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

Size: px

Start display at page:

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

Jared Hicks
5 years ago
Views:

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

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, elementar vs advanced.

1 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 each problem on this ear 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, elementar vs advanced. These solutions are b no means the onl ones possible, nor are the superior to others the reader ma devise. We hope that teachers will inform their students about these solutions, both as illustrations of the kinds of ingenuit 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 10 during the period when students are eligible to participate seriousl jeopardizes the integrit of the results. issemination at an time via copier, telephone, , the World Wide Web or media of an tpe is a violation of the competition rules orrespondence about the problems and solutions for this M 10 should be addressed to: merican Mathematics ompetitions Universit of Nebraska, P.O. ox Lincoln, NE Phone: ; ax: ; amcinfo@unl.edu The problems and solutions for this M 10 were prepared b the M s ommittee on the M 10 and M 1 under the direction of M 10 Subcommittee hair: Prof. ouglas aires, epartment of Mathematics Youngstown State Universit, Youngstown, OH opright 006, The Mathematical ssociation of merica

2 Solutions th M () ive sandwiches cost 5 = 15 dollars and eight sodas cost 8 = 16 dollars. Together the cost = 1 dollars.. () the definition we have h (h h) = h (h h) = h (h h) = h.. () Mar is (/5)(0) = 18 ears old. 4. (E) The largest possible sum of the two digits representing the minutes is 5+9 = 14, occurring at 59 minutes past each hour. The largest possible single digit that can represent the hour is 9. This exceeds the largest possible sum of two digits that can represent the hour, which is 1 + =. Therefore, the largest possible sum of all the digits is =, occurring at 9: () Each slice of plain pizza cost $1. ave paid $ for the anchovies in addition to $5 for the five slices of pizza that he ate, for a total of $7. oug paid onl $ for the three slices of pizza that he ate. Hence ave paid 7 = 4 dollars more than oug. 6. () Take the seventh root of both sides to get (7x) = 14x. Then 49x = 14x, and because x 0 we have 49x = 14. Thus x = /7. 7. () Let E represent the end of the cut on, and let represent the end of the cut on. or a square to be formed, we must have E = = and E + + = 18, so = 6. The rectangle that is formed b this cut is indeed a square, since the original rectangle has area 8 18 = 144, and the rectangle that is formed b this cut has a side of length 6 = 1 = 144. E E E

3 Solutions th M (E) Substitute (, ) and (4, ) into the equation to give = 4 + b + c and = b + c. Subtracting corresponding terms in these equations gives 0 = 1 + b. So b = 6 and c = 4 ( 6) = 11. OR The parabola is smmetric about the vertical line through its vertex, and the points (, ) and (4, ) have the same -coordinate. The vertex has x-coordinate ( + 4)/ =, so the equation has the form = (x ) + k for some constant k. Since = when x = 4, we have = 1 + k and k =. onsequentl the constant term c is ( ) + k = 9 + = () irst note that, in general, the sum of n consecutive integers is n times their median. If the sum is 15, we have the following cases: if n =, then the median is 7.5 and the two integers are 7 and 8; if n =, then the median is 5 and the three integers are 4, 5, and 6; if n = 5, then the median is and the five integers are 1,,, 4, and 5. ecause the sum of four consecutive integers is even, 15 cannot be written in such a manner. lso, the sum of more than five consecutive integers must be more than = 15. Hence there are sets satisfing the condition. Note: It can be shown that the number of sets of two or more consecutive positive integers having a sum of k is equal to the number of odd positive divisors of k, excluding (E) Suppose that k = 10 x is an integer. Then 0 k 10, and because k is an integer, we have 0 k 10. Thus there are 11 possible integer values of k. or each such k, the corresponding value of x is ( 10 k ). ecause ( 10 k ) is positive and decreasing for 0 k 10, the 11 values of x are distinct. 11. () The equation (x + ) = x + is equivalent to x + x + = x +, which reduces to x = 0. Thus the graph of the equation consists of the two lines that are the coordinate axes.

4 Solutions th M () The regions in which the dog can roam for each arrangement are shaded in the figure. or arrangement I, the area of this region is 1 π 8 = π square feet. The area of the shaded region in arrangement II exceeds this b the area of a quarter-circle of radius 4 feet, that is, b 1 4 π 4 = 4π square feet. I 8 8 II () Let x represent the amount the plaer wins if the game is fair. The chance of an even number is 1/, and the chance of matching this number on the second roll is 1/6. So the probabilit of winning is (1/)(1/6) = 1/1. Therefore (1/1)x = $5 and x = $ () The top of the largest ring is 0 cm above its bottom. That point is cm below the top of the next ring, so it is 17 cm above the bottom of the next ring. The additional distances to the bottoms of the remaining rings are 16 cm, 15 cm,..., 1 cm. Thus the total distance is 0 + ( ) = 0 + OR = = 17 cm. The required distance is the sum of the outside diameters of the 18 rings minus a -cm overlap for each of the 17 pairs of consecutive rings. This equals ( ) 17 = ( ) 4 = = 17 cm. 15. () Since Odell s rate is 5/6 that of Kershaw, but Kershaw s lap distance is 6/5 that of Odell, the each run a lap in the same time. Hence the pass twice each time the circle the track. Odell runs ( (0 min) 50 m ) ( 1 min 100π ) laps = 75 laps.87 laps, m π as does Kershaw. ecause.5 <.87 < 4, the pass each other (.5) = 47 times.

5 Solutions th M () Let O and O denote the centers of the smaller and larger circles, respectivel. Let and be the points on that are also on the smaller and larger circles, respectivel. Since O and O are similar right triangles, we have s a consequence, O 1 = O = O +, so O =. = O O = 9 1 =. O 1 O' ' Let be the midpoint of. triangles, we have Since O and are similar right So the area of is 1 = = O + OO + O = = = 16. =.

6 Solutions th M () irst note that since points and trisect, and points G and trisect HE, we have HG = G = E = = = = 1. lso, HG is parallel to and HG =, so GH is a parallelogram. Similarl, is parallel to E and = E, so E is a parallelogram. s a consequence, W XY Z is a parallelogram, and since HG = = = E, it is a rhombus. H G E W X Z Y Since H = =, the rectangle H is a square of side length. Its diagonals and H have length and form a right angle at X. s a consequence, W XY Z is a square. In isosceles HX we have HX = X =. In addition, HG = 1 H. So XW = 1 X = 1, and the square W XY Z has area XW = 1/. 18. () Since the two letters have to be next to each other, think of them as forming a two-letter word w. So each license plate consists of 4 digits and w. or each digit there are 10 choices. There are 6 6 choices for the letters of w, and there are 5 choices for the position of w. So the total number of distinct license plates is () Let n d, n, and n + d be the angles in the triangle. Then 180 = n d + n + n + d = n, so n = 60. ecause the sum of the degree measures of two angles of a triangle is less than 180, we have 180 > n + (n + d) = 10 + d, which implies that 0 < d < 60. There are 59 triangles with this propert. 0. (E) Place each of the integers in a pile based on the remainder when the integer is divided b 5. Since there are onl 5 piles but there are 6 integers, at least one of the piles must contain two or more integers. The difference of two integers in the same pile is divisible b 5. Hence the probabilit is 1. We have applied what is called the Pigeonhole Principle. This states that if ou have more pigeons than boxes and ou put each pigeon in a box, then at least one of the boxes must have more than one pigeon. In this problem the pigeons are integers and the boxes are piles.

7 Solutions th M (E) There are 9000 four-digit positive integers. or those without a or, the first digit could be one of the seven numbers 1, 4, 5, 6, 7, 8, or 9, and each of the other digits could be one of the eight numbers 0, 1, 4, 5, 6, 7, 8, or 9. So there are = 5416 four-digit numbers with at least one digit that is a or a.. () If a debt of dollars can be resolved in this wa, then integers p and g must exist with = 00p + 10g = 0(10p + 7g). s a consequence, must be a multiple of 0, and no positive debt less than $0 can be resolved. debt of $0 can be resolved since 0 = 00( ) + 10(). This is done b giving goats and receiving pigs.. () Radii and are each perpendicular to. the Pthagorean Theorem, E = 5 = 4. ecause E and E are similar, Therefore E E =, so E = E = 4 8 =. = E + E = 4 + = () Two pramids with square bases form the octahedron. The upper pramid is shown. Since the length of is /, the base area of the pramid is ( /) = 1/. The altitude of the pramid is 1/, so its volume is = 1 1. The volume of the octahedron is (1/1) = 1/6.

8 Solutions th M () t each vertex there are three possible locations that the bug can travel to in the next move, so the probabilit that the bug will visit three different vertices after two moves is /. Label the first three vertices that the bug visits as to to, as shown in the diagram. In order to visit ever vertex, the bug must travel from to either G or. H G E The bug travels to G with probabilit 1/, and from there the bug must visit the vertices, E, H, in that order. Each of these choices has probabilit 1/ of occurring. So the probabilit that the path continues in the form G E H is ( 1 5. ) lternativel, the bug could travel from to with probabilit 1/, and then travel to H, which also occurs with probabilit 1/. rom H the bug could go either to G or to E, with probabilit /, and from there to the two remaining vertices, each with probabilit 1/. So the probabilit that the path continues in one of the forms E G H G E is ( ( 1 4. ) ) Hence the bug will visit ever vertex in seven moves with probabilit ( ) [ ( 1 ) 5 + ( ) ( 1 ) 4 ] OR = ( ) ( 1 + ) ( 1 ) 4 = 4. rom a given starting point there are 7 possible walks of seven moves for the bug, all of them equall likel. If such a walk visits ever vertex exactl once, there are three choices for the first move and, excluding a return to the start, two choices for the second. Label the first three vertices visited as,, and, in that order, and label the other vertices as shown. The bug must go to either G or on its third move. In the first case it must then visit vertices, E, H, and in order. In the second case it must visit either H, E,, and G or H, G,, and E in order. Thus there are = 18 walks that visit ever vertex exactl once, so the required probabilit is 18/ 7 = /4.

9 The merican Mathematics ompetitions are Sponsored b The Mathematical ssociation of merica Universit of Nebraska-Lincoln The kamai oundation ontributors merican Mathematical ssociation of Two Year olleges merican Mathematical Societ merican Societ of Pension ctuaries merican Statistical ssociation rt of Problem Solving anada/us Mathcamp anada/us Mathpath asualt ctuarial Societ la Mathematics Institute Institute for Operations Research and the Management Sciences L. G. alfour ompan Mu lpha Theta National ouncil of Teachers of Mathematics National ssessment & Testing Pedagoguer Software Inc. Pi Mu Epsilon Societ of ctuaries U.S.. Math Talent Search W. H. reeman and ompan Wolfram Research Inc.

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

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