE, ∠BAD = ∠ECD and ∠ACB = ∠EBA. Prove that B, C and E are collinear. T–4: Let n be a positive integer. Prove that if the sum of all positive divisors of n is a perfect power of 2, then the number of these divisors is also a perfect power of 2.
Each problem is worth 8 points. The order of the problems does not depend on their difficulty. Time: 5 hours Time for questions: 45 min
language: English
3rd Middle European Mathematical Olympiad Individual Competition 26th September, 2009
ProblemI-1. Find all functions f : R → R such that f (xf (y)) + f (f (x) + f (y)) = yf (x) + f (x + f (y)) for all x, y ∈ R, where R denotes the set of real numbers. ProblemI-2. Suppose that we have n > 3 distinct colours. Let f (n) be the greatest integer with the property that every side and every diagonal of a convex polygon with f (n) vertices can be coloured with one of n colours in the following way: • at least two distinct colours are used, and • any three vertices of the polygon determine either three segments of the same colour or of three different colours. Show that f (n) 6 (n − 1)2 with equality for infinitely many values of n. ProblemI-3. Let ABCD be a convex quadrilateral such that AB and CD are not parallel and AB = CD. The midpoints of the diagonals AC and BD are E and F . The line EF meets segments AB and CD at G and H, respectively. Show that AGH = DHG. ProblemI-4. Determine all integers k > 2 such that for all pairs (m, n) of different positive integers not greater than k, the number nn−1 − mm−1 is not divisible by k .
Time: 5 hours Time for questions: 30 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
language: English
3rd Middle European Mathematical Olympiad Team Competition 27th September, 2009
Problem T-1. Let x, y, z be real numbers satisfying x2 + y 2 + z 2 + 9 = 4(x + y + z). Prove that x4 + y 4 + z 4 + 16(x2 + y 2 + z 2 ) > 8(x3 + y 3 + z 3 ) + 27 and determine when equality holds. Problem T-2. Let a, b, c be real numbers such that for every two of the equations x2 + ax + b = 0,
x2 + bx + c = 0,
x2 + cx + a = 0
there is exactly one real number satisfying both of them. Determine all the possible values of a2 + b2 + c2 . Problem T-3. The numbers 0, 1, 2, . . . , n (n > 2) are written on a blackboard. In each step we erase an integer which is the arithmetic mean of two different numbers which are still left on the blackboard. We make such steps until no further integer can be erased. Let g(n) be the smallest possible number of integers left on the blackboard at the end. Find g(n) for every n. Problem T-4. We colour every square of the 2009 × 2009 board with one of n colours (we do not have to use every colour). A colour is called connected if either there is only one square of that colour or any two squares of the colour can be reached from one another by a sequence of moves of a chess queen without intermediate stops at squares having another colour (a chess queen moves horizontally, vertically or diagonally). Find the maximum n, such that for every colouring of the board at least one colour present at the board is connected. Problem T-5. Let ABCD be a parallelogram with BAD = 60◦ and denote by E the intersection of its diagonals. The circumcircle of the triangle ACD meets the line BA at K 6= A, the line BD at P 6= D and the line BC at L 6= C. The line EP intersects the circumcircle of the triangle CEL at points E and M . Prove that the triangles KLM and CAP are congruent.
Problem T-6. Suppose that ABCD is a cyclic quadrilateral and CD = DA. Points E and F belong to the segments AB and BC respectively, and ADC = 2 EDF . Segments DK and DM are height and median of the triangle DEF , respectively. L is the point symmetric to K with respect to M . Prove that the lines DM and BL are parallel. Problem T-7. Find all pairs (m, n) of integers which satisfy the equation (m + n)4 = m2 n2 + m2 + n2 + 6mn. Problem T-8. Find all non-negative integer solutions of the equation 2x + 2009 = 3y 5z .
Time: 5 hours Time for questions: 45 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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4th Middle European Mathematical Olympiad Individual Competition 11th September, 2010
Problem I-1. Find all functions f : R → R such that for all x, y ∈ R, we have f (x + y) + f (x)f (y) = f (xy) + (y + 1)f (x) + (x + 1)f (y).
Problem I-2. All positive divisors of a positive integer N are written on a blackboard. Two players A and B play the following game taking alternate moves. In the first move, the player A erases N . If the last erased number is d, then the next player erases either a divisor of d or a multiple of d. The player who cannot make a move loses. Determine all numbers N for which A can win independently of the moves of B. Problem I-3. We are given a cyclic quadrilateral ABCD with a point E on the diagonal AC such that AD = AE and CB = CE. Let M be the center of the circumcircle k of the triangle BDE. The circle k intersects the line AC in the points E and F . Prove that the lines F M , AD, and BC meet at one point. Problem I-4. Find all positive integers n which satisfy the following two conditions: (i) n has at least four different positive divisors; (ii) for any divisors a and b of n satisfying 1 < a < b < n, the number b − a divides n.
Time: 5 hours Time for questions: 45 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition 12th September, 2010
STREČNO SLOVAKIA 2010
Problem T-1. Three strictly increasing sequences a1 , a2 , a3 , . . . ,
b1 , b2 , b3 , . . . ,
c1 , c2 , c3 , . . .
of positive integers are given. Every positive integer belongs to exactly one of the three sequences. For every positive integer n, the following conditions hold: (i) can = bn + 1; (ii) an+1 > bn ; (iii) the number cn+1 cn − (n + 1)cn+1 − ncn is even. Find a2010 , b2010 , and c2010 . Problem T-2. For each integer n ≥ 2, determine the largest real constant Cn such that for all positive real numbers a1 , . . . , an , we have 2 a1 + · · · + an a21 + · · · + a2n ≥ + Cn · (a1 − an )2 . n n Problem T-3. In each vertex of a regular n-gon there is a fortress. At the same moment each fortress shoots at one of the two nearest fortresses and hits it. The result of the shooting is the set of the hit fortresses; we do not distinguish whether a fortress was hit once or twice. Let P (n) be the number of possible results of the shooting. Prove that for every positive integer k ≥ 3, P (k) and P (k + 1) are relatively prime. Problem T-4. Let n be a positive integer. A square ABCD is partitioned into n2 unit squares. Each of them is divided into two triangles by the diagonal parallel to BD. Some of the vertices of the unit squares are colored red in such a way that each of these 2n2 triangles contains at least one red vertex. Find the least number of red vertices.
Problem T-5. The incircle of the triangle ABC touches the sides BC, CA, and AB in the points D, E, and F , respectively. Let K be the point symmetric to D with respect to the incenter. The lines DE and F K intersect at S. Prove that AS is parallel to BC. Problem T-6. Let A, B, C, D, E be points such that ABCD is a cyclic quadrilateral and ABDE is a parallelogram. The diagonals AC and BD intersect at S and the rays AB and DC intersect at F . Prove that ∠AF S = ∠ECD. Problem T-7. For a nonnegative integer n, define an to be the positive integer with decimal representation . . 0} 2 |0 .{z . . 0} 1. 1 |0 .{z . . 0} 2 |0 .{z n
n
n
Prove that an /3 is always the sum of two positive perfect cubes but never the sum of two perfect squares. Problem T-8. We are given a positive integer n which is not a power of 2. Show that there exists a positive integer m with the following two properties: (i) m is the product of two consecutive positive integers; (ii) the decimal representation of m consists of two identical blocks of n digits.
Time: 5 hours Time for questions: 45 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
language: English 5th
Middle European Mathematical Olympiad Individual Competition 3rd September 2011
Problem I-1.
Initially, only the integer 44 is written on a board. An integer a on the board can be replaced with four pairwise dierent integers a1 , a2 , a3 , a4 such that the arithmetic mean 1 (a1 + a2 + a3 + a4 ) of the four new integers is equal to the number a. In a step we simulta4 neously replace all the integers on the board in the above way. After 30 steps we end up with n = 430 integers b1 , b2 , . . . , bn on the board. Prove that b21 + b22 + . . . + b2n ≥ 2011 . n Problem I-2.
Let n ≥ 3 be an integer. John and Mary play the following game: First John labels the sides of a regular n-gon with the numbers 1, 2, . . . , n in whatever order he wants, using each number exactly once. Then Mary divides this n-gon into triangles by drawing n − 3 diagonals which do not intersect each other inside the n-gon. All these diagonals are labeled with number 1. Into each of the triangles the product of the numbers on its sides is written. Let S be the sum of those n − 2 products. Determine the value of S if Mary wants the number S to be as small as possible and John wants S to be as large as possible and if they both make the best possible choices. Problem I-3.
In a plane the circles K1 and K2 with centers I1 and I2 , respectively, intersect in two points A and B . Assume that ∠I1 AI2 is obtuse. The tangent to K1 in A intersects K2 again in C and the tangent to K2 in A intersects K1 again in D. Let K3 be the circumcircle of the triangle BCD. Let E be the midpoint of that arc CD of K3 that contains B . The lines AC and AD intersect K3 again in K and L, respectively. Prove that the line AE is perpendicular to KL. Problem I-4.
Let k and m, with k > m, be positive integers such that the number km(k2 − m2 ) is divisible by k3 − m3 . Prove that (k − m)3 > 3km.
Time: 5hours Time for questions: 45min Each problem is worth 8 points. The order of the problems does not depend on their di culty.
language: English
5th
Middle European Mathematical Olympiad Team Competition 4th September 2011
Problem T-1.
Find all functions f : R → R such that the equality y 2 f (x) + x2 f (y) + xy = xyf (x + y) + x2 + y 2
holds for all x, y ∈ R, where R is the set of real numbers. Problem T-2.
Let a, b, c be positive real numbers such that a b c + + = 2. 1+a 1+b 1+c
Prove that
Problem T-3.
√
a+
√ √ 1 1 b+ c 1 >√ +√ +√ . 2 a c b
For an integer n ≥ 3, let M be the set {(x, y) | x, y ∈ Z, 1 ≤ x ≤ n, 1 ≤ y ≤ n} of points in the plane. (Z is the set of integers.) What is the maximum possible number of points in a subset S ⊆ M which does not contain three distinct points being the vertices of a right triangle?
Problem T-4.
Let n ≥ 3 be an integer. At a MEMO-like competition, there are 3n participants, there are n languages spoken, and each participant speaks exactly three dierent languages. 2n of the spoken languages can be chosen in such a way that no participant Prove that at least 9
speaks more than two of the chosen languages. (dxe is the smallest integer which is greater than or equal to x.) Problem T-5.
Let ABCDE be a convex pentagon with all ve sides equal in length. The diagonals AD and EC meet in S with ∠ASE = 60◦ . Prove that ABCDE has a pair of parallel sides. Problem T-6.
Let ABC be an acute triangle. Denote by B0 and C0 the feet of the altitudes from vertices B and C , respectively. Let X be a point inside the triangle ABC such that the line BX is tangent to the circumcircle of the triangle AXC0 and the line CX is tangent to the circumcircle of the triangle AXB0 . Show that the line AX is perpendicular to BC . Problem T-7.
Let A and B be disjoint nonempty sets with A ∪ B = {1, 2, 3, . . . , 10}. Show that there exist elements a ∈ A and b ∈ B such that the number a3 + ab2 + b3 is divisible by 11. Problem T-8.
We call a positive integer n amazing if there exist positive integers a, b, c such that the equality n = (b, c)(a, bc) + (c, a)(b, ca) + (a, b)(c, ab)
holds. Prove that there exist 2011 consecutive positive integers which are amazing. (By (m, n) we denote the greatest common divisor of positive integers m and n.)
Time: 5hours Time for questions: 45min Each problem is worth 8 points. The order of the problems does not depend on their di culty.
Individual competition 8th september 2012
Language: English Problem I-1. Let R+ denote the set of all positive real numbers. Find all functions f : R+ → R+ such that f (x + f (y)) = yf (xy + 1) holds for all x, y ∈ R+ . Problem I-2. Let N be a positive integer. A set S ⊆ {1, 2, . . . , N } is called allowed if it does not contain three distinct elements a, b, c such that a divides b and b divides c. Determine the largest possible number of elements in an allowed set S. Problem I-3. In a given trapezium ABCD with AB parallel to CD and AB > CD, the line BD bisects the angle ^ADC. The line through C parallel to AD meets the segments BD and AB in E and F , respectively. Let O be the circumcentre of the triangle BEF . Suppose that ^ACO = 60◦ . Prove the equality CF = AF + F O. Problem I-4. The sequence {an }n≥0 is defined by a0 = 2, a1 = 4 and an+1 =
an an−1 + an + an−1 for all positive integers n. 2
Determine all prime numbers p for which there exists a positive integer m such that p divides the number am − 1.
Time: 5 hours Time for questions: 45 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
Team competition 9th september 2012
Language: English
Problem T-1. Find all triplets (x, y, z) of real numbers such that 2x3 + 1 = 3zx, 2y 3 + 1 = 3xy, 2z 3 + 1 = 3yz.
Problem T-2. Let a, b, and c be positive real numbers with abc = 1. Prove that p p p 9 + 16a2 + 9 + 16b2 + 9 + 16c2 ≥ 3 + 4(a + b + c). Problem T-3. Let n be a positive integer. Consider words of length n composed of letters from the set {M, E, O} . Let a be the number of such words containing an even number (possibly 0) of blocks M E and an even number (possibly 0) of blocks M O. Similarly, let b be the number of such words containing an odd number of blocks M E and an odd number of blocks M O. Prove that a > b.
Problem T-4. Let p > 2 be a prime number. For any permutation π = (π(1), π(2), . . . , π(p)) of the set S = {1, 2, . . . , p}, let f (π) denote the number of multiples of p among the following p numbers: π(1), π(1) + π(2), . . . , π(1) + π(2) + · · · + π(p). Determine the average value of f (π) taken over all permutations π of S.
Problem T-5. Let K be the midpoint of the side AB of a given triangle ABC. Let L and M be points on the sides AC and BC, respectively, such that ^CLK = ^KM C. Prove that the perpendiculars to the sides AB, AC, and BC passing through K, L, and M , respectively, are concurrent.
Problem T-6. Let ABCD be a convex quadrilateral with no pair of parallel sides, such that ^ABC = ^CDA. Assume that the intersections of the pairs of neighbouring angle bisectors of ABCD form a convex quadrilateral EF GH. Let K be the intersection of the diagonals of EF GH. Prove that the lines AB and CD intersect on the circumcircle of the triangle BKD.
Problem T-7. Find all triplets (x, y, z) of positive integers such that xy + y x = z y , xy + 2012 = y z+1 . Problem T-8. For any positive integer n, let d(n) denote the number of positive divisors of n. Do there exist positive integers a and b, such that d(a) = d(b) and d(a2 ) = d(b2 ), but d(a3 ) 6= d(b3 )?
Time: 5 hours Time for questions: 45 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
Language:
English
7th Middle European Mathematical Olympiad Individual Competition 24th August 2013 Problem I-1. Let a, b, c be positive real numbers such that a+b+c=
1 1 1 + 2 + 2. 2 a b c
Prove that 2(a + b + c) ≥
√ 3
7a2 b + 1 +
√ √ 3 3 7b2 c + 1 + 7c2 a + 1.
Find all triples (a, b, c) for which equality holds. Problem I-2. Let n be a positive integer. On a board consisting of 4n × 4n squares, exactly 4n tokens are placed so that each row and each column contains one token. In a step, a token is moved horizontally or vertically to a neighbouring square. Several tokens may occupy the same square at the same time. The tokens are to be moved to occupy all the squares of one of the two diagonals. Determine the smallest number k(n) such that for any initial situation, we can do it in at most k(n) steps. Problem I-3. Let ABC be an isosceles triangle with AC = BC. Let N be a point inside the triangle such that 2∠AN B = 180◦ + ∠ACB. Let D be the intersection of the line BN and the line parallel to AN that passes through C. Let P be the intersection of the angle bisectors of the angles CAN and ABN . Show that the lines DP and AN are perpendicular. Problem I-4. Let a and b be positive integers. Prove that there exist positive integers x and y such that ! x+y = ax + by. 2
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
Language:
English
7th Middle European Mathematical Olympiad Team Competition 25th August 2013 Problem T-1. Find all functions f : R → R such that f (xf (x) + 2y) = f (x2 ) + f (y) + x + y − 1 for all x, y ∈ R. Problem T-2. Let x, y, z, w ∈ R \ {0} such that x + y 6= 0, z + w 6= 0, and xy + zw ≥ 0. Prove the inequality x+y z+w + z+w x+y
!−1
1 x z + + ≥ 2 z x
−1
w y + + w y
!−1
.
Problem T-3. There are n ≥ 2 houses on the northern side of a street. Going from the west to the east, the houses are numbered from 1 to n. The number of each house is shown on a plate. One day the inhabitants of the street make fun of the postman by shuffling their number plates in the following way: for each pair of neighbouring houses, the current number plates are swapped exactly once during the day. How many different sequences of number plates are possible at the end of the day? Problem T-4. Consider finitely many points in the plane with no three points on a line. All these points can be coloured red or green such that any triangle with vertices of the same colour contains at least one point of the other colour in its interior. What is the maximal possible number of points with this property?
Problem T-5. Let ABC be an acute triangle. Construct a triangle P QR such that AB = 2P Q, BC = 2QR, CA = 2RP , and the lines P Q, QR, and RP pass through the points A, B, and C, respectively. (All six points A, B, C, P , Q, and R are distinct.) Problem T-6. Let K be a point inside an acute triangle ABC, such that BC is a common tangent of the circumcircles of AKB and AKC. Let D be the intersection of the lines CK and AB, and let E be the intersection of the lines BK and AC. Let F be the intersection of the line BC and the perpendicular bisector of the segment DE. The circumcircle of ABC and the circle k with centre F and radius F D intersect at points P and Q. Prove that the segment P Q is a diameter of k. Problem T-7. The numbers from 1 to 20132 are written row by row into a table consisting of 2013 × 2013 cells. Afterwards, all columns and all rows containing at least one of the perfect squares 1, 4, 9, . . . , 20132 are simultaneously deleted. How many cells remain? Problem T-8. The expression ± ± ± ± ± ± is written on the blackboard. Two players, A and B, play a game, taking turns. Player A takes the first turn. In each turn, the player on turn replaces a symbol by a positive integer. After all the symbols are replaced, player A replaces each of the signs ± by either + or −, independently of each other. Player A wins if the value of the expression on the blackboard is not divisible by any of the numbers 11, 12, . . . , 18. Otherwise, player B wins. Determine which player has a winning strategy.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
Individual Competition Sept. 20, 2014 English version Problem I–1 Determine all functions f : R ÝÑ R such that ` ˘ ` ˘ xf pyq ` f xf pyq ´ xf f pyq ´ f pxyq “ 2x ` f pyq ´ f px ` yq holds for all x, y P R. Problem I–2 We consider dissections of regular n-gons into n ´ 2 triangles by n ´ 3 diagonals which do not intersect inside the n-gon. A bicoloured triangulation is such a dissection of an n-gon in which each triangle is coloured black or white and any two triangles which share an edge have different colours. We call a positive integer n ě 4 triangulable if every regular n-gon has a bicoloured triangulation such that for each vertex A of the n-gon the number of black triangles of which A is a vertex is greater than the number of white triangles of which A is a vertex. Find all triangulable numbers. Problem I–3 Let ABC be a triangle with AB ă AC and incentre I. Let E be the point on the side AC such that AE “ AB. Let G be the point on the line EI such that >IBG “ >CBA and such that E and G lie on opposite sides of I. Prove that the line AI, the perpendicular to AE at E, and the bisector of the angle >BGI are concurrent. Problem I–4 ˆˆ ˙˙ n For integers n ě k ě 0 we define the bibinomial coefficient by k ˆˆ ˙˙ n n!! “ . k k!!pn ´ kq!! Determine all pairs pn, kq of integers with n ě k ě 0 such that the corresponding bibinomial coefficient is an integer. Remark. The double factorial n!! is defined to be the product of all even positive integers up to n if n is even and the product of all odd positive integers up to n if n is odd. So e.g. 0!! “ 1, 4!! “ 2 ¨ 4 “ 8, and 7!! “ 1 ¨ 3 ¨ 5 ¨ 7 “ 105.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition Sept. 21, 2014 English version Problem T–1 Determine the lowest possible value of the expression 1 1 1 1 ` ` ` , a`x a`y b`x b`y where a, b, x, and y are positive real numbers satisfying the inequalities 1 1 ě , a`x 2
1 1 ě , a`y 2
1 1 ě , b`x 2
and
1 ě 1. b`y
Problem T–2 Determine all functions f : R ÝÑ R such that xf pxyq ` xyf pxq ě f px2 qf pyq ` x2 y holds for all x, y P R. Problem T–3 Let K and L be positive integers. On a board consisting of 2K ˆ 2L unit squares an ant starts in the lower left corner square and walks to the upper right corner square. In each step it goes horizontally or vertically to a neighbouring square. It never visits a square twice. At the end some squares may remain unvisited. In some cases the collection of all unvisited squares forms a single rectangle. In such cases, we call this rectangle memorable. Determine the number of different memorable rectangles. Remark. Rectangles are different unless they consist of exactly the same squares. Problem T–4 In Happy City there are 2014 citizens called A1 , A2 , . . . , A2014 . Each of them is either happy or unhappy at any moment in time. The mood of any citizen A changes (from being unhappy to being happy or vice versa) if and only if some other happy citizen smiles at A. On Monday morning there were N happy citizens in the city. The following happened on Monday during the day: citizen A1 smiled at citizen A2 , then A2 smiled at A3 , etc., and, finally, A2013 smiled at A2014 . Nobody smiled at anyone else apart from this. Exactly the same repeated on Tuesday, Wednesday and Thursday. There were exactly 2000 happy citizens on Thursday evening. Determine the largest possible value of N .
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition Sept. 21, 2014 English version Problem T–5 Let ABC be a triangle with AB ă AC. Its incircle with centre I touches the sides BC, CA, and AB in the points D, E, and F respectively. The angle bisector AI intersects the lines DE and DF in the points X and Y respectively. Let Z be the foot of the altitude through A with respect to BC. Prove that D is the incentre of the triangle XY Z. Problem T–6 Let the incircle k of the triangle ABC touch its side BC at D. Let the line AD intersect k at L ‰ D and denote the excentre of ABC opposite to A by K. Let M and N be the midpoints of BC and KM respectively. Prove that the points B, C, N , and L are concyclic. Problem T–7 A finite set of positive integers A is called meanly if for each of its nonempty subsets the arithmetic mean of its elements is also a positive integer. In other words, A is meanly if k1 pa1 ` . . . ` ak q is an integer whenever k ě 1 and a1 , . . . , ak P A are distinct. Given a positive integer n, determine the least possible sum of the elements of a meanly n-element set. Problem T–8 Determine all quadruples px, y, z, tq of positive integers such that 20x ` 142y “ px ` 2y ` zqzt .
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Individual Competition 27th August, 2015 English version Problem I–1 Find all surjective functions f : N Ñ N such that for all positive integers a and b, exactly one of the following equations is true: f (a) = f (b), f (a + b) = mintf (a), f (b)u. Remarks: N denotes the set of all positive integers. A function f : X Ñ Y is said to be surjective if for every y P Y there exists x P X such that f (x) = y. Problem I–2 Let n ě 3 be an integer. An inner diagonal of a simple n-gon is a diagonal that is contained in the n-gon. Denote by D(P ) the number of all inner diagonals of a simple n-gon P and by D(n) the least possible value of D(Q), where Q is a simple n-gon. Prove that no two inner diagonals of P intersect (except possibly at a common endpoint) if and only if D(P ) = D(n). Remark: A simple n-gon is a non-self-intersecting polygon with n vertices. A polygon is not necessarily convex. Problem I–3 Let ABCD be a cyclic quadrilateral. Let E be the intersection of lines parallel to AC and BD passing through points B and A, respectively. The lines EC and ED intersect the circumcircle of AEB again at F and G, respectively. Prove that points C, D, F , and G lie on a circle. Problem I–4 Find all pairs of positive integers (m, n) for which there exist relatively prime integers a and b greater than 1 such that am + bm an + bn is an integer.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition 28th August, 2015 English version Problem T–1 Prove that for all positive real numbers a, b, c such that abc = 1 the following inequality holds: b c a2 + b2 + c2 a + + ď . 2b + c2 2c + a2 2a + b2 3 Problem T–2 Determine all functions f : Rzt0u Ñ Rzt0u such that f (x2 yf (x)) + f (1) = x2 f (x) + f (y) holds for all nonzero real numbers x and y. Problem T–3 There are n students standing in line in positions 1 to n. While the teacher looks away, some students change their positions. When the teacher looks back, they are standing in line again. If a student who was initially in position i is now in position j, we say the student moved for |i ´ j| steps. Determine the maximal sum of steps of all students that they can achieve. Problem T–4 Let N be a positive integer. In each of the N 2 unit squares of an N ˆ N board, one of the two diagonals is drawn. The drawn diagonals divide the N ˆ N board into K regions. For each N , determine the smallest and the largest possible values of K. 2
4 3
1 6
5
7
Example with N = 3, K = 7
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition 28th August, 2015 English version Problem T–5 Let ABC be an acute triangle with AB ą AC. Prove that there exists a point D with the following property: whenever two distinct points X and Y lie in the interior of ABC such that the points B, C, X, and Y lie on a circle and =AXB ´ =ACB = =CY A ´ =CBA holds, the line XY passes through D. Problem T–6 Let I be the incentre of triangle ABC with AB ą AC and let the line AI intersect the side BC at D. Suppose that point P lies on the segment BC and satisfies P I = P D. Further, let J be the point obtained by reflecting I over the perpendicular bisector of BC, and let Q be the other intersection of the circumcircles of the triangles ABC and AP D. Prove that =BAQ = =CAJ. Problem T–7 Find all pairs of positive integers (a, b) such that a! + b! = ab + ba . Problem T–8 Let n ě 2 be an integer. Determine the number of positive integers m such that m ď n and m2 + 1 is divisible by n.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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English version
¨ VOCKLABRUCK AUSTRIA 2016
Problem I–1 Let n ě 2 be an integer and x1 , x2 , . . . , xn be real numbers satisfying (a) xj ą ´1 for j “ 1, 2, . . . , n and (b) x1 ` x2 ` ¨ ¨ ¨ ` xn “ n. Prove the inequality
n ÿ
n ÿ 1 xj ě 1 ` xj 1 ` x2j j“1 j“1
and determine when equality holds. Problem I–2 There are n ě 3 positive integers written on a blackboard. A move consists of choosing three numbers a, b, c on the blackboard such that they are the sides of a non-degenerate non-equilateral triangle and replacing them by a ` b ´ c, b ` c ´ a and c ` a ´ b. Show that an infinite sequence of moves cannot exist. Problem I–3 Let ABC be an acute-angled triangle with >BAC ą 45˝ and with circumcentre O. The point P lies in its interior such that the points A, P , O, B lie on a circle and BP is perpendicular to CP . The point Q lies on the segment BP such that AQ is parallel to P O. Prove that >QCB “ >P CO. Problem I–4 Find all functions f : N Ñ N such that f paq ` f pbq divides 2pa ` b ´ 1q for all a, b P N. Remark: N “ t1, 2, 3, . . .u denotes the set of positive integers.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition 25 August 2016
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¨ VOCKLABRUCK AUSTRIA 2016
Problem T–1 Determine all triples pa, b, cq of real numbers satisfying the system of equations a2 ` ab ` c “ 0, b2 ` bc ` a “ 0, c2 ` ca ` b “ 0.
Problem T–2 Let R denote the set of real numbers. Determine all functions f : R Ñ R such that f pxqf pyq “ xf pf py ´ xqq ` xf p2xq ` f px2 q holds for all real numbers x and y. Problem T–3 A tract of land in the shape of an 8 ˆ 8 square, whose sides are oriented north–south and east– west, consists of 64 smaller 1 ˆ 1 square plots. There can be at most one house on each of the individual plots. A house can only occupy a single 1 ˆ 1 square plot. A house is said to be blocked from sunlight if there are three houses on the plots immediately to its east, west and south. What is the maximum number of houses that can simultaneously exist, such that none of them is blocked from sunlight? Remark: By definition, houses on the east, west and south borders are never blocked from sunlight. Problem T–4 A class of high school students wrote a test. Every question was graded as either 1 point for a correct answer or 0 points otherwise. It is known that each question was answered correctly by at least one student and the students did not all achieve the same total score. Prove that there was a question on the test with the following property: The students who answered the question correctly got a higher average test score than those who did not.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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English version
Problem T–5 Let ABC be an acute-angled triangle with AB ‰ AC, and let O be its circumcentre. The line AO intersects the circumcircle ω of ABC a second time in point D, and the line BC in point E. The circumcircle of CDE intersects the line CA a second time in point P . The line P E intersects the line AB in point Q. The line through O parallel to P E intersects the altitude of the triangle ABC that passes through A in point F . Prove that F P “ F Q. Problem T–6 Let ABC be a triangle with AB ‰ AC. The points K, L, M are the midpoints of the sides BC, CA, AB, respectively. The inscribed circle of ABC with centre I touches the side BC at point D. The line g, which passes through the midpoint of segment ID and is perpendicular to IK, intersects the line LM at point P . Prove that >P IA “ 90˝ . Problem T–7 A positive integer n is called a Mozartian number if the numbers 1, 2, . . . , n together contain an even number of each digit (in base 10). Prove: (a) All Mozartian numbers are even. (b) There are infinitely many Mozartian numbers. Problem T–8 We consider the equation a2 ` b2 ` c2 ` n “ abc, where a, b, c are positive integers. Prove: (a) There are no solutions pa, b, cq for n “ 2017. (b) For n “ 2016, a must be divisible by 3 for every solution pa, b, cq. (c) The equation has infinitely many solutions pa, b, cq for n “ 2016.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Individual Competition Aug. 23, 2017 English version
Problem I–1
Determine all functions f : R Ñ R satisfying f px 2
f pxqf py qq xf px
yq
for all real numbers x and y. Problem I–2
Let n ¥ 3 be an integer. A labelling of the n vertices, the n sides and the interior of a regular n-gon by 2n 1 distinct integers is called memorable if the following conditions hold: (a) Each side has a label that is the arithmetic mean of the labels of its endpoints. (b) The interior of the n-gon has a label that is the arithmetic mean of the labels of all the vertices. Determine all integers n ¥ 3 for which there exists a memorable labelling of a regular n-gon consisting of 2n 1 consecutive integers.
Problem I–3 Let ABCDE be a convex pentagon. Let P be the intersection of the lines CE and BD. Assume that =P AD =ACB and =CAP =EDA. Prove that the circumcentres of the triangles ABC and ADE are collinear with P . Problem I–4 Determine the smallest possible value of
|2m 181n|, where m and n are positive integers.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition Aug. 24, 2017 English version
Problem T–1
Determine all pairs of polynomials pP, Qq with real coefficients satisfying P px
Qpy qq Qpx
P py qq
for all real numbers x and y. Problem T–2 Determine the smallest possible real constant C such that the inequality
|x3
y3
z3
1| ¤ C |x5
holds for all real numbers x, y, z satisfying x
y
z
y5
z5
1|
1.
Problem T–3
There is a lamp on each cell of a 2017 2017 square board. Each lamp is either on or off. A lamp is called bad if it has an even number of neighbours that are on. What is the smallest possible number of bad lamps on such a board? (Two lamps are neighbours if their respective cells share a side.)
Problem T–4
Let n ¥ 3 be an integer. A sequence P1 , P2 , . . . , Pn of distinct points in the plane is called good if no three of them are collinear, the polyline P1 P2 . . . Pn is non-self-intersecting and the triangle Pi Pi 1 Pi 2 is oriented counterclockwise for every i 1, 2, . . . , n 2. For every integer n ¥ 3 determine the greatest possible integer k with the following property: there exist n distinct points A1 , A2 , . . . , An in the plane for which there are k distinct permutations σ : t1, 2, . . . , nu Ñ t1, 2, . . . , nu such that Aσp1q , Aσp2q , . . . , Aσpnq is good. (A polyline P1 P2 . . . Pn consists of the segments P1 P2 , P2 P3 , . . . , Pn1 Pn .)
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Team Competition Aug. 24, 2017 English version
Problem T–5
Let ABC be an acute-angled triangle with AB ¡ AC and circumcircle Γ. Let M be the midpoint of the shorter arc BC of Γ, and let D be the intersection of the rays AC and BM . Let E C be the intersection of the internal bisector of the angle ACB and the circumcircle of the triangle BDC. Let us assume that E is inside the triangle ABC and there is an intersection N of the line DE and the circle Γ such that E is the midpoint of the segment DN . Show that N is the midpoint of the segment IB IC , where IB and IC are the excentres of ABC opposite to B and C, respectively.
Problem T–6
Let ABC be an acute-angled triangle with AB AC, circumcentre O and circumcircle Γ. Let the tangents to Γ through B and C meet each other at D, and let the line AO intersect BC at E. Denote the midpoint of BC by M and let AM meet Γ again at N A. Finally, let F A be a point on Γ such that A, M , E and F are concyclic. Prove that F N bisects the segment M D.
Problem T–7
Determine all integers n ¥ 2 such that there exists a permutation x0 , x1 , . . . , xn1 of the numbers 0, 1, . . . , n 1 with the property that the n numbers x0 ,
x0
x1 ,
...,
x0
x1
...
xn1
are pairwise distinct modulo n. Problem T–8
For an integer n ¥ 3 we define the sequence α1 , α2 , . . . , αk as the sequence of exponents in the prime factor decomposition of n! pα1 1 pα2 2 . . . pαk k , where p1 p2 pk are primes. Determine all integers n ¥ 3 for which α1 , α2 , . . . , αk is a geometric progression.
Time: 5 hours Time for questions: 60 min Each problem is worth 8 points. The order of the problems does not depend on their difficulty.
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Individual Problems and Solutions
I–1 Let (an )∞ n=1 be a sequence of positive integers such that an < an+1 for all n > 1. Suppose that for all quadruples of indices (i, j, k, l) such that 1 6 i < j 6 k < l and i + l = j + k, the inequality ai + al > aj + ak is satisfied. Determine the least possible value of a2008 . Solution (Jaromír Šimša, Czech Republic). Since a2 − a1 > 1 and an+2 − an+1 > (an+1 −an )+1 (by applying the quadruple (n, n+1, n+1, n+2) for each n), induction yields an+1 − an > n for all n > 1. Thus an+1 > n + an (and a1 > 1), hence induction again yields an > 12 (n2 − n + 2). Since the sequence an = 21 (n2 − n + 2) is as required (transform ai + al > aj + ak to i2 + l2 > j 2 + k 2 and substitute i = d − y, l = d + y, j = d − x, k = d + x, where 0 6 x < y), the smallest value of a2008 is 2,015,029. I–2 Consider a chessboard n × n where n > 1 is a positive integer. We select the centers of 2n − 2 squares. How many selections are there such that no two selected centers lie on a line parallel to one of the diagonals of the chessboard? Solution. By a k-diagonal we mean any chessboard diagonal formed by k squares, where 1 6 k 6 n. Since the number of stones is 2n−2, while the number of chessboard 1
diagonals in one direction is 2n−1 and two of them, which are 1-diagonals, must not be occupied by stones simultaneously, we can conclude that each k-diagonal with k > 1 contains exactly 1 stone and that exactly two of the 4 corner squares (1-diagonals) are occupied (and lie on the same border side). Let us call two different directions of diagonals as A and B. Now let us consider the set P of all the pairs (s, f ), for which the stone s lies on the same diagonal as the unoccupied (“free”) square f . There are exactly n2 − 2n + 2 free squares on the chessboard, two of them are corner, hence for each of the n2 − 2n free squares f which lie on two k-diagonals with k > 1, we have (s, f ) ∈ P for exactly two stones s. Thus the total number p of the pairs in P is given by the formula p = 2(n2 − 2n) + 2 = 2n2 − 4n + 2, where +2 stands for the two free corner squares. If a stone s lies on the intersection of a k1 -diagonal and a k2 -diagonal with k1 , k2 > 1, then the number of pairs (s, f ) ∈ P with this s equals k1 + k2 − 2. The same holds also for the two other stones with {k1 , k2 } = {1, n}. Obviously, for any stone we have k1 + k2 > n + 1 with equality iff the stone lies on a border square. Thus for each stone s, the number of pairs (s, f ) ∈ P is at least n − 1, and therefore p > (2n − 2)(n − 1) = 2n2 − 4n + 2. Since we have the equality, all the stones must lie on the boarder squares of the chessboard. If we put some stones (even no stone) on the first horizontal row in any way, then the border squares for the other stones are determined in exactly one way. To see this, consider separately the four corner squares and then, for each k, 1 < k < n, the pair of k-diagonals one direction together with the pair of (n + 1 − k)-diagonals in the other direction. Hence there are exactly 2n possibilities how to distribute the stones on the chessboard as required. Comment. The proof of the fact that all the stones must lie on some of the border squares from the preceding solution can be presented in the following algebraic form without counting the pairs (s, f ). Consider the chessboard n × n as the grid {0, 1, . . . , m} × {0, 1, . . . , m} with m = n − 1, in which the occupied squares are represented by points (ai , bi ) with i = 1, 2, . . . , 2m. Since ai −bi are 2m distinct integers from {−m, −m+1, . . . , m−1, m} and the boundary values ±m are not reached simultaneously, the values of |ai − bi | (in nondecreasing order) are the numbers 0, 1, 1, 2, 2, . . . , m − 1, m − 1, m whose sum equals m2 . Thus we have i=2m X
|ai − bi | = m2
(1)
|ai + bi − m| = m2 .
(2)
i=1
and, similarly, i=2m X i=1
2
Summing (1) and (2) and taking into account the identity |a − b| + |a + b − m| = max(|2a − m|, |2b − m|)
for any a, b, m ∈ R ,
we obtain the equality i=2m X
max(|2ai − m|, |2bi − m|) = 2m2 .
(3)
i=1
Since |2ai − m| 6 m and |2bi − m| 6 m for each i, the following inequality i=2m X
max(|2ai − m|, |2bi − m|) 6 2m · m = 2m2
i=1
holds and hence the equality (3) implies that max(|2ai − m|, |2bi − m|) = m for any i = 1, 2, . . . , 2m. This means that any (ai , bi ) is a boundary point of the grid and the proof is complete. Another solution (Bernd Mulansky, Germany). The first paragraph is identical with that from the first solution. It follows that each satisfactory distribution of the 2n − 2 stones can be derived as a result of the following procedure in n steps: ⊲ Step 1: One stone is placed on one of the two 1-diagonals of direction A. ⊲ Step k (where 2 6 k 6 n − 1): Two stones are placed, each on one of the two k-diagonals of direction A. ⊲ Step n: One stone is placed on the n-diagonal of direction A. Notice that for each m = 1, 2, . . . , n − 1 the following conclusion clearly holds: after m steps of our procedure, well-done in the sense that no two stones were placed on the same diagonal (of direction B), all the 2m−1 longest k-diagonals of direction B (those with k > n + 1 − m) are occupied by stones. Consequently, if in addition m < n − 1, in the next step m + 1 the two stones must be placed on the border squares of the two (m + 1)-diagonals of direction A (their other squares lie on the occupied diagonals of direction B) and there are exactly two ways in which this can be well-done. Analogously for the case m + 1 = n. Thus we have two possibilities in each of the n steps and the number of all satisfactory distributions equals 2n . Another solution (Pavol Novotný, Slovakia). Let us colour the chessboard squares as usual, with the black square in the left upper corner. It is easy to show that n − 1 stones must be placed on the black squares (let us call them black stones), analogously for the n − 1 white stones. The number sn of all satisfactory stone distributions on the chessboard n × n is equal to the product bn · wn , where bn and wn are the numbers of satisfactory distributions of black and white stones, respectively. We have w1 = 1, w2 = w3 = 2, b1 = 1, b2 = 2 and b3 = 4. Easy arguments show that for each n > 3, wn = 2wn−2 ,1 and bn = 2wn−1 ,2 hence sn = bn wn = 4wn−2 wn−1 = 2bn−1 wn−1 = 2sn−1 and the result sn = 2n follows. 1
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Remove two white 2-diagonals of one direction and two white (n − 1)-diagonals of the other direction; the remaining white squares form the same diagonals as white squares of the chessboard (n − 2) × (n − 2). Remove one black n-diagonal; the remaining black squares form the same diagonals as the white squares of the chessboard (n − 1) × (n − 1).
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I–3 Let ABC be an isosceles triangle with |AC| = |BC|. Its incircle touches AB and BC at D and E, respectively. A line (different from AE) passes through A and intersects the incircle at F and G. The lines EF and EG intersect the line AB at K and L, respectively. Prove that |DK| = |DL|. Solution. In view of symmetry, suppose that AF < AG, and, in addition, that G is on the smaller arc DE (for the other case see the last two sentences below). If the incircle touches AC at J, then 6 CAB = 6 CJE = 6 JDE = 6 JF E (Fig. 1), hence AJF K is a cyclic quadrilateral. Thus 6 AJK = 6 AF K = 6 EF G = 6 LEB, which implies that AJK and BEL are congruent triangles. Since K and L are inner points of the segment AB, AK = BL means that DK = DL. If G is on the larger arc DE (between E and J), then K, A, B, L is the order of these collinear points and the cyclic quadrilateral is AKJF . The rest of the proof is the same. C
C
E
J
E X
G
G
F
F A
K
D
L
B
A
K
Fig. 1
D
L
B
Fig. 2
Another solution (Tomáš Pavlík, Czech Republic). Let us denote X the intersection of line AF with side BC of the given triangle (Fig. 2). The power of the point X with regard to the incircle of ABC gives XE 2 = XF · XG which means that XG XE = . XE XF
(1)
Let us write Menelaos theorem for triangle ABX and lines EG and EF , respectively: AL BE XG · · =1 LB EX GA
and
AK BE XF · · = 1. KB EX F A
With help of (1) we can rewrite both the last equalities as XE AL · BE · = 1 and XF LB · GA or
XE KB · F A · =1 XF AK · BE
KB · F A AL · BE = LB · GA AK · BE 4
which gives AK · AL · BE 2 = 1, KB · LB · F A · GA hence AK · AL = KB · LB as AF · AG = AD2 = BD2 = BE 2 clearly holds. Depending on the position of point G, the points K and L lie inside or outside the segment AB simultaneously, according to that we choose plus or minus sign in AK · (AB ± BL) = AK · AL = KB · LB = (AB ± AK) · BL which results into AK = BL in both cases. This is equivalent to the wanted equality DK = DL. I–4 Find all integers k such that for every integer n, the numbers 4n + 1 and kn + 1 are relatively prime. Solution. Since 4n + 1 is odd, the identity k − 4 = k(4n + 1) − 4(kn + 1) shows that 4n + 1 and kn + 1 are relatively prime if k − 4 has not any odd divisor p > 1, i.e. if k − 4 = ±2k with any nonnegative integer k. On the other hand, if k − 4 has got an odd divisor p > 1, then we can easily find a multiple of p of the form 4n + 1 (for example, the number p2 or simply one of the numbers p, 3p). For any number 4n + 1 being a multiple of p, the above identity implies that p | kn + 1, hence 4n + 1 and kn + 1 are not relatively prime. Answer : k = 4 ± 2k , where k = 0, 1, 2, . . .
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Team Problems and Solutions
T–1
Let
R denote the set of real numbers. Find all functions f : R → R such that xf (x + xy) = xf (x) + f (x2 )f (y)
for all x, y ∈ R . Solution. Setting x = y = 0 in xf (x + xy) = xf (x) + f (x2 )f (y)
(0)
we get f (0) = 0. Using this in (0) with y = −1 we obtain xf (x) + f (x2 )f (−1) = 0.
(1)
Let us distinguish the cases f (−1) = 0 and f (−1) 6= 0. The case f (−1) = 0. It follows from (1) that f (x) = 0 for all x 6= 0. As we already know, f (0) = 0. Thus we get the zero function f (x) = 0, which is obviously a solution. 1
The case f (−1) 6= 0. Setting x = −1 in (1) yields f (1) = 1. Using this in (1) with x = 1, we get f (−1) = −1 and hence (1) can be transformed to xf (x) = f (x2 ).
(2)
xf (x2 ) = xf (x) + f (x2 )f (x − 1).
(3)
Put now y = x − 1 in (0) to get
Summing up (2) and (3) we obtain the equation f (x2 )(f (x − 1) − (x − 1)) = 0.
(4)
Assume that f (a) = 0 for some a 6= 0. Then f (a2 ) = 0 by (2) and hence (0) with x = a implies that af (a + ay) = 0, i.e. f (a + ay) = 0. Since y is arbitrary here, we get f (−1) = 0, which is not the case. Therefore, for any x 6= 0 we have f (x) 6= 0, and hence f (x2 ) 6= 0 as well. Thus (4) leads to the conclusion that f (x − 1) = x − 1 for any x 6= 0, i.e. f (x) = x for any x 6= −1. Since we already know that f (−1) = −1, we get the identity function f (x) = x, which is obviously a solution. T–2 Let n > 2 be an integer. There are n positive integers written on a blackboard. In each step we choose two of the numbers on the blackboard and replace each of them by their sum. Determine all values of n for which it is always possible to get n identical integers in a finite number of steps. Solution. Starting from the n-tuple (2, 2, 1, 1, . . . , 1) with any n > 3, we get always an n-tuple in which the number of maximal values is even. Hence no odd n > 3 is as required. Let us show by induction that any even n > 2 is satisfactory, which is obvious if n = 2. For an even n > 4, by the induction hypothesis, we can transform any initial n-tuple to (a, a, . . . , a, b, b). If a 6= b, we apply repeatedly some of the following series of steps, which always lead to an n-tuple of type (a, . . . , a, b, . . . , b) in which the | {z } | {z } k
n−k
number k may differ from the initial value k = n − 2 (remaining to be even): series α: (a, . . . , a, b, . . . , b) → (2a, . . . , 2a, b, . . . , b), | {z } | {z } | {z } | {z } k
series β:
n−k
k
n−k
(a, . . . , a, b, . . . , b) → (a, . . . , a, 2b, . . . , 2b), | {z } | {z } | {z } | {z } k
n−k
k
n−k
series γ1 (if k 6 n − k): (a, . . . , a, b, . . . , b) → (a + b, . . . , a + b, b, . . . , b), {z } | {z } | | {z } | {z } k
n−k
n−2k
2k
series γ2 (if k > n − k): (a, . . . , a, b, . . . , b) → (a, . . . , a, a + b, . . . , a + b). {z } | {z } | | {z } | {z } k
n−k
2k−n
2(n−k)
To describe our procedure, we introduce the notation c = 2P (c) N (c) for any positive integer c, where P (c) > 0 and N (c) is odd. To each n-tuple (a, . . . , a, b, . . . , b) with | {z } | {z } k
a 6= b, let us apply
2
n−k
⊲ series α if P (a) < P (b), ⊲ series β if P (a) > P (b), ⊲ series γ1 or γ2 if P (a) = P (b) (and hence N (a) 6= N (b)). Using the series α and β, the numbers N (a), N (b) do not change, while the series γ1 and γ2 cause the changes exactly one of them, namely N (b) →
N (a) + N (b) , 2m
or N (b) →
N (a) + N (b) 2m
respectively,
where m = P (N (a) + N (b)) > 1 and hence N (a) + N (b) N (a) + N (b) 6 < max(N (a), N (b)) m 2 2 (recall that N (a) 6= N (b)). Consequently, throughout our procedure, the value of max(N (a), N (b)) is a nonincreasing variable, and hence constant after a finite numbers of series. From this moment, we must still have either N (a) > N (b), or N (a) 6 N (b). This excludes either series γ1 , or series γ2 from future applications, in which, therefore, all possible changes of the parameter k are either k → 2k, or (n−k) → 2(n−k). Since this can repeat only r times, where 2r 6 n, at the end we always we get an n-tuple (a, . . . , a, b, . . . , b) for which (if a 6= b) the continuation of our procedure reduces only to the series α and β. Applying now either α, or β exactly |P (a) − P (b)| times, we get an n-tuple (a′ , . . . , a′ , b′ , . . . , b′ ) with P (a′ ) = P (b′ ). Since γ1 , γ2 are already excluded, we have a′ = b′ , which completes the induction proof. Another solution (German team, adapted). We show without induction on n that any even n = 2k is satisfactory. At the beginning in the initial 2k-tuple (a1 , . . . , a2k ) we replace every pair (a2i−1 , a2i ) (for i = 1, . . . , k) by the pair (a2i−1 + a2i , a2i−1 + a2i ). From now on, we shall have always identical numbers on the (2i − 1)th and (2i)th position. Hence because of brevity we shall work with k-tuples (x, y, z, . . . ) instead of 2k-tuples (x, x, y, y, z, z, . . . ). We are allowed to do the following transformations on the k-tuples: ⊲ choose two of the numbers x, y and replace each of them by their sum (this corresponds with two steps (. . . , x, x, . . . , y, y, . . . ) → (. . . , x + y, x, . . . , x + y, y, . . . ) → (. . . , x + y, x + y, . . . , x + y, x + y, . . . ) performed on the 2k-tuple); ⊲ choose one number x and multiply it by 2 (this corresponds with one step (. . . , x, x, . . . ) → (. . . , x + x, x + x, . . . ); ⊲ divide all numbers by 2 (this obviously does not affect anything; formally we could remember how many times we have performed this dividing and multiply all the numbers by the proper power of two at the end). Our aim is to obtain k identical numbers. We reach it by iterating the following algorithm: 1. While there are at least two odd numbers, find the minimum and the maximum odd number and replace each of them by their (even) sum. 2. If there is one odd number left after finishing the first step, multiply it by two. 3. Divide all numbers by 2. Clearly, after each iteration, the maximum number among all k numbers either decreases or does not change. As this maximum is permanently a positive integer, after a finite number of iterations, it fixes at the value M and does not change anymore. From now on, look at the number N of M ’s in the k-tuple. 3
Obviously M is odd (otherwise it would decrease in the third step in the next iteration). If N < k, then there is at least one number m with m < M . If m is odd, after the next iteration N decreases. As it is impossible to increase N in the iterations, it must be constant after a finite number of steps and there must be only even m with m < M . But every even m is divided by 2 in each iteration and after some iterations some odd number less than M must appear. So there are no numbers less than M , which completes the proof. T–3 An acute-angled triangle ABC is given. Let E be a point such that B and E lie on different sides of the line AC, and let D be an interior point of the segment AE. Suppose that 6 ADB = 6 CDE, 6 BAD = 6 ECD and 6 ACB = 6 EBA. Prove that B, C and E are collinear. Solution. Condition 6 ADB = 6 CDE motivates us to reflect B over AE to B ′ (Fig. 3). Then C, D and B ′ are collinear and 6 EAB ′ = 6 EAB = 6 ECD = 6 ECB ′ , so B ′ ACE is a cyclic quadrilateral. This implies that 6 ECA = π − 6 EB ′ A = π − 6 EBA = π − 6 ACB, hence 6 ECA + 6 ACB = π and thus B, C, E are collinear. E
B′ D C
A
B Fig. 3
Comment. With the same success, we can reflect C over AE to get a point C ′ collinear with B, D and such that ABEC ′ is cyclic, hence 6 ECA = 6 EC ′ A = π − 6 EBA = π − 6 ACB, i.e. 6 ECA + 6 ACB = π again. Because of the proved collinearity of B, C, E, the condition 6 ACB = 6 EBA implies that AB = AC, while the condition 6 BAD = 6 ECD implies that ABCD must be a cyclic quadrilateral. The last fact serves as a good motivation for another solution. Before we present it, let us note that the situations described in the statement of the problem do exist and all of them are of the following form: ABC is an isosceles triangle with AB = AC, points B, C, E are collinear (C is between B and E) and AE cuts the circumcircle of ABC at D. Another solution. Suppose that B, C, E are not collinear. The line through B, which is parallel to CE, meets the lines CD and AD at C ′ and E ′ , respectively. Since 6 E ′ C ′ D = 6 ECD = 6 BAD, the quadrilateral ABC ′ D is cyclic (Fig. 4). Denote its 4
circumcircle by K. We have 6 AC ′ B = 6 ADB = 6 CDE = 6 C ′ DE = 6 ABC ′ , i.e. 6 AC ′ B = 6 ABC ′ (so ABC ′ is an isosceles triangle). Suppose that C lies inside the segment C ′ D. Then C lies inside K (on the same side of the line AB as C ′ ), therefore 6 ACB > 6 AC ′ B = 6 ABC ′ = 6 ABE ′ > 6 ABE (because E lies between A and E ′ ), which contradicts to 6 ACB = 6 EBA. Similarly, if C does not lie on the segment C ′ D, then C lies outside K (on the same side of the line AB as C ′ ), therefore 6 ACB < 6 AC ′ B = 6 ABC ′ = 6 ABE ′ < 6 ABE (because E ′ lies between A and E), which again contradicts to 6 ACB = 6 EBA. E′ E
D C
A
E
D
C′
B
C
A
Fig. 4
B Fig. 5
Another solution (Karel Horák, Czech Republic). From the given equalities of angles it follows that triangles ABD and CED are similar (Fig. 5). From that similarity we immediately get that triangles ACD and BED are similar (by sas, same angles at common vertex D, and proportional sides). From the equal angles BED and ACD if follows that the sum of three angles BCA, ACD and DCE is equal to the sum of angles in the triangle ABE, so E, C, and B are collinear. T–4 Let n be a positive integer. Prove that if the sum of all positive divisors of n is a perfect power of 2, then the number of these divisors is also a perfect power of 2. Solution. Suppose that n = ps11 ps22 . . . pskk , where p1 , . . . , pk are distinct primes and si > 1 for each i, and that the sum of all positive divisors of n, which is given by (1 + p1 + p21 + · · · + ps11 )(1 + p2 + p22 + · · · + ps22 ) . . . (1 + pk + p2k + · · · + pskk ), is a perfect power of 2. Then each of the factors fi = 1 + pi + p2i + · · · + psi i is also a perfect power of 2 greater than 1 and hence both pi and si are odd. Suppose that si > 1. In this case we have fi = (1 + pi )(1 + p2i + p4i + · · · + psi i −1 ). 5
Since fi has no odd divisor greater than 1, the even integer si − 1 (which is supposed to be positive) must be of the form 4k + 2 and thus we can make another factorization fi = (1 + pi )(1 + p2i )(1 + p4i + p8i + · · · + psi i −3 ). Consequently, both 1 + pi and 1 + p2i are powers of 2, hence 1 + pi | 1 + p2i , which contradicts to 1 + p2i = (1 + pi )(pi − 1) + 2 (as 1 + pi | 2 is impossible). This means that si = 1 for each i and thus the number of divisors of n equals 2k . Note that the above solution can be finished without observing the fact that 1+pi and 1 + p2i cannot be powers of 2 at the same time. Indeed, repeating the procedure of factorization we get finally fi = (1 + pi ) 1 +
p2i
1+
p4i
2ti . . . 1 + pi ,
hence si = 2ti +1 − 1 with some ti > 0 for each i and thus the number of divisors of n equals 2k+t1 +t2 +···+tk . (As we know from the original solution, ti = 0 for each i.)
6
L
ICA AT
EM TH MA
MI DD LE
EUROPEAN
TH R U FO
O L YM P IAD
STREČNO SLOVAKIA 2010
Contest Solutions 4th MEMO, Streˇ cno, Slovakia 9 to 15 September 2010
Problem I-1. Find all functions f : R → R such that for all x, y ∈ R, we have f (x + y) + f (x)f (y) = f (xy) + (y + 1)f (x) + (x + 1)f (y).
Solution. Setting y = 0 yields 0 = f (0)(f (x) − x − 2). It is easy to check that f (x) = x + 2 is not a solution, so f (0) = 0. Setting x = 1, y = −1, we obtain 0 = f (−1)(f (1) − 3), so f (−1) = 0 or f (1) = 3. Let us assume f (−1) = 0. Putting x = 2, y = −1 in the original functional equation, we get f (−2) = f (1). On the other hand, setting x = −2, y = 1 gives f (−2)f (1) = 3f (−2)−f (1) which together with f (−2) = f (1) gives f (1) ∈ {0, 2}. So we have f (1) = a ∈ {0, 2, 3}. Setting y = 1 yields
f (x + 1) = (3 − a)f (x) + a(x + 1).
(1)
for all real x. Now we set y = 1 + 1/x for arbitrary x 6= 0, we obtain f x + x1 + 1 + f (x)f x1 + 1 = f (x + 1) + x1 + 2 f (x) + f
Applying (1) yields (3 − a) f x +
1 x
+ f (x)f
1 x
1 x
− (x + 1)f
1 x
+ 1 (x + 1).
= f (x) 5 − 2a − (a − 1) x1 + 2a + ax.
From this and the original functional equation with y = 1/x we have (3 − a) a + f (x) 1 + x1 = f (x) 5 − 2a − (a − 1) · x1 + 2a + ax, which is equivalent to
f (x) −2 + a +
2 x
= a2 + ax − a.
Using a ∈ {0, 2, 3} we get f (x) = 0, f (x) = x2 + x, f (x) = 3x for all real x (with some exceptions when −2 + a + x2 = 0, but this cases can be handled easily for example by using (1)). We can also easily check that the functions f (x) = 0, f (x) = x2 + x, f (x) = 3x are solutions of the original functional equation. Solution 2. As in the first solution we obtain f (0) = 0, f (1) = a ∈ {0, 2, 3} and f (x + 1) = (3 − a)f (x) + a(x + 1)
for x ∈ R.
(2)
If a = 3, then (2) implies that f (x + 1) = 3(x + 1) for all x ∈ R (and therefore f (x) = 3x for all x). It is easy to verify that this is a solution of the functional equation.
In the sequel we will assume that a ∈ {0, 2} and therefore f (−1) = 0. We will compute f (xyz) in two different ways. Setting yz for y into the original functional equation we obtain f (xyz) = f (x + yz) + f (x)f (yz) − (yz + 1)f (x) − (x + 1)f (yz) =
= f (x + yz) + f (x)f (y + z) + f (x)f (y)f (z) − (z + 1)f (x)f (y) −
− (y + 1)f (x)f (z) − (yz + 1)f (x) − (x + 1)f (y + z) − (x + 1)f (y)f (z)+
+ (x + 1)(z + 1)f (y) + (x + 1)(y + 1)f (z).
On the other hand, setting xy for x and z for y into the original functional equation we obtain f (xyz) = f (xy + z) + f (xy)f (z) − (z + 1)f (xy) − (xy + 1)f (z) =
= f (xy + z) + f (x + y)f (z) + f (x)f (y)f (z) − (y + 1)f (x)f (z) −
− (x + 1)f (y)f (z) − (z + 1)f (x + y) − (z + 1)f (x)f (y) + (y + 1)(z + 1)f (x) +
+ (x + 1)(z + 1)f (y) − (xy + 1)f (z). Therefore
f (x + yz) + f (x)f (y + z) − (yz + 1)f (x) − (x + 1)f (y + z) + (x + 1)(y + 1)f (z) =
= f (xy + z) + f (x + y)f (z) − (z + 1)f (x + y) + (y + 1)(z + 1)f (x) − (xy + 1)f (z). In particular, for x = −1 and y = z we obtain f (z 2 − 1) = f (z − 1)f (z) − (z + 1)f (z − 1) + (z − 1)f (z). On the other hand, setting x = z + 1 and y = z − 1 into the original equation we obtain f (2z) + f (z + 1)f (z − 1) = f (z 2 − 1) + (z + 2)f (z − 1) + zf (z + 1). Therefore f (2z) + f (z + 1)f (z − 1) = f (z − 1)f (z) + f (z − 1) + (z − 1)f (z) + zf (z + 1). Since (2) implies that f (2) = 5a − a2 and since for x = 2 and y = z the original functional equation implies f (z + 2) + f (2)f (z) = f (2z) + (z + 1)f (2) + 3f (z), it follows that f (z + 2) + (5a − a2 )f (z) + f (z + 1)f (z − 1) =
= f (z − 1)f (z) + f (z − 1) + (z + 2)f (z) + zf (z + 1) + (5a − a2 )(z + 1). The equation (2) implies that f (z) = (3 − a)f (z − 1) + az,
f (z + 1) = (3 − a)2 f (z − 1) + (4a − a2 )z + a and f (z + 2) = (3 − a)3 f (z − 1) + (a3 − 7a2 + 13a)z + 5a − a2 .
From the last four equations we obtain that (3 − a)(2 − a)f (z − 1)2 − 2(a − 3)(a − 2)zf (z − 1) + (a2 − 9a + 20)f (z − 1) = = (5a − a2 )z 2 + (a2 − 5a)z.
For a = 2 we get 6f (z − 1) = 6z 2 − 6z, therefore f (z) = z 2 + z and it is easy to verify that this is a solution of the functional equation. For a = 0 we get 6f (z − 1)2 − 12zf (z − 1) + 20f (z − 1) = 0, which implies that for each z ∈ R one of the equalities f (z) = 0 and f (z) = 2z − 43 is satisfied. Assume that f (z) = 2z − 34 for some z ∈ R. Then the original functional equation for x = 1 and y = z implies that f (z + 1) = 3f (z) = 6z − 4. Therefore either 6z − 4 = 0 or 6z − 4 = 2(z + 1) − 43 . The first equation implies that z = 23 and f (z) = 2z − 34 = 0, and the second equation implies that z = 76 and therefore f (z) = 2z − 43 = 1, f (z + 1) = 3z = 3 = 2(z + 1) − 43 and f (z + 2) = 3f (z + 1) = 9 6= 5 = 2(z + 2) − 43 . The contradiction shows that f (z) = 0 for each z ∈ R.
Problem I-2. All positive divisors of a positive integer N are written on a blackboard. Two players A and B play the following game taking alternate moves. In the first move, the player A erases N . If the last erased number is d, then the next player erases either a divisor of d or a multiple of d. The player who cannot make a move loses. Determine all numbers N for which A can win independently of the moves of B. Solution. Let N = pa11 pa22 . . . pakk be the prime factorization of N . In an arbitrary move the players writes down a divisor of N , which we can represent as a sequence (b1 , b2 , . . . , bk ), where bi ≤ ai (such a sequence represents the number pb11 pb22 . . . pbkk ). The rules of the game say that the sequence (b1 , b2 , . . . , bk ) can be followed by a sequence (c1 , c2 , . . . , ck ) with either ci ≤ bi for each i, or ai ≥ ci ≥ bi for each i (obviously, if such a sequence is not on the sheet).
If one of the numbers ai is odd, then the player B posses the winning strategy. Indeed, let for simplicity a1 be odd. Then the response for the move (b1 , b2 , . . . , bk ) should be (a1 − b1 , b2 , . . . , bk ). One can easily check that this is a winning strategy for B: All the legal sequences split up into pairs and when A writes down one sequence from a pair, player B responds with the second one from the same pair (a1 − b1 6= b1 because of a1 is odd).
If all ai are even, then the player A has a winning strategy. Let the move of player B be (b1 , b2 , . . . , bk ), where one of bi is strictly less than ai ((b1 , b2 , . . . , bk ) 6= (a1 , a2 , . . . , ak ), as it was the first move of A). Let j be the smallest index such that bj < aj . Then the response of A can be (b1 , b2 , . . . , bj−1 , aj − bj − 1, bj+1 , . . . , bk )
(symmetric reflection of bj ).
Again, all legal sequences (except for (a1 , a2 , . . . , ak )) split up into pairs and when B writes down one sequence from a pair, player A can respond with the second one (aj − bj − 1 6= bj because of aj is even). Obviously the condition ”all ai are even” means that ”N is a square”. For N ∈ [2000, 2100] it is possible only for N = 2025. The answer for the alternative question is 20101005 .
Points A, B, C and D lie on a circle in this order. Let E be a point on the segment AC such that AD = AE and CB = CE holds. Let k be the circumcircle of triangle BDE let M the center of k. Let F be the second intersection point of k and AC. Prove that the lines F M , AD and BC meet in a point. 1. Solution Assume A lies between C and F (the case when C lies between A and F can be handled in Problem. The problem statement to be inserted. :) the same way). Let the lines BC and AD meet in P . Using |M B| = |M E|, |BC| = |CE| Solution. Assume A lies between C and F (the case when C lies between A and F can and |M E| =be|M F | we getsame way). Let the lines BC and AD meet in P . Using |MB| = |ME|, handled in the |BC| =We |CE|are andgiven |ME| = |MF | wequadrilateral get Problem cyclic ABCD with a=point E on=the ∠M BPI-3. = π − ∠M BE −a∠EBC = π − ∠BEM − ∠CEB ∠F EM ∠Mdiagonal F E. AC such that ∠MBP AD ==AE and CB = CE. Let M be the center of the circumcircle k of the π − ∠MBE − ∠EBC = π − ∠BEM − ∠CEB = ∠F EM = ∠MF E. Therefore ∠M F CThe + ∠CBM = ∠M F Ethe + line π − AC ∠M in BPthe=points π, which implies that the triangle BDE. circle k intersects E and F . Prove that points the Therefore ∠MF C + ∠CBM = ∠MF E + π − ∠MBP = π, which implies that the points Mlines , B, FCMand F are , AD, andconcyclic. BC meet at one point. M, B, C and F are concyclic.
D b
F
A b
E b
b
b
C
b
B
b
M b
P Using |ME| = |MD| and |AE| = |AD| we get
Using |M E| = |M D| and |AE| = |AD| we get
∠AEM = ∠DEM − ∠DEA = ∠EDM − ∠EDA = ∠MDA, Solution. Assume that A lies between C and F (the case when C lies between A and F ∠AEM = ∠DEM − ∠DEA = ∠EDM − ∠EDA = P∠M DA, can be handled in the same way). Let the lines BC and AD meet in . Using M B = M E, hence ∠MDP = ∠MBP and the quadrilateral MP BD is cyclic. This gives (using that ∼ BC =∠M CEDP and=M∠M E =BP M F and we 4M BCcyclic EC = 4M hence the quadrilateral M Pand BDtherefore is cyclic. This gives (using that quadrilaterals ABCD andget F MBC are as well)
quadrilaterals ABCD and F M BC are cyclic as well) ◦
∠P BC MB = DB = ∠ADB ∠ACB = ∠F= CB = ◦π − − ∠F ∠M =∠P ∠M EC = 180= − ∠M EF 180 ∠MMB, F C,
∠P M B = ∠P DB = ∠ADB = ∠ACB = ∠F CB = π − ∠F M B,
which means that the points F , M and P are collinear, therefore the lines AD, BC and which implies that the M , B, C, and F are concyclic. Using M E = M D and AE F Mthat meetthe in a points point. which means points F , M and P are collinear, therefore the lines AD,
= AD BC and (which means 4M EA ∼ = 4M DA) we get ∠AEM = ∠ADM , hence ∠M DP = ∠M BP and F M meet inAlternative a point. solution. Assume A lies between C and F (the case when C lies between the quadrilateral M P BD is cyclic. This gives (using that quadrilaterals ABCD and F M BC A and F can be handled in the same way). Using |MB| = |ME|, |BC| = |CE| and are cyclic as well) that |ME| = |MF | we get
2. Solution Assume A lies∠P between and case C=lies between A◦ and FMcan ∠MBP − ∠MBE ∠EBC ==πwhen − ∠BEM −∠F ∠CEB EM ∠MF E. M B ==πC ∠P DB F=−(the ∠ADB ∠ACB CB ==∠F 180 −=∠F B, be handled in the same way). Using |M B| = |M E|, |BC| = |CE| and |M E| = |M F | we get Therefore ∠MF C + ∠CBM = ∠MF E + π − ∠MBP = π, which implies that the points which means that the points F , M and P are collinear, therefore the lines AD, BC and F M M,=B,πC−and F are ∠M ∠M BEconcyclic. − ∠EBC = π − ∠BEM − ∠CEB = ∠F EM = ∠M F E. meet in BP a point.
Therefore ∠M F C + ∠CBM = ∠M F E + π − ∠M BP = π, which implies that the points Solution 2. Like in the first solution, F M BC is cyclic. Since the points M and A lie Mon , B, C and F are concyclic. the perpendicular bisector of the segment DE, we have ∠M DA = ∠M EA. Moreover, 1 Using E|F= |M D|∠M andF|AE| = |AD| we∠M get M E |M = M , hence A = ∠M FE = EF = ∠M EA. Therefore, the quadrilateral M ADF is cyclic. Then the radical axes of = circumcircles of cyclic=quadrilaterals F M BC, ∠AEM = ∠DEM − ∠DEA ∠EDM − ∠EDA ∠M DA, BCDA and ADF M are the lines F M , AD and BC, which meet in the radical center of the hence DA Thus, = ∠Mthe F Alines andFthe M ADF is cyclic. three∠M circles. M , quadrilateral AD and BC meet in a point.
The radical axes of circumcircles of cyclic quadrilaterals F M BC, BCDA and ADF M are lines F M , AD and BC, which meet in the radical center of the three circles. Thus, the lines F M , AD and BC meet in a point.
Problem I-4. Find all positive integers n which satisfy the following two conditions: (i) n has at least four different positive divisors; (ii) for any divisors a and b of n satisfying 1 < a < b < n, the number b − a divides n. Solution. Clearly primes, squares of primes and the number 1 have the given property. We will exclude these numbers from further considerations. First assume that n is even; thus n = 2x for some integer x. Then x − 2 divides n. Any divisor of n smaller then x = n/2 is at most n/3. Therefore, x − 2 ≤ n/3, yielding x ≤ 6. Checking all possibilities for x, we arrive to three new satisfactory values of n, namely, 6, 8, and 12. Next assume that n is odd. Let n = px, where p is the smallest divisor of n greater than one. Obviously p is an odd prime. Note that p + 1 - n because p + 1 is even; thus x 6= p + 1. Since 1 < p < x < n, we have x − p | px. If p - x then x − p and x are coprime; hence x − p | p. Then x − p ≤ p. On the other hand, x 6= p + 1; therefore, x − p ≥ p since p is the smallest nontrivial divisor. Hence x = 2p, contradicting the assumption that n is odd. If p | x, then x = py for some integer y greater than one. The choice of p implies that y ≥ p. Moreover, py − p | p2 y; thus y − 1 | py. Since y − 1 and y are coprime, y ≤ p + 1. If y = p + 1, then y is an even divisor of n; this contradicts the assumption that n is odd. Otherwise y = p since y ≥ p. This contradicts the condition y − 1 | py. All the solutions are primes, squares of primes and the numbers 1, 6, 8 and 12.
Solution 2. If n = 1, n is prime or n is a square of a prime, there are no divisors a and b satisfying n > a > b > 1, thus the condition is trivially satisfied. Otherwise, for some integers k and `, n = ` · k with ` > k > 1. Then ` − k also divides n = ` · k. If ` and k are coprime, ` − k is coprime to ` and k, thus ` − k = 1. Hence n = k(k + 1). Let p be a prime divisor of k. Since k + 1 − p is coprime to p(k + 1), the condition implies that k + 1 − p divides k. But k k k + 1 − p = (p − 1) −1 + p p
divides k if and only if (p − 1)(k/p − 1) = 0; thus k = p and n = p(p + 1). Clearly p = 2 gives a solution n = 6. Otherwise p + 1 = q · r for some prime q and integer r greater than 1. Since p − q = qr − 1 − q = (q − 1)(r − 1) − 2 + r is a divisor of r, we have (q − 1)(r − 1) ≤ 2. This gives only three possibilities: q = r = 2 or q = 2, r = 3 or q = 3, r = 2. The first one yields a solution n = 12, while the other two give n = 30, which fails to satisfy the conditions: 6 − 2 - 30. It remains to consider the case when n cannot be written as a product of two coprime numbers greater than 1. Then n = pa , where a ≥ 3 (for a ≤ 2, we obtain the solutions we have
already described). This implies that p and p2 are proper divisors of n, hence p2 −p = p(p−1) divides n = pa . Since p and p − 1 are coprime, this is only possible when p − 1 = 1; thus p = 2. However, 23 − 2 = 6 is not a divisor of 2a ; hence there are no solutions for a ≥ 4. Only the number 8 satisfies the condition in this case. Solution 3. Clearly 1, p, p2 are solutions. For the other prime powers, n = pk is possible only for p = 2 and k < 4 (p2 − p is even for odd p, 8 − 2 = 6 does not divide 16).
Now, n is not a prime power, then it has two (or more) prime factors p, q. Then q − p | n. If p, q are both odd, then 2 | n. (Otherwise also 2 | n.) Therefore, if n is not a prime power, one of its factors is 2. Let p be the smallest divisor of n larger than 2; 1 < 2 < p < n. From p − 2 | n follows p = 3, so 6 | n. Clearly, n = 6 is a solution. Let n = 6a, a > 1. Then 1 < 3 < 3a < 6a = n, therefore
(3a − 3) = 3(a − 1) | n = 6a, a − 1 | 2a.
Since gcd(a, a − 1) = 1, we have a − 1 | 2, hence a − 1 = 1 or a − 1 = 2, which yields n = 12 or n = 18. It is easy to check that n = 12 is a solution, and n = 18 is not (e. g., 7 = 9 − 2 is not a divisor of 18).
Problem T-1. Three strictly increasing sequences a1 , a2 , a3 , . . . ,
b1 , b2 , b3 , . . . ,
c1 , c2 , c3 , . . .
of positive integers are given. Every positive integer belongs to exactly one of the three sequences. For every positive integer n, the following conditions hold: (i) can = bn + 1; (ii) an+1 > bn ; (iii) the number cn+1 cn − (n + 1)cn+1 − ncn is even. Find a2010 , b2010 , and c2010 . Solution. Since {cn } is a strictly increasing sequence of positive integers, it is clear that cn ≥ n, n ∈ N. Hence, can ≥ an , n ∈ N. However, the given sequences do not contain equal terms, so can > an and bn = can − 1 > an , n ∈ N. Similarly, from (ii) and (iii), an+1 > bn + 1 = can , n ∈ N. It is also easy to see that bn < can < bn+1 . Let us for any n ∈ N count the number of terms in all three sequences that are less or equal to can . There are n such terms in the first sequence (that is, a1 , a2 , . . . , an ), n such terms in the second sequence (b1 , b2 , . . . , bn ) and an such terms in the third sequence (c1 , c2 , . . . , can ). It is 2n + an terms in total. By (i) any positive integer less or equal can must appear among these terms exactly once; thus, the total number of these terms equals 2n + an = can .
(1)
Now we take an instead of n in (iv): can +1 can − (an + 1)can +1 − an can = can +1 (an + 2n) − (an + 1)can +1 − an (an + 2n) = = can +1 (2n − 1) − a2n − 2an n ≡ can +1 − an − 2n ≡ can +1 − can ≡ 0
(mod 2).
This means that can +1 6= can + 1 and the number can + 1 has to belong to either of the first two sequences. The inequalities an < bn < can < an+1 < bn+1 imply that can + 1 = an+1 , n ∈ N, and, by (1), an+1 = an + 2n + 1, n ∈ N. (2) Next we prove that a1 = 1. Indeed, number 1 has to belong to one of the given sequences, and if a1 > 1 then c1 = 1 or b1 = 1. The latter case is impossible because b1 > a1 . Then we must have c1 = 1, and either c2 = 2, or a1 = 2 and ca1 = c2 = a1 + 2 = 4. In both cases we obtain a contradiction by setting n = 1 in (iv). This proves that a1 = 1, and, together with (2), defines a unique sequence {an }: an = an−1 +(2n−1) = an−2 +(2n−3)+(2n−1) = · · · = a1 +3+5+...+(2n−1) = n2 , Hence, a2010 = 20102 , b2010 = ca2010 − 1 = a2010 + 2 · 2010 − 1 = 20112 − 2,
c1936 = c442 = ca44 = a44 + 2 · 44 = 442 + 88 = 2024, a45 = 452 = 2025,
n ∈ N.
and all the integers between a45 and b45 = ca45 − 1 = a45 + 2 · 45 − 1 = a45 + 89 belong to the sequence {cn }. Hence, these integers have the form c1936+k = a45 + k,
k = 1, 2, ..., 88,
and c2010 = c1936+74 = a45 + 74 = 2099. Answer. a2010 = 20102 , b2010 = 20112 − 2, c2010 = 2099. Solution 2. Denote by (∗) the trivial fact an < can derived at the beginning of the first solution. One can easily fill the sequences inductively. In fact, like in the first solution, we have a1 = 1. Now we will find the place for number 2. If a2 = 2, then by (iii) 2 = a2 > b1 , which is impossible. If c1 = 2, then by (ii) we have 2 = c1 = ca1 = b1 + 1, hence b1 = 1 which is also impossible. So the only way is to put b2 = 2. Then by (ii) c1 = ca1 = b1 + 1 = 3. n an bn cn
1 1 2 3
2
3
4
5
...
Now, because of (iv), we have c2 6= 4. Also, b2 6= 4, because otherwise by (∗) and (ii) a2 < ca2 = b2 + 1 = 5 and there is no number left for a2 . So we have a2 = 4. Then by (iii) a3 6= 5. Also, b2 6= 5, because otherwise by (ii) c4 = ca2 = b2 + 1 = 6 and there are no numbers left for c2 , c3 . So we have c2 = 5. Using the same arguments we derive a3 6= 6 and b2 6= 6, hence, c3 = 6. Now, a3 6= 7 (by (iii)). Also, c4 6= 7, because otherwise by (ii) 7 = c4 = ca2 = b2 + 1, and this leads to b2 = 6, which is not true. Hence, b2 = 7. Then c4 = ca2 = b2 + 1 = 8. n 1 2 3 4 5 ... an 1 4 bn 2 7 cn 3 5 6 8 Now, we can repeat the arguments from the last paragraph: Because of (iv) we have c5 6= 9. By (∗) and (ii) we have b3 6= 9 (otherwise a3 < ca3 = b3 + 1 = 10 and there is no number left for a3 ). So we have a3 = 9. By (iii), a4 6= 10. By (ii), c9 = ca3 = b3 + 1, therefore b3 6= 10 (otherwise there are no numbers left for c5 , . . . , c8 ). So we have c5 = 10. Similarly a4 6= 11, b3 6= 11
=⇒
c6 = 11,
=⇒
c7 = 12,
a4 6= 13, b3 6= 13
=⇒
c8 = 13.
a4 6= 12, b3 6= 12
Finally, a4 6= 14 (by (iii)), c9 6= 14 (otherwise by (ii) 14 = c9 = ca3 = b3 + 1, and this leads to b3 = 13, which is not true). Hence, b3 = 14 and c9 = ca3 = b3 + 1 = 15. n an bn cn
1 1 2 3
2 4 7 5
3 9 14 6
4
5
6
7
8
9
8
10
11
12
13
15
...
We formulate the claim which can be easily proved by induction. (We will skip the formal proof. However, it is just an obvious generalization of the last two paragraphs.) For k ∈ N and i = 1, 2, . . . , 2k − 2, we have ak = k 2 , bk = k 2 + 2k − 1,
c(k−1)2 +i = k 2 + i, ck2 = k 2 + 2k. The rest is straightforward: a2010 = 20102 ,
b2010 = 20102 + 2 · 2010 − 1,
c2010 = c442 +74 = 452 + 74 = 2099.
Problem T-2. For each integer n ≥ 2, determine the largest real constant Cn such that for all positive real numbers a1 , . . . , an , we have a21 + · · · + a2n ≥ n
a1 + · · · + an n
2
+ Cn · (a1 − an )2 .
Solution. Define xij = ai − aj for 1 ≤ i < j ≤ n. After multiplication with n2 , the difference of the squares of quadratic and arithmetic mean equals n2 (QM2 − AM2 ) = n(a21 + · · · + a2n ) − (a1 + · · · + an )2 = (n − 1) =
X i
x2ij = x21n +
n−1 X
(x21i + x2in ) +
i=2
X
n X i=1
a2i −
X
2ai aj =
i
x2ij .
1
Here the last sum is clearly non-negative. By the AM-QM inequality the sum in the middle is at least n−1 n−1 X n−2 2 1X 2 2 · x1n . (x1i + xin )2 = (x1i + xin ) ≥ 2 2 i=2
i=2
Hence we finally get
n2 (QM2 − AM2 ) ≥ x21n +
n−2 2 n · x1n = · (a1 − an )2 2 2
with equality if and only if a2 = · · · = an−1 = 21 (a1 + an ). The largest such constant is therefore 1 C= . 2n
Problem T-3. In each vertex of a regular n-gon there is a fortress. At the same moment each fortress shoots at one of the two nearest fortresses and hits it. The result of the shooting is the set of the hit fortresses; we do not distinguish whether a fortress was hit once or twice. Let P (n) be the number of possible results of the shooting. Prove that for every positive integer k ≥ 3, P (k) and P (k + 1) are relatively prime. Solution. Let us denote each hit fortress by a black dot and each undamaged one with a white dot. Then P (n) is the number of colourings of n dots distributed on the circle with black and white colours in such a way, that no two white dots have exactly one dot in between them. The proof of this bijectivity is straightforward: If there are two white dots with exactly one dot in between, then obviously the fortress in between can not shoot, which is not permitted. On the other hand, if there are no such two white dots, then each fortress can shoot at least one black dot and to ensure that every black dot will be hit, we can force the one in the clockwise direction to shoot at it. If n is odd, then P (n) is equal to the number K(n) of colourings of n dots on a circle with black and white colours in such a way, that no two neighbouring dots have white colour (we define the neighbouring dots to be the dots which have exactly one other dot in between them). For n even, with the same definition of neighbours, the circle splits into two circles with n/2 dots, and we have P (n) = K(n/2)2 . For K(n) it is easy to derive a recurrence formula K(n) = K(n − 1) + K(n − 2). In fact, the number of legal colourings with n-th dot being black is equal to the number of legal colourings of n − 1 dots (just put the black dot in between the first dot and the (n − 1)-th dot) plus the number of colourings of n − 1 dots with no two neighbouring white dots except for the first and (n − 1)-th (we can put the black dot in between two white dots to obtain legal colouring). The latter case gives the same number as the number of legal colouring with n − 2 dots having the first dot white (just span two white dots into one white). On the other hand, the number of legal colourings with n-th dot being white is equal to the number of colourings of n − 1 dots with no two neighbouring white dots and with the first and (n − 1)-th dot black (we can put the white dot only in between to black dots), which is equal to the number of legal colouring with n − 2 dots having the first dot black (again, span two black dots into one black). Together, we have K(n) = K(n − 1) + Kw (n − 2) + Kb (n − 2) = K(n − 1) + K(n − 2), where Kw and Kb stands for the number of legal colourings with first dot white and black respectively. Moreover we can directly count K(2) = 3, K(3) = 4, K(4) = 7, which suggests K(2) = F (4) − F (0),
K(3) = F (5) − F (1),
K(4) = F (6) − F (2)
and we can easily prove by the induction K(n) = F (n + 2) − F (n − 2), where F (k) stands for the k-th term of the Fibonacci sequence (F (0) = 0, F (1) = F (2) = 1, . . . ). Further (K(2), K(3)) = 1, and for n ≥ 3 we have (K(n), K(n − 1)) = (K(n) − K(n − 1), K(n − 1)) = (K(n − 2), K(n − 1)) = · · · = 1.
Similarly we show that for each even n = 2a the number P (n) = K(a)2 is relatively prime both to P (n + 1) = K(2a + 1) and P (n − 1) = K(2a − 1): (K(a), K(2a + 1)) = (K(a), F (2)K(2a) + F (1)K(2a − 1)) =
= (K(a), F (3)K(2a − 1) + F (2)K(2a − 2)) = . . .
· · · = (K(a), F (a + 1)K(a + 1) + F (a)K(a)) = (K(a), F (a + 1)) = = (F (a + 2) − F (a − 2), F (a + 1)) =
= (F (a + 2) − F (a + 1) − F (a − 2), F (a + 1)) =
= (F (a) − F (a − 2), F (a + 1)) = (F (a − 1), F (a + 1)) = = (F (a − 1), F (a)) = 1
(K(a), K(2a − 1)) = (K(a), F (2)K(2a − 2) + F (1)K(2a − 3)) =
= (K(a), F (3)K(2a − 3) + F (2)K(2a − 4)) = . . .
· · · = (K(a), F (a)K(a) + F (a − 1)K(a − 1)) = (K(a), F (a − 1)) =
= (F (a + 2) − F (a − 2), F (a − 1)) = (F (a + 2) − F (a), F (a − 1)) = = (F (a + 2) − F (a + 1), F (a − 1)) = (F (a), F (a − 1)) = 1,
which finishes the proof.
Problem T-4. Let n be a positive integer. A square ABCD is partitioned into n2 unit squares. Each of them is divided into two triangles by the diagonal parallel to BD. Some of the vertices of the unit squares are colored red in such a way that each of these 2n2 triangles contains at least one red vertex. Find the least number of red vertices. Solution. The least number of red vertices is (n + 1)2 . 3 First, we define a colouring and count the number of red vertices. In what follows it will be shown that the number of red vertices, obtained by this colouring, is indeed minimal. It is convenient to replace the square by rhombus ABCD as in this case the isosceles rightangled triangles are equilateral. Cover the rhombus by regular unit hexagons in such a way, that A lies in the vertex of a hexagon (figure 1). Colour the center of each hexagon red. Clearly each equilateral unit triangle belongs to one of the hexagons and thus contains a red point. Hence this colouring satisfies the required conditions. D
A
C
B
Figure 1: Covering with regular hexagons. Denote by an the number of vertices which are coloured red. Let A1 , A2 , . . . , An−1 be the points on AB such that AA1 = A1 A2 = · · · = An−2 An−1 = An−1 B = 1. Similarly define points B1 , . . . , Bn−1 on BC, points C1 , . . . , Cn−1 on CD, and D1 , . . . , Dn−1 on DA. Each of the n points on the line A1 Bn−1 is red (figure 1 and 2). The parallel lines A2 Bn−2 and A3 Bn−3 have no red points while line A4 Bn−4 contains 3 points less than A1 Bn−1 , that is n − 3. All of them are red. Similarly, red points lie on lines A7 Bn−7 , A10 Bn−10 and so on. The number of red points each time decreases by 3. On the other side of the diagonal AC, we have n − 1 red points on the line C2 Dn−2 , n − 4 points on C5 Dn−5 and so on. Hence the number of red points equals an = n + (n − 3) + (n − 6) + . . . + (n − 1) + (n − 4) + (n − 7) + . . . .
When n is of the form n = 3k+1, we have an = 31 n(n+2), for n = 3k+2 we get an = 13 (n+1)2 and n = 3k + 3 implies an = 13 n(n + 2). In general, an = b 13 (n + 1)2 c.
D
C Bn−1 Bn−2
D b
b
b
b
b
b
b
A1
Bn−2
b
b
A4
b
B
An−2 An−1
b
b
B1
b
A2 A3
b
C
Bn−1
b
b
A
b
b
b
b
b
Figure 2: The value of an . b
A
b
b
b
B1 b
b
...
B
An−2Aneeded. A1 number A2 A3of red points n−1 Let bn denote the least Clearly, b1 = 1. Consider a triangle with side length 2 (the first picture on figure 3). The four unit triangles do not have a common vertex, hence at least two vertices beofcoloured. Each of the small marked Slika 2: Themust value an disjoint triangles in the second and third picture (figure 3) must contain at least one red vertex and the bigger marked triangle at least two. This implies that b ≥ 2 + 1 = 3 and vertice and the bigger marked triangle at least two. This implies that2b2 ≥ 2 + 1 = 3 and b3 ≥ 1 + 1 + 1 + 2 = 5. b3 ≥ 1 + 1 + 1 + 2 = 5.
Slika 3: 3:The vertices for fornn==2 2and andn n= = Figure Theminimum minimumnumber number of of red red vertices 3. 3 It has thus been shown that for n = 1, 2 and 3, bn ≥ an , which means that bn = an . The Itrest by induction. The if bthat an−3 then haswill thusfollow been shown that for n = 1, 2,next and step 3, bn will ≥ anshow , whichthat means an ., The n−3 b= n = bnrest = will an . follow by induction. The next step will show that if bn−3 = an−3 , then bn = an . Let in figure figure at least least +(2k (2k+ 1)b22red red verticesare are n−3+ Letnn= = 3k 3k+ +2. 2. As As demonstrated demonstrated in at h 4,4, i bbn−3 h+1)b i vertices 2 −4n+4 2 +2n+1 n n needed. That That is, needed. is, bn ≥ an−3 + (2k + 1) · 3 = + 2n − 1 = = an , hence 3 3 bn = an . (n − 2)2 (n + 1)2 bn ≥ bn−3 + (2k + 1) · 3 =
+ 2n − 1 =
3
= an ,
3
hence bn = an . If n = 3k + 3, we can estimate that
(n − bn ≥ bn−3 + 2kb2 + 2 + 1 + 1 + 1 = 3
b
b
b
2)2
b
(n + 1)2 + 2(n − 3) + 5 = 3 b
b
= an .
Slika 3: The minimum number of red vertices for n = 2 and n = 3 It has thus been shown that for n = 1, 2 and 3, bn ≥ an , which means that bn = an . The rest will follow by induction. The next step will show that if bn−3 = an−3 , then bn = an . Let n = 3k + 2. As demonstrated in figure at least h 4, i bn−3 + (2kh+ 21)b2 red i vertices are 2 needed. That is, bn ≥ an−3 + (2k + 1) · 3 = n −4n+4 + 2n − 1 = n +2n+1 = an , hence 3 3 bn = an .
b
b
b
b
h
b
b
i Slika 4: Case n = 3k + 2 . Finally, taking n = 3k + 1, the+ last2.picture demonstrates Figure 4: Case h 2 thati i hn2 = 3k n −4n+4 n −4n+4 If n = 3k + 3, we can estimate that b ≥ b + 2kb + 2 + 1 + 1 + 1 = +2 2n − 1 = an . In all the cases it + bn ≥ bn−3 + (2(k − 1) + 1)b2 + 1 + 1 + 1 +n1 = n−3 3 3 follows that bn = an . 2(n − 3) + 5 =
n2 +2n+1 3
6
b
b
b
b
b
b
Slika 5: Case n = 3k + 3
Figure 5: Case n = 3k + 3.
b
i Finally, taking n2(n =−3k + 1,h the last picture demonstrates that 3) + 5 = n +2n+1 . Finally, taking n = 3k + 1, the last picture demonstrates that 3 h i + 2n − 12 = an . In all the cases it bn ≥ bn−3 + (2(k − 1) + 1)b2 + 1 + 1 + 1 + 1 = n −4n+4 3 (n − 2) (n + 1)2 follows bn = an .2 + 1 + 1 + 1 + 1 = bn ≥ bn−3 + (2(k −that 1) + 1)b + 2n − 1 = = an . 3 3 b
2
b
2
In all the cases it follows that bn = an . b
b
b
b
b
Slika 6: Case n = 3k + 1 b
b
b
b
Slika 5: Case n = 3k + 3
b
b
b
b
b
b
7
Slika 6: Case n = 3k + 1
Figure 6: Case n = 3k + 1.
Problem T-5. The incircle of the triangle ABC touches the sides BC, CA, and AB in the points D, E, and F , respectively. Let K be the point symmetric to D with respect to the incenter. The lines DE and F K intersect at S. Prove that AS is parallel to BC. Solution. Let S 0 be the intersection point of the line F K and the line parallel to BC passing through A. We need to show that S 0 , D and E are collinear. Let the tangent line to the given circle at the point K intersect AB at Q (it is parallel to BC). Then ∠AS 0 F = ∠QKF = ∠QF K and from that follows that AS 0 = AF = AE. But DC = EC and BC k AS 0 . Thus ∠CDE = ∠CED = ∠AES 0 = ∠AS 0 E = ∠AES and S 0 , D and E are collinear. C
D E S0
K A
Q
F
B
Solution 2. Let α = ∠BAC, β = ∠ABC and let I be the center by the incircle. Then ∠IDF = ∠IF D = β/2 = ∠AF S since ∠KF D = ∠AF I. Because ∠F DS = ∠F IE/2 = 90◦ − α/2 (AF IE is cyclic) and ∠AIF = 90◦ − α/2 we have 4AF I ∼ 4SF D. The ratio of similitude gives AF : SF = IF : DF and using ∠AF S = ∠IF D yields to 4AF S ∼ 4IF D which means that AF = AS = AE. Finally 4ASE ∼ 4CDE gives ∠SAE = 180◦ − α − β and consequently ∠BAS + ∠ABC = 180◦ . Solution 3. Let α = ∠BAC, then ∠F IE = 180◦ − α, ∠F DE = 90◦ − α/2. Because KD is a diameter of the incircle we have ∠DSF = α/2 and ∠KED = 90◦ . If γ = ∠BCA then ∠KDE = γ/2. We want to prove that AE : EC = SE : ED because then ∠CDE = ∠CED = ∠AES = ∠ASE which implies AS k CD. Calculation of these length in terms of the angles of the triangle ABC and its inradius gives α 2 γ EC = r cot 2 α γ α SE = EK cot = ED tan · cot 2 2 2 AE = r cot
and we are done.
Solution 4. We will use complex coordinates. Let I = (0), and let the incircle be a unit circle. Let D = (−i), E = (e), F = (f ). Then K = (i). The tangents in E, F are (w ¯ is the complex conjugate of w) te : z + e2 z¯ = 2e, z + f 2 z¯ = 2f
tf :
and the coordinates of A (intersection of te , tf ) a=
2ef , e+f
The lines DE : FK :
a ¯=
2 . e+f
z − ie¯ z = e − i, z + if z¯ = f + i
intersect in S = (z), z¯ =
(−i)(f − e + 2i) , e+f
z=
i(e − f − 2ief ) . e+f
To prove AS k BC, we calculate the slope1 of AS: t=−
a−z = · · · = −1 a ¯ − z¯
which indeed equals the slope of BC (= d2 = (−i)2 = −1).
1
Slope of the line with the equation z + t¯ z = s is t.
Problem T-6. Let A, B, C, D, E be points such that ABCD is a cyclic quadrilateral and ABDE is a parallelogram. The diagonals AC and BD intersect at S and the rays AB and DC intersect at F . Prove that ∠AF S = ∠ECD. Author: Adrian Satja Kurdija, Croatia
Solution. Let M and N be the feet of perpendicular from S to AB and CD, respectively. Then SM F N is cyclic since it has two opposite right angles. Therefore ∠AF S = ∠M F S = be aNcyclic Point is chosen so that ABDE of is atriangles ∠M N S.Problem. We need Let to ABCD prove ∠M S = quadrilateral. ∠ECD. This willE follow from similarity parallelogram, F is the intersection of lines AB and DC (B is between A and F , C is M SN and EDC. Since ABCD is cyclic, triangles ABS and DCS are similar. Lines SM between D and F ) and S is the intersection of the diagonals AC and BD. Prove that and SN ^AF are the altitudes, so SM : SN = AB : CD = ED : CD. Also, S = corresponding ^ECD. Solution.
∠M SN = 180◦ − ∠AF D = ∠EDF = ∠EDC
and therefore, 4M SN ∼ 4EDC as claimed.
SolutionLet2.M , Let S intersect the lines in the points X and Z respectively N beFfeet of perpendiculars fromAD S toand AB DE and CD, respectively. and denote α = ∠BAD, δ = ∠ADF , AB = a, and CD = c. It is sufficient to prove that Then SM F N is cyclic since it has two opposite right angles. Therefore ^AF S = the triangles ^M F CDE S = ^Mand N S.ZDF are similar, because then ∠ECD = ∠DZF = ∠AF S. These triangles have a common angle, we have to prove that ZD : F D = c : a. We need to prove ^M N S = ^ECD. This will follow from similarity of triangles M SN and EDC. Z E D Since ABCD is cyclic, triangles ABS and DCS are similar. SM and SN are corresponding altitudes, so |SM | : |SN | = |AB| : |CD| = |ED| : |CD|. C Also, XN = 180S◦ − ^AF D = ^EDF = ^EDC ^M SN = 180◦ − ^M F and therefore, △M SN ∼ △EDC as claimed.
A
B
F
The sine law in the triangle BF C gives CF : BF = sin δ : sin α, since ∠F BC = 180◦ − ∠ABC = ∠CDA = δ
and ∠F CB = 180◦ − ∠BCD = ∠BAD = α.
The Ceva’s theorem for the triangle AF D and the point S then gives 1=
DX AB CF DX a sin δ · · = · · . AX BF DC AX c sin α
(1)
From the similitude of triangles AF X and DZX we have ZD = AF · DX/AX, and from the sine law in the triangle AF D we have AF = F D · sin δ/ sin α. Therefore the desired ratio ZD : F D equals ZD AF DX sin δ DX c = · = · = , FD F D AX sin α AX a where (1) is used in the last equality. Solution 3. Let G be the intersection of lines AE and CD and let T be such point on AG that ∠BF S = ∠T F D. Then we have to prove that F T k CE or equivalently, that |CG|/|F G| = |EG|/|T G|. Since ABDE is a parallelogram, the triangles EDG and AF G are similar, therefore |EG| =
|AG| · |ED| (|BD| + |EG|) · |AB| = , |AF | |AF |
so |EG| =
|BD| · |AB| . |BF |
Similarly, since the triangles BF D and EDG are similar, we obtain |DG| =
|F D| · |AB| . |BF |
Since ABCD is cyclic, the triangles BF C and DF A are similar, so |AF | · |BF | = |CF | · |DF |. Now we compute |F D| · |AB| + |F D| − |F C| = |BF | |F D| · |AF | |AF | · |BF | |DF | |BF | − = |F A| · − = |BF | |F D| |BF | |DF |
|CG| = |DG| + |CD| =
and |F G| = |DG| + |F D| = so |CG| =1− |F G|
|F D| · |AF | , |BF |
|BF | |DF |
2
.
Now we will compute also the ratio |EG|/|T G|. By the construction of T we have ∠AF T = ∠SF C. Moreover, since ABCD is cyclic and AT k BD, we have also ∠T AF = ∠T AD + ∠DAF = ∠ADB + ∠BCF = ∠ACB + ∠BCF = ∠ACF, so the triangles AF T and CF S are similar. Since BF C and DF A are also similar, we get |BC|/|AD| = |CF |/|AF | = |CS|/|AT |. Since the angles ∠DAT and SCB are also equal, the triangles ADT and CBS are similar. Then ∠T DA = ∠SBC = ∠DAC, so DT k AS.
Then ASDT is a parallelogram and the triangles T DE and SAB are congruent. Therefore |T G| = |EG| + |T E| = |EG| + |BS|. The triangles ABS and DCS are similar, therefore |BS| |CS| |AC| − |AS| = = |AB| |CD| |CD|
and |AS| = |AC| −
|BS| · |CD| . |AB|
Since the triangles BF S and DF T are similar, we have |BS| |DT | |AS| |AC| |BS| · |CD| = = = − , |BF | |DF | |DF | |DF | |AB| · |DF | so |BS| =
|AC| · |AB| · |BF | . |AB| · |DF | + |BF | · |CD|
We use also the similarities of AF C and DF B and of AF D and CF B and obtain |BS| =
|BD| · |CF | · |AB| · |DF | |BD| · |AB| · |BF | = 2 2 2 |AB| · |DF | + |BF | · |DF | − |BF | · |AF | |DF |2 − |BF |2
and |T G| = |EG| + |BS| = Finally we can compute
|BD| · |AB| · |DF |2 . |BF | · (|DF |2 − |BF |2 )
|EG| |DF |2 − |BF |2 = =1− |T G| |DF |2 which we had to prove.
|BF | |DF |
2
=
|CG| , |F G|
Problem T-7. For a nonnegative integer n, define an to be the positive integer with decimal representation 1 |0 .{z . . 0} 2 |0 .{z . . 0} 2 |0 .{z . . 0} 1. n
n
n
Prove that an /3 is always the sum of two positive perfect cubes but never the sum of two perfect squares. Solution. First we prove that an /3 is never the sum of two perfect squares. Note that perfect squares give only remainders 0 and 1 when divided by four; therefore, integers expressible as the sum of two squares give only remainders 0, 1, and 2. On the other hand, the number an /3 gives remainder 3 because an gives remainder 1; hence it cannot be expressed as the sum of two perfect squares.2 After some experimentation, one finds the formula an = 3
10n+1 + 2 3
3
+
2 · 10n+1 + 1 3
3
.
This follows from the fact that an = 103n+3 + 2 · 102n+2 + 2 · 10n+1 + 1. Both the numbers in brackets are integers since 10n+1 ≡ 1 (mod 3). Thus an /3 can be expressed as the sum of two perfect cubes.
2
Another way to look at an /3 modulo 4 is to note that it always ends with 67.
Problem T-8. We are given a positive integer n which is not a power of 2. Show that there exists a positive integer m with the following two properties: (i) m is the product of two consecutive positive integers; (ii) the decimal representation of m consists of two identical blocks of n digits. Solution. First we prove a lemma. Lemma. Let x and k be integers greater than 2. If k is odd then the number xk + 1 is the product of two coprime numbers. Proof. Let m = gcd(x + 1, k). There is a polynomial Q(x) ∈ Z[x] such that Then
xk + 1 = (x + 1)(xk−1 − xk−2 + · · · + x2 − x + 1) = (x + 1) (x + 1)Q(x) + k . k
x + 1 = (x + 1)m · gives the required product because
(x + 1)Q(x) k + m m
k xk + 1 x3 + 1 (x + 1)Q(x) + = ≥ > 1. m m (x + 1)m (x + 1)2 The numbers 1 and 2 are powers of two, hence we may assume that n ≥ 3. Since n is not a power of two, it has an odd divisor greater than one; therefore, according to our lemma,3 there are coprime numbers a and b such that 10n + 1 = ab. Our task is to prove that there are numbers t and s such that m = (10n + 1)t = abt = s(s − 1). First, we show that there is a positive integer s divisible by a which satisfies s ≡ 1 (mod b). Consider the numbers 0, a, 2a, . . . , (b − 1)a. These numbers give mutually different remainders modulo b since a and b are coprime. Therefore, one of them gives remainder 1 and we take s to be this number. Similarly we can pick a number s0 divisible by b which satisfies s0 ≡ 1 (mod a). The numbers s and s0 are positive and smaller than 10n . Therefore, s(s − 1) and s0 (s0 − 1) are both divisible by ab and smaller than 102n . Moreover, s + s0 ≡ 1 (mod ab). The number s + s0 is greater than 1 and smaller than 2 · 10n . Hence s + s0 = ab + 1. Therefore, one of the numbers s and s0 is greater than 5 · 10n−1 . Then one of the numbers s(s − 1) or s0 (s0 − 1) is greater than 25 · 102n−2 , thus it has 2n digits. This number has all the required properties. Comment. Instead od using congruences we can also look at Diophantine equations ax = by + 1 and ax = by − 1. Both have solutions with 0 < x < b and 0 < y < a and xy > 10n−1 for one of them. 3
We can avoid using the lemma by exploiting the Mihailescu’s theorem, first known as Catalan’s Conjecture; it was proved in 2002. It says that the only solution of the equation xa − y b = 1 in positive integers greater than one is 32 − 23 . This implies that if 10n + 1 is a power of a prime then it is a prime. This cannot happen since n has an odd divisor.
PROBLEMS AND SOLUTIONS 5th Middle European Mathematical Olympiad Varaˇ zdin, Croatia, September 2011
ALGEBRA
I 1 (Vjekoslav Kovaˇ c, Croatia) Initially, only the integer 44 is written on a board. An integer a on the board can be replaced with four pairwise different integers a1 , a2 , a3 , a4 such that the arithmetic mean 1 4 (a1 + a2 + a3 + a4 ) of the four new integers is equal to the number a. In a step we simultaneously replace all the integers on the board in the above way. After 30 steps we end up with n = 430 integers b1 , b2 , . . . , bn on the board. Prove that b21 + b22 + · · · + b2n > 2011 . n
First solution Let us first prove an auxiliary statement. Lemma. If a1 , a2 , a3 , a4 are four different integers such that their average a = (a1 + a2 + a3 + a4 )/4 is also an integer, then a21 + a22 + a23 + a24 5 − a2 > . 4 2 Proof. Note that the expression on the left hand side can be transformed as a21 + a22 + a23 + a24 − a2 4 a2 + a22 + a23 + a24 − 8a2 + 4a2 = 1 4 a21 + a22 + a23 + a24 − 2a(a1 + a2 + a3 + a4 ) + 4a2 = 4 (a1 − a)2 + (a2 − a)2 + (a3 − a)2 + (a4 − a)2 = . 4 Now, a1 − a, a2 − a, a3 − a, a4 − a are four different integers that add up to 0. We claim that sum of their squares is at least 10. If none of these integers is 0, then that sum is at least 12 + (−1)2 + 22 + (−2)2 = 10. On the other hand, if one of the integers is 0, than the remaining three cannot be only from the set {1, −1, 2, −2}, because no three different elements of that set add up to 0. Therefore, the sum of their squares is at least 32 + 12 + (−1)2 = 11. This completes the proof of the lemma. Returning to the given problem, we denote by Sk the average of squares of the numbers on the board after k steps. More precisely, Sk =
b2k,1 + b2k,2 + · · · + b2k,4k 4k
,
where bk,1 , bk,2 , . . . , bk,4k are the numbers appearing on the board after the operation is performed k times. Applying the above lemma to each of the numbers, adding up these inequalities, and dividing by 4k , we obtain Sk+1 − Sk > 52 , so in particular S30 > S0 + 30 ·
5 5 = 442 + 30 · = 2011 . 2 2
1
Second solution (by Michal Zaj¸ ac, Poland) Let a0,1 = 44 and let ai,1 , ai,2 , . . . , ai,4i be number written on the board after i steps. In (i + 1)-st step we replace the number ai,k with ai+1,4k−3 , ai+1,4k−2 , ai+1,4k−1 and ai+1,4k . We denote 4 P
Si =
a2i,j
j=1
4i
.
We want to prove that Si+1 > Si + 2.5, with equality occuring when each number a is replaced by (a − 2, a − 1, a + 1, a + 2). For a given number a, let (b1 , b2 , b3 , b4 ) be an arbitrary quadruple of integers that satisfy the conditions that b1 + b2 + b3 + b4 = 4a and b1 > b2 > b3 > b4 . We will prove that (b1 , b2 , b3 , b4 ) majorizes (a + 2, a + 1, a − 1, a − 2). First we conclude that b1 > a + 2, otherwise b1 + b2 + b3 + b4 6 (a + 1) + a + (a − 1) + (a − 2) < 4a. Next, it holds that b1 + b2 > (a + 2) + (a + 1) = 2a + 3. Otherwise, it holds that b1 + b2 6 2a + 2 and thus b2 6 a, b3 6 a − 1 and b4 6 a − 2. This implies that b1 + b2 + b3 + b4 6 4a − 1 < 4a, which is false. Finally, in order to prove that b1 + b2 + b3 > 3a + 2, which is equivalent to b4 6 a − 2, we assume otherwise: b4 > a − 1 and we arrive to contradiction in the same way as in the first case (in this case the sum is strictly bigger than 4a). Thus, we have proved that (b1 , b2 , b3 , b4 ) ≻ (a + 2, a + 1, a − 1, a − 2). The function f (x) = x2 is convex (because f ′′ (x) = 2 > 0) and by Karamata inequality it holds that: b21 + b22 + b23 + b24 > (a + 2)2 + (a + 1)2 + (a − 1)2 + (a − 2)2 = 4a2 + 10. Similar to first solution, we conclude that Si+1 > Si +2.5 and finally by inductive argument: S30 > S0 + 30 · 2.5 = 2011.
2
T 1 (Ton´ ci Kokan, Croatia) Find all functions f : R → R such that the equality y 2 f (x) + x2 f (y) + xy = xyf (x + y) + x2 + y 2 holds for all x, y ∈ R, where R is the set of real numbers.
First solution Substituting y = 0 we find that x2 f (0) = x2 holds for all real numbers x which implies f (0) = 1. Let us introduce a new function g : R → R given by g(x) = f (x) − 1. Equation from the problem becomes y 2 g(x) + x2 g(y) = xy g(x + y), (1) while g(0) = 0. Denoting c = g(1) and introducing another function h : R → R defined by h(x) = g(x)−cx, we obviously get h(0) = h(1) = 0, whereas the equation that must be satisfied is now y 2 h(x) + x2 h(y) = xy h(x + y).
(2)
Substituting x = y = 1 in the last equation we get h(2) = 0, while another substitution x = −1, y = 1 gives h(−1) = 0. Let us suppose that there exists a real number y0 such that h(y0 ) ̸= 0. Putting x = 1, y = y0 + 1 in (2) we get: h(y0 + 1) = (y0 + 1)h(y0 + 2),
or
h(y0 + 2) =
h(y0 + 1) . y0 + 1
(3)
On the other hand, substituting x = 2, y = y0 in (2) gives 4h(y0 ) = 2y0 h(y0 + 2),
i.e.
h(y0 + 2) =
2h(y0 ) . y0
(4)
Finally, putting x = 1, y = y0 in (2) leads to: h(y0 ) = y0 h(y0 + 1),
or
h(y0 + 1) =
h(y0 ) . y0
(5)
From (3), (4) and (5) it follows that y0 = − 12 . However, substituting x = y = − 21 in (2) and using h(−1) = 0 we arrive at h(− 12 ) = 0, which is a contradiction. We conclude that h(x) = 0 holds for all x ∈ R and thus f (x) = cx + 1 is the only solution. We check that this really is the solution for every real number c.
3
Second solution (by Matija Baˇsi´ c, coordinator) We define function h as in the first solution of the problem. Hence, we have h(0) = 0, h(1) = 0, y 2 h(x) + x2 h(y) = xyh(x + y). (∗) Substituting y = x :
2x2 h(x) = x2 h(2x), ∀x, or h(2x) = 2h(x) for all x.
Thus h(2) = 0. Substituting y = −x :
x2 (h(x) + h(−x)) = x2 h(0) = 0, ∀x, h(−x) = −h(x),
which gives
∀x.
Thus h(−1) = 0. Put y = 1 in (∗) :
h(x) + x2 h(1) = xh(x + 1)
i.e.
h(x) = xh(x + 1)
(1)
h(x + 1) = (x + 1)h(x + 2)
(2)
In (1) we change x → x + 1
Put y = 2 in (∗) :
4h(x) + x2 h(2) = 2xh(x + 2)
i.e.
2h(x) = xh(x + 2)
(3)
Now we conclude 2(x + 1)h(x) = (3) = x(x + 1)h(x + 2) = (2) = xh(x + 1) = (1) = h(x) Therefore, 2(x + 1)h(x) = h(x),
∀x
so h(x) = 0 or 2(x + 1) = 1 for all x. Obviously, h(x) = 0 for all x ̸= − 21 . Moreover, 2h(− 21 ) = h(2 · (− 12 )) = h(−1) = 0 so h(− 12 ) = 0 holds as well. We have proved that h(x) = 0 for all x ∈ R, hence, g(x) = cx, f (x) = cx + 1. Direct check shows that f (x) = cx + 1 is the solution of the given functional equation for all c ∈ R.
4
ˇ Third solution (by Klemen Sivic, Slovenian leader) As in the first solution we obtain f (0) = 1 and we define g(x) = f (x) − 1. Then g(0) = 0 and y x g(x + y) = g(x) + g(y) for x, y ̸= 0. (1) x y Therefore y+z x y+z xy xz g(x + y + z) = g(x) + g(y + z) = g(x) + g(z) + g(y) x y+z x z(y + z) y(y + z) for all nonzero x, y and z such that z ̸= −y. However, since the left side of the above equation is symmetric in x and z, we obtain that y+z xy xz y+x zy xz g(x) + g(z) + g(y) = g(z) + g(x) + g(y) x z(y + z) y(y + z) z x(y + x) y(y + x) for all nonzero x, y and z such that y ̸= −x and y ̸= −z. In this equation we set y = z = 1 and we obtain 2g(x) 1 x + x g(1) = g(x) + x + 1 + g(1) for all x ̸= 0, −1, x x(x + 1) x+1 i.e.
2x + 1 2x + 1 g(x) = g(1) for all x ̸= 0, −1. x(x + 1) x+1
Therefore
1 g(x) = g(1) x for all x ̸= 0, −1, − . 2 Clearly, the above equation holds also for x = 0. If we set x = 1 and y = −1 into the equation (1), we obtain g(−1) − g(1), and if we set x = y = − 12 , then we obtain −g(1) = g(−1) = 2g − 21 , therefore g − 12 = − g(1) 2 . Hence g(x) = g(1) x for all x ∈ R. f (1) = a can be arbitrary, therefore all solutions are functions g(x) = ax, or equivalently, f (x) = ax + 1 for all x ∈ R, where a ∈ R is arbitrary.
Fourth solution (by team Hungary) Similar to the first solution, we introduce the function g(x) and prove that g(0) = 0 and g(−x) = −g(x). Inserting y = 1 and y = −1 into the equation for g(x) we ge: g(x) + x2 g(1) = x g(x + 1),
(1)
g(x) + x g(−1) = −x g(x − 1).
(2)
2
Inserting x + 1 instead of x into (2) we get: g(x + 1) + (x + 1)2 g(−1) = −(x + 1) g(x).
(3)
From (2) and (3) we get represent g(x + 1) in two ways: g(x + 1) =
g(x) + x2 g(1) = −(x + 1) g(x) − (x + 1)2 g(−1) x
Solving for g(x) and using g(−1) = −g(1) we get:
for x ̸= 0.
g(x) x2 + x + 1 = g(1) x x2 + x + 1 . Since x2 + x + 1 > 0 for all x ∈ R we get g(x) = g(1) x and f (x) = cx + 1. Direct check shows that this is, indeed, the solution of the given functional equation for all c ∈ R.
5
ˇ T 2 (Kristina Ana Skreb, Croatia) Let a, b, c be positive real numbers such that b c a + + = 2. 1+a 1+b 1+c Prove that
√
a+
√ √ b+ c 1 1 1 >√ +√ +√ . 2 a c b
First solution Note that the condition of the problem is equivalent to 1 1 1 + + = 1. 1+a 1+b 1+c
(1)
We want to prove that
⇐⇒
√ √ √ a+ b+ c 1 1 1 >√ +√ +√ 2 a c b √ √ √ 1 1 1 ⇐⇒ a+ b+ c>2 √ + √ + √ √ a1 b 1c 1 √ √ 1 1 1 a+ √ + b+ √ + c+ √ >3 √ + √ + √ a a c b 1 c 1 b a+1 b+1 c+1 1 √ + √ + √ >3 √ +√ +√ ⇐⇒ a c a c b b
(2)
From (1) we see that at most one of the numbers a, b, and c can be strictly smaller than 1 1 1 1. (Otherwise, we would have 1+a + 1+b + 1+c > 12 + 12 = 1.) Without loss of generality we can take a > b > c. Case 1. a > b > c > 1 We have √ √ √ √ a ab − 1 > b ab − 1
=⇒
a+1 b+1 √ > √ , a b
√ √ √ √ b bc − 1 > c bc − 1
=⇒
b+1 c+1 √ > √ . c b
and also
Case 2. a > b > 1, and c < 1 √ > b+1 √ . The same way as in Case 1, we get a+1 a b Since a, b, and c are positive numbers, (1) implies 1 c 1 61− = 1+b 1+c 1+c And this gives √
Ê b
! r
b −1 c
>
1 c
Ê
=⇒
bc > 1
!
b −1 c
6
=⇒
=⇒
1 b> . c
b+1 c+1 √ > √ . c b
We have showed that a>b>c
=⇒
a+1 b+1 c+1 √ > √ > √ a c b
(3)
a>b>c
=⇒
1 1 1 6 6 1+a 1+b 1+c
(4)
and
hold. Now (3), (4) and the Chebyshev inequality imply
a + 1
a+1 b+1 c+1 √ + √ + √ = a c b
√ 1a >3 √ a
b+1 c+1 + √ + √ b c 1 1 +√ +√ , c b
1 1 1 + + 1+a 1+b 1+c
which is exactly (2).
ˇ Second solution (by Klemen Sivic, Slovenian leader) We make a substitution x =
1 a+1 , y
=
1 b+1 , z
=
1 c+1 .
The condition
1 1 1 + + =1 1+a 1+b 1+c is then equivalent to x + y + z = 1, and the original variables can be expressed as a = c = x+y z . The inequality
1 x
−1 =
1−x x
=
y+z x ,
b =
x+z y
and
√ √ √ 1 1 a+ b+ c 1 >√ +√ +√ 2 a c b is then equivalent to
r
x+y + 2z
r
y+z + 2x
Ê
z+x > 2y
Ê
2x + y+z
Ê
2y + z+x
Ê
2z . y+x
We will prove that this inequality holds for all positive numbers x, y and z. We make a substitution p = x + y, q = y + z, r = z + x. Then p, q and r are sides of a triangle and we have to prove that
r
p + q+r−p
r
q + r+p−q
r
r > p+q−r
r
p+q−r + r
Ê
q+r−p + p
Ê
r+p−q . (1) q
Since p, q and r are sides of a triangle, we can write p = 2R sin α, q = 2R sin β and r = 2R sin γ, where R is the circumradius and α, β and γ angles of the triangle with sides p, q and r. Then
r
p = q+r−p
Ì
=
Ê
sin α = sin β + sin γ − sin α
s
sin(β + γ) = sin β + sin γ − sin(β + γ)
β+γ 2 sin β+γ 2 cos 2 β−γ β+γ 2 sin β+γ 2 (cos 2 − cos 2 )
7
Ì
=
sin α2 2 sin β2 sin γ2
.
Similarly we compute the other terms in (1), therefore (1) is equivalent to
Ì
Ì
sin α2
+
2 sin β2 sin γ2
sin β2 + 2 sin γ2 sin α2
Ì >
or equivalently, to
α 2
2 sin sin sin γ2
β 2
Ì
sin γ2 2 sin α2 sin β2
Ì
+
2 sin β2 sin γ2 + sin α2
Ì
2 sin γ2 sin α2
sin β2
,
α β γ α β α γ β γ sin + sin + sin > 2 sin sin + sin sin + sin sin 2 2 2 2 2 2 2 2 2 α β γ 2 2 α 2 β 2 γ = sin + sin + sin ) − (sin + sin + sin . 2 2 2 2 2 2 Since sin x is concave function on (0, π), Jensen’s inequality implies that sin
α β γ α+β+γ π 3 + sin + sin 6 3 sin = 3 sin = . 2 2 2 6 6 2
Therefore
β γ 2 α β γ 2 α sin + sin + sin sin + sin + sin > 2 2 2 3 2 2 2 α α β γ 2 β γ > sin + sin + sin − sin2 + sin2 + sin2 , 2 2 2 2 2 2 where at the end we used the arithmetic-quadratic mean. Therefore the inequality is proved.
Third solution (by team Croatia) Let a = 2x, b = 2y, c = 2z. Then our condition is equivalent to : x y z + + =1 1 + 2x 1 + 2y 1 + 2z
1 1 1 + + = 2. 1 + 2x 1 + 2y 1 + 2z
⇐⇒
and we need to prove that √ √ 1 1 1 √ x+ y+ z > √ + √ + √ , x y z which is equivalent to :
Xx−1 cyc
√ >0 x
⇐⇒
X cyc
x − 1 2x + 1 · √ > 0. 2x + 1 x
Since this inequality is symmetric, we can assume x > y > z. We prove that then:
and
x−1 y−1 z−1 > > 2x + 1 2y + 1 2z + 1
(1)
2x + 1 2y + 1 2z + 1 √ > √ > √ . x y z
(2)
In order to prove (1) we note that: y−1 x−1 > 2x + 1 2y + 1
⇐⇒
which holds. The same argument holds for y and z. 8
3x > 3y,
In order to prove (2) we factor the inequality in the following equivalent way: √ √ √ ( x − y)(2 xy − 1) > 0. √ √ √ By the assumption, x − y > 0 thus we need to prove that 2 xy − 1 > 0. Assume the opposite, ie. that 4xy < 1. Then: 1 1 2(1 + x + y) 1 − 4xy + = =1+ > 1, 1 + 2x 1 + 2y 1 + 2(x + y) + 4xy (1 + 2x)(1 + 2y) which contradicts the condition. We have proven that triplets
x−1
y−1 z−1 , , 2x + 1 2y + 1 2z + 1
and
2x + 1 2y + 1 2z + 1 √ , √ , √ x y z
are ordered in the same way thus by Chebyshev inequality we have:
X x − 1 cyc
2x + 1 √ 2x + 1 x ·
>
1 X x − 1 X 2x + 1 √ · = 0. 3 cyc 2x + 1 cyc x
9
COMBINATORICS
I 2 (Tomislav Pejkovi´ c, Croatia) Let n > 3 be an integer. John and Mary play the following game: First John labels the sides of a regular n-gon with the numbers 1, 2, . . . , n in whatever order he wants, using each number exactly once. Then Mary divides this n-gon into triangles by drawing n − 3 diagonals which do not intersect each other inside the n-gon. All these diagonals are labeled with number 1. Into each of the triangles the product of the numbers on its sides is written. Let S be the sum of those n − 2 products. Determine the value of S if Mary wants the number S to be as small as possible and John wants S to be as large as possible and if they both make the best possible choices.
Solution (by Rudi Mrazovi´ c, coordinator) For n = 3 the answer is 6. Suppose n > 4. It is obvious that in each triangulation there are at least two triangles that share two sides with the polygon. We will prove that it is always best for Mary to choose a triangulation for which there is no more than two triangles of this kind. We call a triangle in a triangulation bad if all of its sides are diagonals of the polygon. First we prove that Mary can choose an optimal triangulation that contains no bad triangles. Assume on the contrary that every optimal triangulation contains a bad triangle. For an optimal triangulation T let d(T ) be the length of the smallest side of all bad triangles in T . Among all optimal triangulations with minimal number of bad triangles let T0 be such that d(T0 ) is minimal. Consider a bad triangle ABC in T0 such that |AB| = d(T0 ). Let ABD be the other ø of the triangle of T0 that contains AB as one of its sides. Since D lies on the arc AB circumcircle of ABC that does not contain C and ^ACB is acute, we have |AD| < |AB| and |BD| < |AB|. Let T1 be the triangulation obtained from T0 by replacing AB with CD. If the sides AD and BD have labels a and b respectively, then S(T1 ) − S(T0 ) = a + b − ab − 1 = −(a − 1)(b − 1) 6 0. Because T0 is optimal triangulation, we conclude that T1 is also optimal. Since T0 has the minimal number of bad triangles at least one of the segments AD and BD should be a diagonal, but then d(T1 ) is less than d(T0 ) what is a contradiction. Now that we know that Mary can choose an optimal triangulation that contains no bad triangles, we easily conclude that in a such triangulation there are exactly two triangles that share two sides with the polygon. If we denote by x1 (respectively x2 ) the number of triangles that have exactly one (respectively two) of their sides being the sides of the polygon, then x1 + x2 = n − 2 and x1 + 2x2 = n, so x2 = 2. Mary’s strategy is to choose these two triangles so that the side of the polygon labeled with 1 is contained in one of these triangles and the side labeled with 2 is contained in the other.
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By this strategy Mary makes sure that
(
S 6 max
=
n(n + 1) − (1 + 2 + n + n − 1) + 1 · n + 2 · (n − 1), 2
n(n + 1) − (1 + 2 + n + n − 1) + 1 · (n − 1) + 2 · n 2 n2 + 3n − 6 . 2
)
On the other hand, John can force Mary to achieve exactly this bound by labeling the sides of the polygon in the following order 1, n − 1, 4, n − 3, 5, . . . , n − 2, 3, n, 2. Thus, the answer to our problem is S =
n2 + 3n − 6 , for each n > 3. 2
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T 3 (Viktor Harangi, Hungary) For an integer n > 3, let M be the set {(x, y) | x, y ∈ Z, 1 6 x 6 n, 1 6 y 6 n} of points in the plane. (Z is the set of integers.) What is the maximum possible number of points in a subset S ⊆ M which does not contain three distinct points being the vertices of a right triangle?
Solution We will prove that the maximal cardinality of S is 2n − 2. The set S = {1} × {2, . . . , n} ∪ {2, . . . , n} × {1} has cardinality 2n − 2 and it does not contain three distinct points that form a right triangle. We will show that any subset S ⊂ M which does not contain three distinct points that form a right triangle can have at most 2n − 2 points. For such set S consider its subsets: • Sx consists of those points P = (x, y) in S that have unique x coordinate, that is, there exists no y ′ ̸= y such that (x, y ′ ) ∈ S. • Sy consists of those points P = (x, y) in S that have unique y coordinate, that is, there exists no x′ ̸= x such that (x′ , y) ∈ S. We claim that S = Sx ∪ Sy . We prove this by contradiction. Assume ther exists a point P ∈ S \(Sx ∪Sy ). Since P ∈ / Sx , there exists Px ̸= P in S with the same x coordinate as P . Similarly, there exists Py ̸= P in S with the same y coordinate as P . Hence P, Px , Py ∈ S and ^Px P Py = 90◦ , a contradiction. Clearly, |Sx | 6 n, and if |Sx | = n, then S = Sx . The same holds for Sy . So, |S| = n or |Sx |, |Sy | 6 n − 1 and |S| 6 |Sx | + |Sy | 6 2n − 2. It follows that the cardinality of S is at most max(n, 2n − 2) = 2n − 2.
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T 4 (Vjekoslav Kovaˇ c, Croatia) Let n > 3 be an integer. At a MEMO-like competition, there are 3n participants, there are n languages spoken, and each participant speaks exactly three different languages. ¡ 2n ¤ Prove that at least of the spoken languages can be chosen in such a way that no 9 participant speaks more than two of the chosen languages. (⌈x⌉ is the smallest integer which is greater than or equal to x.)
First solution Consider the classifications of the set of n available languages into easy, medium, and hard languages. There are 3n possible classifications in total and we denote by S the set of all possible classifications. For each classification s ∈ S, let A(s) be the number of easy languages and let B(s) be the number of students who speak 3 easy languages. If we add up quantities A(s) over all possible classifications s ∈ S, the resulting sum will P be s∈S A(s) = n3n−1 . In order to verify that, we realize that the result should be the same for medium and hard languages too, but all three of these sums add up to 3
X
A(s) = number of classifications × number of languages = 3n · n .
s∈S
On the other hand, we use double counting to compute the sum of quantities B(s) over all possible classifications s ∈ S. For each student there are 3n−3 classifications for which he speaks 3 easy languages, as we only have the choice to classify each of the n − 3 languages that the student does not speak. In two ways, we count the cardinality of the set {(X, s) : for a classification s student X speaks 3 easy languages} to get the identity
X
B(s) = 3n · 3n−3 = n3n−2 .
s∈S
We claim that there exists a classification s ∈ S such that A(s) − B(s) > 2n 9 . If we assume on the contrary that A(s) − B(s) < 2n for all classifications s ∈ S, then summing over all 9 n 3 of them would give n3n−1 − n3n−2 =
X
A(s) −
s∈S
X s∈S
B(s) < 3n ·
2n , 9
i.e. 2n3n−2 < 2n3n−2 , which is a contradiction. Let us consider any classification s ∈ S of languages satisfying A(s) − B(s) > 2n 9 . We can first choose all A(s) easy languages. Then we find all B(s) students who can speak 3 of these languages, and for each of them we remove one of the languages the student speaks. This leaves us with a choice of at least 2n 9 languages. Remark: Classification of languages simply as easy or hard would not give the desired bound. It would lead to a choice of at least n8 languages only. Taking more than three language classes would not be a better strategy either.
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Solution (by Rudi Mrazovi´ c, coordinator) In this proof we will use probabilistic method. Let p ∈ [0, 1]. For each language, suppose we choose it with probability p and we make these decisions independently. 1 Let A be the number of chosen languages (i.e. the number of 1s in ω) and B the number of students whose all three languages are among chosen ones. Lets calculate the expectations 2 of these random variables.
2 X EA = E 4 X
3 1we have chosen the language l 5 =
language l
=
X
E [1we have chosen the language l ]
language l
P (we have chosen the language l) = np.
language l
"
X
EB = E
X
# 1student’s s languages are all chosen =
student s
=
X
E [1student’s s languages are all chosen ]
student s
P (student’s s languages are all chosen) = 3np3 .
student s
We will use the following obvious (and easily proved inequality). For arbitrary random variable X we have P(X > EX) > 0. For X = A − B we get
P(A − B > np − 3np3 ) > 0.
In this way we have proved that there is a choosing of languages such that A − B > np − 3np3 . For this choosing for each student that speaks three chosen languages remove one of them. In the end we are left with at least A − B (and thus np − 3np3 ) languages that do the job. Taking p = 31 we get what we need, i.e. we can choose at least ⌈ 2n 9 ⌉ such that no student speaks more than two of them.
Alternative approach (based on the solution by team Poland) We choose ⌈ n3 ⌉ languages uniformly and randomly. Similarly to the previous probabilistic solution we show that with positive probability the number of students that speak three of the chosen languages is less or equal to ⌊ n9 ⌋. Again, use the same trick of removing some of the languages to obtain at least ⌈ 2n 9 ⌉ of them such that no student speaks three of them.
1
Formally, we consider probability space ({0, 1}n , P({0, 1}n ), P) where P(ω) = pk(ω) (1 − p)n−k(ω) ,
for each ω ∈ {0, 1}n
where k(ω) is the number of 1s in ω. 2 The expectation of integer random variable X is the number EX =
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Pn k=0
kP(X = k).
GEOMETRY
I 3 (Nik Stopar, Slovenia) In a plane the circles K1 and K2 with centers I1 and I2 , respectively, intersect in two points A and B. Assume that ^I1 AI2 is obtuse. The tangent to K1 in A intersects K2 again in C and the tangent to K2 in A intersects K1 again in D. Let K3 be the circumcircle of the triangle BCD. Let E be the midpoint of that arc CD of K3 that contains B. The lines AC and AD intersect K3 again in K and L, respectively. Prove that the line AE is perpendicular to KL.
First solution (by Tomislav Pejkovi´ c, coordinator)
K1 K2
A
E B
L
C
D
K
K3 Since AD is tangent to K2 , it follows that ^ACB = ^DAB. Similarly, ^ADB = ^BAC. From this we have ^DBC = (^ADB+^DAB)+(^BAC +^ACB) = 2(^DAB+^BAC), hence ^DBC = 2^DAC.
ø we denote the angle ^XZY where Z is a point on the circle K3 such that X, Y, Z By XY are ordered counterclockwise. Since E is the midpoint of the arc CD and the points C, E, D, K are concyclic we have 1ø 1 ø = 1 (180◦ − ^CBD) = 90◦ − ^DAC. ^AKE = CD = (180◦ − DC) 2 2 2 This means that KE and AL are perpendicular. Analogously, LE and AK are perpendicular and E is the orthocenter of the triangle AKL. Hence AE and KL are perpendicular.
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ø and DC ø because it provides a convenient way of writRemark: We use the notation CD ing the solution in all cases regardless of the mutual position of the points A, D, L, C, K.
Second solution Since AD is tangent to K2 , it follows that ^ACB = ^DAB. Similarly, ^ADB = ^BAC. From this we have ^DBC = (^ADB+^DAB)+(^BAC +^ACB) = 2(^DAB+^BAC), hence ^DEC = ^DBC = 2^DAC. Since |ED| = |EC|, the point E is the circumcenter of ACD. Therefore |EC| = |EA| = |ED|. Because the points C, B, D, K are concyclic we have ^KDB = ^ACB. From this and the first arguments of this solution we have that |DK| = |AK|. Since we proved |EA| = |ED|, we conclude that the line KE is the bisector of the segment AD and therefore perpendicular to it. Analogously, LE and AK are perpendicular and E is the orthocenter of the triangle AKL. Hence AE and KL are perpendicular. Remark: The identity ^KDB = ^ACB holds in all cases regardless of the mutual position of the points A, D, L, C, K.
Third solution (by Karol Kaszuba, Poland) Let us apply inversion with respect to a circle with the center A and radius r. Denote the image of point X with X ′ . From the assumptions of the problem and well known facts about the inversion directly follows that AD′ B ′ C ′ is a parallelogram. From the definition of the image of the point by inversion we have |E ′ C ′ | = |EC|
r2 , |AE||AC|
|E ′ D′ | = |ED|
r2 . |AE||AD|
ø we obtain Dividing these two identities and using that E is the midpoint of the arc CD |E ′ C ′ | |EC| |AD| |AD| |AC ′ | |D′ B ′ | = · = = = . |E ′ D′ | |ED| |AC| |AC| |AD′ | |C ′ B ′ | We consider all points X with the property |D′ B ′ | |XC ′ | = . |XD′ | |C ′ B ′ | These points form the Apollonius circle and hence there are exactly two such points ù ′ D ′ . One of these is the point E ′ . intersecting the image of K3 , each on different arc C Since the point symmetric to B ′ with respect to the line C ′ D′ also lies on the same arc ù ′ D ′ as E ′ and lies on the mentioned Apollonius circle we conclude that E ′ is symmetric C to B ′ .
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This implies |E ′ C ′ | = |B ′ D′ | = |C ′ A′ | (first equality holds because of the symmetry the second because AD′ B ′ C ′ is a parallelogram) and similarly |E ′ D′ | = |D′ A′ |. Hence AC ′ E ′ D′ is a deltoid so AE ′ ⊥ C ′ D′ . This means that AE ′ contains the orthocenter of the triangle AC ′ D′ . It is well know that the orthocenter and circumcenter are isogonal conjugates (lying on the lines which are symmetric with respect to the angle bisector). On the other hand triangles AC ′ D′ and AL′ K ′ are inversely similar, so the circumcenter of AK ′ L′ lies on the same line through A as the orthocenter of AC ′ D′ . All of this shows that AE ′ pass through the circumcenter of AK ′ L′ , so AE is perpendicular to KL.
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T 5 (Michal Szabados, Slovakia) Let ABCDE be a convex pentagon with all five sides equal in length. The diagonals AD and EC meet in S with ^ASE = 60◦ . Prove that ABCDE has a pair of parallel sides.
First solution Let F be such that DEF is an equilateral triangle and the points B and F lay in the opposite half-planes determined by DE. Denote ^DAE = α. Then ^ADE = α.
A
F E
S
B
D
C Since ^ESD = 120◦ , we have ^DEC = 60◦ − α. Then ^SCD = ^ECD and ^ADC = ^SDC = 180◦ − ^SCD − ^DSC = 60◦ + α. Obviously ^ADF = 60◦ + α and because |F D| = |CD| we conclude that ADF ≃ ADC. Similarly, ^AEC = ^F EC = 120◦ − α, so ACE ∼ = F CE. From these two pairs of equal triangles we conclude |AF | = |AC| = |F C|, so both triangles DEF and ACF are equilateral. If E lies on the line AF or D lies on the line F C then |AC| = 2|ED| = |AB| + |BC| and B lies on AC, which is not possible. Therefore exactly one of the points D and E lays inside the triangle ACF . Without loss of generality, let it be the point E. The triangles AEF and ABC have their corresponding sides equal therefore AEF ∼ = ABC ◦ and this yields 60 = ^F AC = ^EAB, so |EB| = |AB|. Hence BCDE is a rhombus, i. e., ED ∥ BC.
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Second solution (by Matija Baˇsi´ c, coordinator) Let α be as in the first solution. In the same way we prove ^AEC = 120◦ − α and ^CED = 60◦ − α. Let F be the symmetric image of A with respect to CE. We get ^DEF = 120◦ − α − (60◦ − α) = 60◦ . Since |DE| = |AE| = |EF |, triangle DEF is equilateral. Because |AB| = |BC| = |DF | = |CD| the triangles ABC and CDF are congruent. If the point D is outside the triangle ACF then this implies that B and D are symmetric with respect to CE, so |BE| = |DE|. Hence BCDE is a rhombus and DE∥BC. If the point D is inside the triangle ACF then the point E is outside that triangle and we see in the similar way that F and C are symmetric with respect to AD and also B and E are symmetric with respect to AD. Hence |BD| = |DE| and ABDE is a rhombus, so DE ∥ AB.
Third solution (by Gerd Baron, Austrian leader) Define the point B ′ such that B ′ CDE is rhombus. If the pentagon AB ′ CDE is convex, denote ^EAD = ^EDA = α. Similarly to other solution we have ^AEB ′ = ^AED−^B ′ EC −^CED = 180◦ −2α−(60◦ −α)−(60◦ −α) = 60◦ . Since |AE| = |B ′ E|, we conclude that AB ′ E is equilateral. Points B and B ′ are on the same side of the line AC, so we conclude that B = B ′ , so DE ∥ AB. If the pentagon AB ′ CDE is not convex, denote the intersection of B ′ E and AD by F and ^DEC = ^DCE = β. Similarly to other solutions we have AEB ′ = 180◦ − ^EAF − ^EF A = 180◦ − ^EDA − (^F EC + ^F SE) = 180◦ − (60◦ − β) − (β + 60◦ ) = 60◦ . Since |AE| = |B ′ E|, we conclude that AB ′ E is equilateral. Let B ′′ be the symmetric image of B ′ with respect to AC. Then AB ′′ CB ′ is a rhombus and B = B ′′ , so we conclude B ′ C ∥ AB ′′ and hence DE ∥ AB.
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Fourth solution (by team Slovakia) Denote ^DEC = ^DCE = α and suppose that all five sides of the pentagon have length a. As in the previous solutions we see that ^SEA = 60◦ + α, ^SDC = 120◦ − α. Applying the law of sines to the triangles ASE and CSD implies |SA| =
a sin(60◦ + α) a sin(120◦ − α) = = |SC|. sin 60◦ sin 60◦
The triangle ASC is isosceles and ^ACS = ^CAS = 30◦ and we have |AC| =
√ 3 · |AS|.
The law of cosines applied to the triangle ABC gives √ a2 = a2 + 3|AS|2 − 2 3a · |AS| · cos(^ACB) 3|AS| from where we get cos(^ACB) = √ = sin(60◦ + α) = cos(30◦ − α). 2 3a ◦ Since 0 < ^ACB < 90 we have two possibilities. The first possibility is that ^ACB = 30◦ − α, so ^BCE = α = ^CED and hence BC ∥ ED. The second possibility is that ^ACB = α − 30◦ , so ^BAD = 60◦ − α = ^ADE and hence AB ∥ ED.
Fifth solution (by team Germany) We construct a point Q on the line SE such that ASQ is the equilateral triangle. As in the previous solutions it is easily seen that ^EAQ = ^DCS and since ^AQS = 60◦ = ^CSD and |AE| = |DC| we have that the triangle AEQ and SCD are congruent, so |AS| = |AQ| = |CS|. This shows that the quadrilateral ABCS is a deltoid, so ^ASB = ^BSC = 60◦ and the point S is the Fermat’s point of the triangle BDE. Let point X be such that BEX is equilateral and that S and X lie on different sides of the line EB. It is well know that the property of the Fermat’s point S is that X, S and D are collinear. Also, since |BX| = |EX|, X lies on the bisector of the segment BE. We have two cases. In the first case, the segment bisector of BE coincides with the line DS, so ABDE is a rhombus and AB ∥ ED. In the second case, the segment bisector of BE intersects the line AS at exactly one point. From the remarks we have given, that point must be X and also A, so A = X. Then the triangle BEA is equilateral, so ABCD is a rhombus and BC ∥ ED.
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T 6, (Michal Rol´ınek, Josef Tkadlec, Czech Republic) Let ABC be an acute triangle. Denote by B0 and C0 the feet of the altitudes from vertices B and C, respectively. Let X be a point inside the triangle ABC such that the line BX is tangent to the circumcircle of the triangle AXC0 and the line CX is tangent to the circumcircle of the triangle AXB0 . Show that the line AX is perpendicular to BC.
First solution
C
A0
H X
A
B
C0
Let A0 be the foot of the altitude from A. The quadrilateral ACA0 C0 is cyclic because ^AA0 C = ^AC0 C = 90◦ . By the power of the point B with respect to that circle we have |BA||BC0 | = |BA0 ||BC|. The power of the point B with respect to the circumcircle of AXC0 gives |BX|2 = |BA||BC0 |. Similarly, we have |CX|2 = |CA||CB0 | = |CA0 ||BC|. Summing these two results we have |BX|2 + |CX|2 = |BA0 ||BC| + |CA0 ||BC| = |BC|2 . The converse of Pythagora’s theorem implies ^BXC = 90◦ . Moreover, from |BX|2 = |BA0 ||BC|, i.e. |BX| : |BC| = |BA0 | : |BX| we have that the triangles BXA0 and BCX are similar. It follows that ^BA0 X = ^BXC = 90◦ = ^BA0 A, so A0 , X and A are collinear, so AX and BC are perpendicular.
21
Second solution (by Tomislav Pejkovi´ c, coordinator) Let H be the orthocenter of the triangle ABC. Because BX is tangent to the circumcircle of AXC0 we have ^BXC0 = ^BAX (the tangent chord angle theorem). Hence the triangles BAX and BXC0 are similar. Analogously, the triangle CAX and CXB0 are similar.
C
B0 H
X
A
B
C0
Observe that the quadrilateral AC0 HB0 is cyclic because ^AC0 H = ^AB0 H = 90◦ . The power of the point B with respect to circumcircles of AC0 X and AC0 HB0 gives |BB0 ||BH| = |BA||BC0 | = |BX|2 . From this we conclude that the triangles BXH and BB0 X are similar and ^BXH = ^XB0 H = ^XB0 C − 90◦ . Since CAX and CXB0 are similar we have ^XB0 C = ^AXC. We obtained ^BXH = ^AXC − 90◦ and analogously ^CXH = ^AXB − 90◦ . Summing up these results we get ^BXC = ^BXH + ^CXH = ^AXC + ^AXB − 180◦ = 180◦ − ^BXC and so ^BXC = 90◦ . Hence, the points B, C0 , X, B0 , C all lie on the same circle and we have ^AXB = ^BC0 X = 180◦ − ^XB0 B = 180◦ − ^BXH which means that A, X and H are collinear. So AX and BC are perpendicular.
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Third solution (by teams Croatia, Hungary and Poland) By power of the point we have |CX|2 = |CA||CB0 |,
|BX|2 = |BA||BC0 |,
È
so the point X is the intersection of the circle with center C and radius
È
|CA||CB0 | and
the circle with center B and radius |BA||BC0 |. There are two such points, but only one is in the interior of the triangle ABC, so we conclude that the point X is unique. On the other hand we will prove that the point Y which is the intersection of the circle with diameter BC and the altitude from the point A has the same properties as the point X, from which we conclude that X and Y are the same point and hence X lies on the line perpendicular to BC. Since ^BB0 C = 90◦ , the quadrilateral BCB0 Y is cyclic and hence ^CBB0 = ^CY B0 . On the other hand ^CAY = 90◦ − ^ACB = ^CBB0 = ^CY B0 , so by the tangent-chord theorem the line CY is tangent to the circumcircle of the triangle AY B0 . Analogously, the line BY is tangent to the circumcircle of the triangle AY C0 . Hence, X = Y .
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NUMBER THEORY
I 4 (Kamil Duszenko, Poland) Let k and m, with k > m, be positive integers such that the number km(k 2 − m2 ) is divisible by k 3 − m3 . Prove that (k − m)3 > 3km.
First solution Let d be the greatest common divisor of k and m. Write k = da, m = db. Then a and b are relatively prime. Moreover, a > b.
The number km k 2 − m2 = d4 ab a2 − b2 = d4 ab(a − b)(a + b) is divisible by k 3 − m3 = d3 (a3 − b3 ) = d3 (a − b)(a2 + ab + b2 ), so we have a2 + ab + b2 | dab(a + b). However, since the numbers a and b are relatively prime, the number a2 +ab+b2 is relatively prime to a, b, and a+b. (For example, in case of a+b we note that a2 +ab+b2 = (a+b)a+b2 , and a + b is relatively prime to b and hence to b2 .) Thus a2 + ab + b2 | d. This, in particular, yields d > a2 + ab + b2 = (a − b)2 + 3ab > 3ab. Therefore (k − m)3 = d3 (a − b)3 > d3 = d2 · d > d2 · 3ab = 3km.
Second solution (by Wojciech Nadara, Poland) Since k 2 + km + m2 divides km(k + m) and (k 2 + km + m2 )(k + m) we have that it divides their difference (k + m)(k 2 + m2 ) = k 3 + k 2 m + km2 + m3 . From this we conclude that k 2 + km + m2 also divides k 3 + k 2 m + km2 + m3 − k(k 2 + km + m2 ) = m3 . Analogously, we conclude that k 2 + km + m2 divides k 3 . Multiplying the second power of k 2 + km + m2 | k 3 with k 2 + km + m2 | m3 we conclude (k 2 + km + m2 )3 | k 6 m3 . Hence k 2 + km + m2 also divides k 2 m and analogously km2 . Adding all the results we have obtained we conclude that k 2 + km + m2 divides k 3 − 3k 2 m + 3km2 − m3 = (k − m)3 . Because k > m, i.e. k − m > 0, we have k 2 + km + m2 6 (k − m)3 . Since (k − m)2 is equivalent to k 2 + km + m2 > 3km, we obtain 3km < (k − m)3 .
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T 7 (Mariusz Skaba, Poland) Let A and B be disjoint nonempty sets with A ∪ B = {1, 2, 3, . . . , 10}. Show that there exist elements a ∈ A and b ∈ B such that the number a3 + ab2 + b3 is divisible by 11.
Solution For each n = 0, 1, 2, . . . the numbers 2n , 2n+1 , 2n+2 , . . . , 2n+9 have different remainders when divided by 11. Suppose that for every b ∈ B there is no a ∈ A such that a ≡ 2b (mod 11). From the above statement there exists n ∈ {0, 1, . . . , 9} such that b ≡ 2n (mod 11) and we conclude that elements of B give ten different remainders when divided by 11, so B has 10 elements. That is a contradiction with the fact that A is nonempty. Therefore there exist b ∈ B and a ∈ A such that a ≡ 2b (mod 11), and we have a3 + ab2 + b3 ≡ 8b3 + 2b3 + b3 = 11b3 ≡ 0 (mod 11).
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T 8 (Aivaras Novikas, Lithuania) We call a positive integer n amazing if there exist positive integers a, b, c such that the equality n = (b, c)(a, bc) + (c, a)(b, ca) + (a, b)(c, ab) holds. Prove that there exist 2011 consecutive positive integers which are amazing. (By (m, n) we denote the greatest common divisor of positive integers m and n.)
Solution We may choose such positive integers x1 , x2 , . . . , x2011 that the numbers y1 = x21 (x1 + 2), y2 = x22 (x2 + 2), . . . ,
y2011 = x22011 (x2011 + 2)
are pairwise coprime. For example, we may choose x1 = 1 and xi = y1 y2 . . . yi−1 − 1 for every consecutive i. This choice guarantees that for every integer 2 6 i 6 2011 both xi and xi + 2 (hence, yi as well) are coprime with any of the numbers y1 , y2 , . . . , yi−1 . If a positive integer n is divisible by any of the numbers y1 , y2 , . . . , y2011 then it is amazing. Indeed, if, say, n = yi m = x2i (xi + 2)m for some positive integers m and 1 6 i 6 2011 then n = (b, c)(a, bc) + (c, a)(b, ca) + (a, b)(c, ab) for a = mx2i , b = mxi , c = xi . Since the numbers y1 , y2 , . . . , y2011 are pairwise coprime, the Chinese remainder theorem implies that there exists a positive integer k satisfying the equalities k ≡ −i (mod yi ),
i = 1, 2, . . . , 2011.
This means that k +i is divisible by yi for any 1 6 i 6 2011. Thus, the consecutive positive integers k + 1, k + 2, . . . , k + 2011 are all amazing, and the statement of the problem is proved.
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Problems and Solutions MEMO 2012 Individual Competition
I1.
Find all functions f : R+ → R+ such that equality
f (x + f (y)) = yf (xy + 1) holds for all x, y ∈ R+ . (R+ denotes the set of all positive real numbers.) Assume that there exists a real number t such that t > 1 and f (t) > 1. If we f (t) − 1 , y = t into the given equality, we get: put x = t−1 f (t) − 1 f (t) − 1 f( + f (t)) = tf ( · t + 1), t−1 t−1 which can be written as tf (t) − 1 tf (t) − 1 ) = tf ( ). f( t−1 t−1 From the previous equality, we conclude that t = 1, which contradicts our assumption that t > 1. 1 − f (t) Therefore, we conclude that such t doesn't exist. In the same way, by putting x = , y = t, 1−t we can prove that there doesn't exist a real number t < 1 such that f (t) < 1. Solution.
Let y > 1 be an arbitrary real number and let x = equality, we get:
f(
y−1 . By plugging them into the given y
y−1 + f (y)) = yf (y). y
y−1 If f (y) > y1 then f ( y−1 y + f (y)) > 1 and thus y + f (y) 6 1. However, the last inequality 1 implies that f (y) 6 y , which is a contradiction. In the same way, if we assume that f (y) < y1 , we also get to a contradiction. Therefore, f (y) = y1 for all y > 1.
Finally, let 0 < a 6 1. Let us take an arbitrary y such that y > x = a − y1 . By plugging this into the given equality, we get:
1 1 f (a) = f (x + ) = f (x + f (y)) = yf (xy + 1) = y · = y xy + 1
1 a
> 1 and let us denote
1 xy+1 y
Therefore, the only solution to the given functional equation is f (x) = 1
=
1 x+
1 . x
1 y
=
1 . a
I2. Let N be a positive integer. A set S ⊂ {1, 2, . . . , N } is called allowed if it does not contain three distinct elements a, b, c such that a divides b and b divides c. Determine the largest possible number of elements in an allowed set S .
The answer is d 34 N e. On one hand, we can reach this optimum by choosing S ∗ = {b 14 N c + 1, b 41 N c + 2, . . . , N }. We call a triplet (a, b, c) forbidden if a divides b and b divides c. For any forbidden triplet we would have 4a 6 c (as if x|y and x 6= y then 2x 6 y ), hence there are no forbidden triplets in S ∗ which means that it is allowed. On the other hand, there are dN/2e odd numbers in {1, 2, . . . , N }. For such an odd number q , consider the set Hq = {q, 2q, 4q, . . . 2iq q}, where iq is the largest index i such that 2i q 6 N . We partitioned {1, . . . , N } to these Hq sets. Any three elements from a single set Hq form a forbidden triplet, hence for any feasible set S we have |Hq ∩ S| 6 2 (for all q odd numbers). If q > N/2, then |Hq | = 1. So for all 1 6 q 6 N/2 we can choose maximum 2 elements of Hq , and for all N/2 < q 6 N we can choose only 1 element of Hq . In both cases the number is equal to |Hq ∩S ∗ |. In summary we have |Hq ∩ S| 6 |Hq ∩ S ∗ | for all odd q which implies that |S| 6 |S ∗ | = d3N/4e. Solution.
I3. In a given trapezium ABCD with AB parallel to CD and AB > CD , the line BD bisects the angle ^ADC . The line through C parallel to AD meets the segments BD and AB in E and F , respectively. Let O be the circumcentre of the triangle BEF . Suppose that ^ACO = 60◦ . Prove the equality |CF | = |AF | + |F O|.
Solution.
D
C
O E Y A
F
Let |AB| = a and |CD| = b. Since AB k CD we have
^DBA = ^BDC = ^ADB 2
X
B
and thus the triangle ABD is isosceles with |AB| = |DA| = a. Since quadrilateral AF CD is parallelogram, we have
|AF | = |CD| = b,
|F C| = |DA| = a and |F B| = a − b.
Since AD k CF , we have
^F EB = ^ADB = ^DBA and thus the triangle F BE is isosceles with |F B| = |EF | = a − b. Let X and Y be the midpoints of segments F B and EF . Due to symmetry, |OX| = |OY |. From a−b a+b |AX| = |AF | + |F X| = b + = 2 2 and a−b a+b |CY | = |CF | − |F Y | = a − = 2 2 we can conclude that right-angled triangles AXO and CY O are congruent and |AO| = |CO|. Since ^ACO = 60◦ , the triangle AOC is equilateral and ^OAC = 60◦ . Denoting ^XAO = ^Y CO = ϕ we get ^F AC = ^XAO + ^OAC = ϕ + 60◦ and ^ACF = ^ACO − ^Y CO = 60◦ − ϕ, so ^CF A = 60◦ , ^BF E = 120◦ and ^XF O = 60◦ . Since the angles of the triangle F XO are 30◦ , 60◦ and 90◦ , from |F O| = 2|F X| = a − b, we nally get |AF | + |F O| = b + (a − b) = a = |CF |.
I4.
The sequence {an }n>0 is de ned by a0 = 2, a1 = 4 and
an+1 =
an an−1 + an + an−1 for all n > 0. 2
Determine all prime numbers p for which there exist m > 0 such that p|am − 1. Solution. If an−1 and an are even the number an an−1 /2 is even and therefore an+1 , too. Hence an is even for all n > 0 and p = 2 is not a solution. Since 3|a1 − 1, p = 3 is a solution. Assume from now on p > 5. The equation is equivalent to
an+1 + 2 =
(an + 2)(an−1 + 2) . 2
Let bn = (an + 2)/2, it follows that bn+1 = bn bn−1 with b1 = 2 and b2 = 3. Note that p - bn for all n ∈ N. Hence the sequence modulo p can be constructed backwards and extended for negative n+2 numbers n by bn ≡ bbn+1 modulo p. Note that two consecutive elements bk , bk+1 determine the entire sequence. As there are p2 ordered pairs of residues modulo p it follows by the pigeon hole principle that there exists 0 6 k < l 6 p2 such that bk ≡ bl and bk+1 ≡ bl+1 . The sequence modulo p is therefore periodic with length l − k and we have bl−k−1 ≡ b−1 ≡ p+3 2 . It follows that al−k−1 ≡ 1 modulo p. Hence all prime numbers greater than 2 are solutions. 3
4
Team Competition
T1.
Find all triplets (x, y, z) of real numbers such that
2x3 + 1
=
3zx,
3
=
3xy,
3
=
3yz.
2y + 1 2z + 1
Solution. Note that no variable can be zero. Assume that one variable is positive. WLOG x > 0. It follows that z > 0 and y > 0. Summing up all three equations and using AM-GM yields
2(x3 +y 3 +z 3 )+3 = 3(xy+yz+zx) 6 3(
x3 + y 3 + 1 y 3 + z 3 + 1 z 3 + x3 + 1 + + ) = 2(x3 +y 3 +z 3 )+3. 3 3 3
Therefore we must have equality and it follows x = y = z = 1 which is a solution. Assume from now on that all variables are negative. Put u = −x, v = −y and w = −z . Then u, v, w must be positive and the equations can then be written as
1
=
2u3 + 3uw,
1
=
2v 3 + 3vu,
1
=
2w3 + 3wv.
WLOG let u > v, w. We have
1 = 2u3 + 3uw > 2w3 + 3wv = 1. Hence we must have equality and therefore u = v = w. It follows that
0 = 2u3 + 3u2 − 1 = (u + 1)2 (2u − 1). Hence v = w = u = 1/2 which is also a solution. The system has therefore the solutions (x, y, z) = (1, 1, 1) and (x, y, z) = (−1/2, −1/2, −1/2).
T2.
Let a, b and c be positive real numbers with abc = 1. Prove that p p p 9 + 16a2 + 9 + 16b2 + 9 + 16c2 > 3 + 4(a + b + c). 5
First Solution.
()
If three
Let us start by showing positive reals x, y , and z sum
up to less than
3,
then
9 − x2 9 − y 2 9 − z 2 · · > 512. x y z Proof of (). Using the inequality between the arithmetic mean and the geometric mean of four positive reals we obtain √ 3 + x = 1 + 1 + 1 + x > 4 4 x. √ √ For similar reasons, one has 3 + y > 4 4 y and 3 + z > 4 4 z . Multiplying the three foregoing estimates we get √ (3 + x)(3 + y)(3 + z) > 64 4 xyz. (1)
Using the A.M.G.M. inequality with just three variables, we get
√ 3 > x + y + z > 3 3 xyz, which tells us
1 > xyz, as well as
xy + yz + zx > 3(xyz)2/3 .
Combining these estimates we infer
(3 − x)(3 − y)(3 − z) = 9(3 − x − y − z) + 3(xy + yz + zx) − xyz > 9(xyz)2/3 − xyz > 8(xyz)2/3 . If we nally multiply this by (1), we arrive at
(9 − x2 )(9 − y 2 )(9 − z 2 ) > 512xyz and () follows. Now to attack the problem itself, we de ne three positive reals x, y , and z by p x = 9 + 16a2 − 4a, p y = 9 + 16b2 − 4b, p and z = 9 + 16c2 − 4c. Then
9 − x2 9 − y 2 9 − z 2 · · = 512abc = 512, x y z
which means that the conclusion of () is violated. This is only possible if its assumption does not hold, i.e. if x + y + z > 3. Hence p p p 9 + 16a2 + 9 + 16b2 + 9 + 16c2 > 3 + 4(a + b + c) and the problem is solved. 6
The function x 7→ ex where e is the Euler Constant is a bijection between R and R>0 , hence we can nd u, v, w ∈ R such that a = eu , b = ev and c = ew and the constraint abc = 1 becomes eu ev ew = 1 ⇔ eu+v+w = 1 ⇔ u + v + w = 0. q 9 − ex the inequality can then be written as Let f (x) = e2x + 16
Second Solution.
f (u) + f (v) + f (w) >
3 4
∀u, v, w with u + v + w = 0
With some calculus we compute the rst two derivatives of f
f 0 (x)
f 00 (x)
e2x
=
q
e2x
+
9 16
2e2x (e2x +
=
(e2x
+
9 4x 16 ) − e 3 9 2 16 )
e2x (e2x + 98 )
=
(e2x +
(2)
− ex
9 32 16 )
(3)
− ex
(4)
− ex .
We determine the convex intervals of f :
f 00 (x) > 0 (5) 3 9 9 ⇔ e2x (e2x + ) > ex (e2x + ) 2 (6) 8 16 9 9 ⇔ e4x (e2x + )2 > e2x (e2x + )3 (7) 8 16 3 9 4x 81 2x 9 ⇔ e + e − > 0 (8) 16 256 16 9 4x 81 2x 9 3 9 3 As the function g(x) = 16 e + 256 e − 16 is monotone increasing, limx→−∞ g(x) = − 16 and limx→∞ g(x) = ∞ there is exactly one value m ∈ R with g(m) = 0. Furthermore it follows that f is convex on (m, ∞) and concave on (−∞, m). Lemma 1. For any three real numbers
u, v
with
u + 2v = 0
x1 , x2 , x3
with
x1 + x2 + x3 = 0
there exists real numbers
such that
f (x1 ) + f (x2 ) + f (x3 ) > f (u) + 2f (v). WLOG we can assume x1 6 x2 6 x3 . If x2 < m we have f (x1 ) + f (x2 ) > f (x1 + x2 − m) + f (m) as f is concave on (−∞, m). Hence we can assume x2 > m. Then we have by Jensen's 3 3 ). Hence we can set u = x1 and v = x2 +x . inequality f (x2 ) + f (x3 ) > 2f ( x2 +x 2 2 Proof.
It remains to prove that r r 9 9 3 2 2 a + + b2 + > 2a + b + 16 16 4
for all a, b with a2 b = 1.
By squaring and rearranging the inequality becomes r 9 9 3 9 4 (a2 + )(b2 + ) > 4ab + 3a + b − . 16 16 2 4 7
(9)
Dividing by 4, squaring and rearranging yields
27 2 9 27 27 3 3 b + ab + a + b > a2 b + ab2 . 64 16 32 64 2 4 After multiplying by a4 and using a2 b = 1 we arrive at
27 5 3 4 9 27 3 27 a − a + a3 + a2 − a + > 0 32 2 16 64 4 64 27 3 3 27 ⇔ (a − 1)2 ( a3 + a2 + a + ) > 0 32 16 32 64
(10) (11)
which is obviously true.
T3. Let n be a positive integer. Consider words of length n composed of letters from the set {M, E, O}. Let a be the number of such words containing an even number (possibly 0) of blocks M E and an even number (possibly 0) of blocks M O. Similarly, let b be the number of such words containing an odd number of blocks M E and an odd number of blocks M O. Prove that a > b. First Solution. Let A be the set of words of length n with even number of ME and even number of MO and B be the set of words of length n with odd number of ME and odd number of MO. We construct an injective map f from B to A. Choose the rst place in the tuple, where either is the block 01 or the block 02 (this place must exist since the number of 01-blocks is odd and therefore >0) and interchange the blocks. One of the number of blocks 01 and the number of blocks 02 will increase by 1 and one will decrease by 1. Hence both numbers change its parity from odd to even and the image of the map is therefore in A. Furthermore when the operation above is applied two times on a sequence the we get the original sequence. Hence we have f (f (b)) = b for all b ∈ B which implies that the operation is injective and therefore |B| 6 |A|. To see that |B| < |A| note that all sequences which do not contain an 0 are in A but can not be the image of an element of B under f . As n > 0 there is at least one such sequence.
We call the blocks ME and the blocks MO important and we say that two words are similar if the positions of the important blocks are the same. This similarity divides the set of all sequences into equivalence classes. Let A and B be de ned as in the rst solution. Note that the elements of A and the elements of B contain an even number of important blocks. Therefore an equivalence class with an odd number of important blocks does not contain any elements of A or B P . Consider an equivalence class with k > 2 even number of important P blocks. k k There are neven = m even m elements of the equivalence class in A and nodd = m odd m elements of the equivalence class in B . We have Second Solution.
neven − nodd = (1 − 1)k = 0. which proves that the equivalence class contains the same number of elements from A and B . For k = 0 all elements belong to A and this set is not empty which concludes our proof. 8
Let an be the number of words of length n with even number of ME and even number of MO, bn be the number of words of length n with odd number of ME and odd number of MO and xn be the common number of words such that
Third Solution.
1) start with E, odd number of ME blocks, even number of MO blocks 2) start with O, odd number of ME blocks, even number of MO blocks 3) start with E, even number of ME blocks, odd number of MO blocks 4) start with O, even number of ME blocks, odd number of MO blocks. Bijections 1) ↔ 4) and 2) ↔ 3) : replace all E by O and all O by E Bijections 1) ↔ 2) and 3) ↔ 4) : switch rst number E ↔ O
To compute an+1 we consider the three possible start letters of the word. If it starts with E or O the number of words is an . If it starts with M we distinguish again the three cases for the second letter. If the word starts with ME or MO we get xn words. When the sequence starts with MM we distinguish again three cases. . . . If we continue this argument we arrive at
an+1 = 2an + 2
n X
xk + 1
(12)
k=1
(the plus 1 comes from the sequence M . . . M ). Similarly
bn+1 = 2bn + 2
n X
yk
(13)
k=1
Taking dierences:
an+1 − bn+1 = 2(an − bn ). Since a1 = 3 and b1 = 0 an induction gives the claim
an > bn . (Actually an = bn + 2n+1 − 1)
T4. Let p > 2 be a prime number. For any permutation π = (π(1), π(2), . . . , π(p)) of the set S = {1, 2, . . . , p}, let f (π) denote the number of multiples of p among the following p numbers:
π(1), π(1) + π(2), . . . , π(1) + π(2) + · · · + π(p) What is the average value of f (π) taken over all permutations of S ?
9
Solution.
that:
We call two permutations π 0 and π 00 equivalent if there is an integer d such
π 0 (i) + d ≡ π 00 (i)
(mod p)
for all i = 1, 2, . . . , p. (It's trivial to check that it is an equivalence relation). This equivalence partitions the set P of all permutations into equivalence classes. Each class consists of p permutations, that can be labeled π0 , . . . , πd , . . . , πp−1 such that π0 (i) + d and πd (i) are congruent for all i. Let sk denote the sum π0 (1) + π0 (2) + · · · + π0 (k) mod p. Then πd (1) + πd (2) + · · · + πd (k) ≡ sk + kd (mod p). So for a xed k if we consider the sums πd (1) + πd (2) + · · · + πd (k) for all d-s, we get these: sk , sk + k, sk + 2k, . . . , sk + (p − 1)k If k < p then among these there is exactly one 0. If k = p then all are 0s (sp is the sum of all remainder classes mod p, it is a well-known fact that it's zero if p > 2). So for any equivalence class, consisting of p pieces of permutations, there 2p − 1 zeros among the sums assigned to them. That means an average of 2p−1 p .
T5. Let K be the midpoint of the side AB of a given triangle ABC. Let L and M be points on the sides AC and BC, respectively, such that ^CLK = ^KM C. Prove that perpendiculars to the sides AB, AC and BC passing through K, L and M , respectively, are concurrent.
Solution.
10
Let S be the intersection of perpendiculars through K and L, and T the intersection of perpendiculars through K and M. Observe that AKSL and BM T M are cyclic (they have two right angles). Therefore ^SAK = ^SLK = ^CLK − π2 (or π2 − ^CKL if ^CKL is acute). Analogously, ^T BK = ^T M K = ^CM K − π2 (or π2 − ^CM K if ^CM K is acute). Therefore
^SAK = ^T BK. But K is the midpoint of AB, hence S and T have to coincide. Thus the perpendiculars are concurrent.
T6. Let ABCD be a convex quadrilateral with no pair of parallel sides, such that ^ABC = ^CDA. Assume that the intersections of the pairs of neighbouring angle bisectors of ABCD form a convex quadrilateral EF GH . Let K be the intersection of the diagonals of EF GH . Prove that the lines AB and CD intersect on the circumcircle of the triangle BKD.
Solution.
11
C
G
D
K
H
I
F
E A B
J Let's denote I ≡ AB ∩ CD and J ≡ BC ∩ AD. Now F is the incenter of triangle IBC thus IF is the angle bisector of ^BIC . We can also note that H is the excenter of triangle ADI su IH is the same angle bisector. Thus I, F, H are collinear. similarly we can deduce that J, E, G are collinear as well. Now if we denote α = ^BAC and β = ^ABC = ^CDA we can see that ^JDI = 180◦ − β = ^JBI . Since IK and JK are the angle bisectors of ^DIA and ^AJB we get:
^IKJ = 360◦ − ^KIA − ^KJA − ^JAI ^AID ^AJB = 360◦ − − − (360◦ − α) 2 2 α + β − 180◦ α + β − 180◦ =α− − 2 2 = 180◦ − β = ^JDI = ^JBI.
12
Thus the points I, J, B, K, D are all concyclic which implies the desired result.
T7.
Find all triplets (x, y, z) of positive integers such that ( xy + y x = z y , xy + 2012 = y z+1 .
Solution.
We consider two cases.
(i) Let x be an odd number. Then, clearly, y is odd (the second equation) and z is even (the rst equation). If a and b are odd positive integers then ab ≡ (±1)b ≡ ±1 ≡ a (mod 4). Let us apply this fact to both equations. The second equation implies that y > 1, thus, 4 divides z y and x + y ≡ xy + y x ≡ z y ≡ 0 (mod 4) =⇒ x ≡ −y (mod 4). On the other hand, x ≡ x + 0 ≡ xy + 2012 ≡ y z+1 ≡ y (mod 4). This means that y ≡ −y (mod 4) which is impossible for an odd y . (ii) Let x = 2x1 be even. Then y and z are even too. If y > 2 then 8 divides xy , but only 4 divides 2012, thus, 8 does not divide y z+1 . This implies that z + 1 6 2 =⇒ z = 1 which is impossible in the light of the rst equation. Hence, y 6 2. We already noted that y > 1, thus, y = 2. We rewrite the rst equation:
x2 + 2x = z 2 =⇒ 2x = (z − x)(z + x). Hence, as x and z are both even, there exist such positive integers u < v that z − x = 2u , z + x = 2v and u + v = x. Since v > u we have v > x2 and u 6 v − 1. This implies x that x = 2v−1 − 2u−1 > 2v−1 − 2v−2 = 2v−2 > 2 2 −2 =⇒ x1 > 2x1 −3 . The inequality does not hold for x1 = 6 and the same can be easily proved by induction for the larger values of x1 (as the left-hand side increases by 1, the right-hand side increases by 2x1 −3 > 22 > 1). Hence, we only need to√check the rst ve values 1, 2, 3, 4, 5 (i.e. x = 2, 4, 6, 8, 10). Only for x = 6 the number z = x2 + 2x = 10 is an integer. The obtained triplet (6, 2, 10) satis es both equations. The answer.
(6, 2, 10).
T8. For any positive integer n, let d(n) denote the number of positive divisors of n. Do there exist positive integers a and b, such that d(a) = d(b) and d(a2 ) = d(b2 ), but d(a3 ) 6= d(b3 )? First Solution. The answer is negative. Denote by τ (n) the number of divisors of n ∈ N. First note that if the prime factorization of n
13
is n = pa1 1 · pa2 2 · · · pakk , then τ (n) = (a1 + 1) · · · (ak + 1) and similarly we nd that τ (n2 ) = (2a1 + 1) · · · (2ak + 1) and τ (n3 ) = (3a1 + 1) · · · (3ak + 1). For convenience denote βi = αi + 1 for i ∈ {1, 2, · · · , k}. Then
τ (n) =
k Y i=1
βi , τ (n2 ) =
k Y
(2βi − 1), τ (n3 ) =
i=1
k Y
(3βi − 2)
i=1
Now we oer two approaches to nd a counterexample in terms of the numbers βi for i ∈ {1, 2, · · · , k}. We look at pairs of positive integers of the form (n, 2n−1) namely at (2, 3), (3, 5), (4, 7), (8, 15), (18, 35), (32, 63) and we aim to nd numbers a1 , a2 , · · · a6 ∈ Z (possibly negative) such that
2a1 · 3a2 · 4a3 · 8a4 · 18a5 · 32a6 = 1 and 3a1 · 5a2 · 7a3 · 15a4 · 35a5 · 63a6 = 1. Comparing the primes powers of 2, 3, 5 and 7 in the latter relations produces a system of equations
a1 + 2a3 + 3a4 + a5 + 5a6
=
0
(14)
a2 + 2a5
=
0
(15)
a1 + a4 + 2a6
=
0
(16)
a2 + a4 + a5
=
0
(17)
a3 + a5 + a6
=
0
(18) (19)
with solution (a1 , a2 , a3 , a4 , a5 , a6 ) = (1, −2, . Then β1 = 2, β2 = 8, β3 = Q 0, 1, 1, Q−1) Q we choose Q 0 18, β10 = β20 = 3,Qβ30 = 32. We have ensured β = β and (2β − 1) = (2βi0 − 1). It remains i i i Q to see that 11| (3βi − 2) and 11 - (3βi − 2), thus the numbers are distinct and we can nd the counterexample to the original problem for example as
a = 2 · 37 · 517 , b = 22 · 32 · 531 .
We look for counterexample in the form β1 = pq, β2 = r, β3 = s and β10 = p, β20 = q, β30 = rs for some p, q, r, s ∈ N. In order to ful ll the second condition we need
(2p − 1)(2q − 1) 2r − 1)(2s − 1) = 2pq − 1 2rs − 1 Thus, looking at the expression (2n−1)(2m−1) for positive integers m and n, we need it to attain 2mn−1 the same value twice for two distinct choices of n and m. After subtracting 2 and switching the sign, we obtain 2m + 2n − 3 2mn − 1 . We attempt the common value to be of the form 1/k (the expression seems asymptotically small) for some k ∈ N. After some manipulation, the condition turns into
(m − k)(n − k) =
1 (2k − 1)(k − 1) 2
14
In order to ensure the right-hand side is not a prime, we choose k = 5. Then we need to solve
(m − 5)(n − 5) = 18 We nd two solutions (6, 23), (7, 14) and so we choose p = 6, q = 23, r = 7, s = 14. We continue similarly as in rst approach and nd counterexample
a = 2137 · 36 · 513 , b = 25 · 322 · 597 . It can be easily seen that the counterexample needs to have at least three prime factors. Also, the set of pairs (2, 3), (3, 5), (4, 7), (8, 15), (18, 35), (32, 63) from the rst approach is the minimal (with smallest maximal element) set for which the system of equations has non-trivial solutions. This leads to the claim that the counterexample from the rst approach is also minimal in this sense.
Second Solution. The problem is immediately reduced to nding nite positive integer sequences (xi ) and (yj ) satisfying Y Y xi = yj (20)
i
Y (2xi − 1)
j
=
i
Y (2yj − 1)
(21)
j
Y (3xi − 2)
6=
i
Y (3yj − 2)
(22)
j
and such that (xi ) is not a permutation of (yj ). We will look for a parametric family of solutions. We base our construction on choosing the y 's such that the generic expression 2y − 1 is decomposable and the decomposition factors can be recombined to generate the x's in (21). The simplest identity of this type seems to be obtained for y = 2z 2 , 2y − 1 = 4z 2 − 1 = (2z − 1)(2z + 1) = (2z − 1)(2(z + 1) − 1). We therefore pick now y0 = 2z 2 , y1 = 2(z + 1)2 , . . . , yn = 2(z + n)2 for some n ∈ N. With this choice it follows that n Y
(2yj − 1)
=
j=0
=
n Y
(4(z + j)2 − 1)
j=0 n Y
(2(z + j) − 1)(2(z + j + 1) − 1)
j=0
=
(2z − 1)(2(z + n + 1) − 1)
n Y
(2(z + j) − 1)2 .
j=1
The obvious choice now for the sequence (xi ) consists in the set of values z , z + n + 1, plus two copies of each number in the sequence (z + j) for j = 1 : n. Note that in this way there 15
are twice as many x's as y 's: 2n + 2 vs. n + 1. Explicitly, x0 = z, x1 = x2 = z + 1, x3 = x4 = z +2, . . . , x2n−1 = x2n = z +n, x2n+1 = z +n+1. With this choice for the sequence (xi ) condition (21) is ful lled by construction. But what about (20) or (22)? Let's start with (20) and calculate Y xi = z(z + n + 1)(z + 1)2 (z + 2)2 · · · (z + n)2 , i
Y
yj = 2n+1 z 2 (z + 1)2 · · · (z + n)2
j
to obtain
Y
yj /
Y
j
xi = 2n+1 z/(z + n + 1).
i
Although the r.h.s. above can not be made equal to 1 as desired, it can be further simpli ed by choosing z = n + 1. With this choice we arrive at Y Y yj / xi = 2n . (23) j
i
Recall that this holds for our current choice y0 = 2n2 , y1 = 2(n + 1)2 , . . . , yn = 2(2n)2 and x0 = n + 1, x1 = x2 = n + 2, x3 = x4 = n + 3, . . . , x2n−1 = x2n = 2n + 1, x2n+1 = 2n + 2. Call now the entire sequence construction above Cn , and consider for arbitrary positive integers n, m the constructions Cn , Cm as well as Cn+m , together with the corresponding sequences (xi , yj ), (x0i , yj0 ) and (x00i , yj00 ) respectively. Consider now the sequences (Xi ) and (Yj ) obtained by collecting (concatenating) the sequences (xi ), (x0i ), (yj00 ) and (yj ), (yj0 ), (x00i ) respectively. We claim that (Xi ) and (Yj ) are in general sequences that satisfy all three conditions (20), (21), (22). Due to the obvious identity 2m 2n = 2m+n we deduce from 23 that the sequences (Xi ) and (Yj ) satisfy, beside (21), also condition (20). It remains to investigate condition (22). There are several ways to do this, for example by analyzing prime decompositions of the l.h.s. and r.h.s. respectively (and formulating a condition on m, n such that the largest prime divisor of the l.h.s. does not divide the r.h.s. anymore). Here however we take a dierent route and estimate the asymptotic behavior of the two sides of equation (22). To this end we use (20) and write ! Y Y Y Y Y Y (3Yj − 2)/ (3Xi − 2) = (3Yj − 2)/ Yj × Xi / (3Xi − 2) . (24) j
Cn ,
i
j
j
i
i
To estimate the two factors on the r.h.s. of (24) we rst observe that in the generic construction
Y
(3yj − 2)/yj = 3n+1
j
with An → 1 as n → ∞, since
n Y
(1 − 2/3yj ) = 3n+1 An ,
j=0
Pn
j=0
1/6(n + j)2 → 0 as n → ∞.
16
(25)
We note now that a similar result holds for the sequence (xi ) in construction Cn , 2n+1 Y Y (3xi − 2)/xi = 32(n+1) (1 − 2/3xi ) = 32(n+1) Bn i
(26)
i=0
with Bn → 2−4/3 as n → ∞, since converging to ln(2) for any real a.
P2n+1 i=0
2/3xi → 4 ln(2)/3 as n → ∞ due to
Pn
j=1
1/(n + j + a)
Using the two facts (25), (26) in (24) and in the context of the construction of (Xi , Yj ) via Cn , Cm , Cn+m we obtain
Y Y (3Yj − 2)/ (3Xi − 2) j
=
3n+1 An · 3m+1 Am · 32(n+m+1) Bn+m ·
i −1 ·3−2(n+1) Bn−1 · 3−2(m+1) Bm · 3−(n+m+1) A−1 n+m
=
−1 −1 (1/3)An Am A−1 n+m Bn Bm Bn+m .
(27)
Since the limit as n, m → ∞ equals 24/3 /3 6= 1 we obtain that the desired conclusion (22) holds for m, n large enough and the proof is complete.
17
Problems and solutions Individual Competition Problem I-1. Let a, b, c be positive real numbers such that a+b+c=
1 1 1 + 2 + 2. 2 a b c
Prove that 2(a + b + c) ≥
√ 3
7a2 b + 1 +
√ √ 3 3 7b2 c + 1 + 7c2 a + 1.
Find all triples (a, b, c) for which equality holds. Solution. From the AM–GM inequality, we obtain that v
u √ u 1 7b 3 3 2 + 2 7a b + 1 = 2 · t a·a· 8 8a
We have analogous upper bounds for inequalities, we obtain that √ 3
7a2 b
√ 3
!
7b2 c + 1 and
√ √ 2 3 3 + 1 + 7b2 c + 1 + 7c2 a + 1 ≤ 3
!
2 7b 1 ≤ a+a+ + 2 . 3 8 8a √ 3
7c2 a + 1. Adding up these three
23(a + b + c) 1 1 1 1 + + 2+ 2 2 8 8 a b c
Using the condition of the problem, we obtain √ √ √ 3 3 3 7a2 b + 1 + 7b2 c + 1 + 7c2 a + 1 ≤ 2(a + b + c). Equality holds if and only if a, b, and c satisfy the system of equations a=
7b 1 + 2, 8 8a
7c 1 + 2, 8 8b 7a 1 c= + 2. 8 8c Note that this systemactuallyimplies the equation stipulated in the problem. Defining f (x) = 87 x − 8x12 , we can rewrite the system as b=
b = f (a), c = f (b), 1
!
.
a = f (c). We prove that f (x) is a non-decreasing function. Let u ≥ v. Then f (u) − f (v) =
8 7
(u − v) +
1 8v 2
=
8 7
(u − v) +
(u−v)(u+v) 8u2 v 2
=
8 (u 7
− v) 1 +
−
u+v 8u2 v 2
1 8u2
≥ 0. Since the system of equations is cyclically symmetric, we may assume that a = max{a, b, c}. Since a ≥ b, we have b = f (a) ≥ f (b) = c, so c = f (b) ≥ f (c) = a. In all, c ≥ a ≥ b ≥ c, so a = b = c. We now have to find the solutions of f (a) = a. 8 7
a−
1 8a2
8a −
= a
1 a2
= 7a
1 a2
= a
1 = a3 Thus, equality holds if and only if a = b = c = 1. Problem I-2. Let n be a positive integer. On a board consisting of 4n × 4n squares, exactly 4n tokens are placed so that each row and each column contains one token. In a step, a token is moved horizontally or vertically to a neighbouring square. Several tokens may occupy the same square at the same time. The tokens are to be moved to occupy all the squares of one of the two diagonals. Determine the smallest number k(n) such that for any initial situation, we can do it in at most k(n) steps. Solution. We shall prove that k(n) = 6n2 . We define the distance from a given square to a given diagonal to be the minimal number of steps needed to get from the square to the diagonal. This equals the minimal number of horizontal steps needed to do that. It also equals the minimal number of vertical steps needed to do that. Given a configuration of tokens, we define the distance from this configuration to a given diagonal to be the sum of distances of the tokens to that diagonal. Choose the coordinate system so that the vertices of the board have coordinates ±2n. Place a token on each of the n squares the coordinates of whose centres satisfy x > 0 and y − x = n. Now complete this configuration of tokens so that it has a rotational symmetry of 90◦ about the origin. Then we have 4n tokens, one in each row, one in each column. The distance from this configuration to either diagonal is 2n · n + 2n · 2n = 6n2 . Therefore, k(n) ≥ 6n2 . Now consider any configuration satisfying the conditions of the problem. We prove that ≤ 6n2 steps suffice even if we only allow horizontal moves. I.e., the smallest of the
2
two distances from the given configuration to the diagonals is ≤ 6n2 . It suffices to prove that the sum of the two distances from the given configuration to the diagonals is ≤ 12n2 . Observe that the sum of the two distances from the square with center (x, y) to the two diagonals is 2 max(|x|, |y|). This number can take values 1, 3, . . . , 4n − 1. The squares where it takes a given value can be covered by two columns and two rows, so we can place at most four tokens there. Thus, the sum of the values for the 4n tokens is ≤ 4((4n − 1) + (4n − 3) + · · · + (2n + 1)) = 4n · 3n = 12n2 . Problem I-3. Let ABC be an isosceles triangle with AC = BC. Let N be a point inside the triangle such that 2∠AN B = 180◦ + ∠ACB. Let D be the intersection of the line BN and the line parallel to AN that passes through C. Let P be the intersection of the angle bisectors of the angles CAN and ABN . Show that the lines DP and AN are perpendicular. Solution. Since AC = BC, there is a circle k such that the lines AC and BC are the tangents to k at the points A and B. The condition defining the point N implies that the point N lies on the circle k. By the tangent-chord theorem, we have ∠BAN = ∠DBC and ∠CAN = ∠ABD. Since DC is parallel to AN , we have ∠CAN = ∠ACD. Hence ∠ACD = ∠ABN , so the quadrilateral ABCD is cyclic. It follows that ∠CAD = ∠CBD = ∠BAN . We can conclude that the angle bisector of ∠CAN is the angle bisector of ∠BAD. Hence the point P is the incenter of the triangle ABD and DP is the angle bisector of ∠ADB. We also note that since CD is parallel to AN , we have ∠AN D = ∠BDC = ∠BAC = ∠BAN + ∠N AC = ∠CAD + ∠N AC = ∠N AD. Hence AD = N D and we conclude that the angle bisector of ∠ADB is the perpendicular bisector of the segment AN . Hence DP is perpendicular to AN . Remark 1. The first part of the solution can easily be deduced also by calculating the angles without noting that AC and BC are tangents to k. Remark 2. It is a well-known fact that the intersection of the perpendicular bisector of the segment AN and the angle bisector of the angle ∠ABN lies on the circumcircle of the triangle ABN , i.e. P also lies on the circle k. This fact is obtained more easily by calculating the angle ∠AP B.
Problem I-4. Let a and b be positive integers. Prove that there exist positive integers x and y such that ! x+y = ax + by. 2 Solution. Denoting A = 2a + 1 and B = 2b + 1, the equation can be translated into (x + y) − A B − (x + y) = . x y For A = B, any integers with x + y = A satisfy the equation. Now suppose that A < B. Let n be the integer in the interval [A, B) which is divisible by d = B − A. Then n 6= A because A is odd and d is even. Now choose n n x = (B − n) , y = (n − A) . d d Here n = x + y, so the equation is satisfied. 3
Team Competition Problem T-1. Find all functions f : R → R such that f (xf (x) + 2y) = f (x2 ) + f (y) + x + y − 1 for all x, y ∈ R. Solution. Putting x = y = 0, we get f (0) = 1. Putting x = 0, y = z, we get f (2z) = f (z) + z.
(1)
Putting x = z, y = −zf (z), we get f (z 2 ) = zf (z) − z + 1.
(2)
Replacing z by 2z in (2) and using (1), we obtain f (4z 2 ) = 2zf (2z) − 2z + 1 = 2z(f (z) + z) − 2z + 1 = 2zf (z) + 2z 2 − 2z + 1.
(3)
Using (1) for 2z 2 and then for z 2 in place of z, and afterwards using (2), we obtain f (4z 2 ) = f (2z 2 ) + 2z 2 = f (z 2 ) + z 2 + 2z 2 = zf (z) − z + 1 + 3z 2 .
(4)
Comparing (3) and (4) we have 2zf (z) + 2z 2 − 2z + 1 = zf (z) − z + 1 + 3z 2 zf (z) − z 2 − z = 0 z(f (z) − z − 1) = 0. For z 6= 0 we obtain f (z) = z + 1. For z = 0, we have f (0) = 1. Thus, for all z ∈ R, we have f (z) = z + 1. This function indeed satisfies the functional equation. Problem T-2. Let x, y, z, w ∈ R \ {0} such that x + y 6= 0, z + w 6= 0, and xy + zw ≥ 0. Prove the inequality x+y z+w + z+w x+y
!−1
1 x z + ≥ + 2 z x
4
−1
y w + + w y
!−1
.
Solution 1. We first subtract 1 on both sides and rewrite the inequality as
(x + y)(z + w) xz yw 1 1 1 ≥ + 2 , − − − 2 2 2 2 2 (x + y) + (z + w) 2 x +z 2 y +w 2
which is equivalent to (x − z)2 (y − w)2 (x + y − z − w)2 + ≥ . x2 + z 2 y 2 + w2 (x + y)2 + (z + w)2 But this is true by the chain of inequalities (x − z)2 (y − w)2 ((x − z) + (y − w))2 (x + y − z − w)2 + ≥ ≥ , x2 + z 2 y 2 + w2 x2 + z 2 + y 2 + w 2 (x + y)2 + (z + w)2 where we first used the Cauchy–Schwarz inequality for the two vectors √
!
(x − z) (y − w) √ ,√ 2 , y + w2 x2 + z 2
x2
+
z2,
q
y2
+
w2
and then the condition xy + zw ≥ 0. Solution 2. The vectors (x, z) and (y, w) form an angle ≤ π/2 containing the sum vector (x + y, z + w). So with suitable angles α ≤ φ ≤ β with β − α ≤ π/2, the desired inequality becomes (tg φ + ctg φ)−1 +
1 ≥ (tg α + ctg α)−1 + (tg β + ctg β)−1 , 2
which is equivalent to 1 + sin 2φ ≥ sin 2α + sin 2β. If sin 2φ ≥ min(sin 2α, sin 2β), then we are done because 1 ≥ max(sin 2α, sin 2β). Otherwise, modulo 2π we have that π/2 < 2α < 3π/2 < 2β and 2β − 2α ≤ π, so sin 2α + sin 2β ≤ 0 ≤ 1 + sin 2φ. Problem T-3. There are n ≥ 2 houses on the northern side of a street. Going from the west to the east, the houses are numbered from 1 to n. The number of each house is shown on a plate. One day the inhabitants of the street make fun of the postman by shuffling their number plates in the following way: for each pair of neighbouring houses, the current number plates are swapped exactly once during the day. How many different sequences of number plates are possible at the end of the day?
5
Solution 1. Let f (n) denote the answer. We shall prove by induction that f (n) = 2n−2 . For n = 2, the answer is clearly 22−2 = 1. We also define f (1) = 1. Now we consider arbitrary n > 2. Let Hi denote the house with number i at the start of the day, and let (i i + 1) denote the swap between Hi and Hi+1 . Let Hk be the house that has plate n at the end of the day. It follows that the swaps (n − 1 n), (n − 2 n − 1), . . . , (k k + 1) must have happened in this order, for otherwise plate n could not have reached the house Hk . In addition, (k − 1 k) must have happened before (k k + 1), otherwise plate n would have ended up somewhere between H1 and Hk−1 . It follows that for any k ≤ i < n the plate i will be on Hi+1 , while plates 1, . . . , k will be on the houses H1 , H2 , . . . , Hk−1 , and Hk+1 in some mixed order. If we restrict our attention to the first k houses: the same happens that would happen if the only houses were H1 , H2 , . . . , Hk ; the only difference is that at the end of the game, Hk should change its plate to n. So there are f (k) different final orderings of the plates such that n is on Hk (for k = 1, . . . , n − 1). Therefore, f (n) =
n−1 X
f (k) = 1 +
n−3 X
2i = 2n−2 .
i=0
k=1
Solution 2. To each pair of neighbouring houses we assign the moment t in time when they swap plates. In this way, we get a sequence t1 , . . . , tn−1 of n − 1 moments in time, such that ti−1 6= ti for all i. This sequence can be split into maximal monotonically increasing subsequences (with each subsequence consisting of consecutive elements). It can also be split into maximal monotonically decreasing subsequences (again, with each subsequence consisting of consecutive elements). If ti−1 > ti , or i = 1, then plate i will finish the day at the eastern end of the maximal increasing subsequence starting at ti . If ti−1 < ti , or i = n, then plate i will finish the day at the western end of the maximal decreasing subsequence ending at ti−1 . Thus, the n − 2 relations < and > between the neighbouring t’s determine the final distribution of the plates. Conversely, given the final distribution of the plates, we can calculate the relations. Indeed, let 2 ≤ i ≤ n − 1. If plate i ends up east of its original position, then ti−1 > ti , whereas if plate i ends up west of its original position, then ti−1 < ti . There is no third possibility. The number of ways to choose these relations is clearly 2n−2 . Each choice can be realized by suitable moments ti in time: we may choose t1 to be any moment during the day, and if t1 , . . . , ti−1 are already given, then we may choose ti to be smaller, resp. greater than ti−1 , as desired. Therefore, the answer is 2n−2 . Problem T-4. Consider finitely many points in the plane with no three points on a line. All these points can be coloured red or green such that any triangle with vertices of the same colour contains at least one point of the other colour in its interior. What is the maximal possible number of points with this property? Solution. The answer is 8. Call a set consisting of red points and green points good if no three points are collinear and any unicoloured triangle contains a point of the other colour. On the one hand, the figure on the left below shows an example of a good set with 8 points. The figure on the right shows the two types of unicololured triangles — all other 6
unicoloured triangles are reflections of those.
On the other hand, we shall prove that a good set can have at most four points of each colour. We give two proofs. First proof. For a proof by contradiction, let S be a counterexample of minimal cardinality. We may assume that S has at least five red points. Let P be any vertex of the convex hull of S. Then P cannot be in the interior of any triangle, so S \ {P } is good. But S was a minimal counterexample, so S \ {P } has at most four points of each colour. Therefore, S has exactly five red points, all vertices of the convex hull of S are red, and S has at most four green points. Consider the convex hull of S. It is a triangle, a quadrilateral, or a pentagon. Case i: The convex hull is a triangle. Let A, B and C denote the vertices of the triangle, and let I and J be the interior red points. Without loss of generality, we may assume that the line IJ intersects sides AB and AC (and not J I BC), and I is nearer to AB than J is. Now ABI, AIJ, AJC, BIJ and BJC are five unired triangles with disjoint interiors, so at least one of them must be empty, because there are at most four C green points. Thus, S is not good, in contradicB tion to our assumptions. A Case ii: The convex hull is a quadrilateral. Let the vertices of the quadrilateral be A, B, C, D, B in this cyclic order, and I be the red point in the I interior. Now ABI, BCI, CDI and DAI are unired triangles with disjoint interiors, each has D a green point inside: denote them by X, Y, Z, W respectively. Then XY Z and ZW X are two uniC green triangles, but both cannot have I (the only possible red point) in their interiors. A
7
A B E
C
D
Case iii: The convex hull is a pentagon. Let A, B, C, D and E denote the vertices of the pentagon, in this cyclic order. Now ABC, ACD and ADE are three unired triangles with disjoint interiors, each must have a green point in its interior, these form a unigreen triangle, which cannot have any red point inside.
Second proof. Lemma. Let a good set of coloured points be given. If the convex hull of some red points contains exactly x red points, with exactly y of them being in its interior, then there are at least x + y − 2 green points in its interior. (The statement is analogous for switched colours.) Proof. If the convex hull is not a polygon (i.e. x ≤ 2), the statement is trivial. Otherwise consider a partition of the convex hull of the red points into triangles, all of which have only red points as vertices, and have no other red points in their interiors. Let N be the number of triangles of the partition. Then the sum of their angles is N π. On the other hand, at each interior point, the sum of the angles of the triangles is always 2π, and at the peripherial points that form a convex (x − y)-gon, the sum is (x − y − 2)π. So the sum of the angles of the triangles in the partition is N π = 2yπ + (x − y − 2)π, which reduces to N = x + y − 2. Each of the disjoint unicoloured triangles must contain a single point in its interior, which proves the lemma. Applying the lemma on all n red points, where m red points are inside their convex hull, gives that there are at least n + m − 2 green points inside the red points’ convex hull. Now applying the lemma again on these green points gives that there are at least (n + m − 2) − 2 red points in their convex hull. But these red points are also interior points of the convex hull of all red points, therefore (n + m − 2) − 2 ≤ m. This reduces to n ≤ 4, and our statement is proven. Remark. The lemma can also be proven by induction on x. Remark. If we allow three colours rather than two, then any set size is possible. For details, see O. Aichholzer. [Empty] [colored] k-gons - Recent results on some Erdős-Szekeres type problems. In Proc. XIII Encuentros de Geometría Computacional, pages 43-52, Zaragoza, Spain, 2009.
Problem T-5. Let ABC be an acute triangle. Construct a triangle P QR such that AB = 2P Q, BC = 2QR, CA = 2RP , and the lines P Q, QR, and RP pass through the points A, B, and C, respectively. (All six points A, B, C, P , Q, and R are distinct.)
8
Solution. There are two possible configurations, but we are giving a proof for only one of them (where P is between A and Q), because it will always give a solution.
Since the angles of triangles P QR and ABC are equal, ∠QAB = ∠RBC = ∠P CA. Let S denote the Brocard point of ABC, the one for which ∠SAB = ∠SBC = ∠SCA. The circle AP C is tangent to AB by the reverse tangent–chord theorem (since ∠P AB = ∠P CA). The same argument holds for S (circle ASC is tangent to AB). It follows that AP SC is cyclic. We intend to prove that triangle AP S is similar to triangle BQS. For this we observe that ∠SAP = ∠SBQ (since ∠SAB = ∠SBC and ∠P AB = ∠QBC), and also ∠ASP = ∠BSQ, because ∠ASP = ∠ACP = ∠BAQ = ∠BSQ. So all the angles are the same in the two triangles, and the same holds for the third triangle CRS. Now considering the twirl with centre S, oriented angle P SA and ratio SA/SP , the image of triangle P QR is triangle ABC. This yields the following construction: • Construct the Brocard point S with the property ∠SAB = ∠SBC = ∠SCA. It is obtained as the intersection point of the three circles ASB, BSC, CSA which are tangent to the lines BC, CA, AB at B, C, A and pass through A, B, C, respectively. • Construct the circle with center S and radius SA/2. • Take the point of intersection of this circle and arc AS of circle ASC not containing C. This point is P . • Take the intersection of line AP and arc BS of circle BSA not containing A: this point is Q. • Take the intersection of line BQ and arc CS of circle CSB not containing B: this is point R.
9
This construction works. We know that the Brocard point S always exists and it is inside the triangle. It is easy to verify that the arcs AS, BS and CS used in the construction are inside the triangle (e.g. arc AS is tangent to side AB and so it is also inside the obtuse triangle ASB). This means that point P is uniquely defined and is inside triangle ABC. Similarly, Q is also unique, and by the tangent–chord theorem ∠P AB = ∠QBC. Now R is also unique, and ∠QBC = ∠RCA. Angle P CA is also the same, because one last time by the tangent–chord theorem ∠P CA = ∠P AB. This means that R, P and C are collinear. We are done. Remarks: • The other point of intersection of the circle around S and circle ASC also works, but in this configuration it is possible for some of the six points (A, B, C, P, Q, R) to coincide. In this case the common angle ∠QAB = ∠RBC = ∠P CA is larger than the Brocard angle, while in the solution it is smaller than the Brocard angle. • The other Brocard point does not work. (It works when P Q passes through B, QR through C and RP through A.) Problem T-6. Let K be a point inside an acute triangle ABC, such that BC is a common tangent of the circumcircles of AKB and AKC. Let D be the intersection of the lines CK and AB, and let E be the intersection of the lines BK and AC. Let F be the intersection of the line BC and the perpendicular bisector of the segment DE. The circumcircle of ABC and the circle k with centre F and radius F D intersect at points P and Q. Prove that the segment P Q is a diameter of k. Solution. The line BC is tangent to the circumcircle of AKC, so the angles BCD and CAK are equal. Analogously, the angles CBE and BAK are equal.
10
Therefore π = |∠KBC| + |∠KCB| + |∠BKC| = |∠DAK| + |∠EAK| + |∠DKE|, so ADKE is a cyclic quadrilateral. Then |∠KBC| = |∠DAK| = |∠DEK|, which implies that DE is parallel to BC. Observe that F is the unique point on line BC for which the angles DF B and CF E are the same. Let F 0 be the point on side BC for which BF 0 KD is cyclic. Then |∠F 0 KE| = 2π−|∠EKD|−|∠DKF 0 | = 2π−(π−|∠BAC|)−(π−|∠CBA|) = π−|∠ACB| which tells us that F 0 CEK is also cyclic, therefore |∠DF 0 B| = |∠DKB| = |∠CKE| = |∠CF 0 E|. This means that F = F 0 and |∠DF B| = |∠CF E| = π − |∠BKC| = |∠BAC|, so the triangles F BD and F EC are both similar to ABC, therefore |AB| |F E| |F B| = = . |AC| |F C| |F D| We get |F B||F C| = |F D|2 . Let Q0 be the intersection of the line P F and the circumcircle of ABC. Using the power of the point F , we get |F D|2 = |F B||F C| = |F P ||F Q0 | = |F D||F Q0 |, therefore |F Q0 | = |F D|, so Q = Q0 . 11
Remark. One can observe that F is Miquel’s point of the cyclic quadrilateral ADKE. Problem T-7. The numbers from 1 to 20132 are written row by row into a table consisting of 2013 × 2013 cells. Afterwards, all columns and all rows containing at least one of the perfect squares 1, 4, 9, . . . , 20132 are simultaneously deleted. How many cells remain? Solution. Let m = 503 and n = 4m + 1 = 2013. Note that (m − 1)n = (m − 1)(4m + 1) < m · 4m = (2m)2 < m(4m + 1) = mn, so the perfect square (2m)2 is in the m-th row. Note that (k + 1)2 − k 2 = 2k + 1 is at most n if k ≤ 2m and is at least n if k ≥ 2m. Therefore, on the one hand, the first 2m + 1 perfect squares never skip a row and the first m + 1 rows are all deleted. On the other hand, the last n − (2m − 1) = 2m + 2 perfect squares are in pairwise distinct rows. Thus, the first m + 1 rows and 2m more are deleted, so m rows remain. The j-th column is deleted if and only if j is a square modulo n = 2013 = 3 · 11 · 61. By the Chinese Remainder Theorem, this is the case if and only if j is a square modulo each of 3, 11, and 61. Since the numbers of squares for these three moduli are 2, 6, and 31 respectively, the number of squares modulo n, again by the Chinese Remainder Theorem, is 2 · 6 · 31 = 372. The number of remaining columns is therefore 2013 − 372 = 1641. The number of remaining cells is 503 · 1641 = 825423. Problem T-8. The expression ± ± ± ± ± ± 12
is written on the blackboard. Two players, A and B, play a game, taking turns. Player A takes the first turn. In each turn, the player on turn replaces a symbol by a positive integer. After all the symbols are replaced, player A replaces each of the signs ± by either + or −, independently of each other. Player A wins if the value of the expression on the blackboard is not divisible by any of the numbers 11, 12, . . . , 18. Otherwise, player B wins. Determine which player has a winning strategy. Solution. We will call a number good if it has a divisor in the set {11, 12, . . . , 18}. We will show that player B has a winning strategy. Namely, B plays 18! in both his first and second move. In his last move he plays a number x (which we will specify later), which ensures that each possible result will be good. When choosing this number, we may, without loss of generality, work with the result mod 18!, thus his first and second moves count as zero. Before he plays the last move, there are eight (= 23 ) possible combinations of signs which give eight results a1 , a2 , . . . , a8 . If we ensure that each of the numbers a1 + x, a2 + x, . . . , a8 + x be good, then also the numbers a1 − x, a2 − x, . . . , a8 − x will be good since for each i ∈ {1 . . . , 8} there exists j ∈ {1, . . . , 8} such that ai − x = −(aj + x). Now observe that all of the numbers a1 , . . . , a8 have the same parity and thus they have at most two different remainders mod 4. Further, if there are two possible remainders, then each appears exactly four times (check it!). At least three of the numbers have the same remainder mod 3, wlog let a1 ≡ a2 ≡ a3 (mod 3). Also, we may assume that a4 ≡ a3 (mod 4). Then, we choose our x from the Chinese Remainder Theorem so that 9 | a1 + x, 5 | a2 + x, 16 | a4 + x, 7 | a5 + x, 11 | a6 + x, 13 | a7 + x, and 17 | a8 + x. With this choice of x we have in fact ensured 18 | a1 + x, 15 | a2 + x, 12 | a3 + x, 16 | a4 + x, 14 | a5 + x, 11 | a6 + x, 13 | a7 + x, and 17 | a8 + x. Hence the result.
13
Contest Problems with Solutions
Jury & Problem Selection Committee
MEMO 2015
I-1
Individual Competition
I-1
A
Find all surjective functions f : N of the following equations is true:
Ñ N such that for all positive integers a and b, exactly one
f pa
f paq f pbq,
bq mintf paq, f pbqu.
Remarks: N denotes the set of all positive integers. A function f : X if for every y P Y there exists x P X such that f pxq y.
Ñ Y is said to be surjective
Solution 1. Each positive integer can be uniquely written as n 2k l where k ¥ 0 and l is odd. We will show that the only function satisfying the conditions is f p2k lq k 1 for all k ¥ 0 and all odd l. Assume that f p1q 1. Since f is surjective, there exists a P N such that f paq 1. Since f p1q 1 f paq, we get f pa 1q mintf paq, f p1qu 1, and inductively we get f pnq 1 for each n ¥ a. However, this contradicts the surjectivity of f . Therefore f p1q 1. Then f p2q mintf p1q, f p1qu 1, and f p3q mintf p1q, f p2qu it easily follows by induction that f pnq 1 if n is odd and f pnq ¡ 1 if n is even.
1.
Now
We will show by induction on k that f p2k lq k 1 for all odd l and f p2k mq ¡ k 1 for all even m. The basis of induction has been proved above. Assume that the statement holds for all k k0 . We define new function g : N Ñ N by g pnq f p2k0 nq k0 . By induction hypothesis g indeed maps to N. In addition, on the set of all integers not divisible by 2k0 , the values of f are smaller than k0 1. Values greater or equal to k0 1 are thus attained by f on the set of integers divisible by 2k0 , making g surjective. A straightforward verification shows that g also satisfies the remaining condition of the initial problem. So g pnq 1 if n is odd and g pnq ¡ 1 if n is even, as we have shown above. Therefore f p2k0 lq k0 1 for odd l and f p2k0 mq ¡ k0 1 for even m, which completes the induction. It is easy to check that this function indeed satisfies the conditions of the problem.
Solution 1a. Like in Solution 1 we prove that f poddq 1,
and
f pevenq ¡ 1.
(1)
We will show by induction on k that f p2k lq k 1 for all odd l and f p2k mq ¡ k 1 for all even m. The basis of induction has been proved above. Assume that the statement holds for all k k0 . Induction step is proved similarly as (1). Suppose f p2k0 q k0 1, meaning that f p2k0 q ¡ k0 1. Surjectivity of f implies, that there exists positive integer b such that
2
MEMO 2015
Individual Competition
I-1
f pbq k0 1. By induction hypothesis b is of the form b 2k0 r for some r (r may be odd or even). Considering the pair p2k0 , bq we get f p2k0 pr 1qq f pb 2k0 q mintf p2k0 q, f pbqu k0 1. By induction we get f p2k0 r1 q k0 1 for all r1 ¥ r, contradicting the surjectivity of f . Hence f p2k0 q k0 1. Conditions of the problem and induction hypothesis imply that f pnq k0 1 iff f pn 2k0 q ¡ k0 1. Therefore it follows inductively that f p2k0 lq k0 1 for odd l and f p2k0 mq ¡ k0 1 for even m, which finishes the induction step. It is easy to check that the function defined by f p2k0 lq conditions of the problem.
k0
1 for odd l indeed satisfies the
Solution 2. Like in Solution 1 we prove that f poddq 1, and f pevenq ¡ 1. Define a sequence of functions gk : N Ñ N by g0 pnq f pnq
and
gk pnq gk1 p2nq 1, for k
P N.
Using first part of solution we prove by induction that all gk satisfy the initial conditions of the problem (they map to N, are surjective and satisfies mutually exclusive equations). It follows from the first part of the solution that gk poddq 1 for all k 0, 1, 2, . . . From gk plq 1 for odd l we inductively obtain f p2k lq k i by backward substitution. This shows that shows that the problem has a unique solution given by f p2k lq k 1 for all k ¥ 0 and all odd l. It is easy to check that this function indeed satisfies the conditions of the problem. Solution 3. Plugging pair pa, aq into the given equations we obtain f p2aq mintf paq, f paqu f paq, in particular f p4aq f p2aq. From pair pa, 2aq we get f p3aq f p2a aq mintf p2aq, f paqu. Suppose f p2aq f paq. Then f p3aq f p2aq f paq. Considering pair pa, 3aq we thus get f p4aq mintf paq, f p3aqu f p2aq, a contradiction. Hence f p2aq ¡ f paq. Next we prove by induction on l that f plaq f paq for all odd l. For l show. We assume that f ppl 2qaq f paq. As f p2aq ¡ f paq, we have
1, there is nothing to
f plaq mintf ppl 2qaq, f p2aqu mintf paq, f p2aqu f paq, which proves the induction step. Let now n 2k l for odd l. By the above we have f pnq f p2k q. Thus we only have to determine f p2k q for k ¥ 0. Since f p2aq ¡ f paq for all a, f p2k q is increasing in k. By surjectivity, the only solution is f p2k q k 1. It is easily seen that f p2k lq k 1, is indeed a solution.
3
MEMO 2015
I-2
Individual Competition
I-2
C
Let n ¥ 3 be an integer. An inner diagonal of a simple n-gon is a diagonal that is contained in the n-gon. Denote by DpP q the number of all inner diagonals of a simple n-gon P and by Dpnq the least possible value of DpQq, where Q is a simple n-gon. Prove that no two inner diagonals of P intersect (except possibly at a common endpoint) if and only if DpP q Dpnq. Remark: A simple n-gon is a non-self-intersecting polygon with n vertices. A polygon is not necessarily convex.
Solution 1. First we prove that for every n-gon P with n ¥ 4 we have DpP q ¥ 1. Let A be one of the vertices of P with inner angle les than 180 . Denote the two vertices adjacent to A by B and C. The segment BC is a diagonal of P , since n ¥ 4. If it lies in P , we are done, so suppose it does not lie in P . Let C 1 be the unique point on the segment AC such that the triangle ABC 1 lies in P and the segment BC 1 contains at least one point in the boundary of P distinct from B and C 1 . Let D C 1 be the point on the segment BC 1 , which lies in the boundary of P and is closest to C 1 . Then D must be a vertex of P and AD is an inner diagonal. Next we prove that Dpnq n 3.
n 1 vertices On the picture diagonals between pairs of points on bottom are clearly outer because that part of polygon is concave. Therefore inner diagonals only exist between upper point and lower points. Number of those diagonals is n 3 therefore Dpnq ¤ n 3. We prove by induction that Dpnq ¥ n 3. The case n 3 is clear. So suppose n ¥ 4 and let P be a n-gon. By the above there exists an inner diagonal of P . This diagonal divides P into two polygons R and S with k and m vertices respectively. Clearly k, m n and k m n 2. By induction we have Dpk q ¥ k 3 and Dpmq ¥ m 3. Note that DpP q ¥ DpRq DpS q 1, hence DpP q ¥ pk 3q pm 3q 1 k m 5 n 3. Since P was arbitrary this shows that Dpnq ¥ n 3. Now we prove the claim by induction. Again the case n 3 is clear. So assume n ¥ 4. As above there exists an inner diagonal d of P and it divides P into polygons R and S with k and
4
Individual Competition
MEMO 2015 m vertices, where k
mn
I-2
2. In addition
DpP q ¥ DpRq
DpS q
1 ¥ Dpk q
Dpmq
1 n 3.
If DpP q Dpnq n 3 then in the above inequality we actually have equalities. In particular DpP q DpRq DpS q 1 which means that the inner diagonals of P are d and those that lie in R or S. In addition DpRq Dpk q and DpS q Dpmq, so by induction the inner diagonals of R and S do not intersect. Thus the inner diagonals of P do not intersect. Conversely, if the inner diagonals of P do not intersect then the inner diagonals of R and S do not intersect and DpP q DpRq DpS q 1 holds. By induction we have DpRq Dpk q k 3 and DpS q Dpmq m 3, thus DpP q pk 3q pm 3q 1 n 3 Dpnq. Solution 2. (using triangulation) Claim 1. Every n-gon P can be triangulated with exactly n 3 inner diagonals. Proof This is well known but can also be proven from DpP q ¥ 1 by induction. Claim 2. For every polygon P and and every inner diagonal ` there exists triangulation that includes `. Proof Diagonal ` divides polygon P into two separate polygons. Both of them can be triangulated. Therefore also P can be triangulated. From claim 1 it follows that Dpnq ¥ n 3. With the same example as in solution 1 we can show that Dpnq ¤ n 3. Therefore Dpnq n 3. Option 1. If inner diagonals in n-gon do not intersect then they must all form one triangulation. Because every triangulation has exactly n 3 diagonals the number of all inner diagonals in P is n 3 and therefore DpP q n 3. Suppose DpP q n 3. If we take a triangulation of P it has exactly n 3 inner diagonals. Therefore all inner diagonals are included in this triangulation and they do not intersect. So no inner diagonals in P intersect. Option 2. We can prove that inner diagonals of n-gon do not intersect if and only if there exists exactly one triangulation of P . Suppose some two inner diagonals would intersect. By claim 2 each of them would be a part of some triangulation and those two triangulations would be different. Hence we have a contradiction. Similarly suppose there would exist at least two different triangulations. Because each of them is produced with n 3 inner diagonals some diagonals must be different. Therefore some of them must intersect otherwise there would exist triangulation with more diagonals. Hence we again get a contradiction and the equivalence is proven.
5
MEMO 2015
Individual Competition
I-2
It remains to prove that P has exactly one triangulation if and only if DpP q n 3. If P has only one triangulation then P has at least n 3 inner diagonals. But there can not be any other inner diagonals otherwise it would follow from claim 2 that there exists a different triangulation with some other diagonals. If it holds DpP q n 3 then there must exist exactly one triangulation with exactly those n 3 inner diagonals.
6
MEMO 2015
I-3
Individual Competition
I-3
G
Let ABCD be a cyclic quadrilateral. Let E be the intersection of lines parallel to AC and BD passing through points B and A, respectively. The lines EC and ED intersect the circumcircle of AEB again at F and G, respectively. Prove that points C, D, F , and G lie on a circle.
Solution 1. The solution uses directed angles. It suffices to show done as follows
=GDC =GF C, which is
=GDC =EDC =EDB =BDC =DEA =BAC =GEA =ABE =GBA =ABE =GBE =GF E =GF C.
7
Individual Competition
MEMO 2015
I-4
I-4
N
Find all pairs of positive integers pm, nq for which there exist relatively prime integers a and b greater than 1 such that am b m an bn is an integer.
Answer. pm, nq integer
pqn, nq where q
Solution 1. If
m n
am an
bm bn
is an odd positive integer and n is an arbitrary positive
q is an odd integer, we have
n q n q pa aqn pbbn q panqq1 panqq2 bn an pbnqq2 pbnqq1,
what is an integer for all positive integers a and b. We will prove that pairs pqn, q q, where q is an odd integer, are the only solutions. Let us assume the opposite, i.e. that there exist pairs pm, nq that are solutions to our problem for which mn is not an odd integer. Among those pairs, let us choose one pair having the minimal sum. Obviously, m ¡ n. Let m n assume a ¡ b. In that case
k for a positive integer k. Without loss of generality, we may
am b m an b k b m ¡ an b n an b n thus there exists a positive integer t such that am an
bm bn
bk
bk ,
t.
This equation can be written as follows: am
pbk tqpan am an bk tpan bm
bn q,
bn q.
Since a and b are relatively prime, an bn and an are relatively prime as well. Therefore, from the last equation we can conclude that t is divisible by an . Let c be a positive integer such that t c an . We have ak b k c an c b n . The right-hand side of the previous equation is greater than an so we conclude that k Previous equation can be written as an pakn cq bn pbkn
¡ n.
cq.
8
Individual Competition
MEMO 2015
I-4
This implies that bkn c is divisible by an , since a and b are relatively prime. Let x be a positive integer such that b k n c x an . The previous equation gives us
ak n c x b n .
Summing the last two equations gives us ak n which means that is an integer. Since pk nq n k sum, we conclude that
m
bk n
x p an
ak n an
bkn bn
bn q,
n and because we have chosen pm, nq to have minimal
kn s n is an odd positive integer. Let r ¥ 0 be an integer such that s 2r k n p2r i.e. k
p2r
1. This implies that
1q n, 2q n.
This means that
m n k p2r 2q n n 2r 3, n n n m which contradicts our assumption that n is not an odd integer. Therefore, the only solutions are pairs pm, nq pqn, nq where q is an odd positive integer and n is an arbitrary positive integer. Solution 2. Clearly m ¡ n. Write m kn am an is integer,
bm apk1qn an bn
r bn
bm bn
apk1qn
r, where k r
¥ 1 and 0 ¤ r n. Since
bm apk1qn r bn an b n
is integer as well. However, since a and b are coprime,
bmn apk1qn an b n
r
apk2qn
r
apk2qn r bn bmn an b n
is again an integer. Proceeding this way we get that an bn divides br p1qk ar . Since |br p1qk ar | an bn, we conclude that br p1qk ar 0. Since a and b are coprime, r has to be zero and k odd. So the only solutions are pkn, nq where k is an odd integer.
9
Individual Competition
MEMO 2015
Solution 3. If m n, then am bm an that m ¥ n. Using long division, we get:
pam
bm q : pan bn q amn am2n bn am bn amn bm bn amn amnbn b2nam2n bm b2n am2n am2n b2n am3n b3n bm am3n b3n .. .
I-4
bn and so there are no solutions. Assume now am3n b2n
The remainders after each step are of the form bm p1qk amkn bkn . For the expression to be an integer, one of these expressions has to be equal to zero. This can only happen when k is odd and m kn. Finally, we check that for pm, nq pkn, nq for k odd we get am
bm
pan
bn qppan qk1 pbn qk1 q.
10
Team Competition
MEMO 2015
T-1
T-1
A
Prove that for all positive real numbers a, b, c such that abc 1 the following inequality holds: a
b c2
2b
a2
2c
b2
2a
a2
¤
c
b2 3
c2
.
Solution 1. Using the given condition abc 1 we get the following: ¸ cyc
c2
a
¸
a b c2 cyc b AM–GM ¸ ?3 a2 2 ¤ cyc 3 b c
2b
¤
GM–AM
¸
cyc
¸ a
3
cyc
a a 3
1
a 3
°
? 3
a2
p
1q
¸ a 2a
9
cyc
°
29
¸
1¸ a. 9 cyc
a2
cyc
°
Now it suffices to prove that 29 cyc a2 91 cyc a ¤ 13 cyc a2 , which is equivalent with ° cyc a and that can be easily proven in the following way: ¸
a2
¥
°
3
QM–AM
cyc
cyc a 3
2
°
¸ A–G cyc a a ¥ 3 cyc
° cyc
a
3
3
? 3
abc
¸
° cyc
a2
¥
a.
cyc
Solution 2. As in the first solution we first show a
b c2
2b
c a2
2c
2a
b2
5
¤
5
a3
5
b3 3
c3
.
Now we use the condition abc 1 and Muirhead’s inequality to get 5
a3
5
b3
a
5
c3
16 9
1
1
b9 c9
1
16
1
1
a9 b 9 c9
1
16
a9 b9 c 9
¤ a2
b2
c2 .
The inequality is proved. Solution 3. Using the given condition abc 1 we get the following: ¸ cyc
a 2b
c2
¸ a2 bc cyc
¤
2b
HM–GM
c2
¸1 cyc
3
¸
a
3
c b
2 cyc c
c
2
a2
b cc c
¸1 cyc
3
¸
a2
1 cyc c
a
2
? 3
c b
1 c
bc
¸1 cyc
3
c
a
2
3
1 a
¸1 cyc
3
5
a3
11
Team Competition
MEMO 2015
T-1
Using the inequality between quadratic mean and mean with coefficient 53 we get the following: 5
a3
5
b3 3
5
c3
Now it is sufficient to prove that
¤
a2
b2 3
c2
15 2 3
a2
b2 3
c2
5 6
.
¥ 1, which follows directly from GM–AM inequality: b2 c 2 ? 2 2 2 ¥ a b c 1 3
a2 b2 c2 3
a2
3
12
Team Competition
MEMO 2015
T-2
T-2
A
Determine all functions f : Rzt0u Ñ Rzt0u such that f px2 yf pxqq
f p1q x2 f pxq
f py q
holds for all nonzero real numbers x and y.
Answer. f pxq
1 x2
and f pxq x12
Solution 1. Let f be any function with the desired property and set α f p1q. Lemma. Let x z 2 3z 2α.
P R0 be arbitrary and put z x2f pxq.
Then f pz q
z, f pz2q 2z α and
Proof. Substituting y 1 and y z into the given functional equation we obtain f pz q z and f pz 2 q α z f pz q, whereby the first two parts of the claim are proved. Applying the first part to z in place of x we infer that z 2 f pz q z 3 is a fixed point of f as well, i.e., f pz 3 q z 3 . On the other hand we may plug y z 2 into the given equation, thus getting f pz 3 q α z f pz 2 q 3z α. Comparing the two previous results we learn indeed z 3 3z 2α. In the particular case x 1 we have z α and the third part of the lemma tells us α3 Since the number α is a value attained by f , it cannot vanish, so α 1.
α.
Let us now return to the situation of the above lemma. The third equation may now be rewritten as pz αq2 pz 2αq z 3 3z 2α3 0. It follows that either z α or z 2α. Assume there were a nonzero real number x such that z x2 f pxq has the property z 2α. Then our lemma yields f pz q 2α, whence z 2 f pz q 8α3 8α R tα, 2αu, which means that z in place of x violates the result from the previous paragraph. This proves that z α holds for all real x 0. In other words we have f pxq xα2 for all nonzero real numbers x. Due to α that f is one of the two functions mentioned in the answer.
1 this shows
It is easy to verify that they do indeed solve the functional equation under consideration both of its sides being equal to α yα2 .
13
MEMO 2015
Team Competition
Solution 2. First we insert y
1 into the equation and we get f px2 f pxqq x2 f pxq
T-2
(1)
for all nonzero real numbers x. In particular, f pf p1qq f p1q. Putting x 1 into the given equation yields f pyf p1qq f py q for each y 0. In particular, inductively we get f pf p1qk q f p1q for each k ¥ 1. On the other hand, (1) for x f p1q yields f pf p1q3 q f p1q3 , so f p1q3 f p1q and f p1q 1. Now we insert y
x2f pxq into the given equation and using (1) we get f px4 f pxq2 q 2x2 f pxq 1
for each x 0. Next, for y
x4f pxq2 we get f px6 f pxq3 q 3x2 f pxq 2
for all x 0. On the other hand, substituting x2 f pxq for x into (1) we get f px6 f pxq3 q x6 f pxq3 for all x 0. Therefore 0 x6 f pxq3 3x2 f pxq 2 px2 f pxq 1q2 px2 f pxq 2q, i.e. f pxq P t x12 , x22 u for each x 0. Assume that f px0 q x22 for some x0 0. Inserting 0 x x0 into the given equation yields f p 2y q f py q 3 for each y 0, in particular, f p 2q 2. However, this is a contradiction, since f p 2q P t 14 , 21 u. Thereofre f pxq x12 for each x 0, i.e., if f p1q 1, then f pxq x12 for each x 0, and if f p1q 1, then f pxq x12 for each x 0. Clearly both functions indeed satisfy the given equation.
14
Team Competition
MEMO 2015
T-3
T-3
C
There are n students standing in line in positions 1 to n. While the teacher looks away, some students change their positions. When the teacher looks back, they are standing in line again. If a student who was initially in position i is now in position j, we say the student moved for |i j | steps. Determine the maximal sum of steps of all students that they can achieve.
Answer.
n2 2
for odd n
n2 1 2
for even n and
°
Solution 1. Let us denote xi the place of student i after switching places. Since ni1 i xi °n °n i1 i i1 xi 0 holds, sum of summands i xi which are negative is the same as sum of ° absolute values of summands i xi which are negative. Therefore to maximize sum ni1 |i xi |, it is enough to maximize sum of positive summands i xi . Let k of summands be positive. ° ° ° Then kj1 ij xij ¤ njnk 1 j kj1 j k pn k q which is maximal when k t n2 u with 2 2 value n2 for even n and n 21 for odd n. This sum of movements can be achieved if students i and n i 1 switch places for i 1, . . . , t n2 u. °n
| xi i |
Solution 2. S pxq
i 1
For maximum define x: xi
n
S pxq
n ¸
1 i, or xi
i n2 , or something similar:
| xi i | . . . 2
i 1
Y
]
To show S pxq ¤ n2 : define di xi i; if di say i moves to the left. So ¸ S p xq 2
0
¸ i
xi °
¸ i
i
¸ i
di
n
2
Y n ]
2
Z
n2 2
^
¥ 0 we say i moves to the right and if di 0 we ¸
di
i;i moves R
and
Yn]
di
i;i moves L
¸
di
i;i moves R
¸
di
i;i moves L
so we need to maximize only i;i moves R di . If i and j (i j) move to the right and xi paths of i and j do not intersect) then
j (the
|xi i| |xj j | |xj i| |xi j | So in order to maximize S pxq the paths of all indices which move to the right intersect (xi On the other hand if i and j (i j) move to the right and xi
¥ j).
¥ j then
|xi i| |xj j | |xj i| |xi j |
15
Team Competition
MEMO 2015
T-3
so the end points of indices moving to the right can be arbitrary permuted. So we can demand n i j ñ xi ¡ xj . SoY the ] sum to the right equals less than pn 1q pn 3q . . . pn 2t 2 u 1q. Times 2 equals . . .
n2 2
Solution 3. S pxq
.
°n
| xi i |
i 1
When we resolve absolute values, we get ¸
S pxq
pxi iq
some i
so
S p xq
¸
pi xiq
other i n ¸
ai
i 1
n ¸
bi
i 1
where the numbers ai and bi are all numbers from 1 to n (twice!). So S p xq ¤ p n
n
pn 1q pn 1q Y
which evaluates to . . . for all i.
n2 2
...
pt n2 u
1qq p1
1
2
2
...
t n 2 1 uq
]
. Equality is attained when i and xi are on different sides of
n 1 , 2
16
Team Competition
MEMO 2015
T-4
T-4
C
Let N be a positive integer. In each of the N 2 unit squares of an N N board, one of the two diagonals is drawn. The drawn diagonals divide the N N board into K regions. For each N , determine the smallest and the largest possible values of K. 2
4
1
3 6
5
7
Example with N
3, K 7 Z
Answer. The smallest K is 2N and the largest is
pN
1q2 2
1
^
.
Solution 1. Minimum A small triangle is a right-angled isosceles triangle whose area is 21 whose hypotenuse is a diagonal of a unit square. A board segment is a horizontal or a vertical segment on the boundary of the board. There are 4N board segments and each of these segments belongs to the boundary of some region. Crucial remark is that each region has either 0 or 2 board segments on its boundary. Indeed, let R be a region that has at least one board segment on its boundary. Let us colour one such board segment in red and then colour the small triangle whose leg is that board segment. In each subsequent step we colour red the unique small triangle which was not coloured so far and which has one of its legs coloured red. This process ends when the other leg of the small triangle is also a board segment. In this way we have exhausted all small triangles of which R consists and shown that R has exactly two board segments on its boundary. It follows that if the number of regions is K, then there is at most 2K board segments. Thus 2K Example with K
¥ 4N
ùñ
K
¥ 2N.
2N :
Maximum The sum of areas of all regions is constant as it is equal to the area of the board.
17
MEMO 2015
Team Competition
T-4
An inner region is a region that has no board segments on its boundary. The boundary of an inner region consists of diagonals and all of them lie on one of two parallel directions. Let us start at some point of the boundary of an inner region and trace the boundary clockwise. In order to return to the same point (i.e. to close the boundary) we need to change direction at least three times, which means that there are at least four diagonals on its boundary. Each diagonal belongs to a different small triangle, so the area of an inner region is at least 2. If a region is not inner, then it has exactly two board segments on its boundary. If these two segments meet at the corner of a board, then the region consists of a single small triangle and has area 21 . We call such regions corner regions. If a region is not inner and not a corner region, we call it outer. An outer region has exactly two board segments on its boundary, which are not legs of the same small triangle, so each such region has an area at least 1. The area is exactly 1 if the two board segments on the boundary are on the same side of the board and share an endpoint. The number of non inner regions is 2N , so their area is at least 4 21
p2N 4q 1 2N 2. Case 1. If N is even, it is possible to make 4 corner regions and 4 21 N 1 2N 4 regions of area 1. So, the area on non inner regions is at least 2N 2 and the area of inner regions is thus at most N 2 2N 2. It follows that there are at most 21 pN 2 2N 2q inner regions, i.e. there are at most 2N
N 2 2N 2
2
pN
1q2 2
1
regions when N is even. Case 2. Let us consider the case when N is odd. If there are exactly 2 corner regions, the total are of outer and corner regions is at least 2 21 p2N 2q 1 2N 1. If there are 3 corner regions, then there are two sides of the board with 2 board segments belonging to corner regions. These sides have an odd number of board segments belonging to outer regions. Hence there must be an outer region which has two board segments on different sides of the boards and its area is at least 32 . We see that in this case the area of all outer and corner regions is at least 3 21 32 p2N 4q 1 2N 1.
18
MEMO 2015
Team Competition
T-4
Also, if there are 4 corner regions, all four sides of the board have an odd number of board segments belonging to outer regions, so at least 2 outer regions have area 23 . The total area (of outer and corner regions) in this case is also at least 4 12 2 32 p2N 6q 1 2N 1. If there would be no corner regions or exactly 1 corner region, then the total area of all outer and corner regions would be at least 1 21 p2N 1q 1 ¡ 2N 1. (We could have argued that these cases are actually impossible, but for the sake of our argument this is sufficient.) So the area of all non inner regions is at least equal to 2N 1. The remaining area is at most N 2 2N 1 pN 1q2 , so there are at most 21 pN 1q2 inner regions. This implies that there are at most p N 1q2 p N 1q2 2N 2 2 regions when N is odd. The following examples show that these numbers of regions can be obtained. Example:
Z
Answer: the smallest K is 2N and the largest is
pN
1q2 2
1
^
.
Remark: every configuration of chosen diagonals determines a set of paths (which may even not be paths but cycles): when you enter into a small square, you leave it on your left or on your right (the chosen diagonal determines that). So if you start in any region, in any square, and follow your path, only one of two possibilities happen: you leave the big square, or you return to the starting point and the path actually is a circle (and, with a chess argument, an even cycle). In the case where you leave the big circle, if you follow the path from the starting point into the opposite dirrection, you cannot return to this point but you also leave the big square so you get a path that starts and ends on the boundary of a big square. Of course every path/circle corresponds to a region. This approach in a way replaces the red-triangle argument from the official solution and the proof that the smallest area of inner regions is 2.
19
Team Competition
MEMO 2015
T-4
Solution 2. We make use of the generalized Euler’s polyhedron formula V
F
E
C
1
Herein V denotes the number of vertices, E the number of edges, F the number of faces (regions) and C the number of connected components of a planar graph. We apply this formula to the graph whose vertices are the pN 1q2 corner points of all the N 2 unit squares. The edges are the 4N segments on the circumference and the N 2 drawn diagonals. Then we get for the number of faces (without the exterior face) K
F 1EV
C
4N
N 2 pN
1q2
C
2N 1
C
Since C ¥ 1 we must have K ¥ 2N . We can easily achieve C 1 and K 2N , for instance by choosing all the diagonals parallel to each other. Hence 2N is the least possible value of K. In order to find an upper bound for C we assign to every corner point its boundary distance, i. e. the smallest distance from the four sides of the N N square. (The corner points with boundary distance d lie on the circumference of a square of side length N 1 2d, there are exactly pN 1 2dq2 pN 1 2dq2 4pN 2dq such points, except for N 2d, in which there is exactly one such point – the midpoint of the board.) Furthermore to a connected component Z of the graph we assign the minimal boundary distance DpZ q of the corner points contained in Z. (Since all the corner points lie in the same connected component, there is exactly one component Z0 such that DpZ0 q 0.) We now consider two neighbouring corner points both having boundary distance d ¥ 1 and we observe that at least one of them must be connected to a point with boundary distance d 1. (Namely these two corner points are two vertices of a unit square whose other two vertices have boundary distance d 1. The diagonal drawn in this unit square is the desired connection.) That means that for 2d N the number of connected components Z with DpZ q d is at most 2pN 2dq, i. e. half the number of corner points with boundary number d. Now it follows that C
¤1
R
pN 1q2 V , 2
that is
K
¤ 2N
R
pN 1q2 V R pN 2
1q2 2
V
.
(Here the ceiling function takes into account the special role of the midpoint in the case 2d N for even N .) In order to prove that this value of K can actually be reached, we consider the board with the corners p0, 0q, pN, 0q, pN, N q, p0, N q and draw in each of the N 2 unit squares that diagonal the both endpoints of which have an odd sum of coordinates. In this case everyQpoint in U the interior p N 1q2 of the board with even sum of coordinates, is isolated. Actually there are such corner 2 points, i. e. z achieved.
¥1
Q
U
pN 1q2 . In this situation the maximal value K 2
Q
U
pN 1q2 is actually 2
20
Team Competition
MEMO 2015
T-5
T-5
G
Let ABC be an acute triangle with AB ¡ AC. Prove that there exists a point D with the following property: whenever two distinct points X and Y lie in the interior of ABC such that the points B, C, X, and Y lie on a circle and
=AXB =ACB =CY A =CBA holds, the line XY passes through D.
Solution. Let D be the point on BC for which AD is a tangent to the circumcircle of ABC. As we will show in the sequel, the point D is as desired.
Solution 1. We assume B, C, X, Y lie on a circle in that order, the other case being similar. We compute the following equality
=AXY =DAY
=AXB =Y XB =DAB =BAY =AXB =Y CB =ACB =BAY =AXB 2=ACB =ACY =BAC =Y AC =AXB 2=ACB =BAC π =CY A =AXB =AY C =CBA =ACB 0.
So AD is tangent to the circumcircle of triangle 4AXY . Consequently AD is the radical axis of the circumcircles of triangles 4AXY and 4ABC. On the other hand BC is the radical axis
21
MEMO 2015
Team Competition
T-5
of the circumcircles of triangles 4BCX and 4ABC. By a well known theorem the radical axis of three circles intersect in one point, so XY passes through D.
Solution 2. Let X and Y be two points satisfying the condition mentioned in the problem and let the line DX meet the circumcircle of triangle BXC for the second time in Y 1 . It suffices to prove that Y Y 1 . Let us assume that the points D, X and Y 1 are collinear in this order, the other case being similar. Due to DX DY 1 DB DC DA2 the triangles ADX and Y 1 DA are similar. Hence
=CY 1A 360 =AY 1D =DY 1C 180 =DAX =CBX p180 =BAX q =DAB =CBX =AXB =XBA =ACB =CBX =AXB =CBA =ACB. Using the condition
=AXB =ACB =AY 1C =CBA
we get =AY C =AY 1 C, so Y and Y 1 lie on the same circle through A and C. On the other hand, Y and Y 1 both lie on the circumcircle of 4BCX, therefore Y Y 1 .
22
Team Competition
MEMO 2015
T-6
T-6
G
Let I be the incentre of triangle ABC with AB ¡ AC and let the line AI intersect the side BC at D. Suppose that point P lies on the segment BC and satisfies P I P D. Further, let J be the point obtained by reflecting I over the perpendicular bisector of BC, and let Q be the other intersection of the circumcircles of the triangles ABC and AP D. Prove that =BAQ =CAJ.
T
C P Q
D I S
B A
J
Solution.
23
Team Competition
MEMO 2015
T-6
Let AI intersect the circumcircle of triangle ABC for the second time at T . It is known that T is the centre of the circumcircle of triangle BIC and due to symmetry the point J lies on this circle as well. Since
=BQP =AQP =AQB π =ADP =ACB =DAC =BAT =BQT, the points T , P , and Q are collinear. Now let T J intersect BC at S. We have IJ k BC and the triangle JT I is isosceles, so ST D is isosceles as well. The same applies to DP I and as their base angles are both equal to π =ACB 2
=CBA
we must have =IT S =IP S as well, meaning that the quadrilateral IP T S is cyclic. It follows that =SP T =SIT . Their angles being equal, the triangles T AB and T BD are similar, whence
|T D | |T B | |T B | |T A| In view of |T D| |T S | and |T B | |T I | |T J | this yields |T S | |T J | |T I | |T A| which proves IS k AJ. It follows that =SIT =JAT , which in combination with the result of our third paragraph proves
=IAQ π =QP D =SP T =JAT . Using
=T AC =BAI we get =JAC =BAQ.
24
Team Competition
MEMO 2015
T-7
T-7
N
Find all pairs of positive integers pa, bq such that b! ab
a!
ba .
Answer. pa, bq P tp1, 1q, p1, 2q, p2, 1qu Solution. If a b, the equation reduces to a! aa . Since aa ¡ a! for a ¥ 2, the only solution in this case is a b 1. If a 1, the equation reduces to b! b, which gives an additional solution a 1, b 2. We prove a b 1; a 1, b 2 and a 2, b 1 are the only solution of the Diophantine equation. Assume a, b is another solution satisfying 1 a b (the case 1 b a is symmetric). This implies a|b! and consequently a|ba . Let p be a prime factor of a. By just argued, also p|b. We compare the exponent of p in prime factorizations of both sides of the equation. LHS of the b! b! b! equation can be rewritten as a!p a! 1q. Since p|b and b ¡ a we have p| a! and hence, a! 1 is coprime to p. Thus, the exponent in prime factorization of LHS equals the exponent of p in prime factorization of a!. It is well know, that this equals
8 ¸
k 1
°8 Y
Z
a pk
^
Z ^
a p
Z
a p2
^
Z
a p3
^
...
]
We have k1 ap pa2 ap p1 1 q ¤ a. The exponent of p in prime factorization of RHS is however at least a since p|a, p|b and b ¡ a. This contradicts the assumption that a, b is a solution. Therefore there are no solutions to the equation, when a, b ¥ 2. a pk
25
Team Competition
MEMO 2015
T-8
T-8
N
Let n ¥ 2 be an integer. Determine the number of positive integers m such that m m2 1 is divisible by n.
¤ n and
Answer. Dp2α0 pα1 1 pα2 2 pαk k q 2k Solution 1. Let Dpnq be the number of positive integers m such that m divisible by n.
¤ n and m2
1 is
No number of the form m2 1 is divisible by 4, so if 4 divides n, we have Dpnq 0. It is also known that Dpnq 0 if n is divisible by some number of the form 4k 3. Furthermore, Dp2q 1. 1. Assume first that n p is an odd prime of the form 4k
1. We show that Dppq 2.
Lemma: If p 4k 1, where k is a positive integer and p is a prime number, and if S tx1 , ..., xp u is a complete residue system modulo p, then there exist exactly two elements x P S for which x2 1 pmod pq. First, we will prove that congruence equation x2 if p 1 pmod 4q.
1 pmod pq has at least one solution
Using Wilson’s theorem, we have
p1 1 2 2
thus x
pp 1qpp 2q p
p 1 2
p 1 ! 2
2
1 pmod pq
! is a solution.
p 1 2
Furthermore, if xi P S is a solution then xj p xi P S is also a solution. If p 2q 1, exactly one of the numbers xi , xj is smaller than or equal to q. We can assume that xi ¤ q. If given congruence equation had another solution xk P S, we could, by the same argument, assume that xk ¤ q. However, x2i x2k 1 pmod pq implies that p divides pxk xiqpxk xiq, which is impossible since xi, xk ¤ q. This completes the proof of lemma. 2. Now let n pk be a prime power where p is congruent to 1 modulo 4. We will prove by induction that Dppk q 2. Induction basis, the case for k
1, is the previous step.
Assume that Dppk q 2 for some positive integer k. Let i and j be those two integers less then pk such that i2 pk .
1 and j 2
1 are divisible by
26
Team Competition
MEMO 2015
Then all numbers less then pk the following numbers: mpk
1
that satisfy congruence equation x2
for m 0, . . . , p 1 and mpk
i,
T-8
j,
1 pmod pk q are
for m 0, . . . , p 1.
k 2 Exactly one of the numbers pmp pkiq 1 (for m 0, . . . , p 1) is divisible by p, i.e. exactly one among numbers pmpk iq2 1 (for m 0, . . . , p 1) is divisible by pk 1 ).
To prove that, assume the opposite – that there are two such numbers, namely pm1 pk iq2 1 and pm2 pk iq2 1. This means that number
pm1pk
iq2
pm2p iq pk pk k k p m1 p i m2 p iqpm1 pk k
1
pm1 m2qppk pm1
pk
m2 q
2
1 m 2 pk
i
iq
2iq
is divisible by p which is impossible because neither m1 m2 nor i are divisible by p. In the same way, we conclude that exactly one of the numbers pmpk 0, . . . , p 1) is divisible by pk 1 . Therefore, Dppk
3.
j q2
1 (for m
q 2. Next, assume that n pa q b where p and q are two distinct prime numbers of the form 4k 1, for positive integers k, then Dppa q b q 4 for all positive integers a and b. According to above, Dppa q 2. 1
Let i and j be those two positive integers smaller than pa such that i2 both divisible by pa . All numbers smaller than pa q b that satisfy congruence equation x2 the following: mpa
i for m 0, ..., q b 1 and mpa
j
1 and j 2
1 are
1 pmod paq are
for m 0, ..., q b 1.
Since t0, 1, 2, . . . , q b 1u is a complete residue system modulo q b , the same is true for t0, pa, 2pa, . . . , pqb 1qpau (because pa and qb are relatively prime), hence T ti, pa i, 2pa i, . . . , pq b 1qpk iu is a complete residue system modulo q b , as well. Lemma implies that there are exactly two elements of the set T that satisfy the congruence equation x2 1 pmod q b q. In the same way, there are exactly two elements of the tj, pa that satisfy congruence equation x2 1 pmod q b q.
j, 2pa
j, ..., pq b 1qpa
ju
27
Team Competition
MEMO 2015
T-8
Therefore, Dppa q b q 4. Using the previous part inductively, we conclude that D pα1 1 pαnn prime numbers pi , i 1, 2, . . . , n.
4. Finally, we show that D pα1 1 pαnn D 2pα1 1 pαnn , if pi , i odd prime numbers, all congruent to 1 modulo 4.
2n for distinct odd
1, 2, . . . , n are distinct
Let a pα1 1 pαnn . If i1 , i2 , i3 , . . . i2n are all positive integers less than a that satisfy congruence equation x2 1 pmod aq, then all positive integers less than 2a that satisfy that equation are the following: δa ij , j 1, 2, 3, . . . , 2n for δ 0, 1. However, exactly one of the number i2j D pα1 1 pαnn D 2pα1 1 pαnn . Thus we conclude that
D pα1 1 pαnn
D
1 and pa
ij q2
2pα1 1 pαnn
1 is even, which implies that
2n,
for distinct odd prime numbers pi , i 1, 2, . . . , n. Solution 2. No number of the form m2 1 is divisible by 4, so if 4 divides n, we have Dpnq 0. Also Dp2q 1. Write n pα0 0 pα1 1 pαk k with p0
2, α0 P t0, 1u and pi odd and αi ¥ 1 for i ¥ 1. The problem is to find the number of residue classes m modulo n with m2 1 pmod nq. It is clear that, m2 1 pmod nq if and only if m2 1 pmod pαi q for all i. i
We use the following lemma. Lemma: Let p be a prime number and α ¥ 1. Then the number of residue classes m fulfilling m2 equals
$ ' ' 0 ' &
1
' ' ' %2
1 pmod pαq if p 3
pmod 4q,
2, if p 1 pmod 4q. if pα
Proof of lemma: For pα 2, there is nothing to show. It is well-known that 1 is a quadratic residue modulo an odd prime p if and only if p 1 pmod 4q. We now assume p 1 pmod 4q. It is also known (Hensel lifting) that if some b is a quadratic residue modulo some odd prime p, then b is also a quadratic residue modulo pα .
28
MEMO 2015
Team Competition
T-8
Thus there is at least one residue class m with m2 1 pmod pα q. Another residue class r is also a solution r2 1 pmod pα q if and only if m2 r2 pmod pα q or equivalently pα | pm rqpm rq. We have gcdpm r, m rq | 2m which is coprime to pα. Thus r m pmod pαq.
1 pmod pαq has exactly two solutions. This proves the lemma. By the lemma, α0 does not influence the result. For each 1 ¤ i ¤ k, there are two choices for
Thus m2
m modulo pαi i , thus there are 2k choices for m modulo n by the Chinese remainder theorem.
29
EUROPEAN
MI
DD
LE
ICA AT EM TH MA L
H T N
OL Y M
TE
P I A D
¨ VOCKLABRUCK AUSTRIA 2016
Contest Problems with Solutions Jury & Problem Selection Committee Version 7, 21 October 2016
The Note of Confidentiality
IMPORTANT
We would like to remind everybody of the following MEMO regulation:
These exam problems have to be kept strictly confidential until the contest will have been finished.
Jury & Problem Selection Committee selected 12 problems submitted by the following countries: T-1 Croatia T-2 Lithuania I-1 Austria
T-3 Croatia
I-2 Switzerland
T-4 Austria
I-3 Slovakia
T-5 Croatia
I-4 Croatia
T-6 Poland T-7 Slovakia T-8 Austria
The Problem Selection Committee would also like to thank Roger Labahn for providing the LATEX templates.
Contents Individual I-1 . . I-2 . . I-3 . . I-4 . .
Competition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Team Competition T-1 . . . . . . . . T-2 . . . . . . . . T-3 . . . . . . . . T-4 . . . . . . . . T-5 . . . . . . . . T-6 . . . . . . . . T-7 . . . . . . . . T-8 . . . . . . . .
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15 15 19 22 24 25 28 34 37
Individual Competition
MEMO 2016
I-1
I-1
A
Let n ě 2 be an integer and x1 , x2 , . . . , xn be real numbers satisfying (a) xj ą ´1 for j “ 1, 2, . . . , n and (b) x1 ` x2 ` ¨ ¨ ¨ ` xn “ n. Prove the inequality
n ÿ 1 xj ě 1 ` xj j“1 1 ` x2j j“1 n ÿ
and determine when equality holds.
Solution. We have to prove n ÿ
n n ÿ ÿ 1 xj 1 ´ xj ´ “ ě 0. 2 2 1 ` x 1 ` x p1 ` x qp1 ` x q j j j j j“1 j“1 j“1
We use the supporting line method and consider the function f defined by f pxq “
1´x p1 ` xqp1 ` x2 q
for all x ą ´1. The tangent line of f at x “ 1 is given by y “ f pxq “
1´x . 4
We claim that
1´x 1´x ě 2 p1 ` xqp1 ` x q 4
for all x ą ´1 with equality for x “ 1. For x ě 1 we get 4 ď p1 ` xqp1 ` x2 q and for ´1 ă x ď 1 we get 4 ě p1 ` xqp1 ` x2 q. Both inequalities are obviously true. Now we conclude that
n ÿ
n ÿ 1 ´ xj 1 ´ xj ě “ 0, 2 p1 ` x qp1 ` x q 4 j j j“1 j“1
and we are done. Equality occurs if and only if all n numbers are equal to 1.
Solution. Since 1 ` xj ą 0 for j “ 1, 2, . . . , n, Cauchy-Schwarz inequality yields n ÿ
n ÿ 1 ¨ p1 ` xj q ě 1 ` x j j“1 j“1
˜
n ÿ
¸2 1
,
j“1
4
Individual Competition
MEMO 2016 which is equivalent to
I-1
n ÿ
1 n ě . 1 ` xj 2 j“1 It therefore suffices to prove that
n ÿ
n xj ď , 2 1 ` xj 2 j“1 but this last inequality is equivalent to the trivial one n ÿ p1 ´ xj q2 ě0, 1 ` x2j j“1
so the inequation is proven. In the last equation, we have equality if and only if xj “ 1 for j “ 1, 2, . . . , n, and one can easily see that this is indeed a case of equality, so it is the only case of equality.
Solution The inequality is equivalent to n ÿ
1 ´ xi ě0 p1 ` xi qp1 ` x2i q i“1 1 p1`xqp1`x2 q
As the functions f pxq “ 1 ´ x and gpxq “ the Chebychev inequality to obtain: n ÿ
1 ´ xi n¨ ě p1 ` xi qp1 ` x2i q i“1
˜
n ÿ
are both strictly decreasing, we can apply
¸˜ 1 ´ xi
i“1
n ÿ
1 p1 ` xi qp1 ` x2i q i“1
¸ “0
So we’re done.
Solution (via Lagrange multipliers) Let us write f pxq “ that the expression f px1 q ` . . . ` f pxn q in the domain
1 x ´ . We want to show 1 ` x 1 ` x2
D : x1 , . . . , xn ą ´1, x1 ` . . . ` xn “ n attains its minimal value 0 exactly at the point x1 “ . . . “ xn “ 1. We first consider the boundary of D. This means that w. l. o. g. we may assume that x1 “ ´1, in which case the expression attains the value `8, which is not the minimum. We now look for minima in the interior of the domain: The method of Lagrange multipliers gives the Langrange function F px1 , . . . , xn , λq “
n ÿ
pf pxj q ´ λpxj ´ 1qq
j“1
5
Individual Competition
MEMO 2016
I-1
and results in the system of equations f 1 pxj q “ λ, n ÿ xj “ n.
j “ 1, . . . , n,
j“1
Now note that • f 1 pxq “
´1 x2 ´ 1 2px3 ´ x2 ´ x ´ 1q ` “ . p1 ` xq2 p1 ` x2 q2 p1 ` xq2 p1 ` x2 q2
• f 1 pxq ă 0 for ´1 ă x ď 1. This is obvious from the first expression for f 1 pxq. • f 1 pxq ą 0 for x ą 2. This can be seen from the second expression for f 1 pxq, since for x3 x3 x3 ` ` ă x3 . x ą 2 we have 1 ` x ` x2 ă 8 4 2 2xp3 ´ x2 q 2 • f 2 pxq “ ` . 2 3 p1 ` x q p1 ` xq3 • f 2 pxq ą 0 for ´1 ă x ă 2 can be shown by considering the following three cases: – For ´1 ă x ă 0, we have 1 1 1 “ ą . p1 ` xq3 p1 ` xqp1 ` 2x ` x2 q p1 ` xqp1 ` x2 q Thus we get 2xp3 ´ x2 q 2 2xp3 ´ x2 qp1 ` xq ` 2p1 ` x2 q2 f pxq ą ` “ “ p1 ` x2 q3 p1 ` xqp1 ` x2 q p1 ` xqp1 ` x2 q3 p6x ` 6x2 ´ 2x3 ´ 2x4 q ` p2 ` 4x2 ` 2x4 q 2 ` 6x ` 10x2 ´ 2x3 “ “ “ p1 ` xqp1 ` x2 q3 p1 ` xqp1 ` x2 q3 1 p2 ` 3xq2 ` 11 x2 ´ 2x3 2 ą 0. “ 2 p1 ` xqp1 ` x2 q3 2
? – For 0 ď x ď 3, the assertion is obvious. ? – For 3 ă x ă 2, the assertion follows from 2 2 ą , 3 p1 ` xq 27
2xp3 ´ x2 q 2xpx2 ´ 3q 2 ¨ 2p22 ´ 3q 1 2 “ ´ ą ´ ? 2 “´ ą´ . 2 3 2 3 p1 ` x q p1 ` x q 16 27 p1 ` 3 q3
• Hence f 1 is strictly increasing for ´1 ă x ď 2. ř Since x “ n1 nj“1 xj “ 1, we know that xj ď 1 for some j. Therefore λ ă 0. This means xj ď 2 for every j “ 1, . . . , n. From the monotonicity of f 1 in the domain ´1 ă x ď 2, we now conclude x1 “ . . . “ xn . In view of the condition x1 ` . . . ` xn “ n this means x1 “ . . . “ xn “ 1. Since f p1q “ 0, we are done.
6
MEMO 2016
I-2
Individual Competition
I-2
C
There are n ě 3 positive integers written on a blackboard. A move consists of choosing three numbers a, b, c on the blackboard such that they are the sides of a non-degenerate non-equilateral triangle and replacing them by a ` b ´ c, b ` c ´ a and c ` a ´ b. Show that an infinite sequence of moves cannot exist.
Solution 1. We will show that the product of all the numbers on the blackboard can never increase. Indeed, for the three numbers a, b and c we have the inequalities a2 ě a2 ´ pb ´ cq2 “ pa ` b ´ cqpa ` c ´ bq b2 ě b2 ´ pa ´ cq2 “ pb ` a ´ cqpb ` c ´ aq c2 ě c2 ´ pa ´ bq2 “ pc ` a ´ bqpc ` b ´ aq Since by the conditions of the problem all factors on both sides are positive, we can multiply these equations and obtain a2 b2 c2 ě pa ` b ´ cq2 pb ` c ´ aq2 pc ` a ´ bq2 , which is equivalent to abc ě pa ` b ´ cqpb ` c ´ aqpc ` a ´ bq.
(1)
Since every non-increasing sequence of positive integers is eventually constant, we see that the product of the numbers on the blackboard cannot change after a finite number of moves. Furthermore, it is clear that we have equality in (1) if and only if a “ b “ c, in which case the numbers on the blackboard do not change. It is now clear that after a finite number of moves the numbers on the blackboard will not change.
Solution 2. Since pa ` b ´ cq ` pb ` c ´ aq ` pa ` c ´ bq “ a ` b ` c, it is clear that the sum of the numbers on the blackboard is invariant. On the other hand, we learn from pa ` b ´ cq2 ` pb ` c ´ aq2 ` pc ` a ´ bq2 “ a2 ` b2 ` c2 ` pa ´ bq2 ` pa ´ cq2 ` pb ´ cq2 that the sum of squares of the numbers increases in every move, except in the case a “ b “ c, ř ř when nothing at all changes. But in view of the inequality x2i ď p xi q2 , it now becomes evident that the number of possible "moves with effect" is bounded by the constant right-hand side.
7
MEMO 2016
Individual Competition
I-2
Solution 3. Let m be the smallest integer on the blackboard and k the number of times that m is written on the blackboard. The number n of numbers on the blackboard will stay fixed throughout the process, while m and k might change. We first prove that any move involving a number m will either decrease m or keeping m fixed and increasing k or change nothing on the blackboard: • If m “ a ď b ă c, we have a ` b ´ c ă a “ m, so the new minimum is clearly smaller than the original m. • If m “ a ă b “ c, we have a ` b ´ c “ c ` a ´ b “ a “ m, so the number k has increased while m is fixed. • If m “ a “ b “ c, nothing changes on the blackboard. Now, if there are only finitely many moves that change anything and involve a minimal number then we can conclude by induction on n that we can only make finitely many moves that change anything. The most convenient base case is n “ 2 where nothing changes because no three numbers can be chosen. However, if there are infinitely many moves involving a minimal number that change something then we have seen above that either m decreases or m is fixed and k increases. Since k is at most n and m is at least 1 this is impossible, so the process has to terminate as desired. (The same argument works with the maximum because it is bounded by the constant sum of the written integers.)
8
Individual Competition
MEMO 2016
I-3
I-3
G
Let ABC be an acute-angled triangle with >BAC ą 45˝ and with circumcentre O. The point P lies in its interior such that the points A, P , O, B lie on a circle and BP is perpendicular to CP . The point Q lies on the segment BP such that AQ is parallel to P O. Prove that >QCB “ >P CO.
C
O Q1 A
P
Q B
Solution 1. Since >BAC ą 45˝ , we have >BOC ą 90˝ , and the points A, P , O, B lie on a circle in this order. Instead of the equality >QCB “ >P CO, we show the equivalent statement >OCB “ >P CQ. We know that >OCB “ 90˝ ´ >BAC and >P CQ “ 90˝ ´ >CQP . So we just need to show >BAC “ >CQP . By the construction of P and Q, we have >P QA “ >QP O “ >BP O “ >BAO “ >OBA “ 180˝ ´ >AP O “ >QAP, so the triangle AQP is isosceles with P A “ P Q. Thus if Q1 denotes the point obtained by reflecting Q about P , then >QAQ1 “ 90˝ . Moreover, reusing some part of the above calculation and exploiting that O is the circumcentre of ABC, we find >AQ1 B “ >AQ1 Q “ 90˝ ´ >Q1 QA “ 90˝ ´ >P QA “ 90˝ ´ >OBA “ >ACB, which proves that the quadrilateral ABCQ1 is cyclic. Finally, since >QP C “ 90˝ and P Q “ P Q1 , the triangle Q1 QC is isosceles with CQ “ CQ1 , whence >CQP “ >CQQ1 “ >QQ1 C “ >BQ1 C “ >BAC, as we wanted to show.
9
Individual Competition
MEMO 2016
I-3
Another solution of the second part: Let Q1 be the point of intersection of BP with the circumcircle of the traingle ABC. Then we have >BQ1 C “ >BAC “ α and >AQ1 B “ >ACB “ γ. Since >Q1 P A “ 180˝ ´ >AP B “ 180˝ ´ 2γ we get >P AQ1 “ 180˝ ´ >AQ1 B ´ >Q1 P A “ γ and triangle AP Q1 is isosceles with P Q1 “ P A. Therefore we get P Q “ P A “ P Q1 and the point P is the midpoint of QQ1 . Finally, since >QP C “ 90˝ and P Q “ P Q1 , the triangle Q1 QC is isosceles with CQ “ CQ1 . Therefore >CQP “ >QQ1 C “ >BAC, as we wanted to show. C
Z
O
P
Y Q
A
B
Solution 2. Since >BAC ą 45˝ , we have >BOC ą 90˝ , and the points A, P , O, B lie on a circle in this order. Instead of showing the equality >QCB “ >P CO, we prove the equivalent statement >OCB “ >P CQ. Let Y be the point of intersection of AQ with the circle through A, O and B. Due to P O k AY the quadrilateral AY OP is an isosceles trapezoid with P A “ OY . Since AO “ OB, we can conclude OP “ Y B, e.g. by proving the congruence of the triangles AOP and OBY . Therefore, P BY O is an isosceles trapezoid as well and we get that P QY O is a parallelogram. The triangle AQP is isosceles due to P A “ OY “ P Q and, in addition, similar to triangle ABO because of >AP B “ >AOB. Now let Z be the midpoint of BC. Since >BP C “ 90˝ , we know that Z is the center point of the circle through P , B and C. Hence we get ZC “ ZP “ ZB. Since triangle P BZ is
10
MEMO 2016
Individual Competition
I-3
isosceles, we have >ZBP “ >BP Z. Because of >ZBP “ >ZBY ` >Y BP and >BP Z “ >BP O ` >OP Z and >Y BP “ >BP O, we conclude >ZBY “ >OP Z. Now we have the equality of two corresponding sides and the enclosed angle. Therefore, the triangles Y BZ and P OZ are congruent, yielding OZ “ OY . Since >OZY “ 90˝ ´ >Y ZB “ 90˝ ´ >P ZO “ >CZP and ZC “ ZP , we see that the triangles OY Z and CP Z are similar. Therefore, we get CZ CP “ , OY OZ
hence
CP CZ “ . PQ OZ
Now the similarity of the triangles P QC and OZC follows from >QP C “ >CZO “ 90˝ and we get >P CQ “ >OCZ “ >OCB, which completes the proof.
11
MEMO 2016
I-4
Individual Competition
I-4
N
Find all functions f : N Ñ N such that f paq ` f pbq divides 2pa ` b ´ 1q for all a, b P N. Remark: N “ t1, 2, 3, . . .u denotes the set of positive integers.
Answer. The only solutions are the constant function f paq “ 1 for all a P N and the function f paq “ 2a ´ 1 for all a P N. Solution 1. We will first prove that f is either injective or bounded. Assume that we have f pmq “ f pnq “ t for some m and n. If we plug in m and n as b, we get respectively: f paq ` t | 2pa ` m ´ 1q f paq ` t | 2pa ` n ´ 1q. Since the divisor of two numbers must divide their difference, we get f paq ` t | 2m ´ 2n. That means that if f is not bounded, it is injective, because then we must have m “ n. Case 1: f is injective If we put a “ b “ 1 in the given relation, we get f p1q “ 1. Putting a “ b gives us f paq | 2a ´ 1. Since f is injective, we can now prove by induction that f paq “ 2a ´ 1 for a “ 2, 3, . . . since all smaller divisors of 2a ´ 1 are already attained by previous values of f . Case 2: f is bounded If f is bounded, the maximum of f exists and for any a P N we can choose a prime p that is greater than a and at least three times greater than this maximum. We can now choose b P N such that a ` b ´ 1 “ p and we get f paq ` f pbq | 2p. Clearly, f paq ` f pbq ă p because of our choice of p. So we must have f paq ` f pbq “ 2, which gives us f paq “ 1 for all a P N. It is easily checked that the functions f pnq “ 2n ´ 1 and f pnq “ 1 satisfy the condition of the problem.
Solution 2. Setting a “ b gives f paq | 2a ´ 1, which implies that f p1q “ 1 and that f paq is odd for all a. Setting a “ 2 and b “ 1 gives f p2q ` 1 | 4, therefore f p2q “ 1 or f p2q “ 3. Case 1: f p2q “ 1. We choose b “ 1 and b “ 2 to obtain f paq`1 | 2a and f paq`1 | 2a`2, which implies f paq`1 | 2 and therefore f paq “ 1 for all a. The constant function f paq “ 1 is clearly a solution. Case 2: f p2q “ 3.
12
MEMO 2016
Individual Competition
I-4
We first show that f paq “ 2a ´ 1 for a ą 1 implies f pa ` 2q “ 2a ` 3. Setting b “ a ` 2 gives f pa ` 2q ` 2a ´ 1 | 2p2a ` 1q. If f pa ` 2q “ 1, then a | 2a ` 1. This implies a “ 1, which was excluded. Therefore, the odd number f pa ` 2q is greater than 2, so f pa ` 2q ` 2a ´ 1 is a divisor of 2p2a ` 1q that is greater than 2a ` 1 (half of 2p2a ` 1q). Thus f pa ` 2q ` 2a ´ 1 must be equal to 2p2a ` 1q, which gives f pa ` 2q “ 2a ` 3 as desired. Therefore, f p2q “ 3 implies f p4q “ 7. Now, setting a “ 3 and b “ 4 gives f p3q ` 7 | 12, which implies f p3q “ 5. Since we now know that f paq “ 2a ´ 1 holds for 1, 2, 3, 4 and we can use induction in steps of two, we get f paq “ 2a ´ 1 for all a, which is clearly a solution. The two solutions are f paq “ 1 for all a and f paq “ 2a ´ 1 for all a.
Solution. 3. We have f p1q “ 1 and f paq | p2a ´ 1q as in the previous solution. If f p2q “ 1, then f paq “ 1 for all a as in the previous solution. Therefore, we only have to consider f p2q “ 3. We easily check that f paq “ 2a ´ 1 for all a is a solution. Choose k maximally such that f paq “ 2a ´ 1 holds for all 1 ď a ď k. Then setting a “ k and b “ k ` 1 yields 2k ´ 1 ` f pk ` 1q “ f pkq ` f pk ` 1q | 4k, which by maximality of k implies that f pk ` 1q “ 1. Setting a “ k ´ 1 and b “ k ` 1 yields 2k ´ 2 “ f pk ´ 1q ` f pk ` 1q | 2pk ´ 1 ` k ` 1 ´ 1q “ 4k ´ 2, which also implies 2k ´ 2 | pp4k ´ 2q ´ 2p2k ´ 2qq “ 2 and thus k “ 2. We conclude that f p3q “ 1 and f p4q | 7. If f p4q “ 1, then 4 “ f p2q ` f p4q | 10, a contradiction. Thus f p4q “ 7. This leads to the contradiction 8 “ f p3q ` f p4q | 12. Thus there are only the constant solution and the solution f paq “ 2a ´ 1 for all a.
13
Individual Competition
MEMO 2016
I-4
Solution 4. We have f p1q “ 1 and f paq | p2a ´ 1q as in the previous solutions. If f p2q “ 1, then f paq “ 1 for all a as in the previous solutions. Therefore, we only have to consider f p2q “ 3. Let p be a prime with p ” ´1 pmod 4q. Since we already know that f paq | 2a ´ 1, we get ` ˘ ` ˘ f p`1 | p which implies that f p`1 is either 1 or p. 2 2 ` ˘ “ 1 then we choose a “ 2 and b “ p`1 If f p`1 in the original equation and get 4 | p ` 3 which 2 2 ` p`1 ˘ is impossible. Therefore, f 2 “ p for all such primes p. Now we choose b “
p`1 2
in the original equation and get
f paq ` p | 2a ´ 1 ` p ùñ f paq ` p | 2a ´ 1 ´ f paq. Since there exist arbitrarily large primes p with p ” ´1 pmod 4q, the right-hand side has to be 0, so f paq “ 2a ´ 1 which is indeed a solution.
14
Team Competition
MEMO 2016
T-1
T-1
A
Determine all triples pa, b, cq of real numbers satisfying the system of equations a2 ` ab ` c “ 0, b2 ` bc ` a “ 0, c2 ` ca ` b “ 0.
Answer. The solutions are ˙* " ˆ 1 1 1 . pa, b, cq P p0, 0, 0q, ´ , ´ , ´ 2 2 2 Solution. If one of the numbers a, b and c is equal to zero, it is easy to see that the other two numbers also have to be equal to zero, which gives us the solution p0, 0, 0q. Now assume that a, b, c ‰ 0. If all three numbers are positive, then the left-hand side of each equation is positive, while the right-hand sides are equal to zero, which is impossible. Let us assume that only one of the numbers is positive, and without loss of generality let it be a. Since b, c ă 0, it follows that b2 ` bc ` a ą 0, which is a contradiction. It remains to consider the two following cases: (a) All three numbers are negative. We substitute a “ ´x, b “ ´y and c “ ´z, where x, y, z ą 0. The original system transforms into x2 ` xy “ z
(1)
y 2 ` yz “ x
(2)
z 2 ` zx “ y. The system is cyclic, so we can assume that x ď y and x ď z. Now we have x2 ` xy “ z ě x ùñ x ` y ě 1, y 2 ` yz “ x ď y ùñ y ` z ď 1. From the previous two inequalities we conclude that x ` y ě 1 ě y ` z,
i.e. x ě z.
15
Team Competition
MEMO 2016
T-1
On the other hand x ď z, so we get x “ z. Now, from equation (1) it follows that x ` y “ 1, while from equation (2) it follows that x “ y 2 ` yz “ y 2 ` yx “ ypy ` xq “ y. Thus x “ y “ z and from x ` y “ 1 we see that x “ y “ z “ 1{2. ˘ ` We easily verify that pa, b, cq “ ´ 21 , ´ 12 , ´ 12 is indeed a solution. (b) Exactly one of the numbers is negative. Without loss of generality we can assume that c is negative, while a and b are positive. From the second equation we conclude that bpb ` cq “ ´a ă 0, thus b ` c ă 0. The third equation yields cpa ` cq “ ´b ă 0, thus a ` c ą 0. Adding a ` b to the first equation and cyclic permutation yields a ` b ` c “ p1 ´ aqpa ` bq “ p1 ´ bqpb ` cq “ p1 ´ cqpc ` aq. The last product is positive. This implies that 1 ´ a ą 0 and 1 ´ b ă 0 by our above considerations. Therefore 0 ă a ` c ă 1 ` c ă b ` c ă 0, a contradiction.
Solution by symmetric functions. We set p :“ a ` b ` c,
q :“ ab ` ac ` bc,
r “ abc.
(3)
Our strategy will be to determine p, q and r by considering equations of the form ÿ
f pa, b, cqpa2 ` ab ` cq “ 0.
cyc
By setting f pa, b, cq “ 1, we find p2 ´ q ` p “ 0. By setting f pa, b, cq “ b, we find pq ´ 3r ` q “ 0. ř (Here we use the general identity cyc pa2 b ` ab2 q “ pq ´ 3r.) By setting f pa, b, cq “ c2 , we find pq 2 ´ 2prq ` pr ` pp3 ´ 3pq ` 3rq “ 0. By elimination of q “ p2 ` p and r “ pp2 ` pq2 ´ p ¨
pq`q 3
“
ppp`1q2 3
we find
ppp ` 1q2 ` p3 ´ 3ppp2 ` pq ` ppp ` 1q2 “ 0, 3
which is equivalent to 0 “ 2p4 ` p3 ´ p2 ` 3p “ pp2p ` 3qpp2 ´ p ` 1q.
16
Team Competition
MEMO 2016
T-1
Now we see that either p “ 0 or p “ ´ 23 . In the case p “ 0 we find that also q “ 0 and r “ 0, whence a “ b “ c “ 0. In the case p “ ´ 32 we find q “ 43 and r “ ´ 18 , hence a, b, c are the solutions to the cubic equation 3 3 1 0 “ x ` x2 ` x ` “ 2 4 8
ˆ
3
1 x` 2
˙3 .
This gives a “ b “ c “ ´ 12 . Solution. As in the first solution, we prove that as soon as one of the variables is 0, all three variables have to be 0. Obviously, the triple p0, 0, 0q is a solution. From now on, we may therefore assume that all three variables are non-zero. We can rewrite the system of equations as ´c “ apa ` bq, ´a “ bpb ` cq, ´b “ cpc ` aq. Multiplying these equations and dividing by abc ‰ 0 gives (4)
pa ` bqpb ` cqpc ` aq “ ´1. On the other hand by summing up the equations we get ´ a ´ b ´ c “ a2 ` b2 ` c2 ` ab ` bc ` ac.
(5)
Now we substitute x “ a ` b,
y “ b ` c,
z “ c ` a.
which transforms equation (4) and (5) into xyz “ ´1,
´
x2 ` y 2 ` z 2 x`y`z “ 2 2
(6)
Now we calculate 3px2 ` y 2 ` z 2 q ě p|x| ` |y| ` |z|q2 ě |x ` y ` z|2 “ px2 ` y 2 ` z 2 q2 ě 32
a 2 3 x2 y 2 z 2 “ 9,
where the first inequality comes from Cauchy-Schwarz (or QM-AM), the second one from the triangle inequality and the last one from AM-GM. The equalities come from (6). Now, if we denote S “ x2 ` y 2 ` z 2 , we have the inequalities 3S ě S 2 ě 9,
17
MEMO 2016
Team Competition
T-1
and because we trivially have S ą 0 (note that S “ 0 would imply x “ y “ z “ 0 and hence a “ b “ c “ 0, which has already been excluded), we can split it into the inequalities 3 ě S and S ě 3, so we have equality and actually all the inequalities are equalities. The case of equality for the triangle equality is when all nonzero x, y, z have the same sign and, in view of equations 6, the only possibility is that x, y, z are all negative. Moreover, in the last inequality, we have equality exactly when x2 “ y 2 “ z 2 and, because they have the same sign, it means x “ y “ z. Finally, in view of xyz “ ´1, the only possibility is x “ y “ z “ ´1. By definition of x, y, z the values of a, b, c are then ˆ pa, b, cq “
´1 ´1 ´1 , , 2 2 2
˙
and this is indeed a solution.
18
Team Competition
MEMO 2016
T-2
T-2
A
Let R denote the set of real numbers. Determine all functions f : R Ñ R such that f pxqf pyq “ xf pf py ´ xqq ` xf p2xq ` f px2 q holds for all real numbers x and y.
Answer. There are two solutions: f : R Ñ R : x ÞÑ 0,
g : R Ñ R : x ÞÑ 3x.
Solution. We set x “ 0 and get f p0qf pyq “ f p0q, so f p0q “ 0 or f pyq “ 1 for all y. The latter leads to a contradiction. We set y “ x ` z and get f pxqf px ` zq “ xf pf pzqq ` xf p2xq ` f px2 q
(1)
for all x and z. Setting z “ 0 yields f pxq2 “ xf p2xq ` f px2 q.
(2)
We set C “ f p1q. For x “ 1 and x “ 2, we get f p2q “ CpC ´ 1q and f p4q “ C 2 pC ´ 1q2 {3, respectively. If C “ 0, so (1) with x “ 1 yields f pf pzqq “ f p2q “ 0, so (1) reads f pxqf px`zq “ xf p2xq`f px2 q for all z, which implies that f pxq “ 0 for all x. Setting x “ 1 and x “ 2 in (1) leads to Cf pz ` 1q “ f pf pzqq ` C 2 , CpC ´ 1qf pz ` 2q “ 2f pf pzqq ` C 2 pC ´ 1q2 . Eliminating f pf pzqq and division by C ‰ 0 leads to pC ´ 1qf pz ` 2q ´ 2f pz ` 1q “ CpC 2 ´ 2C ´ 1q
(3)
for all z. Setting z “ ´1 leads to CpC ´ 1q “ CpC 2 ´ 2C ´ 1q. In view of C ‰ 0, this implies C “ 3. Inserting this in (3), dividing by 2 and shifting z leads to f pz ` 1q ´ f pzq “ 3 for all z.
19
Team Competition
MEMO 2016
T-2
We set z “ 1 in (1) and get f pxqpf pxq ` 3q “ 9x ` xf p2xq ` f px2 q. Together with (2), we get 3f pxq “ 9x, i.e., f pxq “ 3x for all x. Both f pxq “ 3x and f pxq “ 0 are solutions.
Alternative Solution. Setting x “ 0 in the original equation gives f p0qf pyq “ f p0q. If f p0q ‰ 0 then f pxq “ 1, x P R, but this function does not satisfy the original equation. Hence, f p0q “ 0. Setting y “ 0 and y “ x in the original equation we get 0 “ xf pf p´xqq ` xf p2xq ` f px2 q,
f pxq2 “ xf p2xq ` f px2 q.
(4)
In combination this gives ´xf pf p´xqq “ f pxq2
and
xf pf pxqq “ f p´xq2 , x P R.
Multiplying the original equation by y ´ x gives py ´ xqf pxqf pyq “ xpy ´ xqf pf py ´ xqq ` py ´ xqpxf p2xq ` f px2 qq “ xf px ´ yq2 ` py ´ xqf pxq2 , which gives xf px ´ yq2 “ py ´ xqf pxqpf pyq ´ f pxqq.
(5)
Now setting x “ 2y ‰ 0 in (5) we get 2yf pyq2 “ ´yf p2yqpf pyq ´ f p2yqq and consequently 0 “ f p2yq2 ´ f pyqf p2yq ´ 2f pyq2 “ pf p2yq ` f pyqqpf p2yq ´ 2f pyqq. Thus, for any y ‰ 0 (and for y “ 0 as well) we have f p2yq “ ´f pyq or f p2yq “ 2f pyq. Setting y “ 2x ‰ 0 in (5) we get xf p´xq2 “ xf pxqpf p2xq ´ f pxqq and consequently f p´xq2 ` f pxq2 “ f pxqf p2xq.
20
MEMO 2016
Team Competition
T-2
If f p2xq “ ´f pxq holds for some x P R, then f p´xq2 ` 2f pxq2 “ 0 implies f p´xq “ f pxq “ 0 “ f p2xq. Hence, the equality f p2xq “ 2f pxq holds for all x P R. Replacing y by y ` x in the original equation and using (4) gives f pxqf px ` yq “ xf pf pyqq ` xf p2xq ` f px2 q “ xf pf pyqq ` f pxq2 .
(6)
Now setting y “ x gives 2f pxq2 “ xf pf pxqq ` f pxq2 , which means f pxq2 “ xf pf pxqq. Multiplying (6) by y yields yf pxqf px ` yq “ xf pyq2 ` yf pxq2 (7) Here we can deduce that f pxq “ 0 implies x “ 0, unless f is identically 0. Now we can interchange x and y and achieve yf pxqf px ` yq “ xf pyqf px ` yq. Setting y “ 1 now gives f pxq “ xf p1q for all x ‰ ´1. However, we also have 1 1 f p´1q “ f p´2q “ ¨ p´2qf p1q “ ´f p1q, 2 2 so f pxq “ xf p1q is valid for all x. Plugging in f pxq “ cx easily gives c “ 0 or c “ 3. Hence these are the two solutions.
21
Team Competition
MEMO 2016
T-3
T-3
C
A tract of land in the shape of an 8 ˆ 8 square, whose sides are oriented north–south and east–west, consists of 64 smaller 1 ˆ 1 square plots. There can be at most one house on each of the individual plots. A house can only occupy a single 1 ˆ 1 square plot. A house is said to be blocked from sunlight if there are three houses on the plots immediately to its east, west and south. What is the maximum number of houses that can simultaneously exist, such that none of them is blocked from sunlight? Remark: By definition, houses on the east, west and south borders are never blocked from sunlight.
Answer. The maximal number of houses is 50.
Solution. Let us represent the tract as an 8 ˆ 8-chessboard, with cells colored black if the corresponding parcel is occupied, and white otherwise. We denote by pi, jq the cell in the i´th row and j´th column (with the first row being the northernmost and the first column being the westernmost). We start by showing that an optimal configuration can be obtained by coloring all the cells along the east, south, and west borders. Assume that there is an optimal configuration in which one of those cells, for example pi, 1q, is left white. Since we have an optimal configuration, this cell cannot be colored black. This means that by coloring pi, 1q, we would block the cell pi, 2q. In other words, we know that the cells pi, 2q, pi, 3q and pi ` 1, 2q are all colored black in this optimal configuration. However, we now see that we can color pi, 1q instead of pi, 2q, keeping the same number of black cells and coloring pi, 1q, without disturbing any of the other cells. We can apply the same reasoning to any of the cells p1, 1q ´ p8, 1q, p8, 1q ´ p8, 8q and, similarly, to p8, 8q ´ p1, 8q, thus showing that there is an optimal configuration in which all the cells along the E, S, W borders are colored black. Those cells being colored, we are left with a 7 ˆ 6 area of the board. We can now show that no more than 28 cells in this area can be colored black. In order to obtain 28, the average number of black cells per row has to be 4. However, if any row contains six black cells, the next row down cannot contain any black cells, since such a black cell would block the cell immediately north of it. Similarly, if a row were to contain 5 black cells, the next row down would be able to contain at most 3 black cells (namely in the cell immediately below the single white one and the two next to it). This shows that the average number of black cells per row in our 7 ˆ 6 area cannot be greater than 4. This gives us an upper bound on the total number of black cells: the 22 border cells plus 28 cells in the remaining 7 ˆ 6 area, i.e., 50 cells in total.
22
MEMO 2016
Team Competition
(a)
T-3
(b)
Figure 1: Alternative Solution An example to show that 50 can indeed be achieved is the following. We color the columns 1, 2, 4, 5, 7, 8 and row 8 black, leaving the other cells white. This coloring clearly satisfies the conditions and contains exactly 50 black cells, completing the proof.
Alternative Solution. By building a house on each dark grey plot in Figure 1(a), we see that 50 houses can be built accordingly. We will prove that no more than 50 houses can be built, or, equivalently, that at least 14 plots remain empty. Consider the 14 4 ˆ 1 rectangles in Figure 1(b) marked by their thick boundary. We will uniquely assign one empty plot to each of these 14 rectangles as follows: • if the rectangle contains at least one empty plot, assign to it the easternmost such plot; • if the rectangle contains no empty plots, assign to it the westernmost empty plot in the rectangle directly to the south of it. In order to see that this assignment is feasible, note that if some rectangle contains no empty plots, then its two central houses are blocked from sunlight from the east and west. Therefore, the two central plots of the rectangle directly to the south of it must be empty, showing that we can indeed assign its westernmost empty plot to the original rectangle while leaving its easternmost empty plot unassigned. The above assignment is thus feasible and shows that there are at least 14 empty plots, concluding the proof.
23
Team Competition
MEMO 2016
T-4
T-4
C
A class of high school students wrote a test. Every question was graded as either 1 point for a correct answer or 0 points otherwise. It is known that each question was answered correctly by at least one student and the students did not all achieve the same total score. Prove that there was a question on the test with the following property: The students who answered the question correctly got a higher average test score than those who did not.
Solution. Let n be the number of the students in the class and a their average score. Denote by P and S the set of all problems, and the set of all students resp., and let Sppq be the non-empty set of students who solved problem p P P . For any student s, let scpsq be the score of s. For any proposition A, let rAs “ 1 if A is true and 0 if A is false. We will prove the assertion by contradiction. Assume that on all questions the average test score of solvers was at most the general average a, that is
aě
ÿ ÿ 1 scpsq ô a|Sppq| ě scpsq. |Sppq| sPSppq sPSppq
We now sum these inequalities over all problems p P P to get a
ÿ
ÿ ÿ
pPP
ô
a
ÿÿ
scpsq
|Sppq| ě pPP sPSppq
rs solved ps ě
pPP sPS
ÿÿ
rs solved ps
pPP sPS
a
ô
scpsq ě
sPS
ô
1 n
rs solved qs
qPP
˜ ÿ
ÿ
¸ ˜
ÿ
ÿ
sPS
pPP
ÿ
scpsq2 .
rs solved ps
¸ ÿ
¨
rs solved qs
qPP
¸2
˜ ÿ sPS
scpsq
ě sPS
However, this is the reverse of the inequality between the arithmetic and the quadratic mean. Since the case of equality, namely that scpsq is the same for all s P S, is excluded by the problem statement, we arrive at the desired contradiction.
24
Team Competition
MEMO 2016
T-5
T-5
G
Let ABC be an acute-angled triangle with AB ‰ AC, and let O be its circumcentre. The line AO intersects the circumcircle ω of ABC a second time in point D, and the line BC in point E. The circumcircle of CDE intersects the line CA a second time in point P . The line P E intersects the line AB in point Q. The line through O parallel to P E intersects the altitude of the triangle ABC that passes through A in point F . Prove that F P “ F Q.
P
C ω
F “ F1 E
D
O A Q
B
Solution 1. Let us denote >ABC by β and >BCA by γ. Without loss of generality, AB ą AC, or equivalently β ă γ, as in the figure. Segment AD is a diameter of ω, so by Thales’ theorem we have >DCA “ 90˝ . Since the quadrilateral CEDP is cyclic, we get >P ED “ 90˝ , which immediately gives >AEQ “ 90˝ . Since >EAQ “ >OAB “ 90˝ ´ γ, we also get >AQP “ >AQE “ γ. Since CEDP is cyclic, we get >ADP “ >EDP “ 180˝ ´ >P CE “ >ACB “ γ. This means that the quadrilateral AQDP is cyclic. Let us denote the circumcentre of AQDP by F 1 . We show that F “ F 1 . We have >AP Q “ 180˝ ´ >CAB ´ >AQP “ β. Hence >F 1 AQ “ 90˝ ´ β, which implies that F 1 lies on the altitude of ABC that passes through A. Moreover, by definition F 1 must lie on the perpendicular bisector of AD, which is the line through O parallel to P E. So we get F 1 “ F , and consequently F P “ F Q.
25
Team Competition
MEMO 2016
T-5
A
O
Q1
F
B
E C
D
P1
Solution 2. Let α, β and γ denote the angles of ABC in the natural way. Point F is defined as the intersection point of the perpendicular bisector of diameter AD with the altitude of triangle ABC through A. We define points P 1 and Q1 as the intersection points of the circle with center F passing through A (and therefore also through D) with sides AC and AB, respectively. First we show P “ P 1 . We calculate >EDP 1 “ >ADP 1 “ >AQ1 P 1 “ >AQ1 F ` >F Q1 P 1 “ 180˝ ´ >Q1 F P 1 >AQ F ` “ 90˝ ´ β ` 90˝ ´ α “ γ “ 180˝ ´ >ECP 1 . 2 1 So we have that P is the intersection of the circumcircle of triangle EDC with AC and therefore we have P “ P 1 . 1
Now we prove Q “ Q1 . We have >ABC “ >ADC “ >EDC “ >EP C “ >Q1 P A “ >Q1 DA “ β and therefore quadrilateral BDEQ1 is cyclic. Since >DBQ1 “ 90˝ we get Q1 E K AD. Together with >DEP 1 “ >DEP “ 90˝ we have that Q1 lies on the line P E and thus Q1 “ Q. Therefore we have proven F P “ F Q.
26
Team Competition
MEMO 2016
T-5
Solution 3. Let us denote =ABC by β and =BCA by γ and the foot of the altitude from A by G. Without loss of generality let AB ą AC, or equivalently β ă γ, as in the figure.
P ω2 B1 ω1 C
G ω F O1 D
E
O C1 A
Q
B
Let ω1 be the reflection of ω about the angle bisector of =BAC. Since =EAB “ =OAB “ =CAG “ 90˝ ´ γ, the center O1 of ω1 lies on the altitude from A and AO “ AO1 . The circle ω1 intersects AB for the second time at C1 with AC “ AC1 and AC for the second time at B1 with AB “ AB1 . Now, if we can prove AC1 : AQ “ AB1 : AP “ AO1 : AF, then there is a homothety with center A which maps C1 Ñ Q, B1 Ñ P and O1 Ñ F . Hence P and Q lie on a circle ω2 with center F and and the problem is solved. Therefore it remains to prove AC1 : AQ “ AB1 : AP “ AO1 : AF. The triangles AOF and AGE are similar, so we have AF “ AO¨AE . Due to segment AD being AG ˝ a diameter of ω, we have =DCA “ 90 by Thales’ theorem. Since the quadrilateral CEDP is cyclic, we get =P ED “ 90˝ , which immediately gives =AEQ “ 90˝ . Since =EAQ “ =OAQ “ AG AE 90˝ ´ γ, we have AQ “ sin and with AC “ sin , we get γ γ AC1 : AQ “ AC : AQ “
AG AE AO ¨ AE : “ AG : AE “ AO : “ AO1 : AF. sin γ sin γ AG
Similarly we can prove AB1 : AP “ AO1 : AF and we are ready.
27
Team Competition
MEMO 2016
T-6
T-6
G
Let ABC be a triangle with AB ‰ AC. The points K, L, M are the midpoints of the sides BC, CA, AB, respectively. The inscribed circle of ABC with centre I touches the side BC at point D. The line g, which passes through the midpoint of segment ID and is perpendicular to IK, intersects the line LM at point P . Prove that >P IA “ 90˝ .
Solution 1. Let pXY Zq denote the circumcircle of a triangle XY Z. We use the following well-known lemma: Lemma. The centre of the circle pBICq is the midpoint of arc BC of circle pABCq and therefore lies on the angle bisector of >BAC.
A
g L
M
P
I Y k H
K B D
C
X
Now assume without loss of generality that AB ă AC, as in the figure. Let the circle pBICq cut AB and AC at X and Y respectively. From the lemma it follows that X is symmetric to C, and Y is symmetric to B with respect to the angle bisector of >BAC. We have ˆ ˙2 AC ´ AB p2AC ´ ABq ¨ AB AC 2 M X¨M B “ pAX´AM q¨M B “ pAC´AM q¨M B “ “ ´ , 4 4 2
28
Team Competition
MEMO 2016 and since
AC 2
T-6
“ M K and AC ´ AB BC AB ` BC ´ AC “ ´ “ BK ´ BD “ DK, 2 2 2
we get M X ¨ M B “ M K 2 ´ DK 2 . So M lies on the radical axis of the circle pBICq and the circle k, which is the circle with midpoint K and radius DK. An analogous calculation shows that L lies on it as well, so the line through M and L is the radical axis of circles pBICq and k. Obviously g is the radical axis of circle k and point I (regarded as a degenerate circle). Thus P is the radical centre of pBICq, k, and I. It follows that P also lies on the radical axis of pBICq and I, which is the line perpendicular to AI passing through I. This shows that >P IA “ 90˝ .
T
A
L
M
P
Y I
X
B
K D
S
C
Solution 2. Let X be the midpoint of ID, and let Y be the midpoint of AD (which is also the intersection of AD with LM ). Clearly, XY is parallel to AI. Next we show that IK passes through Y . To this end, let S be the intersection of AI with BC, and let T be the intersection of KI with the line through A that is parallel to BC. Note that triangles AT I and SKI are similar. By the angle bisector theorem, we have AT AI AB “ “ , KS IS BS
29
Team Competition
MEMO 2016
T-6
so
AB ¨ pBK ´ BSq AB ¨ BK AB ¨ KS “ “ ´ AB. BS BS BS Using the angle bisector theorem once again, we get AT “
AT “
pAB ` ACq ¨ BK AB ` AC BC AB ` BC ´ AC ´AB “ ´AB “ ´ “ BK´BD “ DK. BC 2 2 2
Thus DKAT is a parallelogram, which means that AD and KT meet at Y , the midpoint of AD. Now we know that KI passes through Y , and by definition it is perpendicular to XP . Hence it is an altitude in XY P . Moreover, DI is clearly also an altitude in XY P (it passes through X and is perpendicular to P Y ). Thus I is the orthocentre of XY P , which means that P I is perpendicular to XY and thus also to AI. This proves the desired statement.
C
D
K
L I
Y
B
X A
M
P
Z
Solution 3. Let X, Y , Z be reflections of D about I, K, P respectively. We shall prove that A, X, Y are collinear. Since K is the midpoint of segments BC, DY , we have BD “ CY . Therefore Y is the common point of segment BC and the A-excircle. Consider
30
Team Competition
MEMO 2016
T-6
homothety centered at A mapping incircle to A-excircle. It’s easy to see that this homothety maps X to Y . Therefore A, X, Y are collinear. Consider homothety with centre D and ratio 2. We easily see that AZ k LM . Since IX K BC k LM , we have IX K AZ. Moreover IZ k g K KI k XY . Since A, X, Y are collinear, we have IZ K AX. Thus X is the orthocentre of triangle AIZ. Therefore ZX K AI. Since ZX k P I, we conclude that P I K AI. In other words, =P IA “ 90˝ . the half-perimeter, ρ the radius Solution 4. Let a, b, c be the sides of the triangle, s “ a`b`c 2 of the inscribed circle, α, β, γ the angles and A the area of the triangle. In the coordinate plane, let B “ p0, 0q be the origin and C “ pa, 0q. Then A “ pc cos β, c sin βq. We get: ´ ´a ¯ ρ¯ I “ ps ´ b, ρq, D “ ps ´ b, 0q, H “ s ´ b, (midpoint of DI), K “ ,0 . 2 2 The line g that passes through H and is perpendicular to IK has the following equation: ´ ´ a¯ a¯ ρ2 s´b´ x ` ρy “ s ´ b ´ ps ´ bq ` . 2 2 2 This is equivalent to pc ´ bqx ` 2ρy “ pc ´ bqps ´ bq ` ρ2 . On the other hand, the line h “ LM that is parallel to the x-axis has the equation y“
c sin β. 2
We deduce that P (the intersection of g and h) has the coordinates ´ ¯ ρ2 ´ ρc sin β c , sin β . P “ s´b` c´b 2 Now we have to prove that ÝÑ ÝÑ AI ¨ IP “ 0,
31
Team Competition
MEMO 2016
T-6
which is equivalent to ps ´ b ´ c cos βq ¨
´c ¯ ρ2 ´ ρc sin β ` pρ ´ c sin βq sin β ´ ρ “ 0. c´b 2
Since ρ ´ c sin β ‰ 0, we can cancel the factor ρ ´ c sin β to obtain the equation ¯ ´c ps ´ b ´ c cos βqρ ` pc ´ bq sin β ´ ρ “ 0, 2 which is equivalent to 2ρps ´ cq ´ 2ρc cos β ` cpc ´ bq sin β “ 0. Now we use the well-known identities ρ “
A s
and ac sin β “ 2A to end up with the equation
aps ´ cq ´ ac cos β ` spc ´ bq “ 0. Since this is equivalent to b2 “ a2 ` c2 ´ 2ac cos β, which holds by the law of cosines, the proof is complete.
Solution 5. Let S be the point of intersection of the interior angle bisector of =BAC and BC, H the midpoint of ID, R the point of intersection of LM and ID, Q the point of intersection of LM and the altitude from A, r the inradius and h the length of the altitude from A. Since KD and P R are perpendicular to RD and IK is perpendicular to P H, we have =IKS “ =IHP and =RP H “ =DIK. So the triangles RP H and DIK are similar. We have P H : and KD “ a2 ´ a`b´c “ c´b we conclude HR “ IK : KD and since HR “ h´r 2 2 2 PH “
IKph ´ rq . pc ´ bq
Now we want to show that the triangles IKS and P HI are similar. Since =IKS “ =P HI, it suffices to prove that IK : KS “ P H : HI.
32
Team Competition
MEMO 2016
T-6 A
P
M
R
L Q
g I
H
B
K
With the angle bisector theorem we get SC “ ab KS “ a2 ´ b`c “ apc´bq . 2pb`cq
ab b`c
S
D
C
and we deduce
Now we have P H : HI “ IK : KS ðñ
IKph ´ rq r apc ´ bq : “ IK : ðñ rpa ` b ` cq “ ah pc ´ bq 2 2pb ` cq
and we are done, because rpa ` b ` cq “ ah = twice the area of the triangle ABC. Hence the triangles IKS and P HI are similar and we conclude that the triangles P IR und ISD are similar too. Now we have =QP I “ =RP I “ =DIS “ =QAI and it follows that AP IQ is cyclic, and consequently 90˝ “ =P QA “ =P IA.
33
Team Competition
MEMO 2016
T-7
T-7
N
A positive integer n is called a Mozartian number if the numbers 1, 2, . . . , n together contain an even number of each digit (in base 10). Prove: (a) All Mozartian numbers are even. (b) There are infinitely many Mozartian numbers.
Solution 1. (a) Note that we need an even number of digits alltogether if every digit occurs an even number of times. There is an odd number of numbers with one digit. For k ą 1, there are 9 ¨ 10k´1 numbers with k digits, which is an even number. Thus we need to end after a segment of odd length of numbers with an odd number of digits, i.e., we end on an even number, so a Mozartian number is indeed even. (b) The numbers n “ lo 2o.mo . .o2n 0 are Mozartian numbers for all natural numbers `: There are an 2`
even number of least significant digits 0, 1, . . . , 9; and all other digits at higher positions except for those in n are repeated 10 times in a row which does not change the parities of occurrences. The leading 2` digits 2 of n do not change parities, either.
Solution 2. (a) Let k be any integer ě 0. In the pairing p2, 3q, p4, 5q, ..., p2k, 2k ` 1q, the members of each pair need the same number of digits, so each pair needs an even number of digits together, so alltogether the numbers from 1 to 2k ` 1 need an odd number of digits. Therefore, any Mozartian number has to be even because the total number of digits used up to a Mozartian number has to be even. (b) We will show that 102k ` 22 are Mozartian numbers for all natural numbers k. We first note that by the proof of the first part, we know that we need an odd number of digits up to 102k ` 21, and therefore an even number of digits up to 102k ` 22. So it is sufficient to check that the digits 1, 2, . . . , 9 occur an even number of times because the condition for 0 will be automatically satisfied. Now, we will consider the numbers from 0 to 102k ´ 1 as numbers with 2k ` 1 digits with leading zeros where necessary. Clearly, each digit must occur equally often. Since the number of all digits in this list is divisible by 100, this quantity is still divisible by 10,
34
MEMO 2016
Team Competition
T-7
therefore even. This proves that nonzero digits occur an even number of times in this interval. It remains to consider the numbers 102k , 102k ` 1, . . . , 102k ` 22. Clearly, the leading ones occur an odd number of times. Since the list 1, 2, . . . , 22 contains an odd number of ones and an even number of the other digits, the proof is finished.
Solution 3. (only Part (b)) We will first show that for any k ě 1 the numbers from 0 to 20k ´ 1 together contain an even number of each digit from 0 to 9. The units digits clearly run from 0 to 9 an even number of times, so they contribute an even number to each digit count. For any possible fixed choice of all digits except the units digits, there are 10 numbers that satisfy this condition, so again, they contribute an even number to each digit count which proves the assertion. Consider now the numbers from 1 to M “ 20k where M has a decimal representation that contains an odd number of zeros and an even number of each digit from 1 to 9. Since the odd number of zeros compensates for the missing zero that was counted in the above assertion, we find that M is a Mozartian number. There are clearly infinitely many such numbers, for example all numbers of the form 22 . . . 20 that contain an even number of 2s. Comment. The argument of Solution 3 shows that Mozartian numbers that are multiples of 20 are exactly those multiples that contain an odd number of 0s and an even number of all other digits. By an analogous argument, one can now find all Mozartian numbers. The following table lists the parity restrictions on the digit counts for each possible even residue modulo 20 where e stands for even and o stands for odd. The rows list the different possible residues and the columns lists the digits from 0 to 9. For example, 10198 is a Mozartian number because it has residue 18 modulo 20 and the digits that occur an odd number of times are 0, 8 and 9. These conditions are the only restrictions on Mozartian numbers.
35
Team Competition
MEMO 2016
0 2 4 6 8 10 12 14 16 18
0
1
2
3
4
5
6
7
8
9
o e e e e e o o o o
e o o o o o e e e e
e e o o o o o e e e
e e o o o o o e e e
e e e o o o o o e e
e e e o o o o o e e
e e e e o o o o o e
e e e e o o o o o e
e e e e e o o o o o
e e e e e o o o o o
T-7
36
Team Competition
MEMO 2016
T-8
T-8
N
We consider the equation a2 ` b2 ` c2 ` n “ abc, where a, b, c are positive integers. Prove: (a) There are no solutions pa, b, cq for n “ 2017. (b) For n “ 2016, a must be divisible by 3 for every solution pa, b, cq. (c) The equation has infinitely many solutions pa, b, cq for n “ 2016.
Solution 1. (a) We distinguish cases depending on the parity of a, b, c: • If all three are odd, we have a2 ` b2 ` c2 ` 2017 ” 0 pmod 2q and abc ” 1 pmod 2q. • If exactly one of them is even, we have a2 ` b2 ` c2 ` 2017 ” 1 pmod 2q and abc ” 0 pmod 2q. • If exactly two of them are even, we have a2 ` b2 ` c2 ` 2017 ” 2 pmod 4q (recalling that squares are either congruent to 0 or 1 modulo 4) and abc ” 0 pmod 4q. • If all three are even, we have a2 ` b2 ` c2 ` 2017 ” 1 pmod 2q and abc ” 0 pmod 2q. In each of the four cases, we see that the two sides of the equation cannot be equal. (b) Note that m2 ” 0 pmod 3q if m is divisible by 3 and m2 ” 1 pmod 3q otherwise, and note also that 2016 is divisible by 3. We consider two cases: • If none of the three numbers a, b, c is divisible by 3, then neither is abc, while on the other hand a2 ` b2 ` c2 ` 2016 ” 1 ` 1 ` 1 ` 0 ” 0 pmod 3q. Hence we get a contradiction. • Otherwise, abc is divisible by 3, so a2 ` b2 ` c2 must be divisible by 3 as well. If exactly one of the three variables is divisible by 3, we have a2 ` b2 ` c2 ” 2 pmod 3q, and if exactly two of them are divisible by 3, we have a2 ` b2 ` c2 ” 1 pmod 3q. In both cases, we see that there cannot be a solution. This leaves us with the only possibility that a, b, c are all divisible by 3. (c) We know from the previous part that we must have a “ 3x, b “ 3y, c “ 3z for certain positive integers x, y, z. We plug these into the given equation and divide by 9 to obtain x2 ` y 2 ` z 2 ` 224 “ 3xyz.
37
MEMO 2016
Team Competition
T-8
Note that 225 “ 152 is a perfect square, so we try to find solutions with x “ 1: y 2 ` z 2 ` 225 “ 3yz. Indeed, y “ z “ 15 is a solution, and we find further solutions by means of “Vieta jumping”. Suppose that py0 , z0 q is a solution, i.e., y02 ` z02 ` 225 “ 3y0 z0 , where y0 ě z0 . The second solution to the quadratic equation z 2 ´ 3y0 z ` p225 ` y02 q “ 0 is z1 “ 3y0 ´ z0 ě 2y0 ą y0 , giving us a new solution pair pz1 , y0 q that has a greater first component than the previous one. Repeating the procedure, we obtain infinitely many solutions.
Solution 2. The third part can also be solved by means of the theory of Pellian equations. Let us return to the equation y 2 ` z 2 ` 225 “ 3yz. We multiply by 4 and complete the square: 4y 2 ´ 12yz ` 4z 2 ` 900 “ p2y ´ 3zq2 ´ 5z 2 ` 900 “ 0. For odd k, we have
? ? p2 ` 5qk ¨ p2 ´ 5qk “ ´1. ? ? ? We can write p2 ` 5qk as u ` v 5 for certain positive integers u and v, so that p2 ´ 5qk “ ? u ´ v 5 and thus u2 ´ 5v 2 “ ´1. Now simply set z “ 30v and y “ 15u ` 45v (so that 2y ´ 3z “ 30u) to obtain p2y ´ 3zq2 ´ 5z 2 ` 900 “ 0, as desired. Since we obtain a solution for every odd k in this way, there must be infinitely many.
38
Contest problems with solutions
Jury & Problem Selection Committee
VILNIUS UNIVERSITY FACULTY OF MATHEMATICS AND INFORMATICS
Selected problems Countries Austria
Algebra
Combinatorics
T-2
T-3
Croatia
Geometry
Number theory T-8
T-5
Czech Republic
I-2
Germany
I-4
Poland
T-1
Slovakia
I-1
T-7
T-4 I-3, T-6
Contents Individual Competition . . . . . . . . . . . . . . . . . . . . . . .
4
I-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
I-2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
I-3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
I-4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
Team Competition . . . . . . . . . . . . . . . . . . . . . . . . .
10
T-1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
T-2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
T-3
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T-4
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T-5
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T-6
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T-7
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T-8
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25
I-1
Individual Competition
MEMO 2017
Individual Competition
I-1 Determine all functions f : R → R satisfying f (x2 + f (x)f (y)) = xf (x + y) for all real numbers x and y.
Answer: f (x) ≡ 0, or f (x) ≡ x, or f (x) ≡ −x. Solution. Put x := 0. Then f (f (0)f (y)) = 0, so there is at least one real number a such that f (a) = 0. Let z be an arbitrary real number, let’s put x := a, y := z − a. Then we get f (a2 ) = a · f (z). If a 6= 0, we’ll get that f is constant, i. e. f (x) ≡ c. The equation (1) is then equivalent to c = cx for all x, which gives us c = 0. The function f (x) ≡ 0 is really a solution. So we’ve proved that f (x) = 0 if and only if x = 0. Therefore after putting y := −x we get x2 + f (x)f (−x) = 0
(1)
f (x2 ) = xf (x)
(2)
Now let’s put y := 0, which gives us
From (3) for x := −x we get f (x2 ) = −xf (−x), so x(f (x) + f (−x)) = 0, which for x 6= 0 means f (x) = −f (−x) and for x = 0 it holds too, so f is an odd function. By putting f (−x) = −f (x) to (2) we get x2 = f (x)2 , which means that for every x is f (x) = x or f (x) = −x. We’ll prove that f (x) ≡ x or f (x) ≡ −x. Assume there are a, b such that f (a) = a and f (b) = −b. Let’s put x := a, y := b to (1). Then we have f (a2 − ab) = a · f (a + b) This relation together with f (x) = x or f (x) = −x means that e1 (a2 − ab) = e2 (a2 + ab) where e1 , e2 ∈ {1, −1}. After checking options (if we WLOG suppose e1 = 1, there’ll be only two) we’ll find out it must hold a = 0 or b = 0. If a = 0, then for all x 6= 0 we have f (x) = −x, 4
I-1
Individual Competition
MEMO 2017
which is true even for x = 0, so f (x) ≡ −x, in that case. Analogously, if b = 0, then f (x) ≡ x. These two solutions really satisfy (1). The equation has 3 solutions: f (x) ≡ 0, f (x) ≡ x, f (x) ≡ −x. J Alternative solution 1. Suppose, f is injective; that is, f (y0 ) = f (y1 ) implies y0 = y1 . Plugging x = 1 gives f (1 + f (1)f (y)) = f (1 + y). From injectivity, this gives 1 + f (1)f (y) = 1 + y. We see that f (1) = c 6= 0, and f (y) = c−1 y. For y = 1 this gives c2 = 1. Thus, f (y) = y or f (y) = −y, and both are the solutions. Assume, f is not injective, and f (y0 ) = f (y1 ) for a certain pair y0 6= y1 . Plugging y 7→ y0 , and y 7→ y1 , and comparing, gives xf (x + y0 ) = xf (x + y1 ). Thus, if x 6= 0, then f (x + y0 ) = f (x + y1 ). This is also valid for x = 0. So, for P = y0 − y1 6= 0, we get f (x + P ) = f (x). Put in the initial equation x = 0. We see that f (f (y)f (0)) = 0. Let a be such that f (a) = 0, which does exist. Put x = a, y = z − a. This gives f (a2 ) = af (z). If a 6= 0, we get that f (x) = c, and only c = 0 does satisfy this. So, if f is not identically 0, a can be only 0. But this now contradicts to f (P ) = f (0) = 0. J Alternative solution 2. Obviously, the functions f (x) = 0 and f (x) = ±x are solutions. Let us prove that there are no more solutions. Let f be any other solution. Denote f (0) = c. Taking x = 0 we get f (cf (y)) = 0 for all y. Hence f (a) = 0 for a = cf (1) and therefore 0 = f (cf (a)) = f (0) = c. Now take y = 0; then f (x2 ) = xf (x) for all x and therefore xf (x) = −xf (−x), i.e. f (−x) = −f (x) for all x 6= 0. Hence f is an odd function. Next prove that 0 is the only point, where f (x) = 0. Let f (c) = 0 for c 6= 0. Then f (c2 ) = cf (y) for all y, i.e. f (y) = c1 for all y. Obviously, such f is not a solution; we got a contradiction. Now take y = −x; then f (x2 − f 2 (x)) = 0 and therefore x2 = f 2 (x) for all x. Suppose f (1)f (y) = −y for some y 6= 0; then f (1 − y) = f (1 + y), which yields 1 − y = ±(1 + y), a contradiction.
J 5
I-2
Individual Competition
MEMO 2017
I-2 Let n > 3 be an integer. A labelling of the n vertices, the n sides and the interior of a regular n-gon by 2n + 1 distinct integers is called memorable if the following conditions hold: 1. Each side has a label that is the arithmetic mean of the labels of its endpoints. 2. The interior of the n-gon has a label that is the arithmetic mean of the labels of all the vertices. Determine all integers n > 3 for which there exists a memorable labelling of a regular n-gon consisting of 2n + 1 consecutive integers. Solution. We prove that the desired n’s are precisely those divisible by 4. Fix n and assume such labelling exists. Without loss of generality, the labels form a set {0, 1, . . . , 2n}. A maximum can’t be obtained by averaging, so number 2n labels a vertex. In order for the side labels to be integers, the vertex labels have to have the same parity, hence all of them are even. Since the label of the interior is the average of the vertex as well as edge labels, we see that the interior label is the average of all labels. Thus the interior label is equal to n. There are n + 1 even labels and n of them label vertices. Denote by e the one that doesn’t. We have n(n + 1) − e 0 + 2 + · · · + 2n − e = n= n n which gives n = e, i.e. n has to be even. In that case, the vertex labels form a set V = {0, 2, . . . , n − 2, n + 2, . . . , 2n} and hence the side labels form a set S = {1, 3, . . . , 2n − 1}. Now assume n = 4k + 2 for some integer k > 1. Then V contains n/2 + 1 numbers divisible by four, hence two such labels are used on neighbouring vertices which contradicts the fact that all edges get odd label. Therefore, n is divisible by 4. Finally, for any n = 4k we construct a satisfying labelling: Label the vertices by numbers 0, 2, 4, . . . , 4k − 2, 4k + 4, 4k + 2, 4k + 8, 4k + 6, . . . , 8k, 8k − 2 in this order. Then all the even labels but n = 4k are used for vertices, n itself is used for the interior, and the side labels are 1, 3, . . . , 4k − 3, 4k + 1, 4k + 3, . . . , 8k − 1, 4k − 1 in this order. Remark. Another construction of a satisfying labelling: Label the vertices by numbers 0, 2, . . . , 2k − 2, 6k, 6k − 2, . . . , 4k + 2, 8k, 8k − 2, . . . , 6k + 2, 2k, 2k + 2, . . . , 4k − 2 in this order. 6
J
I-3
Individual Competition
MEMO 2017
I-3 Let ABCDE be a convex pentagon. Let P be the intersection of the lines CE and BD. Assume that ∠P AD = ∠ACB and ∠CAP = ∠EDA. Prove that the circumcentres of the triangles ABC and ADE are collinear with P . Solution. Simple angle chasing gives us: ∠BCD + ∠EDC = ∠ACB + ∠ACD + ∠EDA + ∠ADC = = ∠P AD + ∠ACD + ∠CAP + ∠ADC = 180◦ , so BC k DE. Therefore there exists a homothety H centered in P that maps BC to DE. Let A0 be the image of A under this homothety. Then simply ∠A0 ED = ∠ACB = ∠A0 AD, so quadrilateral A0 DEA is cyclic. This means that the circumcircle of triangle AED is the same as the circumcircle of triangle DA0 E. But triangle ABC maps to the triangle A0 DE, so their circumcenters are collinear with the center of homethety P , which concludes the proof. J
7
I-4
Individual Competition
MEMO 2017
I-4 Determine the smallest possible value of |2m − 181n |, where m and n are positive integers. Answer: |215 − 1812 | = 7. Solution. Calculating 1812 = 32, 761 one should get the idea that this may be close to 215 = 32, 768 so taking the difference of both we arrive at the minimum possible value 7. As we can clearly see that the difference must be positive and odd, we only need to eliminate the possibilities 1, 3 and 5 for the given difference. First consider the difference modulo 15 since this will lead to a short period of basis 2 and an even shorter one of basis 181. Since 2m ≡ 1, 2, 4, 8 modulo 15 and 181n ≡ 1 modulo 15 we get that 2m − 181n ≡ 0, 1, 3, 7 modulo 15 thus leaving us with possible residues 0, 1, 3, 7, 8, 12, 14 for the absolute of the difference. Therefore we can eliminate the minimum 5 for the difference, leaving us with 1 and 3 as possible values less than 7. We can easily see that indeed m > 4 is clearly required for the difference to be anywhere near 7 or less. So now let us consider the following equations: 2m − 181n = −1 ⇒ 2m ≡ 181n − 1 ≡ 0 modulo 3, which is impossible; 2m − 181n = 1 ⇒ 2m ≡ 181n + 1 ≡ 2 modulo 4, which is impossible for m > 2; 2m − 181n = −3 ⇒ 2m ≡ 181n − 3 ≡ 2 modulo 4, which is still impossible. Therefore one last equation needs to be considered: 2m − 181n = 3 ⇔ 2m = 181n + 3. By looking at the period of the values obtained modulo 15 we can see that m ≡ 2 modulo 4 is required, so m = 4k + 2. 8
I-4
Individual Competition
MEMO 2017
But then we can look at the equation modulo 13 and see that: 2m − 181n = 3 ⇔2m = 181n + 3 ⇒24k+2 ≡ (−1)n + 3 ⇔4 · 16k ≡ (−1)n + 3 ⇔4 · 3k ≡ (−1)n + 3
modulo 13 modulo 13 modulo 13
We can clearly see that (−1)n + 3 ≡ 2, 4 modulo 13 and 4 · 3k ≡ 12, 10, 4 modulo 13 periodically. So the only possible solution would be if (−1)n + 3 ≡ 4 but that requires n to be even, thus n = 2q and hence |2m − 181n | = |24k+2 − 1812q | = |(22k+1 − 181q ) · (22k+1 + 181q )| > 183 so this will not lead to any solution less than 7, which proves 7 to be the minimum possible value. J
9
T-1
Team Competition
MEMO 2017
Team Competition T-1 Determine all pairs of polynomials (P, Q) with real coefficients satisfying P (x + Q(y)) = Q(x + P (y)) for all real numbers x and y. Answer: Either P ≡ Q or P (x) = x + a and Q(x) = x + b for some real numbers a, b. Solution. If either P or Q is constant then clearly P ≡ Q. Suppose neither of P , Q is constant. Write P (x) = axn + bxn−1 + R(x) and Q(x) = cxm + dxm−1 + S(x) with n, m > 1, a 6= 0 6= c, deg R < n − 1, deg S < m − 1. Then the degree of P (x + Q(y)) (with respect to x) is n and the leading coefficient is equal to a. Similarly, the degree of Q(x + P (y)) is m and the leading coefficient is c. It follows that m = n and a = c. Now, the xn−1 -coefficient of the polynomial P (x+Q(y)) is anQ(y)+b and the corresponding coefficient of Q(x + P (y)) is equal to anP (y) + d. It follows that anQ(y) + b = anP (y) + d for every y ∈ R. we have P (x) = Q(x)+t for every x. If t = 0 then P ≡ Q. Such polynomials Putting t = b−d an satisfy required conditions. Suppose that t 6= 0. Substituting P (x) = Q(x) + t to P (x + Q(y)) = Q(x + P (y)) we get Q(x + Q(y)) + t = Q(x + Q(y) + t). Putting x := x − Q(y) we have Q(x) + t = Q(x + t). This implies that Q(kt) = Q(0) + kt for k ∈ Z and therefore Q(x) = Q(0) + x for every x. It follows that P (x) = x + Q(0) + t. Those polynomials clearly satisfy required conditions. J Alternative solution. Substitute x = −P (y). We get P (Q(y)−P (y)) = Q(0). If Q(y)−P (y) is not a constant, then Q(y) − P (y) is a polynomial which takes infinitely many values, which would imply thatP (x) = Q(0) with infinitely many x, hence P (x) is a constant. In this case P ≡ Q ≡ d, where d is a constant, which is a valid solution. Therefore Q(y) − P (y) = c, where c is a constant. If c = 0, then we get P ≡ Q, which is also a valid solution. If c 6= 0, we get P (x + P (y) + c) = P (x + P (y)) + c. Let’s call z = x + P (y), z is any real number and P (z + c) = P (z) + c. Substituting z = 0 and z = c gives P (c) = P (0) + c, P (2c) = P (c) + c = P (0) + 2c and it is easy to see by induction that P (kc) = P (0) + kc for every positive integer k. 10
T-1
Team Competition
MEMO 2017
Let’s call G(x) = P (x) − x − P (0), then G is a polynomial which has infinitely many roots and so G ≡ 0 and P (x) = x + a for all real x and a constant a. Then Q(x) = P (x) + c = x + b with a constant b. Those polynomials clearly satisfy required conditions. J
11
T-2
Team Competition
MEMO 2017
T-2 Determine the smallest possible real constant C such that the inequality |x3 + y 3 + z 3 + 1| 6 C|x5 + y 5 + z 5 + 1| holds for all real numbers x, y, z satisfying x + y + z = −1.
Answer: The smallest constant C is
9 . 10
Solution. The key for our solution is the replacement of 1 by −(x + y + z)3 and −(x + y + z)5 on the LHS and RHS, resp., of the inequality under consideration. Thus we have to deal with the equivalent inequality |x3 + y 3 + z 3 − (x + y + z)3 | 6 C · |x5 + y 5 + z 5 − (x + y + z)5 |. Now, for instance z = −x shows that the expressions on either side of our inequality are 0. Therefore, both expressions have (x + z) as a factor. Similarly, (x + y) and (y + z) are factors, too. A bit of moderate algebra yields |3(x + y)(x + z)(y + z)| 6 C · |5(x + y)(x + z)(y + z)(x2 + y 2 + z 2 + xy + xz + yz)|. Therefore, (x + y)(x + z)(y + z) = 0 certainly implies equality in our inequality. We thus let further on be (x + y)(x + z)(y + z) 6= 0 and thus have to deal with 3 6 5C · |x2 + y 2 + z 2 + xy + xz + yz|. Because of |x2 + y 2 + z 2 + xy + xz + yz| = x2 + y 2 + z 2 + xy + xz + yz, this is equivalent to
⇔ ⇔
3 6 x2 + y 2 + z 2 + xy + xz + yz, 5C 6 6 x2 + y 2 + z 2 + (x + y + z)2 , 5C 6 − 1 6 x2 + y 2 + z 2 . 5C
Now, the arithmetic-square root-inequality implies x2 + y 2 + z 2 > 12
(x + y + z)2 1 = 3 3
T-2
Team Competition
MEMO 2017
with equality iff x = y = z = − 31 . Thus, a constant that works for all values of x, y, z satisfies 6 1 −16 , 5C 3 that is finally C> As a summary, we have Cmin =
9 10
9 . 10
and there occurs equality in our inequality iff
1 (x + y)(x + z)(y + z) = 0 or x = y = z = − . 3
J
Alternative solution. We replace z = −1 − x − y and factor both sides to obtain |3(x + 1)(y + 1)(x + y)| 6 5C|(x + 1)(y + 1)(x + y)|(x2 + xy + y 2 + 1 + x + y). If x = −1, y = −1 or x = −y, both sides are zero; otherwise we may cancel the corresponding factors. We have 1 2 2 y 1 2 3 y+ (x2 + xy + y 2 + 1 + x + y) = x + + + + , 2 2 4 3 3 so the right hand side is at least 23 . Thus 10 ·
C 3
6 3, i.e., C =
9 10
is the optimal value.
J
13
T-3
Team Competition
MEMO 2017
T-3 There is a lamp on each cell of a 2017 × 2017 square board. Each lamp is either on or off. A lamp is called bad if it has an even number of neighbours that are on. What is the smallest possible number of bad lamps on such a board? (Two lamps are neighbours if their respective cells share a side.) Answer: The smallest possible number of lamps with an even number of neighbours that are on is 1. Solution. Please consult the figures at the end of this solution. We divide the square in 1 × 1-squares and color the square in checkerboard fashion such that the corners are black and we call lamps on black and white squares black and white lamps, respectively. We assign the number 1 to a lamp that is on, and the number 0 to a lamp that is off. If we assign coordinates (0, 0) to the lamp in the center, we see that the black lamps are exactly the lamps with the coordinates (i, j) where i + j is even. Now we assume that the minimum number is 0 that is, there is a configuration where every lamp has an odd number of neighbours that are on, and we try to get a contradiction. For every black lamp with coordinates (i, j), i and j even, we add the numbers associated to its neighbours, and add all these numbers. The parity of this sum S can be determined in the following two ways: On the one hand, we know that every lamp has an odd number of neighbours with value 1, so we simply have to determine the number modulo 2 of lamps with i and j even. Since we can group lamps at (i, j) with lamps at (−i, −j) and the lamp in the center is the only one left, we get that S is odd. On the other hand, every white lamp enters the sum as often as it has neighbours with i and j even. But there are exactly two such lamps because exactly one of the coordinate of the white lamp is odd and can be modified with plus or minus 1 to get a neighbour with two even coordinates. There are no problems at the boundary because this process will not change the coordinate ±1008 so we will stay inside the square. Therefore, S is even, which is clearly a contradiction. So, it is impossible that all lamps have an odd number of neighbours that are on. Now, we will provide a concrete arrangement where all lamps except for the lamp at the center have an odd number of neighbours that are on. For the black lamps, i.e. i + j even, we choose the values: 0, if max(|i|, |j|) ≡ 0, 1 mod 4, f (i, j) = 1, if max(|i|, |j|) ≡ 2, 3 mod 4. 14
T-3
Team Competition
MEMO 2017
For the white lamps, i.e. i + j odd, we choose the values: 0, if max(|i|, |j| − 1) ≡ 0, 1 f (i, j) = 1, if max(|i|, |j| − 1) ≡ 2, 3
mod 4, mod 4.
(This assignment can be found by replacing 2017 with a small number, say 17, starting with a row of zeros, using the assumptions to determine the rest and then notice that the zeros and ones for black or white lamps only form frames of depth 2 around the center.) It is now easily checked that the condition is satisfied for all non-central lamps: For a white lamp we assume without loss of generality |i| < |j| (equality is impossible because they have different parity). Then, for the neighbours (i ± 1, j) and (i, j ± 1), the bigger coordinates are |j − 1|, |j|, |j| and |j + 1| and we can check easily that an odd number of them are ≡ 2, 3 mod 4. For a black lamp with j > 0 or j < 0, we argue anlogously. If j = 0, then i 6= 0 for a non-central lamp, therefore the maximum is |i| and wie have again the values |i − 1|, |i|, |i|, |i + 1| to check which contain and odd number of values ≡ 0, 1 mod 4. Therefore, we have found an arrangement with exactly one lamp with an even number of neighbours that are on as desired.
The images show the discussed optimal arrangement for n = 77. Lamps that are on are yellow, lamps that are off are blue. The first image shows all lamps, the second image shows the lamps with i + j even and the third image shows the lamps with i + k odd. J Alternative solution. We color the board as a chess board in such a way that the four corners are white. An active lamp on a black field has no influence on the number of active neighbours of any lamp on a black field, and vice versa an active lamp on a white field has no influence on the number of active neighbours of a lamp on a white field. Therefore, we can optimize the number of lamps with an even number of active neighbours separately for lamps on black and white fields. 15
T-3
Team Competition
MEMO 2017
For the black fields, it is easy to find an arrangement in which all black fields have exactly one active neighbour, by turning on the following lamps: In the 1st, 5th, 9th, 13th, ... row the 1st, 5th, 9th, 13th, ... lamp (so all lamps with x ≡ y ≡ 1 mod 4, assuming that the corner has coordinates (1, 1)), and in the 3rd, 7th, 11th, ... row the 3rd, 7th, 11th, ... lamp (so all lamps with x ≡ y ≡ 3 mod 4). An example for n = 77 with the same color coding as in the previous solution:
For white fields, we first show that at least one white field has to have an even number of active neighbours. To do so, we colour all white fields in the 1st, 3rd, 5th, ... row and in the 1st, 3rd, 5th, ... column red. Each black field is neighbour to exactly two red fields, and red fields have only black neighbours. If x lamps on black fields are active, then all red fields together have exactly 2x active neighbours. Since the number of red fields is odd, at least one of them has to have an even number of active neighbours. All that is left to do is finding an arrangement in which all white fields except for one have an odd number of active neighbours. To do so, we separate the board into four “triangles” roughly as follows: 16
T-3
Team Competition
MEMO 2017
In the upper triangle we turn on the following lamps: In the 1st, 3rd, 5th, ... row always the lamp on the second field from the left, and then every 4th lamp, like this:
For filling the other triangles, we rotate the same pattern by 90◦ . Each lamp is only neighboured to fields from the own triangle, and each field within the triangle has exactly one active neighbour. Only the field in the middle of the board is left.
17
T-3
Team Competition
MEMO 2017
An example for n = 77:
Altogether, the pattern looks like this:
J
18
T-4
Team Competition
MEMO 2017
T-4 Let n > 3 be an integer. A sequence P1 , P2 , . . . , Pn of distinct points in the plane is called good if no three of them are collinear, the polyline P1 P2 . . . Pn is non-self-intersecting and the triangle Pi Pi+1 Pi+2 is oriented counterclockwise for every i = 1, 2, . . . , n − 2. For every integer n > 3 determine the greatest possible integer k with the following property: there exist n distinct points A1 , A2 , . . . , An in the plane for which there are k distinct permutations σ : {1, 2, . . . , n} → {1, 2, . . . , n} such that Aσ(1) , Aσ(2) , . . . , Aσ(n) is good. (A polyline P1 P2 . . . Pn consists of the segments P1 P2 , P2 P3 , . . . , Pn−1 Pn .) Answer: n2 − 4n + 6. Solution. Fix n points on a plane, no three of which are collinear. Let P be their convex hull. Let the vertices of P be A1 , A2 , . . . , Am (lying in this order on the boundary of P counterclockwise). We denote Am+1 = A1 . Also, let I be the set of our fixed points other than A1 , . . . , Am , i.e. the points lying in the interior of P. Lemma. Every good polyline contains all but one side of P. Proof. The key observation is that if a segment Ai Ai+1 is not a part of our polyline, then the point Ai+1 appears in the polyline before Ai . This is clear if Ai is the last vertex of the polyline. Otherwise there is a segment Ai X in the polyline, where X 6= Ai+1 . Observe that all segments appearing after Ai X are located in the halfplane determined by the line Ai X which does not contain the point Ai+1 . This is because the polyline always turns left, has no self-intersections, and Ai is a vertex of P. This implies that the point Ai+1 must appear in the polyline before Ai . It is clear that at least one side of P does not appear in the polyline. Suppose now that Ai1 Ai1 +1 , Ai2 Ai2 +1 , . . . , Aij Aij +1 are all segments on the boundary of P that do not appear in the polyline (where 1 6 i1 < i2 < . . . < ij 6 n and j > 2). Using the observation we know that Ai1 appears after Ai1 +1 , which is followed by Ai1 +2 , Ai1 +3 , . . . , Ai2 . Since Ai2 6= Ai1 , this means that Ai1 appears after Ai2 . Analogously, Ai2 appears after Ai3 , and so on, and Aij appears after Ai1 . Thus Ai1 appears after Ai1 , which is absurd. Therefore there is exactly one i such that Ai Ai+1 does not belong to the polyline. J Lemma. For every good polyline there is a line which intersects exactly one segment of the polyline. Proof. Using the previous lemma we know that there is exactly one i such that Ai Ai+1 does not belong to the polyline. Thus the polyline is of the form B1 B2 . . . Bj Ai+1 Ai+2 . . . Ai−1 Ai C1 C2 . . . Cl . (It may happen that there are no points before Ai+1 and/or no points after Ai .) 19
T-4
Team Competition
MEMO 2017
It is quite clear that B1 B2 . . . Bj Ai+1 and Ai C1 C2 . . . Cl can be separated by a line. Again, this follows form the fact that polyline has no self-intersections and always turns left, and the fact that you can separate two non-intersecting convex polygons by a line. It is clear that this separating line must intersect exactly one segment of the polyline Ai+1 Ai+2 . . . Ai−1 Ai . J Lemma. Each good polyline is uniquely determined by an i such that Ai Ai+1 is not in the polyline and by the partition I = B ∪ C such that there is a line intersecting Ai Ai+1 separating B from C. Proof. This is easy to see. We use the previous lemma and the fact that the polyline only turns left and has no self-intersections. J Lemma. Consider the n−m lines determined by the points of I. Suppose that j of them 2 intersect segment Ai Ai+1 . Then there are exactly n − m + j + 1 good polylines not containing Ai Ai+1 . Proof. We will move a point X along the segment Ai Ai+1 , starting from Ai , and count how many good partitions of I are there. In the beginning of our journey there are n − m + 1 possible partitions of I by a line passing through X. Every time we cross a line determined by some two points of I we get exactly one new partition. Since we cross j such lines, the total number of good partitions is equal to n − m + 1 + j. This corresponds to n − m + j + 1 good polylines. J Lemma. There are exactly (n − m + 1)m + 2 n−m good polylines. 2 Proof. For each i there are n − m + ji + 1 good polylines. Summing up yields m X
n − m + ji + 1 = m(n − m + 1) +
i=1
because there are of P.
m X i=1
n−m 2
n−m ji = (n − m + 1)m + 2 2
lines determined by points in I and each of them intersects two sides J
Since m 7→ 2 n−m + (n − m + 1)m is decreasing, it follows that the greatest possible number 2 of good polylines is achieved for the smallest possible value of m, i.e. for m = 3. Therefore the answer is 2 n−3 + 3(n − 2) = n2 − 4n + 6. J 2
20
T-5
Team Competition
MEMO 2017
T-5 Let ABC be an acute-angled triangle with AB > AC and circumcircle Γ. Let M be the midpoint of the shorter arc BC of Γ, and let D be the intersection of the rays AC and BM . Let E 6= C be the intersection of the internal bisector of the angle ACB and the circumcircle of the triangle BDC. Let us assume that E is inside the triangle ABC and there is an intersection N of the line DE and the circle Γ such that E is the midpoint of the segment DN . Show that N is the midpoint of the segment IB IC , where IB and IC are the excentres of ABC opposite to B and C, respectively. Solution. Consider the following implications: Let us denote by P the other point of intersection of internal bisector from C and the circle Γ. BDCE is cyclic =⇒ ∠BDC = ∠BEP =⇒ 4BEP ∼ 4BDA BD BE = =⇒ BP BA BE BP =⇒ = BD BA =⇒ 4BDE ∼ 4BAP =⇒ EB = ED
(∠BP E = ∠BP C = ∠BAC = ∠BAD)
1 (∠DBE = ∠ABP = ∠ACB) 2 (4BAP is isosceles)
=⇒ ∠M BN = ∠DBN = 90◦ =⇒ M and N are diametrically opposite on Γ =⇒ N B = N C. The quadrilateral IB IC BC is cyclic, as IB B ⊥ IC B and IC C ⊥ IB C. Let IA be the excentre of 4ABC opposite to A. Now consider the nine point circle of 4IA IB IC . We can clearly see that this circle is Γ, as 4ABC is the orthic triangle of 4IA IB IC . So Γ passes through the midpoint of IB IC and this midpoint is also equidistant from B and C, so N must be the described midpoint. J
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T-6
Team Competition
MEMO 2017
T-6 Let ABC be an acute-angled triangle with AB 6= AC, circumcentre O and circumcircle Γ. Let the tangents to Γ through B and C meet each other at D, and let the line AO intersect BC at E. Denote the midpoint of BC by M and let AM meet Γ again at N 6= A. Finally, let F 6= A be a point on Γ such that A, M , E and F are concyclic. Prove that F N bisects the segment M D.
Solution. We may suppose AB < AC. It is known that AD is a symmedian of triangle ABC. Let Q be its second point of intersection with k. The triangles ABM and and AQC are similar to each other, since their corresponding angles are equal, and it follows that ∠AF Q = ∠ACQ = ∠AM B = ∠AF E. Hence the points Q, E, and F are collinear. Now let R 6= F denote the point where F D intersects k again. As before, F D is a symmedian of triangle F BC, the triangles CF M and RF B are similar, ∠EAF = ∠EM F = ∠RBF = ∠RAF, and the points A, O, and R are collinear as well. ∠AF D = 90◦ .
So AR is a diameter of k and thus
Now let X be the midpoint of AD and put J = M D ∩ F N . As we have just seen, the triangle AXF is isosceles at X, the angle at its base being ∠XAF = ∠QN J. But since AM is a median in triangle ABC and AQ the corresponding symmedian, we have QN ||BC ⊥ OD, so the triangle QJN is likewise isosceles. As we have just seen, it has the same angle at its base. We thus have two similar isosceles triangles and looking at their remaining angles we learn ∠F XA = ∠N JQ. This implies that the quadrilateral XQJF is cyclic. Thus ∠QXJ = ∠QF J = ∠QF N = ∠QAN or in other words, the lines XJ and AM are parallel. Since X was defined to be the midpoint of AD, this tells us that J is the midpoint of M D, as desired. J Alternative solution. As in the first solution we restrict our attention to the case AB < AC, define the points Q and J, and remark that the points Q, E, and F are collinear. Our next step is to prove the similarity of the triangles JF M and JM N . Evidently their angles at J coincide and in view of ∠M F J = ∠M F Q + ∠QF N = ∠M AE + ∠QAM = ∠QAO 22
T-6
Team Competition
MEMO 2017
it remains to be shown that ∠QAO = ∠JM N . To this end we notice that because of DM ·DO = DB 2 = DQ · DA the quadrilateral AQM O is cyclic, and deduce that ∠QAO = ∠OQA = ∠OM A = ∠JM N is indeed the case. This concludes the proof of 4JF M ∼ 4JM N . Now JM 2 = JN · JF has become clear. Writing r for the radius of k and considering the power of J with respect to this circle we obtain JN · JF = JO2 − r2 = JO2 − OM · OD, so altogether we have OM · OD = JO2 − JM 2 = (JO − JM )(JO + JM ) = OM · (2JM + OM ). This simplifies immediately to 2JM = M D, whereby the problem is solved.
J
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T-7
Team Competition
MEMO 2017
T-7 Determine all integers n > 2 such that there exists a permutation x0 , x1 , . . . , xn−1 of the numbers 0, 1, . . . , n − 1 with the property that the n numbers x0 ,
x0 + x1 ,
...,
x0 + x1 + . . . + xn−1
are pairwise distinct modulo n. Answer: All even numbers. Solution. Suppose that x0 , . . . , xn−1 is such a permutation. Note that x0 = 0. Indeed, if xi = 0 for some i > 0 then x0 + · · · + xi−1 = x0 + · · · + xi−1 + xi , which is a contradiction. On the other hand x0 + x1 + · · · + xn−1 = 0 + 1 + 2 + · · · + n − 1 = n ·
n−1 . 2
This means that if n is odd then x0 + x1 + · · · + xn−1 ≡ 0 (mod n). This gives a contradiction if n > 1, because x0 = 0. If n is even then we put xi = i if i is even and xi = n − i if i is odd. Then x0 + x1 + · · · + x2m = 0 + (n − 1) + 2 + (n − 3) + · · · + 2m ≡ m (mod n) and x0 + x1 + · · · + x2m+1 = x0 + x1 + · · · + x2m + (n − 2m − 1) ≡ n − m − 1
(mod n).
Thus the numbers x0 + x1 + · · · + xi , i = 0, 1, . . . , n − 1, are pairwise distinct modulo n.
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J
T-8
Team Competition
MEMO 2017
T-8 For an integer n > 3 we define the sequence α1 , α2 , . . . , αk as the sequence of exponents in the prime factor decomposition of n! = pα1 1 pα2 2 . . . pαk k , where p1 < p2 < · · · < pk are primes. Determine all integers n > 3 for which α1 , α2 , . . . , αk is a geometric progression. Answer: The solutions are n = 3, 4, 6, 10. Solution. Let pi be the ith prime number and let αi be the exponent of pi in n!. It is wellknown that n n n αi = + 2 + 3 + .... pi pi pi For n > 9, we have 4n − 14 n 2 n 8 − + − = , 3 3 9 9 9 n−6 α4 > , 7 n n n + ··· = . α3 < + 5 25 4 α2 >
In the geometric sequence, we have α2 α4 = α32 . Since all terms in the above inequalities are positive, we get:
4n − 14 9
n−6 7
< α2 α4 = (α3 )2 <
n2 . 16
After simplification, we get: n2 − 608n + 1344 6 0. This is certainly false for n > 608 and it remains to check the small cases. For two primes p and q with p < q < 2p, we know that for q 6 n 6 2p − 1, they will both have exponent 1 in the prime factor decomposition of n! which is impossible because the ratio of the geometric sequence is bigger than 1 for n > 3. We will now give a list of appropriate primes p and q such that the intervals [q, 2p − 1] cover most of our interval. p q 2p − 1
3 5 7 11 17 29 47 83 157 311 5 7 11 13 19 31 53 89 163 313 5 9 13 21 33 57 93 165 313 621
It remains to check n = 3, 4, 6, 10 which give the sequences of exponents (1, 1), (3, 1), (4, 2, 1) and (8, 4, 2, 1), respectively, which clearly work. J Alternative solution. Let pi be the ith prime number, that is p1 = 2, p2 = 3, p3 = 5, etc. We check the small cases up to n = 11 and find the solutions 3!, 4!, 6! and 10!. From now on, let n > 11. 25
T-8
Team Competition
MEMO 2017
Claim. The smallest and last exponent αk in the prime factor decomposition of n! is always 1. Proof. By Bertrand’s postulate, we know pk+1 < 2pk . Therefore, for all n ∈ [pk , pk+1 − 1], the largest prime number that occurs in n! is pk and it occurs exactly once. J Therefore, α1 = f m for some f ∈ N and m + 1 is the number of primes in the prime factor decomposition of n!. Case 1: f = 2. The exponent of 2 in the prime factor decomposition of (2m )! is 2m−1 +2m−2 +· · ·+1 = 2m −1. Therefore, α1 = 2m holds exactly for n = 2m + 2 and 2m + 3, and since n > 11, we have m > 4. Let π(x) be the number of primes 6 x. We check easily that π(16) = 6. Bertrand’s postulate implies π(2x) > π(x) + 1, therefore, π(n) > π(2m ) > π(2m−1 ) + 1 > . . . > π(24 ) + m − 4 = 6 + m − 4 = m + 2 which is not compatible with the fact that there are m + 1 primes in the prime factor decomposition of n!. Case 2: f > 2. Since the exponent of 2 is now even bigger than 2m , n must be bigger than 2m + 3, so as before, the number of primes in the prime factor decomposition of n! must be bigger than m + 1 which is again a contradiction. We have seen, that the solutions 3, 4, 6, 10 are indeed all the solutions. J
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