The MATHEMATICAL ASSOCIATION of AMERICA American Mathematics Competitions. AMC 12 Contest B. Solutions Pamphlet. Wednesday, February 27, 2008

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1 The MATHEMATICAL ASSOCIATION of AMERICA American Mathematics Competitions 59 th Annual American Mathematics Contest 1 AMC 1 Contest B Solutions Pamphlet Wednesday, February 7, 008 This Pamphlet gives at least one solution for each problem on this year s contest and shows that all problems can be solved without the use of a calculator. When more than one solution is provided, this is done to illustrate a significant contrast in methods, e.g., algebraic vs geometric, computational vs conceptual, elementary vs advanced. These solutions are by no means the only ones possible, nor are they superior to others the reader may devise. We hope that teachers will inform their students about these solutions, both as illustrations of the kinds of ingenuity needed to solve nonroutine problems and as examples of good mathematical exposition. However, the publication, reproduction or communication of the problems or solutions of the AMC 1 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, , World Wide Web or media of any type during this period is a violation of the competition rules. After the contest period, permission to make copies of problems in paper or electronic form including posting on webpages for educational use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear the copyright notice. Correspondence about the problems/solutions for this AMC 1 and orders for any publications should be addressed to: American Mathematics Competitions University of Nebraska, P.O. Box 81606, Lincoln, NE Phone: ; Fax: ; amcinfo@maa.org The problems and solutions for this AMC 1 were prepared by the MAA s Committee on the AMC 10 and AMC 1 under the direction of AMC 1 Subcommittee Chair: Prof. David Wells, Department of Mathematics Penn State University, New Kensington, PA Copyright 008, The Mathematical Association of America

2 Solutions th AMC 1 B 1. Answer (E: The number of points could be any integer between 5 = 10 and 5 = 15, inclusive. The number of possibilities is = 6.. Answer (B: The two sums are = 50 and = 5, so the positive difference between the sums is 5 50 =. Query: If a different block of dates had been chosen, the answer would be unchanged. Why? Answer (C: A single player can receive the largest possible salary only when the other 0 players on the team are each receiving the minimum salary of $15,000. Thus the maximum salary for any player is $700,000 0 $15,000 = $00,000.. Answer (D: The measure of COD is = 105. Therefore the ratio of the area of the sector to the area of the circle is = Answer (C: The total cost of the carnations must be an even number of dollars. The total number of dollars spent is the even number 50, so the number of roses purchased must also be even. In addition, the number of roses purchased cannot exceed 50. Therefore the number of roses purchased must be one of the even integers between 0 and 16, inclusive. This gives 9 possibilities for the number of roses purchased, and consequently 9 possibilities for the number of bouquets. 6. Answer (A: During the year Pete takes = steps. At 1800 steps per mile, the number of miles Pete walks is = = Answer (A: Note that (y x = (x y, so (x y $(y x = (x y $(x y = ( (x y (x y = 0 = Answer (C: Because AB + BD = AD and AB = BD, it follows that BD = 1 5 AD. By similar reasoning, CD = 1 10 AD. Thus BC = BD CD = 1 5 AD 1 10 AD = 1 10 AD.

3 Solutions th AMC 1 B 9. Answer (A: Let O be the center of the circle, and let D be the intersection of OC and AB. Because OC bisects minor arc AB, OD is a perpendicular bisector of chord AB. Hence AD =, and applying the Pythagorean Theorem to ADO yields OD = 5 =. Therefore DC = 1, and applying the Pythagorean Theorem to ADC yields AC = + 1 = 10. A 5 C D O B 10. Answer (B: Let n be the number of bricks in the chimney. Then the number of bricks per hour Brenda and Brandon can lay working alone is n 9 and n 10, respectively. Working together they can lay ( n 9 + n bricks in an hour, or ( n n bricks in 5 hours to complete the chimney. Thus ( n n = n, and the number of bricks in the chimney is n = 900. OR Suppose that Brenda can lay x bricks in an hour and Brandon can lay y bricks in an hour. Then the number of bricks in the chimney can be expressed as 9x, 10y, or 5(x + y 10. The equality of these expressions leads to the system of equations x 5y = 50 5x + 5y = 50. It follows that x = 100, so the number of bricks in the chimney is 9x = Answer (A: The portion of the mountain that is above the water forms a cone that is similar to the entire mountain. The ratio of the volumes of the cones is the cube of the ratio of their heights. Let d be the depth of the ocean, in feet. Then the height of the mountain above the water is 8000 d feet, and (8000 d 8000 = 1 8. Taking cube roots on both sides gives 8000 d = , from which 16,000 d = 8000, and d = Answer (B: Because the mean of the first n terms is n, their sum is n. Therefore the nth term is n (n 1 = n 1, and the 008th term is = 015.

4 Solutions th AMC 1 B 1. Answer (B: Draw a line parallel to AD through point E, intersecting AB at F and intersecting CD at G. Triangle AEF is a triangle with hypotenuse AE = 1, so EF =. Region R consists of two congruent trapezoids of height 1 6, shorter base EG = 1, and longer base the weighted average EG + 1 AD = ( 1 C B = 1. G F E A D Therefore the area of R is (( ( ( = OR = Place ABCD in a coordinate plane with B = (0, 0, A = (1, 0, and C = (0, 1. Then the equation of the line BE is y = x, so E = ( 1,, and the point of R closest to B is ( 1,. Thus the region R consists of two congruent trapezoids with height 1 6 and bases 1 and 1. Then proceed as in the first solution. 1. Answer (C: The given information implies that π log 10 (a = log 10 (b or, equivalently, that π log 10 a = log 10 b. Thus log a b = log 10 b log 10 a = π. 15. Answer (C: The region inside S but outside R consists of four triangles, each of which has two sides of length 1. The angle between those two sides is = 0. Thus the area of each triangle is so the required area is 1 = sin 0 = 1, 16. Answer (B: Because the area of the border is half the area of the floor, the same is true of the painted rectangle. The painted rectangle measures a by b feet. Hence ab = (a (b, from which 0 = ab a b + 8. Add 8 to each side of the equation to produce 8 = ab a b + 16 = (a (b.

5 Solutions th AMC 1 B 5 Because the only integer factorizations of 8 are 8 = 1 8 = = ( ( = ( 8 ( 1, and because b > a > 0, the only possible ordered pairs satisfying this equation for (a, b are (1, 8 and (,. Hence (a, b must be one of the two ordered pairs (5, 1, or (6, Answer (C: Let A = (a, a and C = (c, c. Then B = ( a, a. If either A or B is 90, then c = ±a, but this is impossible because A, B, and C must have distinct x-coordinates. Thus C = 90, so AC BC. Consequently c a c a c a c + a = 1, from which 1 = a c, which is the length of the altitude from C to AB. Because ABC has area 008, it follows that AB = 016, a = 008 and a = 008 = 006. Therefore c = a 1 = 006 and the sum of the digits of c is Answer (E: Square ABCD has side length 1. Let F and G be the feet of the altitudes from E in ABE and CDE, respectively. Then F G = 1, EF = = 15 and EG = 1 = 1. Because EF G is perpendicular to the plane of ABCD, the altitude to F G is the altitude of the pyramid. By Heron s Formula, the area of EF G is (1(6(7(8 = 8, so the altitude to F G is 8 1 = 1. Therefore the volume of the pyramid is ( 1 (196(1 = Answer (B: Let α = a + bi and γ = c + di, where a, b, c, and d are real numbers. Then f(1 = ( + a + c + (1 + b + di, and f(i = ( b + c + ( 1 + a + di. Because both f(1 and f(i are real, it follows that a = 1 d and b = 1 d. Thus α + γ = a + b + c + d = (1 d + ( 1 d + c + d = + d + c + d. The minimum value of α + γ is consequently, which is achieved when c = d = 0. In this case we also have a = 1 and b = 1.

6 Solutions th AMC 1 B 6 0. Answer (B: Number the pails consecutively so that Michael is presently at pail 0 and the garbage truck is at pail 1. Michael takes 00/5 = 0 seconds to walk between pails, so for n 0 he passes pail n after 0n seconds. The truck takes 0 seconds to travel between pails and stops for 0 seconds at each pail. Thus for n 1 it leaves pail n after 50(n 1 seconds, and for n it arrives at pail n after 50(n 1 0 seconds. Michael will meet the truck at pail n if and only if 50(n 1 0 0n 50(n 1 or, equivalently, 5 n 8. distance (ft truck Michael time (sec Hence Michael first meets the truck at pail 5 after 00 seconds, just as the truck leaves the pail. He passes the truck at pail 6 after 0 seconds and at pail 7 after 80 seconds. Finally, Michael meets the truck just as it arrives at pail 8 after 0 seconds. These conditions imply that the truck is ahead of Michael between pails 5 and 6 and that Michael is ahead of the truck between pails 7 and 8. However, the truck must pass Michael at some point between pails 6 and 7, so they meet a total of five times. 1. Answer (E: Circles A and B both have radius 1, b so they intersect if and only if the distance between (0, their centers is no greater than. Let the centers of the circles be (a, 0 and (b, 1. The distance between these points is (a b + 1, so the circles intersect(0, if and only if (a b + 1. This condition is equivalent to (a b, or a b. a Points in the square correspond to ordered pairs (a, b (0,0 (,0 (,0 with 0 a and 0 b. The shaded region corresponds to the points that satisfy a b. Its area is (. The requested probability is the area of the shaded region divided by the area of the square, which is ( =.

7 Solutions th AMC 1 B 7. Answer (E: The four vacant spaces can be located in any of ( 16 = 180 combinations of positions. The arrangements in which Auntie Em is unable to park may be divided into two cases. If the rightmost space is occupied, then every vacant space is immediately to the left of an occupied space. Let X denote the union of a vacant space and the occupied space immediately to its right, and let Y denote a single occupied space not immediately to the right of a vacant space. The arrangement of cars and spaces can be represented by a sequence of four X s and eight Y s in some order, and there are ( 1 = 95 possible orders. If the rightmost space is vacant, the arrangement in the remaining 15 spaces can be represented by a sequence of three X s and nine Y s in some order, and there are ( 1 = 0 possible orders. Therefore there are = 1105 arrangements in which Auntie Em can park, and the requested probability is = OR Let O denote an occupied space, and let V denote a vacant space. The problem is equivalent to finding the probability p that in a string of 1 O s and V s, there are at least two consecutive V s. Then 1 p is the probability that no two V s are consecutive. In a string of 1 O s, there are 1 spaces in which to insert V s to create a string in which no two V s are consecutive. Thus p = 1 ( 1 ( 16 = Answer (A: Because the prime factorization of 10 is 5, the positive divisors of 10 n are the numbers a 5 b with 0 a n and 0 b n. Thus 79 = a=0 b=0 ( log 10 a 5 b = = a=0 b=0 b=0 a=0 (a log 10 + b log 10 5 (a log 10 + a=0 a=0 b=0 (b log 10 5 = (n + 1 (log 10 a + (n + 1 (log 10 5 b = (n + 1 (log 10 + log 10 5 ( 1 n(n + 1 = 1 n(n + 1 (log = 1 n(n + 1. b=0 Hence n(n + 1 = 79 = 11 7 = 11 1, so n = 11. OR Let d(m denote the number of divisors of a positive integer M. The sum of the logs of the divisors of M is equal to the log of the product of its divisors.

8 Solutions th AMC 1 B 8 If M is not a square, its divisors can be arranged in pairs, each with a product of M. Thus the product of the divisors is M d(m/. A similar argument shows that this result is also true if M is a square. Therefore ( 79 = log (10 n d(10n / = 1 d(10n n = 1 d(n 5 n n = 1 (n + 1 n, and the conclusion follows as in the first solution.. Answer (C: For n 0, let A n = (a n, 0, and let c n+1 = a n+1 a n. Let B 0 = A 0, and let c 0 = 0. Then for n 0, ( so B n+1 = ( cn+1 a n + c n+1, cn+1 = a n + c n+1, from which c n+1 c n+1 a n = 0. For n 1, B n = ( a n c n, cn,, so ( cn = a n c n, from which c n + c n a n = 0. Hence c n+1 c n+1 = a n = c n + c n, and (c n+1 + c n = (c n+1 c n = (c n+1 + c n (c n 1 c n. Thus c n+1 = c n + for n 0. It follows that a n = n = Solving n(n + 1/ 100 gives n 17. n(n + 1 = n(n Answer (B: Let M and N be the midpoints of sides AD and BC. Set BAD = y and ADC = x. We have x + y = 90, from which it follows that AP D = 90. Hence in triangle AP D, MP is the median to the hypotenuse AD, so AM = MD = MP and MP A = MAP = P AB. Thus, MP AB. Likewise, QN AB. It follows that M, P, Q, and N are collinear, and P Q = MN MP QN = AB + CD AD BC = 9.

9 Solutions th AMC 1 B 9 The area of ABQCDP is equal to the sum of the areas of two trapezoids ABQP and CDP Q. Let F be the foot of the perpendicular from A to CD. Then the area of ABQCDP is equal to AB + P Q AF + CD + P Q AF = 1AF. Let E lie on DC so that AE BC. Then AE = BC = 5 and DE = CD CE = CD AB = 8. We have AD DF = AF = AE EF = AE (DE DF, or 9 DF = 5 (8 DF. Solving the last equation gives DF = 11. Thus AF = 5 and the area of ABQCDP is 1AF = 0. D M x x A y y y x P A Q B N C B D F E C OR As in the first solution, conclude that AE = 5 and DE = 8. Apply the Law of Cosines to ADE to obtain cos( AED = = 1. Therefore AED = 60, so AF = 5 /, and the area of ABQCDP is 0.

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