The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions Presented by The Akamai Foundation. AMC 12 - Contest B. Solutions Pamphlet

Size: px

Start display at page:

Download "The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions Presented by The Akamai Foundation. AMC 12 - Contest B. Solutions Pamphlet"

Bennett Bennett
5 years ago
Views:

The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions Presented by The Akamai Foundation 54 th Annual American Mathematics Contest 2 AMC 2 - Contest B Solutions Pamphlet

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.

1 The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions Presented by The Akamai Foundation 54 th Annual American Mathematics Contest 2 AMC 2 - Contest B Solutions Pamphlet Wednesday, FEBRUARY 26, 200 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 2 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Duplication at any time via copier, phone, , the Web or media of any type is a violation of the copyright law. Correspondence about the problems and solutions should be addressed to: Prof. David Wells, Department of Mathematics Penn State University, New Kensington, PA 5068 Orders for prior year Exam questions and Solutions Pamphlets should be addressed to: Titu Andreescu, AMC Director American Mathematics Competitions University of Nebraska-Lincoln, P.O. Box 8606 Lincoln, NE Copyright 200, Committee on the American Mathematics Competitions The Mathematical Association of America

2 Solutions th AMC 2 B 2. (C) We have ( ) = ( ) = (D) The cost of each day s pills is 546/4 = 9 dollars. If x denotes the cost of one green pill, then x + (x ) = 9, so x = 20.. (A) To minimize the cost, Rose should place the most expensive flowers in the smallest region, the next most expensive in the second smallest, etc. The areas of the regions are shown in the figure, so the minimal total cost, in dollars, is ()(4) + (2.5)(6) + (2)(5) + (.5)(20) + ()(2) = (C) The area of the lawn is =,500 ft 2. Moe cuts about two square feet for each foot he pushes the mower forward, so he cuts 2(5000) = 0,000 ft 2 per hour. Therefore, it takes about,500 0,000 =.5 hours. 5. (D) The height, length, and diagonal are in the ratio : 4 : 5. The length of the diagonal is 27, so the horizontal length is 4 (27) = 2.6 inches (B) Let the sequence be denoted a, ar, ar 2, ar,..., with ar = 2 and ar = 6. Then r 2 = and r = or r =. Therefore a = 2 or a = (D) Let n represent the number of nickels in the bank, d the number of dimes, and q the number of quarters. Then n + d + q = 00 and 5n + 0d + 25q = 85. Further, n, d, and q must all be nonnegative integers. Dividing the second equation by 5 yields n + 2d + 5q = 67. Subtracting the first equation from this gives d+4q = 67. Since q cannot be negative, d is at most 67, and we check that

3 Solutions th AMC 2 B 67 dimes and nickels indeed produces $8.5. On the other hand, d cannot be 0,, or 2 because then q would not be an integer. Thus the smallest d can be is, leaving q = 6. We check that 6 quarters, dimes, and 8 nickels also produces $8.5. Thus the largest d can be is 67, the smallest is, and the difference is (E) Let y = (x). Since x 99, we have y 8. Thus if (y) =, then y = or y = 2. The values of x for which (x) = are 2, 2, and 0, and the 7 values of x for which (x) = 2 are 9, 48, 57, 66, 75, 84, and 9. There are 0 values in all. 9. (D) Since f is a linear function, its slope is constant. Therefore f(6) f(2) 6 2 and f(2) f(2) = 0. = f(2) f(2), so f(2) f(2) =, 0 Since f is a linear function, it has a constant rate of change, given by f(6) f(2) 6 2 Therefore f(2) f(2) = (2 2) = 0. If f(x) = mx + b, then so m =. Hence = 2 4 =. 2 = f(6) f(2) = 6m + b (2m + b) = 4m, f(2) f(2) = 2m + b (2m + b) = 0m = (B) We may assume that one of the triangles is attached to side AB. The second triangle can be attached to BC or CD to obtain two non-congruent figures. If the second triangle is attached to AE or to DE, the figure can be reflected about the vertical axis of symmetry of the pentagon to obtain one of the two already counted. Thus the total is two.. (C) When it is :00 PM it is 60 minutes after noon, but Cassandra s watch has recorded only 57 minutes and 6 seconds, that is, 57.6 minutes. Thus when her watch has recorded t minutes since noon, the actual number of minutes passed is t = t.

4 Solutions th AMC 2 B 4 In particular, when her watch reads 0:00 PM it has recorded 600 minutes past noon, so the actual number of minutes past noon is 25 (600) = 625, 24 or 0 hours and 25 minutes. Therefore the actual time is 0:25 PM. 2. (D) Among five consecutive odd numbers, at least one is divisible by and exactly one is divisible by 5, so the product is always divisible by 5. The cases n = 2, n = 0, and n = 2 demonstrate that no larger common divisor is possible, since 5 is the greatest common divisor of 5 7 9, 5 7 9, and (B) Let r be the radius of the sphere and cone, and let h be the height of the cone. Then the conditions of the problem imply that ( ) 4 4 πr = πr2 h, so h = r. Therefore, the ratio of h to r is :. 4. (D) Let H be the foot of the perpendicular from E to DC. Since CD = AB = 5, F G = 2, and F EG is similar to AEB, we have and EH = 2. Hence EH EH + = 2, so 5EH = 2EH + 6, 5 Area( AEB) = 2 (2 + ) 5 = E D F H G 2 C A 5 B Let I be the foot of the perpendicular from E to AB. Since EIA is similar to ADF and EIB is similar to BCG,

5 Solutions th AMC 2 B 5 we have AI EI = E and 5 AI EI = 2. D F G 2 C A AI I 5 2AI B Adding gives 5/EI =, so EI = 5. The area of the triangle is = (D) Let O be the intersection of the diagonals of ABEF. Since the octagon is regular, AOB has area /8. Since O is the midpoint of AE, OAB and BOE have the same area. Thus ABE has area /4, so ABEF has area /2. A B H G O C D F E Let O be the intersection of the diagonals of the square IJKL. Rectangles ABJI, JCDK, KEF L, and LGHI are congruent. Also IJ = AB = AH, so the right isosceles triangles AIH and JOI are congruent. By symmetry, the area in the center square IJKL is the sum of the areas of AIH, CJB, EKD, and GLF. Thus the area of rectangle ABEF is half the area of the octagon.

6 Solutions th AMC 2 B 6 A B H I J C G L O K D F E 6. (E) The area of the larger semicircle is 2 π (2)2 = 2π. The region deleted from the larger semicircle consists of five congruent sectors and two equilateral triangles. The area of each of the sectors is and the area of each triangle is 6 π ()2 = π 6 2 so the area of the shaded region is 2π 5 π = 4, 4 = 7 6 π (D) We have = log(xy ) = log x + log y and = log(x 2 y) = 2 log x + log y. Solving yields log x = 2 5 and log y = 5. Thus log(xy) = log x + log y = 5.

7 Solutions th AMC 2 B 7 The given equations imply that xy = 0 = x 2 y. Thus y = 0 ( ) 0 x 2 and x x 2 = 0. It follows that x = 0 2/5 and y = 0 /5, so log(xy) = log(0 /5 ) = /5. Since log(xy ) = log(x 2 y), we have xy = x 2 y, so x = y 2. Hence So log(xy) = log(y ) = /5. 8. (B) We have = log(xy ) = log(y 5 ) = 5 log y, and log y = 5. y = 7x 5 = 7(a c b d ) 5 = 7a 5c b 5d. The minimum value of x is obtained when neither x nor y contains prime factors other than 7 and. Therefore we may assume that a = 7 and b =, so x = 7 c d and 7x 5 = 7 5c+ 5d. Letting y = 7 m n we obtain y = 7 m n+. Hence 7 5c+ 5d = 7 m n+. The smallest positive integer solutions are c = 5, d = 8, m = 2, and n =. Thus a + b + c + d = =. 9. (E) Since the first term is not, the probability that it is 2 is /4. If the first term is 2, then the second term cannot be 2. If the first term is not 2, there are four equally likely values, including 2, for the second term. Thus the probability that the second term is 2 is = 6, so a + b = + 6 = 9. The set S contains (4)(4!) = 96 permutations, since there are 4 choices for the first term, and for each of these choices there are 4! arrangements of the remaining terms. The number of permutations in S whose second term is 2 is ()(!) = 8, since there are choices for the first term, and for each of these choices there are! arrangements of the last terms. Thus the requested probability is 8/96 = /6, and a + b = (B) We have 0 = f( ) = a + b c + d and 0 = f() = a + b + c + d, so b + d = 0. Also d = f(0) = 2, so b = 2. The polynomial is divisible by (x + )(x ) = x 2, its leading term is ax, and its constant term is 2, so f(x) = (x 2 )(ax 2) = ax 2x 2 ax + 2 and b = 2.

8 Solutions th AMC 2 B 8 2. (D) Let β = π α. Apply the Law of Cosines to ABC to obtain Thus AC < 7 if and only if (AC) 2 = (8)(5) cos β = cos β cos β < 49, that is, if and only if cos β > 2. Therefore we must have 0 < β < π π/, and the requested probability is π =. 22. (C) Let O be the point of intersection of AC and BD. Then AOB is a right triangle with legs OA = 8 and OB = 5. Quadrilateral OP NQ is a rectangle because it has right angles at O, P, and Q. Thus P Q = ON, because the diagonals of a rectangle are of equal length. The minimum value of P Q is the minimum value of ON. This is achieved if and only if N is the foot of the altitude from O in triangle AOB. Writing the area of AOB in two different ways yields 2 AB ON = OA OB. 2 Hence the minimum value of P Q is ON = OA OB AB = OA OB OA2 + OB = 8 5 = A N B P O Q D C 2. (A) The intercepts occur where sin(/x) = 0, that is, where x = /(kπ) and k is a nonzero integer. Solving yields < kπ < π 0, 000 < k <. π Thus the number of x intercepts in (0.000, 0.00) is 0, = 8 8 = 2865, π π which is closest to 2900.

9 Solutions th AMC 2 B (C) Since the system has exactly one solution, the graphs of the two equations must intersect at exactly one point. If x < a, the equation y = x a + x b + x c is equivalent to y = x + (a + b + c). By similar calculations we obtain x + (a + b + c), if x < a x + ( a + b + c), if a x < b y = x + ( a b + c), if b x < c x + ( a b c), if c x. Thus the graph consists of four lines with slopes,,, and, and it has corners at (a, b + c 2a), (b, c a), and (c, 2c a b). On the other hand, the graph of 2x + y = 200 is a line whose slope is 2. If the graphs intersect at exactly one point, that point must be (a, b + c 2a). Therefore 200 = 2a + (b + c 2a) = b + c. Since b < c, the minimum value of c is (D) We can assume that the circle has its center at (0, 0) and a radius of. Call the three points A, B, and C, and let a, b, and c denote the length of the counterclockwise arc from (, 0) to A, B, and C, respectively. Rotating the circle if necessary, we can also assume that a = π/. Since b and c are chosen at random from [0, 2π), the ordered pair (b, c) is chosen at random from a square with area 4π 2 in the bc-plane. The condition of the problem is met if and only if 0 < b < 2π, 0 < c < 2π, and b c < π. This last inequality is equivalent to b π < c < b + π. c 2p 2p 2p 2p b The graph of the common solution to these inequalities is the shaded region shown. The area of this region is ( ) ( ) 2 6 2π = π 2 /, 8 so the requested probability is π 2 / 4π 2 = 2.

10 The American Mathematics Contest 2 (AMC 2) Sponsored by Mathematical Association of America The Akamai Foundation University of Nebraska Lincoln Contributors American Mathematical Association of Two Year Colleges American Mathematical Society American Society of Pension Actuaries American Statistical Association Canada/USA Mathpath & Mathcamp Casualty Actuarial Society Clay Mathematics Institute Consortium for Mathematics and its Applications Institute for Operations Research and the Management Sciences Kappa Mu Epsilon Mu Alpha Theta National Association of Mathematicians National Council of Teachers of Mathematics Pi Mu Epsilon School Science and Mathematics Association Society of Actuaries

The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions Presented by The Akamai Foundation. AMC 12 - Contest A. Solutions Pamphlet

The MATHEMATICAL ASSOCIATION OF AMERICA American Mathematics Competitions Presented by The Akamai Foundation 53 rd Annual American Mathematics Contest 1 AMC 1 - Contest A Solutions Pamphlet TUESDAY, FEBRUARY