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

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1 The MATHEMATICAL ASSOCIATION OF AMERICA Americn Mthemtics Competitions Presented by The Akmi Foundtion 53 rd Annul Americn Mthemtics Contest AMC - Contest B Solutions Pmphlet WEDNESDAY, FEBRUARY 7, 00 This Pmphlet gives t lest one solution for ech problem on this yer s contest nd shows tht ll problems cn be solved without the use of clcultor. When more thn one solution is provided, this is done to illustrte significnt contrst in methods, e.g., lgebric vs geometric, computtionl vs conceptul, elementry vs dvnced. These solutions re by no mens the only ones possible, nor re they superior to others the reder my devise. We hope tht techers will inform their students bout these solutions, both s illustrtions of the kinds of ingenuity needed to solve nonroutine problems nd s exmples of good mthemticl exposition. However, the publiction, reproduction, or communiction of the problems or solutions of the AMC during the period when students re eligible to prticipte seriously jeoprdizes the integrity of the results. Dupliction t ny time vi copier, phone, emil, the Web or medi of ny type is violtion of the copyright lw. Correspondence bout the problems nd solutions should be ddressed to: Prof. Dvid Wells, Deprtment of Mthemtics Penn Stte University, New Kensington, PA 5068 Orders for prior yer Exm questions nd Solutions Pmphlets should be ddressed to: Titu Andreescu, AMC Director Americn Mthemtics Competitions University of Nebrsk-Lincoln, P.O. Box 8606 Lincoln, NE Copyright 00, Committee on the Americn Mthemtics Competitions The Mthemticl Assocition of Americ

2 . (A) The number M is equl to Solutions rd AMC B ( ,999,999) = ,, = 3,456, The number M does not contin the digit 0.. (D) Since (3x )(4x + ) (3x )4x + = (3x )(4x + 4x) + when x = 4 we hve the vlue 3 4 =. 3. (B) If n 4, then n 3n + = (n )(n ) = (3x ) + = 3x, is the product of two integers greter thn, nd thus is not prime. For n =,, nd 3 we hve, respectively, ( )( ) = 0, ( )( ) = 0, nd (3 )(3 ) =. Therefore, n 3n + is prime only when n = (E) The number n is greter thn 0 nd less thn <. Hence, n = n is n integer precisely when it is equl to. This implies tht n = 4, so the nswer is (E). 5. (D) A pentgon cn be prtitioned into three tringles, so the sum of the degree mesures of the ngles of the pentgon is v + w + x + y + z = 540. The rithmetic sequence cn be expressed s x d, x d, x, x+d, nd x+d, where d is the common difference, so Thus, x = 08. (x d) + (x d) + x + (x + d) + (x + d) = 5x = 540.

3 Solutions rd AMC B 3 6. (C) The given conditions imply tht so x + x + b = (x )(x b) = x ( + b)x + b, + b = nd b = b. Since b 0, the second eqution implies tht =. The first eqution gives b =, so (, b) = (, ). 7. (B) Let n, n, nd n + denote the three integers. Then (n )n(n + ) = 8(3n). Since n 0, we hve n = 4. It follows tht n = 5 nd n = 5. Thus, (n ) + n + (n + ) = = (D) Since July hs 3 dys, Mondy must be one of the lst three dys of July. Therefore, Thursdy must be one of the first three dys of August, which lso hs 3 dys. So Thursdy must occur five times in August. 9. (C) We hve b = + r, c = + r, nd d = + 3r, where r is positive rel number. Also, b = d yields ( + r) = ( + 3r), or r = r. It follows tht r = nd d = + 3 = 4. Hence d = (A) Ech number in the given set is one more thn multiple of 3. Therefore the sum of ny three such numbers is itself multiple of 3. It is esily checked tht every multiple of 3 from = through = 38 is obtinble. There re 3 multiples of 3 between nd 48 inclusive.. (E) The numbers A B nd A + B re both odd or both even. However, they re lso both prime, so they must both be odd. Therefore, one of A nd B is odd nd the other even. Becuse A is prime between A B nd A + B, A must be the odd prime. Therefore, B =, the only even prime. So A, A, nd A + re consecutive odd primes nd thus must be 3, 5, nd 7. The sum of the four primes, 3, 5, nd 7 is the prime number 7.

4 Solutions rd AMC B 4. (D) If n 0 n = k, for some k 0, then n = 0k k +. Since k nd k + hve no common fctors nd n is n integer, k + must be fctor of 0. This occurs only when k = 0,,, or 3. The corresponding vlues of n re 0, 0, 6, nd (B) Let n, n +,...,n + 7 be the 8 consecutive integers. Then the sum is 8n + ( + + 7) = 8n = 9(n + 7). Since 9 is perfect squre, n + 7 must lso be perfect squre. The smllest vlue of n for which this occurs is n = 4, so 9(n + 7) = 9 5 = (D) Ech pir of circles hs t most two intersection points. There re ( 4 ) = 6 pirs of circles, so there re t most 6 = points of intersection. The following configurtion shows tht points of intersection re indeed possible: 5. (D) Let denote the leftmost digit of N nd let x denote the three-digit number obtined by removing. Then N = 000+x = 9x nd it follows tht 000 = 8x. Dividing both sides by 8 yields 5 = x. All the vlues of in the rnge to 7 result in three-digit numbers. 6. (C) The product will be multiple of 3 if nd only if t lest one of the two rolls is 3 or 6. The probbility tht Jun rolls 3 or 6 is /8 = /4. The probbility tht Jun does not roll 3 or 6, but Aml does is (3/4)(/3) = /4. Thus, the probbility tht the product of the rolls is multiple of 3 is =.

5 Solutions rd AMC B 5 7. (B) Let A be the number of squre feet in Andy s lwn. Then A/ nd A/3 re the res of Beth s lwn nd Crlos lwn, respectively, in squre feet. Let R be the rte, in squre feet per minute, tht Crlos lwn mower cuts. Then Beth s mower nd Andy s mower cut t rtes of R nd 3R squre feet per minute, respectively. Thus, nd Andy tkes A minutes to mow his lwn, 3R Beth tkes A/ R = A 4R Crlos tkes A/3 R = A 3R Since A 4R < A, Beth will finish first. 3R minutes to mow hers, minutes to mow his. 8. (C) The re of the rectngulr region is. Hence the probbility tht P is closer to (0, 0) thn it is to (3, ) is hlf the re of the trpezoid bounded by the lines y =, the x- nd y- xes, nd the perpendiculr bisector of the segment joining (0, 0) nd (3, ). The perpendiculr bisector goes through the point (3/, /), which is the center of the squre whose vertices re (, 0), (, 0), (, ), nd Hence, the line cuts the squre into two qudrilterls of equl re /. Thus the re of the trpezoid is 3/ nd the probbility is 3/4. y (w, q) (3, ) 3 x 9. (D) Adding the given equtions gives (b+bc+c) = 484, so b+bc+c = 4. Subtrcting from this ech of the given equtions yields bc = 90, c = 80, nd b = 7. It follows tht b c = = 70. Since bc > 0, we hve bc = 70.

6 0. (B) Let OM = nd ON = b. Then Solutions rd AMC B 6 9 = () + b nd = + (b). X X 9 M O b N Y O b Y Hence It follows tht 5( + b ) = 9 + = 845. MN = + b = 69 = 3. Since XOY is similr to MON nd XO = MO, we hve XY = MN = 6. X M O b 3. (A) Since 00 = 3 4, we hve N 6, if n = 3 4 i, where i =,,...,0; 3, if n = 4 j, where j =,,..., ; n = 4, if n = 3 k, where k =,,..., 3; 0, otherwise. 00 Hence n = = 448. n=. (B) We hve n = log n 00 = log 00 n, so b c =(log 00 + log log log 00 5) b (log log 00 + log 00 + log log 00 4) =log = log = log =. Y

7 Solutions rd AMC B 7 3. (C) Let M be the midpoint of BC, let AM =, nd let θ = AMB. Then cos AMC = cosθ. Applying the Lw of Cosines to ABM nd to AMC yields, respectively, cosθ = nd cosθ = 4. Adding, we obtin 0 = 5, so = / nd BC = =. B M A C OR As bove, let M be the midpoint of BC nd AM =. Put rectngulr coordinte system in the plne of the tringle with the origin t M so tht A hs coordintes (0, ). If the coordintes of B re (x, y), then the point C hs coordintes ( x, y), y A(0, ) B(x,y) M x C(-x,-y) so x + ( y) = nd x + ( + y) = 4. Combining the lst two equtions gives ( x + y ) +8 = 5. But, x +y =, so 0 = 5. Thus, = / nd BC =. 4. (E) We hve Are (ABCD) AC BD,

8 Solutions rd AMC B 8 with equlity if nd only if AC BD. Since 00 = Are(ABCD) AC BD 5 77 (AP + PC) (BP + PD) = = 00, it follows tht the digonls AC nd BD re perpendiculr nd intersect t P. Thus, AB = = 40, BC = = 4 3, CD = = 53, nd DA = = 5. The perimeter of ABCD is therefore ( 3 = ) (E) Note tht nd f(x) + f(y) = x + 6x + y + 6y + = (x + 3) + (y + 3) 6 f(x) f(y) = x y + 6(x y) = (x y)(x + y + 6). The given conditions cn be written s (x + 3) + (y + 3) 6 nd (x y)(x + y + 6) 0. The first inequlity describes the region on nd inside the circle of rdius 4 with center ( 3, 3). The second inequlity cn be rewritten s (x y 0 nd x + y + 6 0) or (x y 0 nd x + y + 6 0). Ech of these inequlities describes hlf-plne bounded by line tht psses through ( 3, 3) nd hs slope or. Thus, the set R is the shded region in the following digrm, nd its re is hlf the re of the circle, which is 8π 5.3. y 6 x (3,3) 6

9 Solutions rd AMC B 9 This pge left intentionlly blnk.

10 The Americn Mthemtics Contest (AMC ) Sponsored by Mthemticl Assocition of Americ The Akmi Foundtion University of Nebrsk Lincoln Contributors Americn Mthemticl Assocition of Two Yer Colleges Americn Mthemticl Society Americn Society of Pension Acturies Americn Sttisticl Assocition Cnd/USA Mthpth & Mthcmp Csulty Acturil Society Cly Mthemtics Institute Consortium for Mthemtics nd its Applictions Institute for Opertions Reserch nd the Mngement Sciences Kpp Mu Epsilon Mu Alph Thet Ntionl Assocition of Mthemticins Ntionl Council of Techers of Mthemtics Pi Mu Epsilon School Science nd Mthemtics Assocition Society of Acturies