The Mathematical Association of America. American Mathematics Competitions AMERICAN INVITATIONAL MATHEMATICS EXAMINATION (AIME)

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1 The Mathematical Association of America American Mathematics Competitions 6 th Annual (Alternate) AMERICAN INVITATIONAL MATHEMATICS EXAMINATION (AIME) SOLUTIONS PAMPHLET Wednesday, April, 008 This Solutions Pamphlet gives at least one solution for each problem on this year s AIME and shows that all the problems can be solved using precalculus mathematics. When more than one solution for a problem is provided, this is done to illustrate a significant contrast in methods, e.g., algebraic vs geometric, computational vs. conceptual, elementary vs. advanced. The solutions are by no means the only ones possible, nor are they necessarily superior to others the reader may devise. We hope that teachers 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. Routine calculations and obvious reasons for proceeding in a certain way are often omitted. This gives greater emphasis to the essential ideas behind each solution. Correspondence about the problems and solutions for this AIME and orders for any of the publications listed below should be addressed to: American Mathematics Competitions University of Nebraska, P.O. Box 8606 Lincoln, NE Phone: ; Fax: ; The problems and solutions for this AIME were prepared by the MAA s Committee on the AIME under the direction of: Steve Blasberg, AIME Chair San Jose, CA 959 USA Copyright 008, The Mathematical Association of America

2 008 AIME SOLUTIONS. (Answer: 00) Reordering the sum shows that N = (00 98 ) + (99 97 ) + (96 94 ) + + (4 ) + ( ), which equals = 4( ) = 4( ) + 4( ) [ ] [ ] 5(99 + ) 5(98 + ) = = 00 (5 + 50) = 000, and the required remainder is 00.. (Answer: 60) Suppose Rudolph bikes at r miles per minute. He takes 49 five-minute breaks in reaching the 50-mile mark, so his total time in minutes is 50 r = 50 r Jennifer bikes at 4r miles per minute and takes 4 fiveminute breaks in reaching the 50-mile mark, so her total time in minutes is 0.75r = r + 0. Setting these two times equal gives r = 5, and hence r = 5. This yields a total time of 50 / = 60 minutes.. (Answer: 79) Let a, b, and c be the dimensions of the cheese after ten slices have been cut off, giving a volume of abc cu cm. Because each slice shortens one of the dimensions of the cheese by cm, a + b + c = ( ) 0 = 7. By the Arithmetic-Geometric Mean Inequality, the product of a set of positive numbers with a given sum is greatest when the numbers are equal, so the remaining cheese has maximum volume when a = b = c = 9. The volume is then 9 = 79 cu cm. Query: What happens if a + b + c is not divisible by? OR The volume of the remaining cheese is greatest when it forms a cube. This can be accomplished by taking one slice from the 0 cm dimension, 4 slices from the cm dimension, and 5 slices from the 4 cm dimension for a total of = 0 slices. The remaining cheese is then 9 cm by 9 cm by 9 cm for a volume of 79 cu cm.

3 008 AIME SOLUTIONS 4. (Answer: 0) Every positive integer has a unique base- representation, which for 008 is 00. Note that k = ( + ( )) k = k+ + ( ) k, so that 00 = = ( 7 + ( ) 6 ) + ( 6 + ( ) 5 ) + ( 4 + ( ) ) = 7 + ( ) ( ) + + 0, and n + n + + n r = =. 5. (Answer: 504) A 7 N 5 D B M C E Extend leg AB past B and leg CD past C, and let E be the point of intersection of these extensions. Then because BM AN = CM DN, line MN must pass through point E. But A = 7 and D = 5 implies that AED = 90. Thus EDA is a right triangle with median EN, and EBC is a right triangle with median EM. The median to the hypotenuse in any right triangle is half the hypotenuse, so EN = 008 = 004, EM = 000 = 500, and MN = EN EM = (Answer: 56) Observe that if x n = x n + x n and y n = x n, then y n = + y n, x n x n so x n = x n xn x = y y y n = y (y + ) (y + n ). x 0 x n x n x 0 In particular, for the first sequence, y = a a 0 =, and so a n = n!. Similarly, for the second sequence, y = b b 0 =, and so b n = (n + )!. The required ratio is then 4!/! = 7 = 56.

4 008 AIME SOLUTIONS 4 7. (Answer: 75) Because the equation is cubic and there is no x term, the sum of the roots is 0; that is, r + s + t = 0. Therefore, (r + s) + (s + t) + (t + r) = ( t) + ( r) + ( s) = (r + s + t ). Because r is a root, 8r + 00r = 0, and similarly for s and t. Therefore, 8(r + s + t ) + 00(r + s + t) = 0, and r + s + t = 00(r + s + t) = = 75, and hence (r + s) + (s + t) + (t + r) = (r + s + t ) = (Answer: 5) The product-to-sum formula for cos(k a) sin(ka) yields sin(k a + ka) sin(k a ka), which equals sin(k(k +)a) sin((k )ka). Thus the given sum becomes sin( a) sin(0) + sin( a) sin( a) + sin(4 a) sin( a) + + sin(n(n + )a) sin((n )na). This is a telescoping sum, which simplifies to sin(n(n + )a) sin(0) = sin(n(n + )a). But sin(x) is an integer only when x is an integer multiple of π/, so n(n + ) must be an integer multiple of 004 = 5. Thus either n or n + is a multiple of 5, because 5 is prime. The smallest such n is 50, but 50 5 is not a multiple of 004. The next smallest such n is 5, and 5 5 is a multiple of 004. Hence the smallest such n is (Answer: 09) Let the coordinate plane be the complex plane, and let z k be the complex number that represents the position of the particle after k moves. Multiplying a complex number by cis θ corresponds to a rotation of θ about the origin, and adding 0 to a complex number corresponds to a horizontal translation of 0 units to the right. Thus z 0 = 5, and z k+ = ωz k + 0, where ω = cis (π/4), for k 0. Then z = 5ω + 0, z = ω(5ω + 0) + 0 = 5ω + 0ω + 0, z = ω(5ω + 0ω + 0) + 0 = 5ω + 0ω + 0ω + 0, and, in general, z k = 5ω k + 0(ω k + ω k + + ).

5 008 AIME SOLUTIONS 5 In particular, z 50 = 5ω (ω 49 + ω ). Note that ω 8 = and ω k+4 = cis ((k + 4)π/4) = cis (kπ/4 + π) = cis (kπ/4) = ω k. Applying the second equality repeatedly shows that z 50 = 5ω 50 +0(ω 49 + ω ) = 5ω 6 +0( ω+( )+ω +ω +ω+) = 5ω 6 +0(ω +ω ). The last expression equals 5cis (π/) + 0(cis (π/4) + cis (π/)) ( ) = 5( i) i + i = 5 + (5 + 5)i Thus (p, q) = ( 5, 5 +5), and p + q = The required value is therefore the greatest integer less than or equal to = 9.4, which is (Answer: 40) The following argument shows that m = 0 and r = 4. Note that the square of any distance between distinct points in the array has the form a + b for some integers a and b in the set {0,,, } with a and b not both equal to 0. There are 9 such possible values, namely,,, 4, 5, 8, 9, 0,, and 8. Hence a growing path can consist of at most 0 points. It remains to show that 0 points can be so chosen, and that there are 4 such paths. Let the points of these paths be labeled P, P,..., P 0. First, P 9 P 0 must be 8; that is, P 9 and P 0 must be a pair of opposite corners in the array. There are 4 equivalent ways to label P 9 and P 0. Without loss of generality, assume that P 9 and P 0 are the bottom right corner and upper left corner, respectively. Then there are symmetrical positions for P 8, both neighboring P 0. (See the left-hand diagram below.) Assume that P 8 is beside P 0, as shown in the middle diagram below. P0 P0 P8 P0 P8 P6 P P4 P5 P P9 P9 P7 P9 Note that the bottom left corner cannot be P 7, because otherwise P 6 = P 9 or P 6 = P 0. The positions of points P 7, P 6, P 5, P 4, P, and P are then fixed. (See the right-hand diagram above.) Finally, there are possible positions for P. Hence r = 4 = 4, and mr = 40.

6 008 AIME SOLUTIONS 6. (Answer: 54) A P Q B C Let q be the radius of circle Q. The perpendicular from A to BC has length = 4 4, and so sin B = sin C = 4 5. Thus tan C = tan B = sin B + cos B = 4. Place the figure in a Cartesian coordinate system with B = (0, 0), C = (56, 0), and A = (8, 96). Note that circles P and Q are both tangent to BC, and their centers P and Q lie on the angle bisectors of angles C and B, respectively. Thus P = (56 4 6/, 6) and Q = (4q/, q). Also, P Q = q + 6, and so ( 4(q + 6) P Q = 56) + (q 6) = (q + 6), which yields (4q 04) = 576q, and thus q 88q = 0. The roots of this equation are 44 ± 6 5, but is impossible because it exceeds the base of ABC. Hence q = , and the requested number is = 54.. (Answer: 0) For convenience, label the flagpoles and, and denote by G n the number of flag arrangements in which flagpole has n green flags. If either flagpole contains all the green flags, then it must also contain at least 8 blue flags to act as separators. The flagpole with no green flags must therefore contain either or blue flags. Hence, G 0 = G 9 =, because there are 0 possible positions in which to place blue flag on the flagpole with all

7 008 AIME SOLUTIONS 7 the green flags (otherwise, the flagpole with no green flags contains blue flags). If 0 < n < 9, then a flagpole that has n green flags must contain at least n blue flags, and the other flagpole has 9 n green flags and must contain at least 8 n blue flags. Thus in any flag arrangement where each flagpole contains at least one green flag, 7 of the blue flags have fixed positions relative to the green flags, and there are remaining blue flags that can be freely distributed. If flagpole has n green flags, then there are n+ possible positions in which blue flag can be placed on that flagpole, and there are 0 n possible positions in which blue flag can be placed on flagpole. Thus, for 0 < n < 9, after placing the green flags and 7 of the blue flags, there remain blue flags with distinguishable locations. This is a standard problem of distributing indistinguishable objects among distinguishable bins. Because bins require 0 separators to divide them, the problem is equivalent to choosing locations out of (the flags and the 0 separators). Thus the number of ways to place the blue flags is ( ). The desired number of flag arrangements is then N = 9 ( ) G i = + 8 = 0, i=0 and the required remainder is 0.. (Answer: 09) One pair of vertices lies at ± i. Express points on the line segment determined by these two vertices in the form z = + yi, where y is real and y. Reciprocals of points on this line segment are then of the form yi with y + y. Because 4 z = yi 4 + y 4 + y 4 + y = ( 4 y ) + y ( 4 + y ) = ( 4 + y ) ( 4 + y ) =, the curve traced by the reciprocals of complex numbers on this line segment is an arc of a circle centered at with radius, running from to + i. The region enclosed by this arc and the lines from the origin to the endpoints can be partitioned into a 0-degree sector of the disk centered at with radius, together with two triangles, each of base and height. Thus it has area π +. Multiply by six to find that the total area is π + = π + 7. Thus a + b = 9. Query: The above argument shows that the reciprocals of the points on a line not through the origin fall on a circle through the origin. What happens to a line through the origin? OR i

8 008 AIME SOLUTIONS 8 Because pairs of parallel sides are one unit apart, one side of the hexagon lies along the line x = Re z =. If w = u + vi = z, where z is on this line, then ( w + ) = w, which is equivalent to w+w = ww. Hence u +v = u, or (u ) +v =, which is the equation of a circle of radius centered at. Because z = is mapped to w =, it follows that Re z > is mapped to the interior of this circle. By symmetry, the five remaining half-planes whose union produces R are mapped to corresponding disks. Thus the points that are in the image of the half-plane Re z > but none of the other five half-planes belong to the disk (u ) +v < and the region bounded by the two rays v = ± u. The resulting set can be partitioned into a circular sector with radius and central angle π and two isosceles triangles with equal sides of length and vertex angle π. There are six such nonoverlapping congruent figures forming S. It follows that the area of S is 6( π + ) = π + 7. Thus a + b = (Answer: 007) Let ABCD be a rectangle such that AB = CD = a, and BC = DA = b. Let E and F be points on the sides AB and BC respectively, such that AE = x and CF = y. Solving the given system of equations is thus equivalent to requiring that DEF be equilateral. Let ADE = α. Then F DC = 0 α, tan α = x b, and EDF = 60. Thus y a = tan(0 α) = x b + x b = b x b + x. () Squaring and adding yields 4(x + b ) (b + x) = y + a a = x + b a. Thus 4a = (b + x), and x 0 implies that x = a b, which is a positive real number because a b. Equation () shows that y 0 if and only if b x = b (a b ) = 4b a 0. It follows that a b, and so ρ =. Hence ρ = 4, and m + n = 7. This value of ρ is achieved when a =, b =, x =, and y = (Answer: 8) Let m be an integer such that (m + ) m = n, which implies that (m + ) = (n )(n + ). Because n and n + are consecutive odd integers, they are relatively prime. Thus can only divide one of n and n +. Therefore either n = k and n + = j, or n = k and n + = j. The first case implies that j k =,

9 008 AIME SOLUTIONS 9 which can be shown to be impossible by examining the equation modulo. The second case implies that 4n = j + k and j k =. Let k = a+ for some integer a. Then j = k + = (a+) +. Therefore 4n = (a + ) + (a + ) + = 8a + 8a + 4, or n = a + a +. Furthermore, because n + 79 = d for some integer d, then n + 79 = d = (a + a + ) + 79 = 4a + 4a + 8. This equation is equivalent to 80 = d (4a + 4a + ) = d (a + ) = (d a )(d + a + ). Both factors on the right side are of the same parity, so they both must be even. Then the two factors on the right are either (, 40), (4, 0), or (8, 0), and (d, a) = (, 9), (, ), (9, 0). The first solution gives n = = 8, and the last solution gives n = 8 79 =. Thus the largest such n is 8 (with m = 04).

10 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.