AMC 10 Contest B. Solutions Pamphlet. Wednesday, FEBRUARY 21, American Mathematics Competitions

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1 The MATHEMATICAL ASSOCIATION of AMERICA Ameican Mathematics Competitions 8 th Annual Ameican Mathematics Contest 10 AMC 10 Contest B Solutions Pamphlet Wednesday, FEBRUARY 21, 2007 This Pamphlet gives at least one solution fo each poblem on this yea s contest and shows that all poblems can be solved without the use of a calculato. When moe than one solution is povided, this is done to illustate a significant contast in methods, e.g., algebaic vs geometic, computational vs conceptual, elementay vs advanced. These solutions ae by no means the only ones possible, no ae they supeio to othes the eade may devise. We hope that teaches will infom thei students about these solutions, both as illustations of the kinds of ingenuity needed to solve nonoutine poblems and as examples of good mathematical exposition. Howeve, the publication, epoduction o communication of the poblems o solutions of the AMC 10 duing the peiod when students ae eligible to paticipate seiously jeopadizes the integity of the esults. Dissemination via copie, telephone, , Wold Wide Web o media of any type duing this peiod is a violation of the competition ules. Afte the contest peiod, pemission to make copies of individual poblems in pape o electonic fom including posting on web-pages fo educational use is ganted without fee povided that copies ae not made o distibuted fo pofit o commecial advantage and that copies bea the copyight notice. Coespondence about the poblems/solutions fo this AMC 10 and odes fo any publications should be addessed to: Ameican Mathematics Competitions Univesity of Nebaska, P.O. Box 81606, Lincoln, NE Phone: ; Fax: ; amcinfo@unl.edu The poblems and solutions fo this AMC 10 wee pepaed by the MAA s Committee on the AMC 10 and AMC 12 unde the diection of AMC 10 Subcommittee Chai: Pof. Douglas Faies, Depatment of Mathematics Youngstown State Univesity, Youngstown, OH Copyight 2007, The Mathematical Association of Ameica

2 Solutions th AMC 10 B 2 1. Answe (E): The peimete of each bedoom is 2( ) = 44 feet, so the suface to be painted in each bedoom has an aea of = 292 squae feet. Since thee ae 3 bedooms, Isabella must paint = 876 squae feet. 2. Answe (E): Since 3 5 = (3+5)5 = 8 5 = 40 and 5 3 = (5+3)3 = 8 3 = 24, we have = = Answe (B): The student used 120/30 = 4 gallons on the tip home and 120/20 = 6 gallons on the tip back to school. So the aveage gas mileage fo the ound tip was 240 miles = 24 miles pe gallon. 10 gallons 4. Answe (D): Since OA = OB = OC, tiangles AOB, BOC, and COA ae all isosceles. Hence Since ABC = ABO + OBC = OR AOC = = 100, the Cental Angle Theoem implies that ABC = 1 2 AOC = 50. = Answe (D): Let A, B, C, and D epesent the following statements about a peson in the land. A : Is an Aog. B : Is a Baf. C : Is a Cup. D : Is a Damp. Then the statement in the fist sentence of the poblem can be expessed as: A = B, C = B, D = A and C = D. The most we can conclude is that C = D = A = B. So the only statement listed that we ae cetain is tue is that Cups ae both Aogs and Bafs.

3 Solutions th AMC 10 B 3 6. Answe (D): Saah will eceive 4.5 points fo the thee questions she leaves unansweed, so she must ean at least = 95.5 points on the fist 22 poblems. Because 15 < 95.5 < 16, 6 she must solve at least 16 of the fist 22 poblems coectly. This would give he a scoe of Answe (E): Because AB = BC = EA and A = B = 90, quadilateal ABCE is a squae, so AEC = 90. A B E C D Also CD = DE = EC, so CDE is equilateal and CED = 60. Theefoe E = AEC + CED = = Answe (D): Once a and c ae chosen, the intege b is detemined. Fo a = 0, we could have c = 2, 4, 6, o 8. Fo a = 2, we could have c = 4, 6, o 8. Fo a = 4, we could have c = 6 o 8, and fo a = 6 the only possibility is c = 8. Thus thee ae = 10 possibilities when a is even. Similaly, thee ae 10 possibilities when a is odd, so the numbe of possibilities is Answe (D): The last s is the 12th appeaance of this lette in the message, so it will be eplaced by the lette that is = 1 (12 13) = lettes to the ight of s. Since the alphabet has 26 lettes, this lette s is coded as s. 10. Answe (A): If the altitude fom A has length d, then ABC has aea (1/2)(BC)d. The aea is 1 if and only if d = 2/(BC). Thus S consists of the two lines that ae paallel to line BC and ae 2/(BC) units fom it, as shown. B C d d

4 Solutions th AMC 10 B Answe (C): Let BD be an altitude of the isosceles ABC, and let O denote the cente of the cicle with adius that passes though A, B, and C, as shown. B 3 O A 1 D C Then BD = = 2 2 and OD = 2 2. Since ADO is a ight tiangle, we have ( 2 = ) 2 2 = , and = = As a consequence, the cicle has aea ( ) π = π. 12. Answe (D): Tom s age N yeas ago was T N. The sum of his thee childen s ages at that time was T 3N. Theefoe T N = 2(T 3N), so 5N = T and T/N = 5. The conditions of the poblem can be met, fo example, if Tom s age is 30 and the ages of his childen ae 9, 10, and 11. In that case T = 30 and N = Answe (D): The two cicles intesect at (0, 0) and (2, 2), as shown. y (2, 2) x

5 Solutions th AMC 10 B 5 Half of the egion descibed is fomed by emoving an isosceles ight tiangle of leg length 2 fom a quate of one of the cicles. Because the quate-cicle has aea (1/4)π(2) 2 = π and the tiangle has aea (1/2)(2) 2 = 2, the aea of the egion is 2(π 2). 14. Answe (C): Let g be the numbe of gils and b the numbe of boys initially in the goup. Then g = 0.4(g + b). Afte two gils leave and two boys aive, the size of the entie goup is unchanged, so g 2 = 0.3(g + b). The solution of the system of equations g = 0.4(g + b) and g 2 = 0.3(g + b) is g = 8 and b = 12, so thee wee initially 8 gils. OR Afte two gils leave and two boys aive, the size of the goup is unchanged. So the two gils who left epesent 40% 30% = 10% of the goup. Thus the size of the goup is 20, and the oiginal numbe of gils was 40% of 20, o Answe (D): Let x be the degee measue of A. Then the degee measues of angles B, C, and D ae x/2, x/3, and x/4, espectively. The degee measues of the fou angles have a sum of 360, so Thus x = (12 360)/25 = = x + x 2 + x 3 + x 4 = 25x Answe (C): Let N be the numbe of students in the class. Then thee ae 0.1N junios and 0.9N senios. Let s be the scoe of each junio. The scoes totaled 84N = 83(0.9N) + s(0.1n), so s = 84N 83(0.9N) 0.1N = 93. Note: In this poblem, we could assume that the class has one junio and nine senios. Then s = = and s = 9(84 83) + 84 = Answe (D): Let the side length of ABC be s. Then the aeas of AP B, BP C, and CP A ae, espectively, s/2, s, and 3s/2. The aea of ABC is the sum of these, which is 3s. The aea of ABC may also be expessed as ( 3/4)s 2, so 3s = ( 3/4)s 2. The unique positive solution fo s is 4 3.

6 Solutions th AMC 10 B Answe (B): Constuct the squae ABCD by connecting the centes of the lage cicles, as shown, and conside the isosceles ight BAD. A D 2 B C Since AB = AD = 2 and BD = 2 + 2, we have 2(2) 2 = (2 + 2) 2. So = 2 2, and = 0. Applying the quadatic fomula gives = Answe (C): The fist emainde is even with pobability 2/6 = 1/3 and odd with pobability 2/3. The second emainde is even with pobability 3/6 = 1/2 and odd with pobability 1/2. The shaded squaes ae those that indicate that both emaindes ae odd o both ae even. Hence the squae is shaded with pobability = Answe (C): Afte one of the 25 blocks is chosen, 16 of the emaining blocks do not shae its ow o column. Afte the second block is chosen, 9 of the emaining blocks do not shae a ow o column with eithe of the fist two. Because the thee blocks can be chosen in any ode, the numbe of diffeent combinations is = = ! 21. Answe (B): Let s be the side length of the squae, and let h be the length of the altitude of ABC fom B. Because ABC and W BZ ae simila, it follows that h s s = h AC = h 5h, so s = h. Because h = 3 4/5 = 12/5, the side length of the squae is s = 5(12/5) /5 = OR

7 Solutions th AMC 10 B 7 Because W BZ is simila to ABC, we have BZ = 4 5 s and CZ = s. Because ZY C is simila to ABC, we have s 4 (4/5)s = 3 5. Thus 5s = s and s = Answe (B): The pobability of the numbe appeaing 0, 1, and 2 times is P (0) = = 9 16, P (1) = = 6 16, and P (2) = = 1 16, espectively. So the expected etun, in dollas, to the playe is P (0) ( 1) + P (1) (1) + P (2) (2) = = Answe (E): Let h be the altitude of the oiginal pyamid. Then the altitude of the smalle pyamid is h 2. Because the two pyamids ae simila, the atio of thei altitudes is the squae oot of the atio of thei suface aeas. Thus h/(h 2) = 2, so h = 2 2 = Answe (C): Since n is divisible by 9, the sum of the digits of n must be a multiple of 9. At least one digit of n is 4, so at least nine digits must be 4, and at least one digit must be 9. Fo n to be divisible by 4, the last two digits of n must each be 4. These conditions ae satisfied by seveal ten-digit numbes, of which the smallest is 4,444,444,944.

8 Solutions th AMC 10 B Answe (A): Let u = a/b. Then the poblem is equivalent to finding all positive ational numbes u such that u u = k fo some intege k. This equation is equivalent to 9u 2 9uk + 14 = 0, whose solutions ae u = 9k ± 81k = k 18 2 ± 6 1 9k2 56. Hence u is ational if and only if 9k 2 56 is ational, which is tue if and only if 9k 2 56 is a pefect squae. Suppose that 9k 2 56 = s 2 fo some positive intege s. Then (3k s)(3k + s) = 56. The only factos of 56 ae 1, 2, 4, 7, 8, 14, 28, and 56, so (3k s, 3k + s) is one of the odeed pais (1, 56), (2, 28), (4, 14), o (7, 8). The cases (1, 56) and (7, 8) yield no intege solutions. The cases (2, 28) and (4, 14) yield k = 5 and k = 3, espectively. If k = 5, then u = 1/3 o u = 14/3. If k = 3, then u = 2/3 o u = 7/3. Theefoe thee ae fou pais (a, b) that satisfy the given conditions, namely (1, 3), (2, 3), (7, 3), and (14, 3). OR Rewite the equation a b + 14b 9a = k in two diffeent foms. Fist, multiply both sides by b and subtact a to obtain 14b 2 9a = bk a. Because a, b, and k ae integes, 14b 2 must be a multiple of a, and because a and b have no common factos geate than 1, it follows that 14 is divisible by a. Next, multiply both sides of the oiginal equation by 9a and subtact 14b to obtain 9a 2 = 9ak 14b. b This shows that 9a 2 is a multiple of b, so 9 must be divisible by b. Thus if (a, b) is a solution, then b = 1, 3, o 9, and a = 1, 2, 7, o 14. This gives a total of twelve possible solutions (a, b), each of which can be checked quickly. The only such pais fo which a b + 14b 9a is an intege ae when (a, b) is (1, 3), (2, 3), (7, 3), o (14, 3).

9 The Ameican Mathematics Competitions ae Sponsoed by The Mathematical Association of Ameica The Akamai Foundation Contibutos Ameican Mathematical Association of Two Yea Colleges Ameican Mathematical Society Ameican Society of Pension Actuaies Ameican Statistical Association At of Poblem Solving Awesome Math Canada/USA Mathcamp Canada/USA Mathpath Casualty Actuaial Society Clay Mathematics Institute Institute fo Opeations Reseach and the Management Sciences L. G. Balfou Company Mu Alpha Theta National Assessment & Testing National Council of Teaches of Mathematics Pedagoguey Softwae Inc. Pi Mu Epsilon Society of Actuaies U.S.A. Math Talent Seach W. H. Feeman and Company Wolfam Reseach Inc.