Wednesday, February 27, th Annual American Mathematics Contest 12 AMC 12

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1 Wednesday, February 7, th Annual American Mathematics Contest 1 AMC 1 Contest B The MATHEMATICAL ASSOCIATION of AMERICA American Mathematics Competitions 1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR GIVES THE SIGNAL TO BEGIN.. This is a 5-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.. Mark your answer to each problem on the AMC 1 Answer Form with a # pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded.. SCORING: You will receive 6 points for each correct answer, 1.5 points for each problem left unanswered, and 0 points for each incorrect answer. 5. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor, and erasers. No calculators are allowed. No problems on the test will require the use of a calculator. 6. Figures are not necessarily drawn to scale. 7. Before beginning the test, your proctor will ask you to record certain information on the answer form. When your proctor gives the signal, begin working the problems. You will have 75 MINUTES to complete the test. 8. When you finish the exam, sign your name in the space provided on the Answer Form. Students who score 100 or above or finish in the top 5% on this AMC 1 will be invited to take the 6 th annual American Invitational Mathematics Examination (AIME) on Tuesday, March 18, 008 or Wednesday, April, 008. More details about the AIME and other information are on the back page of this test booklet. The Committee on the American Mathematics Competitions (CAMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The CAMC also reserves the right to disqualify all scores from a school if it is determined that the required security procedures were not followed. 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 web-pages 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. Copyright 008, The Mathematical Association of America

2 59 th AMC 1 B A basketball player made 5 baskets during a game. Each basket was worth either or points. How many different numbers could represent the total points scored by the player? (A) (B) (C) (D) 5 (E) 6. A block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums? (A) (B) (C) 6 (D) 8 (E) 10. A semipro baseball league has teams with 1 players each. League rules state that a player must be paid at least $15,000, and that the total of all players salaries for each team cannot exceed $700,000. What is the maximum possible salary, in dollars, for a single player? (A) 70,000 (B) 85,000 (C) 00,000 (D) 0,000 (E) 700,000. On circle O, points C and D are on the same side of diameter AB, AOC = 0, and DOB = 5. What is the ratio of the area of the smaller sector COD to the area of the circle? B (A) 9 (B) 1 (C) 5 18 (D) 7 (E) 10 D 5 0 O C A 5. A class collects $50 to buy flowers for a classmate who is in the hospital. Roses cost $ each, and carnations cost $ each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50? (A) 1 (B) 7 (C) 9 (D) 16 (E) Postman Pete has a pedometer to count his steps. The pedometer records up to steps, then flips over to on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to During the year, the pedometer flips from to forty-four times. On December 1 the pedometer reads Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year? (A) 500 (B) 000 (C) 500 (D) 000 (E) 500

3 59 th AMC 1 B For real numbers a and b, define a$b = (a b). What is (x y) $(y x)? (A) 0 (B) x + y (C) x (D) y (E) xy 8. Points B and C lie on AD. The length of AB is times the length of BD, and the length of AC is 9 times the length of CD. The length of BC is what fraction of the length of AD? (A) 1 6 (B) 1 1 (C) 1 10 (D) 5 6 (E) Points A and B are on a circle of radius 5 and AB = 6. Point C is the midpoint of the minor arc AB. What is the length of the line segment AC? (A) 10 (B) 7 (C) 1 (D) 15 (E) 10. Bricklayer Brenda would take 9 hours to build a chimney alone, and bricklayer Brandon would take 10 hours to build it alone. When they work together, they talk a lot, and their combined output is decreased by 10 bricks per hour. Working together, they build the chimney in 5 hours. How many bricks are in the chimney? (A) 500 (B) 900 (C) 950 (D) 1000 (E) A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top 1 8 of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain, in feet? (A) 000 (B) 000( ) (C) 6000 (D) 600 (E) For each positive integer n, the mean of the first n terms of a sequence is n. What is the 008th term of the sequence? (A) 008 (B) 015 (C) 016 (D),00,056 (E),0,06 1. Vertex E of equilateral ABE is in the interior of unit square ABCD. Let R be the region consisting of all points inside ABCD and outside ABE whose distance from AD is between 1 and. What is the area of R? (A) (B) (C) 18 (D) 9 1. A circle has a radius of log 10 (a ) and a circumference of log 10 (b ). What is log a b? (A) 1 π (B) 1 π (C) π (D) π (E) 10 π (E) 1

4 59 th AMC 1 B On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 1 triangles have no points in common. Let R be the region formed by the union of the square and all the triangles, and let S be the smallest convex polygon that contains R. What is the area of the region that is inside S but outside R? (A) 1 (B) (C) 1 (D) (E) 16. A rectangular floor measures a feet by b feet, where a and b are positive integers with b > a. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width 1 foot around the painted rectangle and occupies half the area of the entire floor. How many possibilities are there for the ordered pair (a, b)? (A) 1 (B) (C) (D) (E) Let A, B and C be three distinct points on the graph of y = x such that line AB is parallel to the x-axis and ABC is a right triangle with area 008. What is the sum of the digits of the y-coordinate of C? (A) 16 (B) 17 (C) 18 (D) 19 (E) A pyramid has a square base ABCD and vertex E. The area of square ABCD is 196, and the areas of ABE and CDE are 105 and 91, respectively. What is the volume of the pyramid? (A) 9 (B) (C) 9 (D) 9 (E) A function f is defined by f(z) = ( + i)z + αz + γ for all complex numbers z, where α and γ are complex numbers and i = 1. Suppose that f(1) and f(i) are both real. What is the smallest possible value of α + γ? (A) 1 (B) (C) (D) (E)

5 59 th AMC 1 B Michael walks at the rate of 5 feet per second on a long straight path. Trash pails are located every 00 feet along the path. A garbage truck travels at 10 feet per second in the same direction as Michael and stops for 0 seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet? (A) (B) 5 (C) 6 (D) 7 (E) 8 1. Two circles of radius 1 are to be constructed as follows. The center of circle A is chosen uniformly and at random from the line segment joining (0, 0) to (, 0). The center of circle B is chosen uniformly and at random, and independently of the first choice, from the line segment joining (0, 1) to (, 1). What is the probability that circles A and B intersect? (A) + (B) + 8 (C) 1 (D) + (E). A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires adjacent spaces. What is the probability that she is able to park? (A) 11 0 (B) 7 (C) (D) 5 (E) The sum of the base-10 logarithms of the divisors of 10 n is 79. What is n? (A) 11 (B) 1 (C) 1 (D) 1 (E) 15. Let A 0 = (0, 0). Distinct points A 1, A,... lie on the x-axis, and distinct points B 1, B,... lie on the graph of y = x. For every positive integer n, A n 1 B n A n is an equilateral triangle. What is the least n for which the length A 0 A n 100? (A) 1 (B) 15 (C) 17 (D) 19 (E) 1 5. Let ABCD be a trapezoid with AB CD, AB = 11, BC = 5, CD = 19, and DA = 7. Bisectors of A and D meet at P, and bisectors of B and C meet at Q. What is the area of hexagon ABQCDP? (A) 8 (B) 0 (C) (D) 5 (E) 6

6 008 AMC 1 Contest B DO NOT OPEN UNTIL Wednesday, February 7, 008 **Administration On An Earlier Date Will Disqualify Your School s Results** 1. All information (Rules and Instructions) needed to administer this exam is contained in the TEACHERS MANUAL, which is outside of this package. PLEASE READ THE MANUAL BEFORE February 7, 008. Nothing is needed from inside this package until February 7, Your PRINCIPAL or VICE PRINCIPAL must sign the Certification Form found in the Teachers Manual.. The Answer Forms must be mailed by First Class mail to the AMC no later than hours following the examination.. The publication, reproduction or communication of the problems or solutions of this test during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination during this period via copier, telephone, , World Wide Web or media of any type is a violation of the competition rules. The American Mathematics Competitions are Sponsored by The Mathematical Association of America MAA... The Akamai Foundation... Contributors American Mathematical Association of Two Year Colleges AMATYC... American Mathematical Society AMS... American Society of Pension Actuaries ASPA... American Statistical Association ASA... Art of Problem Solving AoPS... Awesome Math... Canada/USA Mathcamp C/USA MC... Canada/USA Mathpath C/USA MP... Casualty Actuarial Society CAS... Clay Mathematics Institute CMI... IDEA Math... Institute for Operations Research and the Management Sciences INFORMS... L. G. Balfour Company... Mu Alpha Theta MAT... National Assessment & Testing... National Council of Teachers of Mathematics NCTM... Pedagoguery Software Inc Pi Mu Epsilon PME... Society of Actuaries SOA... U. S. A. Math Talent Search USAMTS... W. H. Freeman and Company... www. whfreeman.com/ Wolfram Research Inc....

7 WRITE TO US! Correspondence about the problems and solutions for this AMC 1 and orders for publications should be addressed to: American Mathematics Competitions University of Nebraska, P.O. Box 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 dmw8@psu.edu 008 AIME The 6 th annual AIME will be held on Tuesday, March 18, 008, with the alternate on Wednesday, April, 008. It is a 15-question, -hour, integer-answer exam. You will be invited to participate only if you score 10 or above, or finish in the top 1% of the AMC 10, or if you score 100 or above or finish in the top 5% of the AMC 1. Top-scoring students on the AMC 10/1/AIME will be selected to take the USA Mathematical Olympiad (USAMO) on April 9 and 0, 008. The best way to prepare for the AIME and USAMO is to study previous exams. Copies may be ordered as indicated below. PUBLICATIONS A complete listing of current publications, with ordering instructions, is at our web site:

8 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

9 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.

10 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 10 10) 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.

11 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 ).

12 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, 8). 17. 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 + d)i, and f(i) = ( b + c) + ( 1 + a + d)i. 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.

13 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 ( ) =.

14 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 10 10) = 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.

15 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 + 1). 5. 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.

16 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.

17 The American Mathematics Competitions are Sponsored by The Mathematical Association of America The Akamai Foundation Contributors American Mathematical Association of Two Year Colleges American Mathematical Society American Society of Pension Actuaries American Statistical Association Art of Problem Solving Awesome Math Canada/USA Mathcamp Canada/USA Mathpath Casualty Actuarial Society Clay Mathematics Institute IDEA Math Institute for Operations Research and the Management Sciences L. G. Balfour Company Mu Alpha Theta National Assessment & Testing National Council of Teachers of Mathematics Pedagoguery Software Inc. Pi Mu Epsilon Society of Actuaries U.S.A. Math Talent Search W. H. Freeman and Company Wolfram Research Inc.