PURPLE COMET MATH MEET April 2012 HIGH SCHOOL - SOLUTIONS c Copyright Titu Andreescu and Jonathan Kane
Problem 1 Last month a pet store sold three times as many cats as dogs. If the store had sold the same number of cats but eight more dogs, it would have sold twice as many cats as dogs. How many cats did the pet store sell last month?
Answer: 48 Let x be the number of dogs sold. Then 3x is the number of cats sold, and 2(x + 8) = 3x. Thus, 2x + 16 = 3x and x = 16. It follows that the number of cats sold is 3x = 3 · 16 = 48.
Problem 2 What is the greatest three-digit divisor of 111777?
Answer: 703 Factoring yields 111777 = 111 · 1007 = 3 · 37 · 19 · 53. The product of any three of the prime factors of 111777 exceeds 1000, so consider the product taken two at a time: 3 · 19 = 57, 3 · 37 = 111, 3 · 53 = 159, 19 · 37 = 703, 19 · 53 = 1007, and 37 · 53 is clearly greater than 1000. The greatest of the 3-digit products is 703.
Problem 3 The diagram below shows a large square divided into nine congruent smaller squares. There are circles inscribed in five of the smaller squares. The total area covered by all the five circles is 20π. Find the area of the large square. 1
Answer: 144 Let the circles each have radius r. Then the total area covered by all the five circles is 5πr2 = 20π, so r = 2. Each small square has side length 2r = 4, and the larger square has side length 3 · 4 = 12. It follows that the larger square has area 122 = 144.
Problem 4 The following diagram shows an equilateral triangle and two squares that share common edges. The area of each square is 75. Find the distance from point A to point B.
Answer: 15 Let the squares and triangle have side length s. Let the vertex common to the two squares and the triangle be labeled C. Note that ∠ACB = 120◦ . Then the Law of Cosines gives AB 2 = s2 + s2 − 2 · s · s cos(120◦ ) = 3 · s2 = 3 · 75 = 225. Thus, AB = 15.
2
Problem 5 Find the sum of the squares of the values x that satisfy
1 x
+
2 x+3
+
3 x+6
= 1.
Answer: 33 Multiply the equation by x(x + 3)(x + 6) to get (x + 3)(x + 6) + 2x(x + 6) + 3x(x + 3) = x(x + 3)(x + 6). This simplifies to x3 + 3x2 − 12x − 18 = (x − 3)(x2 + 6x + 6) = 0. The possible values of x are 3, √ √ −3 + 3, and −3 − 3. The requested sum of squares is √ √ √ √ 32 + (−3 + 3)2 + (−3 − 3)2 = 9 + (9 − 6 3 + 3) + (9 + 6 3 + 3) = 33.
Problem 6 Find the least positive integer n so that both n and n + 1 have prime factorizations with exactly four (not necessarily distinct) prime factors.
Answer: 135 Note that either n or n + 1 must be odd. The smallest odd positive integer with four prime factors is 34 = 81, but neither 80 = 24 · 5 nor 82 = 2 · 41 have four prime factors. The second smallest odd positive integer with four prime factors is 33 · 5 = 135. The neighbors of 135 are 134 = 2 · 67 and 136 = 23 · 17, so it follows that n = 135 is the requested number.
Problem 7 Two convex polygons have a total of 33 sides and 243 diagonals. Find the number of diagonals in the polygon with the greater number of sides.
Answer: 189 If a convex polygon has n sides, it has
n(n−1) 2
−n =
n2 −3n 2
diagonals. So if one
convex polygon has n sides and another has 33 − n sides, together the number of diagonals in the two polygons is
n2 −3n 2
+
(33−n)2 −3(33−n) 2
=
2n2 −66n+990 . 2
Setting this equal to 243 yields 2n2 − 66n + 990 = 2 · 243 which simplifies to n2 − 33n + 252 = 0. The polynomial factors to give (n − 12)(n − 21) = 0. Thus, the two polygons have 12 and 21 sides. The number of diagonals in the polygon with 21 sides is, therefore,
212 −3·21 2
3
=
18·21 2
= 9 · 21 = 189.
Problem 8 In the tribe of Zimmer, being able to hike long distances and knowing the roads through the forest are both extremely important, so a boy who reaches the age of manhood is not designated as a man by the tribe until he completes an interesting rite of passage. The man must go on a sequence of hikes. The first hike is a 5 kilometer hike down the main road. The second hike is a 5 14 kilometer hike down a secondary road. Each hike goes down a different road and is a quarter kilometer longer than the previous hike. The rite of passage is completed at the end of the hike where the cumulative distance walked by the man on all his hikes exceeds 1000 kilometers. So in the tribe of Zimmer, how many roads must a man walk down, before you call him a man?
Answer: 73 The total distance walked by a man in n hikes is h Pn 5 + 5 41 + 5 12 + · · · + 19+n = k=1 19+k = 41 19n + 4 4
n(n+1) 2
i . This exceeds 1000
when n2 + 39n > 8000. The smallest natural number n for which this inequality is satisfied is n = 73.
Problem 9 Find the value of x that satisfies log3 (log9 x) = log9 (log3 x).
Answer: 81 Using the change of base formula for logarithms, the logarithms base 9 can be log3 log3 x 3x changed to logarithms base 3 to get log3 log log3 9 = log3 9 . Thus, 2 2 log3 log23 x = log3 log3 x, so log23 x = log3 x. Let u = log3 x. Then u2 4
= u, so u = 0 or u = 4. But u = 0 is not a possible solution since log3 u
must be defined. Therefore, u = 4 and log3 x = 4. Finally, x = 34 = 81.
Problem 10 Consider a sequence of eleven squares that have side lengths 3, 6, 9, 12, . . . , 33. Eleven copies of a single square each with area A have the same total area as the total area of the eleven squares of the sequence. Find A.
4
Answer: 414 Pn Recall that k=1 k 2 =
n(n+1)(2n+1) . 6
The total area of the squares in the sequence is 3 + 6 + 9 + · · · + 33 = 32 12 + 22 + 32 + · · · + 112 = 2
9·
11·12·23 6
2
2
= 11A. Thus, A = 9 ·
2
12·23 6
= 414.
Problem 11 Define f (x) = 2x + 3 and suppose that g(x + 2) = f (f (x − 1) · f (x + 1) + f (x)). Find g(6).
Answer: 259 One has g(6) = g(4 + 2) = f (f (4 − 1) · f (4 + 1) + f (4)) = f (f (3) · f (5) + f (4)) = f (9 · 13 + 11) = f (128) = 259.
Problem 12 Ted flips seven fair coins. There are relatively prime positive integers m and n so that
m n
is the probability that Ted flips at least two heads given that he
flips at least three tails. Find m + n.
Answer: 190 There are k7 equally likely ways to flip k tails with seven flips. Thus, the probability Ted flips at most five tails given that Ted flips at least three tails is (73)+(74)+(75) 35+35+21 = 35+35+21+7+1 = 91 7 99 . The requested sum is 91 + 99 = 190. (3)+(74)+(75)+(76)+(77)
Problem 13 Find the least n for which n!(n + 1)!(2n + 1)! − 1 ends in thirty digits that are all 9’s.
Answer: 34 If n!(n + 1)!(2n + 1)! − 1 ends in thirty digits all equal to 9, then n!(n + 1)!(2n + 1)! must end in thirty 0’s. The number of trailing zeros in n! is n the number of factors of 5 in n! which, for n < 100, is b n5 c + b 25 c. (Here, bxc
refers to the greatest integer less than or equal to x.) For example, when 27 n = 27, the number of trailing zeros in n! is b 27 5 c + b 25 c = 5 + 1 = 6. The least
5
n for which n!(n + 1)!(2n + 1)! ends with thirty 0’s can be found by searching. If n = 27, the number of zeros is 6 + 6 + 13 = 25. If n = 30, the number of zeros is 7 + 7 + 14 = 28. If n = 33, the number of zeros is 7 + 7 + 15 = 29. But if n = 34, the number of zeros is 7 + 8 + 15 = 30, so the least n with the desired property is n = 34.
Problem 14 A circle in the first quadrant with center on the curve y = 2x2 − 27 is tangent to the y-axis and the line 4x = 3y. The radius of the circle is
m n
where m and
n are relatively prime positive integers. Find m + n.
Answer: 11 The distance from a point to the line 4x = 3y is given by the normal form for the equation of the line and is
3y−4x √ 32 +42
=
3y−4x . 5
For the circle to be tangent to
2
the line and the y-axis, its center (x, 2x − 27) must be equidistant from the y-axis and the line, so x =
3·(2x2 −27)−4x 5
or 6x2 − 9x − 81 = 0. This has
solutions x = −3 and x = 92 . Thus, the circle in the first quadrant has radius 9 2,
and the requested sum is 9 + 2 = 11.
Problem 15 Let N be a positive integer whose digits add up to 23. What is the greatest possible product the digits of N can have?
Answer: 4374 Digits 0 and 1 do not increase the product of the digits, so the N with the largest product of digits has no digits of 0 or 1. A digit of 4 can be replaced by two digits of 2 which have the same sum and same product as the one digit of 4. Any digit k ≥ 5 can be replaced by the digits 2 and k − 2 because 2(k − 2) > k for these digits. Hence, an N with the greatest product of digits has all of its digits being 2’s and 3’s. The largest product is then of the form 2a 3b where 2a + 3b = 23. Note that 2 + 2 + 2 = 3 + 3, but 2 · 2 · 2 < 3 · 3, so two 3’s give a larger product than three 2’s. It follows that the largest product is 2 · 37 = 4374.
6
Problem 16 Let a, b, and c be non-zero real numbers such that ca c+a
ab a+b
= 3,
bc b+c
= 4, and
= 5. There are relatively prime positive integers m and n so that
abc ab+bc+ca
=
m n.
Find m + n.
Answer: 167 Taking the reciprocals of each side of the given equations shows 1 3
=
a+b ab
=
1 b
+ a1 ,
1 4
equations to get
2 a
n m
1 a
=
ab+bc+ca abc
=
=
+
2 b
+
1 b
b+c bc
=
1 c
+ 1b , and
+
2 c
=
1 3
+
+
1 c
=
47 120 .
1 4
+
1 5
1 5
=
= 47 60 .
c+a ca
=
1 a
+ 1c . Add these three
Note that
The requested sum is
m + n = 120 + 47 = 167.
Problem 17 How many positive integer solutions are there to w + x + y + z = 20 where w + x ≥ 5 and y + z ≥ 5?
Answer: 781 Assume a solution assigns w + x = k and y + z = 20 − k. There are k − 1 ways to assign positive integer values to w and x so that their sum is k, and 19 − k ways to assign positive integer values to y and z so that y + z = 20 − k. It follows that the number of solutions satisfying the given conditions is 15 X
(k − 1)(19 − k) =
k=5
45 · 11 + 12 ·
11 X
(k + 3)(15 − k) =
k=1
11 X
45 + 12k − k 2 =
k=1
11(12) 11(12)(23) − = 11(45 + 72 − 46) = 11(71) = 781. 2 6
Problem 18 Find the number of three-digit numbers such that its first two digits are each divisible by its third digit.
Answer: 138 If the third digit of a three-digit number divides the other two digits, the third digit clearly cannot be 0.
7
Third Digit
First Digit
Second Digit
Possibilities
1
1,2,3,4,5,6,7,8,9
0,1,2,3,4,5,6,7,8,9
90
2
2,4,6,8
0,2,4,6,8
20
3
3,6,9
0,3,6,9
12
4
4,8
0,4,8
6
5,6,7,8,9
same as third
0 or same as third
10
Thus, the number of such numbers is 90 + 20 + 12 + 6 + 10 = 138.
Problem 19 9
2
5
Find the remainder when 25 + 59 + 92 is divided by 11.
Answer: 8 Fermat’s Little Theorem states that if integer n is not a multiple of prime p, then np−1 ≡ 1 (mod p). Thus, 210 , 510 , and 910 are all congruent to 1 (mod 11). Note that 59 ≡ 5 (mod 10), 92 = 1 (mod 10), and 25 ≡ 2 (mod 10). 9
2
5
It follows that 25 + 59 + 92 ≡ 25 + 51 + 92 ≡ 32 + 5 + 81 ≡ 8 (mod 11).
Problem 20 Square ABCD has side length 68. Let E be the midpoint of segment CD, and let F be the point on segment AB a distance 17 from point A. Point G is on segment EF so that EF is perpendicular to segment GD. The length of √ segment BG can be written as m n where m and n are positive integers, and n is not divisible by the square of any prime. Find m + n.
Answer: 46 −−→ −−→ −−→ −−→ Let vector BC = u and vector BA = v. Then EF = v4 − u. Note that EG is −−→ −−→ the projection of ED = v2 onto EF . This projection is ! −−→ −−→ ! v v v 1 v ED · EF −−→ 1 2 2 · 4 − u 8 EF = − u = − u = v− u. −−→ −−→ 1 v v 4 4 34 17 + 1 − u · − u EF · EF 16 4 4 −−→ −−→ 9 Then BG = u + v2 + EG = 15 17 u + 17 v. This vector has length √ √ √ 68 2 2 17 15 + 9 = 4 306 = 12 34. The requested sum is 12 + 34 = 46.
8
Problem 21 Each time you click a toggle switch, the switch either turns from off to on or from on to off. Suppose that you start with three toggle switches with one of them on and two of them off. On each move you randomly select one of the three switches and click it. Let m and n be relatively prime positive integers so that
m n
is the probability that after four such clicks, one switch will be on
and two of them will be off. Find m + n.
Answer: 142 Without loss of generality assume that at the beginning the first of the three toggle switches is on, and the second and third are off. There are 34 = 81 equally likely ways to select a sequence of four toggle switches to click. After four clicks, there will either be one or three switches in the on position. To get three toggle switches in the on position, one must click the first switch an even number of times and the second and third switches an odd number of times each. This can be done by clicking the first switch twice and the other switches once each, or by clicking either the second or third switch three times and the other of the two switches one time. The number of ways to do one of 4 4 4 4! 4! 4! these is 2,1,1 + 0,1,3 + 0,3,1 + 0!·1!·3! + 0!·3!·1! = 12 + 4 + 4 = 20. It = 2!·1!·1! follows that the desired probability is
81−20 81
=
61 81 .
The requested sum is 61 +
81 = 142. Alternatively, there are four states of the toggle switches indicating whether there are 0, 1, 2, or 3 switches in the on position. Let ai,j be the probability that the system will transition from state i to state j with one random toggle 0 1 0 0 1 3 0 23 0 . switch. The transition matrix with i, j entry ai,j is A = 0 23 0 13 0 0 1 0 1 2 7 20 0 0 0 0 3 27 3 27 7 2 61 20 0 0 0 0 9 9 81 81 2 Then A2 = and A4 = 2 . Since A 7 20 61 9 0 9 0 81 0 81 0 2 1 61 0 3 0 3 0 20 0 27 81 gives the probabilities of transitioning from one state to another with two clicks, and A4 gives the probabilities of transitioning from one state to another 9
with four clicks, it is seen that the probability of transitioning from one toggle on to one toggle on in four clicks is
61 81 .
Problem 22 The diagram below shows circles radius 1 and 2 externally tangent to each other and internally tangent to a circle radius 3. There are relatively prime positive integers m and n so that a circle radius
m n
is internally tangent to the
circle radius 3 and externally tangent to the other two circles as shown. Find m + n.
Answer: 13 Let the unknown radius be r. Let the centers of the circles radius 3, 2, 1, and r be A, B, C, and D, respectively, as shown. Applying the Law of Cosines to 4ABD gives 12 + (2 + r)2 − 2 · 1 · (2 + r)(cos ABD) = (3 − r)2 . Applying the Law of Cosines to 4CBD gives 32 + (2 + r)2 − 2 · 3 · (2 + r)(cos ABD) = (1 + r)2 . Subtracting three times the first equation from the second yields 3(3 − r)2 − (1 + r)2 = −6 + 2(2 + r)2 which reduces to 24 = 28r implying r = 76 . The requested sum is 6 + 7 = 13. Note that this can also be solved as an application of the Descartes’ Kissing Circles Theorem.
10
Problem 23 Find the greatest seven-digit integer divisible by 132 whose digits, in order, are 2, 0, x, y, 1, 2, z where x, y, and z are single digits.
Answer: 2093124 Solution: From the given information it follows that • Since 132 is divisible by 3, 5 + x + y + z ≡ 0 (mod 3). • Since 132 is divisible by 11, 1 + x − y + z ≡ 0 (mod 11). • Since 132 is divisible by 4, z ≡ 0 (mod 4). Thus, z must be 0, 4, or 8. The number would be maximized if x could be 9. In this case, it follows that y ≡ 1 − z (mod 3) and y ≡ z − 1 (mod 11). If z = 0, then the digit y would need to be equivalent to 10 (mod 11) which is not possible. If z = 8, the digit y would need to be both equivalent to 1 − 8 ≡ 2 (mod 3) and equivalent to 8 − 1 ≡ 7 (mod 11) which is also not possible. But when z = 4, y = 3 satisfies both y ≡ 1 − 4 ≡ 0 (mod 3) and y ≡ 4 − 1 ≡ 3 (mod 11). Thus, the number 2093124 satisfies all the conditions and is the greatest such value that does.
Problem 24 There are positive integers m and n so that x = m + √ equation x2 − 10x + 1 = x(x + 1). Find m + n.
√
n is a solution to the
Answer: 55 √ Divide both sides of the given equation by x to get x − 10 + x1 = x + √ y = x + √1x . Then y 2 = x + 2 + x1 . This shows y 2 − 12 = y which is
√1 . x
Let
equivalent to (y − 4)(y + 3) = 0. The solution y = −3 will not work, but y = 4 √ √ √ means x + √1x = 4 or x − 4 x + 1 = 0. This quadratic equation in x has √ √ solutions x = 2 ± 3. The desired solution is, therefore, √ √ √ x = (2 + 3)2 = 7 + 4 3 = 7 + 48. The requested sum is 7 + 48 = 55.
Problem 25 Find the largest prime that divides 1 · 2 · 3 + 2 · 3 · 4 + ... + 44 · 45 · 46. 11
Answer: 47 The sum has terms of the form (k − 1)k(k + 1) = k 3 − k. Then 2 Pn Pn n(n+1) 3 − n(n+1) = k=1 (k − 1)k(k + 1) = k=1 k − k = 2 2 n(n+1)[n(n+1)−2] 4
=
n(n+1)(n2 +n−2) 4
special case when n = 45, so the
n(n+1)(n−1)(n+2) . The given 4 (45)(46)(44)(47) sum equals whose 4
=
sum is the largest prime
factor is 47.
Problem 26 A paper cup has a base that is a circle with radius r, a top that is a circle with radius 2r, and sides that connect the two circles with straight line segments as shown below. This cup has height h and volume V . A second cup that is exactly the same shape as the first is held upright inside the first cup so that its base is a distance of
h 2
from the base of the first cup. The volume of liquid
that will fit inside the first cup and outside the second cup can be written m n
· V where m and n are relatively prime positive integers. Find m + n.
Answer: 93 The volume between the cups does not change if the sides of each cup are extended so that the cups form circular cones with radius 2r and height 2h. The volume of the first cone is then 31 π(2r)2 (2h) = 83 πr2 h. The part of the second cone inside the first is a cone that is
3 4
the height of the first cone, so 3 the volume that it takes up within the first cone is 34 83 πr2 h = 98 πr2 h. Thus, the volume inside the first cup and outside the second cup is 8 9 2 2 − πr h = 37 24 πr h. The volume of the first cup is h3 8 i 37 3 4 7 37 2 2 24 1 − 12 3 πr (2h) = 3 πr h. The desired fraction is 7 = 56 . The requested 3
sum is 37 + 56 = 93.
12
Problem 27 You have some white one-by-one tiles and some black and white two-by-one tiles as shown below. There are four different color patterns that can be generated when using these tiles to cover a three-by-one rectangle by laying these tiles side by side (WWW, BWW, WBW, WWB). How many different color patterns can be generated when using these tiles to cover a ten-by-one rectangle?
Answer: 286 Let Aj refer to the number of color patterns one can obtain by laying these tiles together to tile a j-by-one rectangle. Clearly, A0 = A1 = 1 and A2 = 3. When j > 2, categorize the coloring patterns by the position of the first black square among the j squares. If the coloring pattern begins with a black square in its first position, the pattern must have a two-by-one tile followed by one of the Aj−2 patterns of length j − 2. Similarly, if the coloring pattern has a white square in the first position and a black square in the second position, the first two squares can be followed by any of the Aj−2 patterns of length j − 2. If the first black square appears in the k th position where k > 2, it can be followed by any of the Aj−k patterns of length j − k. Finally, there is one pattern consisting of all white squares. Thus Aj = Aj−2 + Aj−2 + Aj−3 + Aj−4 + Aj−5 + · · · A1 + A0 + 1. 13
From this one can generate A3 = A1 + A1 + A0 + 1 = 4 and A4 = A2 + A2 + A1 + A0 + 1 = 9. The following table is then easy to complete. 0
1
2
3
4
5
6
7
8
9
10
1
1
3
4
9
14
28
47
89
155
286
Problem 28 A bag contains 8 green candies and 4 red candies. You randomly select one candy at a time to eat. If you eat five candies, there are relatively prime m n
positive integers m and n so that
is the probability that you do not eat a
green candy after you eat a red candy. Find m + n.
Answer: 6 The five candies can be eaten in the order GRRRR, GGRRR, GGGRR, GGGGR, or GGGGG. The probabilities of eating the candies in these orders are, respectively, 8 12
·
7 11
·
6 10
·
5 9
8 12
·
4 11
· 48 , and
·
8 12
3 10
·
2 9
·
7 11
·
· 18 , 6 10
4·3·2·1+7·4·3·2+7·6·4·3+7·6·5·4+7·6·5·4 12·11·10·9
·
8 12 5 9
=
·
7 11
·
4 10
·
3 9
· 28 ,
8 12
·
7 11
·
6 10
·
4 9
· 38 ,
· 48 . The sum of these is 12(2+14+42+70+70) 12·11·10·9
=
198 11·10·9
= 51 . The
requested sum is 1 + 5 = 6.
Problem 29 Let A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8, 10}. Let f be a randomly chosen function from the set A ∪ B into itself. There are relatively prime positive integers m and n such that
m n
is the probability that f is a one-to-one function
on A ∪ B given that it maps A one-to-one into A ∪ B and it maps B one-to-one into A ∪ B. Find m + n.
Answer: 253 A one-to-one function f that maps A ∪ B into itself is just a permutation of the set A ∪ B. Thus, there are 10! equally likely such functions. The number of one-to-one functions that map A into A ∪ B is
10 P5
= 10 · 9 · 8 · 7 · 6. Similarly,
there are that many one-to-one functions that map B into A ∪ B. It follows that the desired probability is
10·9·8·7·6·5·4·3·2·1 (10·9·8·7·6)(10·9·8·7·6)
The requested sum is 1 + 252 = 253.
14
=
5·4·3·2·1 10·9·8·7·6
=
1
(10 5)
=
1 252 .
Problem 30 The diagram below shows four regular hexagons each with side length 1 meter attached to the sides of a square. This figure is drawn onto a thin sheet of metal and cut out. The hexagons are then bent upward along the sides of the square so that A1 meets A2 , B1 meets B2 , C1 meets C2 , and D1 meets D2 . The resulting dish is set on a table with the square lying flat on the table. If this dish is filled with water, the water will rise to the height of the corner where the A1 and A2 meet. There are relatively prime positive integers m and p n so that the number of cubic meters of water the dish will hold is m n . Find m + n.
Answer: 67 Let the point where A1 and A2 meet be labeled A, and the point where B1 and B2 meet be labeled B. Let the center of the square be labeled F . Let the corner of the square nearest point A be labeled H, the corner nearest B be labeled G, and the corner diagonally opposite G be labeled E as shown in the figure below. Let J be the projection of the point A in the plane of the square. Because ∠AHE = 120◦ , the Law of Cosines gives √ √ √ AE = AH 2 + HE 2 − 2 · AH · HE cos 120◦ = 1 + 1 + 1 = 3. Note that EF = HF = √12 . Then the Pythagorean Theorem gives q q √ AF = AE 2 − EF 2 = 3 − 12 = 52 . The Law of Cosines then gives cos AHF =
AF 2 −AH 2 −HF 2 2·AH·HF
=
5 1 2 −1− √ 2
− 2
= − √12 . It follows that ∠AHF = 135◦ , 15
and ∠AHJ = 45◦ . Thus, AJ =
√1 2
is the height of A from the plane of the
square. Since the diameter of a regular hexagon is twice the length of one of its sides, it follows that AB = 2. If the vertical edges of the dish AH and BG are extended, they will meet at a point K as shown in the diagram. Note that K is the vertex of a downward pointing square pyramid with base having side √ length 2 and with height 2. The volume of this pyramid is given by 13 the √ area of its base times its height = 31 · 4 · 2. The lower half of the pyramid is √
another square pyramid with base having side length 1 and with height its 4 3
·
volume is 13 √ √ 2 − 13 · 22
√
·1 =
· 22 . The desired volume is q √ 7 2 49 6 = 18 . The requested
16
then sum is 49 + 18 = 67.
2 2
so
