PURPLE COMET MATH MEET April 2012 MIDDLE SCHOOL - SOLUTIONS Copyright c Titu Andreescu and Jonathan Kane Problem 1 Evaluate 5 4 4 3 3 2 2 1 1 0. Answer: 549 The expression equals 625 64 9 2 1 = 549. Problem 2 The diagram below shows rectangle ABDE where C is the midpoint of side BD, and F is the midpoint of side AE. If AB = 10 and BD = 24, find the area of the shaded region. Answer: 90 Parallelogram ACDF has side length AF = 12 and height AB = 10, so it has area 12 10 = 120. Let G be the intersection of CE and DF. Then CDG has base 12 and height 5, so it has area 12 5 2 = 30. It follows that the shaded region has area 120 30 = 90. Alternatively, notice that ACF is half the area of rectangle ABCF, and F CG is a quarter of the area of rectangle CDEF. Then the shaded region 1
has area 1 2 [ABCF ] + 1 4 [CDEF ] = ( 1 2 + ) 1 4 [ABCF ] = 3 8 [ABCE] = 3 8 240 = 90. Problem 3 While Peter was driving from home to work, he noticed that after driving 21 miles, the distance he had left to drive was 30 percent of the total distance from home to work. How many miles was his complete trip from home to work? Answer: 30 If the total distance Peter had to drive from home to work was x miles, then (1 0.30)x = 21. Thus, x = 21 0.7 = 210 7 = 30. Problem 4 How many two-digit positive integers contain at least one digit equal to 5? Answer: 18 There are nine two-digit integers which end in 5, ten two-digit integers which begin with 5, and one two-digit integer that both begins and ends with 5. Thus, there are 9 + 10 1 = 18 two-digit integers that contain at least one digit equal to 5. Problem 5 Meredith drives 5 miles to the northeast, then 15 miles to the southeast, then 25 miles to the southwest, then 35 miles to the northwest, and finally 20 miles to the northeast. How many miles is Meredith from where she started? Answer: 20 Rotate Meredith s trip by 45 degrees. Then she drives 5 miles east, 15 miles south, 25 miles west, 35 miles north, and 20 miles east. The total distance she travels in the east-west direction is 5 25 + 20 = 0. The total distance she travels in the north-south direction is 15 + 35 = 20. Thus, she ends up 20 miles from where she started. 2
Problem 6 Volume A equals one fourth of the sum of the volumes B and C, while volume B equals one sixth of the sum of the volumes A and C. There are relatively prime positive integers m and n so that the ratio of volume C to the sum of the other two volumes is m n. Find m + n. Answer: 35 The relations between the volumes give A = B + C 4 B = C + A 6 Then 4A B = A + 6B and B = 5 7A, so = 4A B = C = A + 6B = C. C = A + 6B = A + 6 5 7 A = 23 7 A. Finally, C requested sum is m + n = 23 + 12 = 35. A+B = 23 7 A = 23 A+ 5 7 A 7+5 = 23 12. The Problem 7 A snail crawls 2 1 2 centimeters in 4 1 4 minutes. At that rate, how many centimeters can the snail crawl in 85 minutes? Answer: 50 The snail is moving at a rate of 2 1 2 4 1 4 85 minutes, the snail can crawl 10 17 = 5 2 17 4 = 10 17 85 = 50 centimeters. centimeters per minute. So, in Problem 8 Seven boys and three girls are playing basketball. In how many different ways can they make two teams of five players so that both teams have at least one girl? 3
Answer: 105 If there must be at least one girl on each team, exactly one team will have two girls. There are ( 3 2) = 3 ways to select the two girls who will be on that team and ( 7 3) = 35 ways to select the three boys who will be on their team. Thus, there are 3 35 = 105 ways to select the team with two girls which determines the two basketball teams. Problem 9 Points E and F lie inside rectangle ABCD with AE = DE = BF = CF = EF. If AB = 11 and BC = 8, find the area of the quadrilateral AEF B. Answer: 32 Note that since E is equidistant from A and D, and F is equidistant from B and C, the segment EF is parallel to AB. Let EF intersect the rectangle at points G and H as shown. Let AE = EF = F B = x, and since AEG is a right triangle with AG = 4, the Pythagorean Theorem gives EG = x 2 4 2 = x 2 16. Then 11 = GH = GE + EF + F H = 2 x 2 16 + x. Thus, 4(x 2 16) = (11 x) 2 = 121 22x + x 2, so 3x 2 + 22x 185 = 0. This factors as (3x + 37)(x 5) = 0. The positive solution is x = 5. It follows that quadrilateral AEF B is a trapezoid with bases with lengths 11 and 5 and height 4. Thus, its area is 4 11+5 2 = 32. 4
Problem 10 Find the least positive multiple of 999 that does not have 9 as a digit. Answer: 111888 For any positive integer k, one has 999k = k(1000 1) = 1000k k which is k less than a multiple of 1000. Note that each of the integers from 889 through 999 contains 9 as a digit. Thus, in order for 999k not to have 9 as a digit, k must be at least 1000 888 = 112. The least positive multiple of 999 that does not have 9 as a digit is 112 999 = 112000 112 = 111888. Problem 11 For some integers a and b the function f(x) = ax + b has the properties that f (f(0)) = 0 and f (f (f(4))) = 9. Find f (f (f (f(10)))). Answer: 10 The given information says f (f(0)) = ab + b = (a + 1)b = 0, so either b = 0 or a = 1. Furthermore, f (f (f(4))) = a ( 4a 2 + ab + b ) + b = 4a 3 + a(ab + b) + b = 4a 3 + b = 9. If b = 0, then a = 3 9 4 which is not an integer. If a = 1, then 4 + b = 9 and b = 13. Conclude that f(x) = x + 13, so that f (f (f (f(10)))) = f (f (f (3))) = f (f (10)) = f (3) = 10. Problem 12 Pentagon ABCDE consists of a square ACDE and an equilateral triangle ABC that share the side AC. A circle centered at C has area 24. The intersection of the circle and the pentagon has half the area of the pentagon. Find the area of the pentagon. Answer: 20 Note that if the circle intersects line AB, its radius would need to be at least 3 times the side length of the pentagon. In this case the circle would contain 2 more than half the area of the both the triangle and the square. The intersection of the square ACDE and the circle is a quarter of the circle. The 5
intersection of the equilateral triangle ABC and the circle is a sixth of the circle. It follows that the area of the pentagon is 2 24 ( 1 4 + 1 6) = 2(6 + 4) = 20. Problem 13 Find the least positive integer N which is both a multiple of 19 and whose digits add to 23. Answer: 779 Any such N must have at least 3 digits in order for the sum of its digits to be 23. If the decimal representation of N is abc, then a + b + c = 23 and 100a + 10b + c = 100a + 10b + (23 a b) = 99a + 9b + 23 0 (mod 19). Thus, 4a + 9b 15 (mod 19) and multiplying this by 17 yields 11a + b 8 (mod 19) and b 8 11a (mod 19). Since b and c are each at most 9, it follows that a 5. If a = 5, then b 8 11 5 10 (mod 19) which is not possible. If a = 6, then b 8 11 6 18 (mod 19) which is not possible. If a = 7, then b 8 11 7 7 (mod 19). This gives the solution a = b = 7 and c = 23 7 7 = 9 and N = 779. Problem 14 At the 4 PM show, all the seats in the theater were taken, and 65 percent of the audience was children. At the 6 PM show, again, all the seats were taken, but this time only 50 percent of the audience was children. Of all the people who attended either of these shows, 57 percent were children although there 6
were 12 adults and 28 children who attended both shows. How many people does the theater seat? Answer: 520 Let s be the number of seats in the theater. Then at the first show, there were 0.65s children, and at the second show, there were 0.5s children. The number of children attending the two shows was 0.65s + 0.5s 28, and the number of people attending the two shows was s + s 28 12. It follows that 0.65s+0.5s 28 s+s 40 s = 2800 57 40 = 520. = 0.57. Thus, 115s 2800 = 57(2s 40), and Problem 15 The top and bottom of a rectangular tank each has area 48 square inches. The four vertical sides of the tank have a total area of 96 square inches. The longest diagonal of the tank is 13 inches. Find the sum of the height, the width, and the length of this tank in inches. Answer: 19 Let the tank have length x, width y, and height z. Then the conditions of the problem give xy = 48, 2(xz + yz) = 96, and x 2 + y 2 + z 2 = 13. It follows that 2xz + 2yz + 2xy = 96 + 2 48 = 192. Then (x + y + z) 2 = x 2 + y 2 + z 2 + 2xz + 2yz + 2xy = ( x 2 + y 2 + z 2) + 2(xz + yz + xy) = 13 2 + 192 = 361 = 19 2. Thus, the requested sum x + y + z = 19. Problem 16 The following sequence lists all the positive rational numbers that do not exceed 1 2 by first listing the fraction with denominator 2, followed by the one with denominator 3, followed by the two fractions with denominator 4 in increasing order, and so forth so that the sequence is 1 2, 1 3, 1 4, 2 4, 1 5, 2 5, 1 6, 2 6, 3 6, 1 7, 2 7, 3 7,. Let m and n be relatively prime positive integers so that the 2012 th fraction in the list is equal to m n. Find m + n. 7
Answer: 61 In the sequence there are two denominators that each appear once (2 and 3), two denominators that each appear twice (4 and 5), and so forth. The number of terms of the sequence through the fraction n 2n+1 is 1 + 1 + 2 + 2 + 3 + 3 + + n + n = 2 n k=1 k = n(n + 1). The least n for which n(n + 1) > 2012 is n = 45 since 44(44+1) = 1980 and 45(45+1) = 2070. Because n = 44 corresponds to fractions with denominators 88 and 89, the 2012 th term of the sequence is the thirty-second fraction with denominator 90. The fraction is 32 90 = 16 45. The requested sum is 16 + 45 = 61. Problem 17 The diagram below shows nine points on a circle where AB = BC = CD = DE = EF = F G = GH. Given that GHJ = 117 and BH is perpendicular to EJ, there are relatively prime positive integers m and n so that the degree measure of AJB is m n. Find m + n. Answer: 29 Note that AJB = BJC = CJD = DJE = EJF = F JG = GJH since all the angles subtend equal sized arcs. Let x represent the degree measure of those angles. Since they subtend equal arcs, GHB = GJB = 5x. Because BH is perpendicular to EJ, it follows that JBH is isosceles. Thus, the sum of the angles in JBH is 2 BHJ + BJH = 2( GHJ GHB) + BJH = 2(117 5x) + 6x = 180. Solving yields 117 2x = 90 and x = 27 2. The requested sum is 27 + 2 = 29. 8
Problem 18 Find the smallest positive integer whose remainder when divided by 5 is 3, when divided by 7 is 5, and when divided by 9 is 1. Answer: 208 Let the answer be n. Because n is 3 more than a multiple of 5, when n is divided by 35, it must have a remainder in {3,8,13,18,23,28,33}. Similarly, because n is 5 more than a multiple of 7, when divided by 35, it must have a remainder in {5,12,19,26,33}. Thus, n must have a remainder of 33 when divided by 35. Thus, when divided by 35 9 = 315, it must have a remainder in {33,68,103,138,173,208,243,278,313}. The only one of these numbers which is 1 more than a multiple of 9 is 208. Thus, the smallest possible value of n is 208. Alternatively, note that n + 2 must be divisible by 5 and 7, so it is divisible by 35. Because 35 1 (mod 9), it follows that n + 2 which is 3 (mod 9) must satisfy n + 2 3 6( 1) 6 35 (mod 9). So n + 2 = 6 35 = 210 and n = 208. Problem 19 A teacher suggests four possible books for students to read. Each of six students selects one of the four books. How many ways can these selections be made if each of the books is read by at least one student? Answer: 1560 If each of six students can select any one of the four choices, there are 4 6 = 4096 ways for them to make choices. For j = 1, 2, 3, 4 let A j be the set of book choices where no student selects book number j. The Inclusion/Exclusion Principle says that the size of A 1 A 2 A 3 A 4 is 4 #(A 1 ) ( ) 4 2 #(A1 A 2 ) + ( 4 3) #(A1 A 2 A 3 ) #(A 1 A 2 A 3 A 4 ) = 4 3 6 6 2 6 + 4 1 0 = 2536. Thus, the number of ways for the students to make their selections so that each book is selected at least once is 4096 2536 = 1560. 9
Problem 20 In the following addition, different letters represent different non-zero digits. What is the 5-digit number ABCDE? Answer: 84269 Note that since each letter is the first digit of at least one number, none of the letters represents the digit zero. From the ones column of the addition, it follows that the number 7B ends in the digit A which implies that A and B are either both even or both odd. Since the top row of the addition is clearly smaller than the bottom row, the digit A must be greater than the digit B. Because there is no carry from the hundred-thousands place to the millions place in the addition, B must be less than 5, and A must be at least 2B. Since the carry into the hundred-thousands place is at most 3, A cannot exceed 2B + 3. The possible values for (B, A) that satisfy 2B A 2B + 3, B < 5, with A and B having the same parity are (1, 3), (1, 5), (2, 4), (2, 6), (3, 7), (3, 9), (4, 8). But the ones column shows that 7B must end in the digit A. This limits the possibilities to (2, 4) and (4, 8). If B = 2, then there is a carry of 1 from the ones column to the tens column. But then 6D + 1 must end with A = 4 which is not possible. Thus, A = 8 and B = 4. Then 6D + 2 must exceed 10 and end in the digit 8 which requires D to be 6. Thus, there will be a carry of 3 from the tens column to the hundreds column, so 5E + 3 ends in the digit 8 implying E is odd. Moreover, the thousands column contains four 6 s and thus, with the carry, must add to 28. The carry from the hundreds column must be 4 implying that E = 9. Finally, the ten-thousands column has 3C + 2 adding to 8, so C = 2. This determines that ABCDE = 84269. 10