The MATHEMATICAL ASSOCIATION OF AMERICA. AMC 12 Contest B. Solutions Pamphlet. Wednesday, February 15, American Mathematics Competitions

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The MTHEMTIL SSOITION OF MERI merican Mathematics ompetitions 57 th nnual merican Mathematics ontest 1 M 1 ontest Solutions Pamphlet Wednesday, February 15, 006 This Pamphlet gives at least one

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 MTHEMTIL SSOITION OF MERI merican Mathematics ompetitions 57 th nnual merican Mathematics ontest 1 M 1 ontest Solutions Pamphlet Wednesday, February 15, 006 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 M 1 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. issemination at any time via copier, telephone, , the World Wide Web or media of any type is a violation of the competition rules orrespondence about the problems and solutions for this M 1 and orders for any of the publications listed below should be addressed to: merican Mathematics ompetitions University of Nebraska, P.O. ox Lincoln, NE Phone: ; Fax: ; amcinfo@unl.edu The problems and solutions for this M 1 were prepared by the M s ommittee on the M 10 and M 1 under the direction of M 1 Subcommittee hair: Prof. avid Wells, epartment of Mathematics Penn State University, New Kensington, P opyright 006, The Mathematical ssociation of merica

2 1. () ecause the sum can be written as Solutions th M 1 ( 1) k = { 1, if k is even, 1, if k is odd, ( 1 + 1) + ( 1 + 1) + + ( 1 + 1) = = 0.. () ecause 4 5 = (4 + 5)(4 5) = 9, it follows that (4 5) = ( 9) = ( + ( 9))( ( 9)) = ( 6)(1) = 7.. () Let c and p represent the number of points scored by the ougars and the Panthers, respectively. The two teams scored a total of 4 points, so c + p = 4. The ougars won by 14 points, so c p = 14. The solution is c = 4 and p = 10, so the Panthers scored 10 points. 4. () The five items cost approximately = 19 dollars, so Mary s change is about $1.00, which is 5 percent of her $ () In order to catch up to John, ob must walk 1 mile farther in the same amount of time. ecause ob s speed exceeds John s speed by 5 = miles per hour, the time required for ob to catch up to John is 1/ hour, or 0 minutes. 6. () Francesca s 600 grams of lemonade contains = 411 calories, so 00 grams of her lemonade contains 411/ = 17 calories. 7. () There are only two possible occupants for the driver s seat. fter the driver is chosen, any of the remaining three people can sit in the front, and there are two arrangements for the other two people in the back. Thus there are = 1 possible seating arrangements.

3 Solutions th M 1 8. (E) Substituting x = 1 and y = into the equations gives 1 = 4 + a and = b. It follows that a + b = ( 1 ) ( + 1 ) = = 9 4. OR ecause a = x y 4 and b = y x 4, we have a + b = (x + y). 4 Since x = 1 when y =, this implies that a + b = 4 (1 + ) = () Let the integer have digits a, b, and c, read left to right. ecause 1 a < b < c, none of the digits can be zero and c cannot be. If c = 4, then a and b must each be chosen from the digits 1,, and. Therefore there are ( ) = choices for a and b, and for each choice there is one acceptable order. Similarly, for c = 6 and c = 8 there are, respectively, ( 5 ) = 10 and ( 7 ) = 1 choices for a and b. Thus there are altogether = 4 such integers. 10. () The sides of the triangle are x, x, and 15 for some positive integer x. y the Triangle Inequality, these three numbers are the sides of a triangle if and only if x + x > 15 and x + 15 > x. ecause x is an integer, the first inequality is equivalent to x 4, and the second inequality is equivalent to x 7. Thus the greatest possible perimeter is = (E) Joe has ounces of cream in his cup. Jonn has drunk ounces of the 14 ounces of coffee-cream mixture in her cup, so she has only 1/14 = 6/7 of her ounces of cream in her cup. Therefore the ratio of the amount of cream in Joe s coffee to that in Jonn s coffee is 6 7 = () parabola with the given equation and with vertex (p, p) must have equation y = a(x p) + p. ecause the y-intercept is (0, p) and p 0, it follows that a = /p. Thus so b = 4. y = p (x px + p ) + p = p x + 4x p,

4 Solutions th M () Since = 60, isosceles is also equilateral. s a consequence, E, E, E, F, F, and F are congruent. These six triangles have equal areas and their union forms rhombus, so each has area 4/6 = 4. Rhombus F E is the union of E and F, so its area is 8. E F OR Let the diagonals of rhombus intersect at O. Since the diagonals of a rhombus intersect at right angles, O is a triangle. Therefore O = O. ecause O and O are half the length of the longer diagonals of rhombi and F E, respectively, it follows that rea(f E) rea() = ( ) O = 1 O. Thus the area of rhombus F E is (1/)(4) = 8. O E F 14. () The total cost of the peanut butter and jam is N(4 + 5J) = 5 cents, so N and 4 + 5J are factors of 5 = 11. ecause N > 1, the possible values of N are 11,, and 5. If N = 5, then 4 + 5J = 1, which is impossible since and J are positive integers. If N =, then 4 + 5J = 11, which also has no solutions in positive integers. Hence N = 11 and 4 + 5J =, which has the unique positive integer solution = and J =. So the cost of the jam is 11()(5 ) = $1.65.

5 Solutions th M () Through O draw a line parallel to intersecting P at F. O F 6 P Then OF is a rectangle and OP F is a right triangle. Thus F =, F P =, and OF = 4. The area of trapezoid OP is 1, and the area of hexagon OP is 1 = 4. OR Lines,, and OP intersect at a common point H. H O 6 4 P ecause P H = OH = 90, triangles P H and OH are similar with ratio of similarity. Thus HO = HP = HO + OP = HO + 6, so HO = 6 and H = HO O = 4. Hence the area of OH is (1/)()(4 ) = 4, and the area of P H is ( )(4 ) = 16. The area of the hexagon is twice the area of P H minus twice the area of OH, so it is () iagonals, E, E,, F, and E divide the hexagon into twelve congruent triangles, six of which make up equilateral E. ecause = = 50, the area of E is ( ) 4 50 = 5. The area of hexagon EF is ( ) 5 = 5. OR Let O be the center of the hexagon. Then triangles, E, and EF are congruent to triangles O, OE, and EO, respectively. Thus the area of the hexagon is twice the area of equilateral E. Then proceed as in the first solution.

6 Solutions th M () On each die the probability of rolling k, for 1 k 6, is k = k 1. There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs (1, 6), (, 5), (, 4), (4, ), (5, ), and (6, 1). Thus the probability of rolling a total of 7 is = 56 1 = () Each step changes either the x-coordinate or the y-coordinate of the object by 1. Thus if the object s final point is (a, b), then a+b is even and a + b 10. onversely, suppose that (a, b) is a lattice point with a + b = k 10. One ten-step path that ends at (a, b) begins with a horizontal steps, to the right if a 0 and to the left if a < 0. It continues with b vertical steps, up if b 0 and down if b < 0. It has then reached (a, b) in k steps, so it can finish with 5 k steps up and 5 k steps down. Thus the possible final points are the lattice points that have even coordinate sums and lie on or inside the square with vertices (±10, 0) and (0, ±10). There are 11 such points on each of the 11 lines x + y = k, 5 k 5, for a total of 11 different points. 19. () The 4-digit number on the license plate has the form aabb or abab or baab, where a and b are distinct integers from 0 to 9. ecause Mr. Jones has a child of age 9, the number on the license plate is divisible by 9. Hence the sum of the digits, (a + b), is also divisible by 9. ecause of the restriction on the digits a and b, this implies that a + b = 9. Moreover, since Mr. Jones must have either a 4-year-old or an 8-year-old, the license plate number is divisible by 4. These conditions narrow the possibilities for the number to 1188, 77, 66, 5544, 66, 77, and The last two digits of 9900 could not yield Mr. Jones s age, and none of the others is divisible by 5, so he does not have a 5-year-old. Note that 5544 is divisible by each of the other eight non-zero digits. 0. () The given condition is equivalent to log 10 x = log 10 4x. Thus the condition holds if and only if n log 10 x < log 10 4x < n + 1 for some negative integer n. Equivalently, This inequality is true if and only if 10 n x < 4x < 10 n n x < 10n+1. 4

7 Solutions th M 1 7 Hence in each interval [ 10 n, 10 n+1), the given condition holds with probability ( 10 n+1 /4 ) 10 n 10 n+1 10 n = 10n ((10/4) 1) 10 n = 1 (10 1) 6. ecause each number in (0, 1) belongs to a unique interval [ 10 n, 10 n+1) and the probability is the same on each interval, the required probability is also 1/6. 1. () Let a and b, respectively, be the lengths of the major and minor axes of the ellipse, and let the dimensions of the rectangle be x and y. Then x + y is the sum of the distances from the foci to point on the ellipse, which is a. The length of a diagonal of the rectangle is the distance between the foci of the ellipse, which is a b. Thus x + y = a and x + y = 4a 4b. The area of the rectangle is 006 = xy = 1 [ (x + y) (x + y ) ] = 1 so b = 100. Thus the area of the ellipse is 006π = πab = πa 100, [ (a) (4a 4b ) ] = b, so a = 100, and the perimeter of the rectangle is (x + y) = 4a = () Note that n is the number of factors of 5 in the product a!b!c!, and 006 < 5 5. Thus 4 ( n = a/5 k + b/5 k + c/5 k ). k=1 ecause x + y + z x + y + z for all real numbers x, y, and z, it follows that n = 4 ( (a + b + c)/5 k ) k=1 4 ( 006/5 k ) k=1 = = 49. The minimum value of 49 is achieved, for example, when a = b = 64 and c = (E) Let, E, and F be the reflections of P about,, and, respectively. Then F = E = 90, and EF = 180. Thus the area of pentagon EF is twice that of, so it is s.

8 Solutions th M 1 8 E F Observe that E = 7, EF = 1, and F = 11. Furthermore, (7 ) + 1 = = 4 = (11 ), so EF is a right triangle. Thus the pentagon can be tiled with three right triangles, two of which are isosceles, as shown. It follows that E F s = 1 ( ) = , so a + b = 17. OR Rotate 90 counterclockwise about, and let and P be the images of and P, respectively. P' P ' Then P = P = 6, and P P = 90, so P P is an isosceles right triangle. Thus P P = 6, and P = P = 11. ecause ( 6 ) + 7 = 11, the converse of the Pythagorean Theorem implies that P P = 90. Hence P = 15. pplying the Law of osines in P gives = cos 15 = ,

9 Solutions th M 1 9 and a + b = () For a fixed value of y, the values of sin x for which sin x sin x sin y+sin y = 4 can be determined by the quadratic formula. Namely, sin y 4(sin y 4 ) sin x = sin y ± ecause cos ( ) π = 1 sin x = cos and sin ( π ) = ( π ) sin y ± sin = 1 sin y ±, this implies that ( π ) cos y = sin ( y ± π cos y. Within S, sin x = sin(y π ) implies x = y π. However, the case sin x = sin(y + π ) implies x = y + π when y π 6, and x = y + π when y π 6. Those three lines divide the region S into four subregions, within each of which the truth value of the inequality is constant. Testing the points (0, 0), ( π, 0), (0, π ), and ( π, π ) shows that the inequality is true only in the shaded subregion. The area of this subregion is ( π ) 1 ( π ) 1 ( π ) π = 6 6. π (0, ) x = y+ π x = y π π π (, ) ). x = y+ π (0,0) π (,0) 5. () The condition a n+ = a n+1 a n implies that a n and a n+ have the same parity for all n 1. ecause a 006 is odd, a is also odd. ecause a 006 = 1 and a n is a multiple of gcd(a 1, a ) for all n, it follows that 1 = gcd(a 1, a ) = gcd( 7, a ). There are 499 odd integers in the interval [1, 998], of which 166 are multiples of, 1 are multiples of 7, and 4 are multiples of 7 = 111. y the Inclusion-Exclusion Principle, the number of possible values of a cannot exceed = 4. To see that there are actually 4 possibilities, note that for n, a n < max(a n, a n 1 ) whenever a n and a n 1 are both positive. Thus a N = 0 for some N If gcd(a 1, a ) = 1, then a N = a N 1 = 1, and for n > N the sequence cycles through the values 1, 1, 0. If in addition a is odd, then a k+ is odd for k 1, so a 006 = 1.

10 The merican Mathematics ompetitions are Sponsored by The Mathematical ssociation of merica University of Nebraska-Lincoln The kamai Foundation ontributors merican Mathematical ssociation of Two Year olleges merican Mathematical Society merican Society of Pension ctuaries merican Statistical ssociation rt of Problem Solving anada/us Mathcamp anada/us Mathpath asualty ctuarial Society lay Mathematics Institute Institute for Operations Research and the Management Sciences L. G. alfour ompany Mu lpha Theta National ouncil of Teachers of Mathematics National ssessment & Testing Pedagoguery Software Inc. Pi Mu Epsilon Society of ctuaries U.S.. Math Talent Search W. H. Freeman and ompany Wolfram Research Inc.

The MATHEMATICAL ASSOCIATION OF AMERICA. AMC 10 Contest A. Solutions Pamphlet. American Mathematics Competitions

The MATHEMATICAL ASSOCIATION OF AMERICA. AMC 10 Contest A. Solutions Pamphlet. American Mathematics Competitions The MTHEMTIL SSOITION O MERI merican Mathematics ompetitions 57 th nnual merican Mathematics ontest 10 M 10 ontest Solutions Pamphlet Tuesda, JNURY 1, 006 This Pamphlet gives at least one solution for