2012AMC12B Solutions

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2012 AMClzB Solutions 21. Answer (C): There are 18-2 : 16 more students than rabbits per classroori.Altogether there are 4.16:64 more students than rabbits.2. Answer (E): The width of the rectangle is the diameter of the circlc, sr: thr:width is 2.5 : 10. The length of the rectangle is 2. 10 : 20. Therefore thc a,roaof the rectangle is 10 .20 :200.3. Answer (D): Let h be the number of holes dug by the chipmunk. Thcrr tlxrchipmunk hid 3h acorns, while the squirrel hid 4(h - 4) acorns. Sirrce thcy hi


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2012 AMCL2B Solutions 412. Answer (E): By symmetry, half of all such sequences end in zero. Of thosl,exactly one consists entirely of zeros. Each of the others contains a single sub-

sequence of one or more consecutive ones beginning at position j and ending atposition k with I < j < k < L9. Thus the number of sequences that meet therequirements is

2012 AMCL2B SolutionsAnswer (C): Each sector forms a cone with slant height 12. The circumferenceof the base of the smaller cone ir *?3 . 2 . 12 . r : 8r. Hence the radius of thebase of the smaller cone is 4 and its height is l@=E : 8rt. Similarly, thecircumference of the base of the larger cone is 162r. Hence the radius of the baseof the larger cone is 8 and its height is lt/i. fne ratio of the volume of thesmaller cone to the volume of larger cone is

*r . +2 .8t/2 _ {LaIr .82 . 4\fr 10

16. Answer (B): There are two cases to consider.Case 1Each song is liked by two of the girls. Then one of the three pairs of girls likesone of the six possible pairs of songs, one of the remaining pairs of girls likes oneof the remaining two songs, and the last pair of girls likes the last song. Thiscase can occur in 3 . 6 .2: 36 ways.Case 2Three songs are each liked by a different pair of girls, and the fourth song isliked by at most one girl. There are 4t : 24 ways to assign the songs to thesefour categories, and the last song can be liked by Amy, Beth, Jo, or no one.This case can occur in 24 . 4: 96 ways.The total number of possibilities is 96 -l 36 : 132.Answer (C): Let A: (3,0), B: (5,0), C : (7,0), D: (13,0), and dbe the acute angle formed by the line PQ and the r-axis. Then S.R : PQ :ABcos9 :2cos0, and SP : QR: CDsin9 : 6sind. Because PQRS isa square, it follows that 2cosd:6sind and tand: {. Therefore lines SPand RQ have slope 3, and lines S.R and PQ have slope -$. l,et the pointsM : (4,0) and 1nr : (10,0) be the respective midpoints of segments AB ar,dCD. Let 11 be the line through M parallel to line SP. Let 12 be the linethrough l/ parallel to line SrB. Lines 1.1 and /2 intersect at the center of thesquare PQRS. Line 14 satisfies the equation U:3(r -4), and Iine 12 satisflesthe equation ,: -*(" - 10). Thus the lines h and /2 intersect at the point(4.6, 1.8), and the required sum of coordinates is 6.4.

/1ek2 (,*tt\ k-r j:r ,) : 2(t+(1 +2+3+ +1e)) -2(t*q4)-382.'/ \ 2 /

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ORLet A be the set of zero-one sequences of length 20 where all the zeros appeartogether, and let B be the equivalent set of sequences where all the ones appea,rtogether. Set A contains one sequence with no zeros and 20 sequences withexactly one zero. Each sequence of A with more than one zero has a positionwhere the flrst zero appears and a position where the last zero appears, so there*" ('r') : 190 such.sequences, and thus lAl: t+20+190 : 211. By symmetrylBl : Ztt. A sequence in A o B begins with zero and contains from 1 to 20zeros, or it begins with one and contains from l- to 20 ones; thus 1,4 n Bl: 49.Therefore the required number of sequences equals

lAu Bl : lAl + lBl - lAn Bl : 2Lr +211 - 40 : 382.Answer (D): The parabolas have no points in common if and only if theequation xz + ar + b : 12 + cr + d has no solution. This is true if and only if thelines with equations A : ar*b and g : cr*d are parallel, which happens if andonly if a : c and b I d. Theprobability that a : c is I and the probability thatb I d is !, so the probability that the two parabolas have a point in common iSr 1 5" 31^ 6 6 36'Answer (A): The smallest initial number for which Bernardo wins afterone round is the smallest integer solution of 2n 150 > 1000, which is 475. Thesmallest initial number for which he wins after two rounds is the smallest integersolution of 2n * 50 > 475, which is 213. Similarly, the smallest initial numbersfor which he wins after three and four rounds are 82 and 16, respectively. Thereis no initial number for which Bernardo wins after more than four rounds. ThusN : 16, and the sum of the digits of l[ is 7.

t7.


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2012 AMCL2B Solutions

18. Answer (B): If a1 :1, then the list must be an increasing sequence. Other-wise let k: ar. Then the numbers 1 through k - 1 must appear in increasingorder from right to left, and the numbers from k through 10 must appear in in-creasing order from left to right. For 2 S k < 10 there are (o1r) *uy. to choosepositions in the list for the numbers from 1 through k - 1, and the positions ofthe remaining numbers are then determined. The number of lists is therefore- 2e - br2.

19. Answer (A): Let s be the length of the octahedron's side, and let Q6 and Q|be the vertices of the octahedron on PiPt and ffi, rcspectively. If Q2 alnd. Q71were opposite vertices of the octahedron, then the midpoint M of Qfi woddbe the center of the octahedron. Because M lies on the plane P1P2P1, the vertexof the octahedron opposite Qa would be outside the cube. Therefore Qz, Qs,and Qa are all adjacent vertices of the octahedron, and by symmetry so are Q'2,Q!, and Q'n. For 2 < i < j < 4, the Pythagorean Theorem applied to LPyQaQigives ,, : (eoe)2 : (he)2 + (pre)z.It follows that P1Q2: PtQs: hQt: *t, and by symmetry, P\QL: P\AS:P\QL: f,s. Co.rsequently Qa and. Ql are opposite vertices of the octahcdron.The Pythagorean Theorem on AQ2P2P! and LQtsPlQ2 gives

Pi,,t

1+E(-1,) :*(;)

(QrP!), : (PrPl), + (QrPz)2 - 1 +s2:(QrA)2-ercL)'

(,- *,)+ (Q,Pt)2 - ('- +,) '*

2and

1+ (r-\ *,) 'Solving for s gives s - M4

Ax2 : s2 - (20 * u)' + (20\fr)2

.t.

2012 AMCL2B Solutions

20. Answer (D): Let ABCD be a trapezoid with AB lleD and. AB < CD. Let.E be the point on CD such that CE: AB. Then ABCE is a parallelogram.Set AB : a, BC :b, CD: c, and DA: d. Then the side lengths of LADE'are b, d,, and c - o. If one of b or d is equal to 11, say b : 11 by symmetry, thenrl* (c:- a) < 7 + (5 - 3) < 11 : d, which contradicts the triangle inequality.Thrn c : 11. There are three cases to consider, namely, o,:3, a: 5, and a : 7.If n : 3, tlrcn LADE has side lengths 5, 7, and 8 and by Heron's formula itsarea is 1

4 (b + T +8)(z+ 8 - b)(8 + b - 7)(5 +T* 8) - 10/5,Thearea of LAEC is $ of thearea of AADE, andtriangles ABC andAEChavc the same area. It follows that the area of the trapezoid is +(35/5).If. o,:5, then LADE has side lengths 3, 6, and 7, and area

- 4Jb.Tlte area of LAEC is f; of the area of AADE, and triangles ABC and AEChave the same area. It follows that the area of the trapezoid is ](:Z/5).If. o, : 7, then LADE has side lengths 3, 4, and 5. Hence this is a righttrapezoid with height 3 and base lengths 7 and 11. This trapezoid has area;(3(z + Lr)):27.Thc sum of the three possible areas is ff rt+ ff ,/ 5127. Hence rr : T, r, : t,t.s:27, nt :3, nz :5, and,111-r21-is1-ntinr: T+T+ZZ+S+B: 63+:.Thus the required integer is 63.

21. Answer (A): Extend EF to H and extend eB to..I so that -FlJ- containsA and is perpendicular to lines EF arrd CB. Let s be the side length of thesquare and let u: BX. Because IABJ:60o, it follows that BJ:20 andAJ :20t/5. Then by the Pythagorean Theorem

14 3)(7 + 6)(3 + 6


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2012 AMCL2B Solutions 8Because ABCDEF is equiangular, it follows that ED llTE and soW llTB."AlsoW ll 7X and thus it follows that IEYZ : IBAX and so LEYZ eABAX. Th,os EZ: u. AIso, IHZA:90" - IYZ.E:90o - IAXJ : IJAX;thus AAX"/ = LZAH and so ZH : 2Ot/5 and HA : 20 * u. Moreover,IHFA:60" andso FII : #: rt,QO+u). But EZ+ZH: EF+FH,and so

u + 2ot/5 : tl\/i- 1) + ']+/ZSolving for z yields u : 2tJ5 -20. Then s2 : Qt\h)2 + (20\/5)2 : 2.292 and,therefore s :29t/5.

22. Answer (E): Label the columns having arrows as cL,c2,cs,. ..tcz accordingto the figure. Call those segments that can be traveled only from left to rightforward segrnents. Call the segments sL, s2, and s3, in columns c2, c4, dfldc6, respectively, which can be traveled only from right to left, back segments.Denote ,S as the set of back segments traveled for a path.First suppose that ^9 : 0. Because it is not possible to travel a segment morethan once, it follows that the path is uniquely determined by choosing oneforward segment in each of the columns cy. There are 2,2, 4, 4, 4,2, and2 choices for the forward segment in columns ct, c2, cs, c4t cbt c6t and c7,respectively. This gives a total of 210 total paths in this case.

B-+

order in which the segments of c2 are traveled is determined. Moreover, thereare only 2 choices for possible segments in ca depending on the last segmenttraveled in c2, either the bottom 2 or the top 2. For the rest of the columns, thepath is determined by choosing any forward segment. Thus the total numberof paths in this case is 2 . l .2 . 4. 4.2.2 :28, and by symmetry this is alsothe total for the number of paths when .9 : {re}. A similar argument gives2 . 1, . 2 . 4 . 2 . | . 2 : 26 trips for the case when ^9 : {s1, s3}.

Suppose ,9 : {rr}. Because s2 is traveled, it follows that 2 forward segments inc4 need to belong to the path, one of them above s2 (2 choices) and the otherbelow it (2 choices). Once these are determined, there are 2 possible choices forthe order in which these segments are traveled: the bottom forward segmentflrst, then s2, then the top forward segment, or vice versa. Next, there areonly 2 possible forward segments that can be selected in ca and also only 2possible forward segments that can be selected in c5. The forward segments incr, c2, c6) and c7 can be freely selected (2 choices each). This gives a total of(2' .2 . 2) . 24 :2e parhs.If S: {sr,sz}, then the analysis is similar, except for the last step, where theforward segments of c1 and c2 are determined by the previous choices. Thusthere are 1Zt .2 . 2) . 22 :27 possibilities, and by symmetry the same numberwhenS:{sz,ss}.Finally, if S : {sr, s21s3}1then in the last step, all forward segments of c1, c2, c6,and c7 zlc determined by the previous choices and hence there are 23 . 2 .2 - 2possible paths. Altogether the total number of paths is 2'0 + 2 . 28 + 26 + 2e2 .27 + 25 - 2400.Answer (E): First note that P(1) : 4* atb-f c+ d> 4,so zs : 1 is not azeroof the polynomial. On the other hand, P(-1) : (a-")+(b-c)td and byassumption 4) a,b ) c, and d > 0. Thus if P(-1) :0, then a:4,b: c, andd:0. Conversely, if a:4,b: c, and d:0, then P(r):4za+4zs tbz2+bz:z(z+t)(422 +b) satisfles the required conditions. The only remaining possibilityis that zs is not real. In this case,

425 - (z - t)P(z) : z4(4 - a) + z3(a - q + zz(b - c) + z(c - d,) + d,.

r.2012 AMCL2B Solutions

23.

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Next suppose that S - {rr}. The two forward segments in c2, together withneed to be part of the path, and once the forward segment froffr ct is chosen,

51rthe

IIII

I IItts

I IIIIt

It i


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2012 AMCLZB SolutionsIf z: zo is a zero of P(z) with lzsl:1, then the triangle inequality yields

+ : lz$(+ - a) + z3@ - b) + zfi(b- c) + zg(c - d,) + d,l1 lroln @ - o) + lrolu (" - b) + lrol' (b - c) + lzsl (c - d) + d: 4 - a -l a - b*b - cl c- dt d : 4.Thus equality must occur throughout. Therefore z$(4 - a) is a positive scalarmultiple ot z$(a - b), which implies that zg(4 - a) is real. Because z0 d lR,it follows that a : 4. Similarly, "3@ - b) : z$(+ - b) is a positive scalarmultiple of zfi(b - c) and thus b -- 4. In the same way, c: d,:4. Conver:scly,the polynomial P(z) : 4za + 423 + 422 -f 4z * 4 satisfies the given conrlitionsbecause zo : cos(2r 15) -f i,sin(%r l5) is a zero of this polynomial. Therofbrc thecomplete list of polynomials is 4za + 423 + 422 + 4z + 4 and 4za + 423 * bz2 I bz for0 < b < 4. The required sum is 20+tb4:0(8+ 2b) : 2g1ag+ (2+4+6+8) : 80.Answer (D): Let ,S1-, : (/r(N), fz(N),"fs(If),...). If I[r


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!.2012 AMC12B Solutions 12of the line OY has the form f with I I r 14, so if the right angle were ai Y,then the vertex Z wot\d. need to be at least 5 horizontal units away from Y,which is impossible. Therefore the right angle is at Z. There are 4 such triangleswith Z on the r-axis, with vertices O,Z: (r,0), and Y: (r,5) for 1 ( r 14.There are two more triangles: with verticesO, Z: (3,3), and Y: (1,5), andwith vertices O, Z : (4,4), and Y : (3,5). The product of the values /(l) overthese six triangles is equal to

72948\/r4\/, 144B 5 E's rrt' ,n:ax'Thus the required product equals

The problems and solutions in this contest were proposed by Bernardo Abrego, BetsyBennett, Barb Currier, Steve Davis, Zuming Feng, Silvia Ferniindez, Peter Gilchrist,Jerrold Grossman, Dan Kennedy, Joe Kennedy, David Wells, LeRoy Wenstrom.

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Solutions PamphletAmerican Mathematics Competitions ;Nn

63'd AnnualAMC 12HAmerican Mathematics Contest 12 BWednesday, Februa ry 22, 2012

This Pamphlet gives at least one solution for each problem on this yeart contest and showsthat all problems can be solved without the use of a calculator. '\Vhen more than one solutionis provided, this is done to illustrate a significant contrast in methods, e.g., algebraic usgeometric, computational as conceptual, elementary us advancr.d. These solutions are by nomeans the only ones possible, nor are they superior to others the reader may devise.'W'e hope that teachers will inform their students about these solutions, both as illustrationsof the kinds of ingenuity needed to solve nonroudne problems and as examples of goodmathematical exp osition- Howeaer, the publication, reproduction or communication ofthe probbmsor solutions of the AMC 12 during the period when studentt are eligiblz to participate seriouslyjeopardizes the integrity of the resuhs. Dissemination ain copier, telephone, e-mail, W'orld\Yid.e \Yebor media of any type during this period is a uiolation of the competition rules.Afier the contest peiod, permission to mahe copies of problems in paper or elcctonic form including po*ing onueb-pages for educational usc is granted without fee proaidzd that copies drc not madz or d*tibuted for proft orcommercial aduantage and that copies bear the copyight notice.Conespondtnce about the probhmslsolutiins for this AMC 12 and ordtrs for ary publications should be addresed to:

American Mathematics CompetitionsUniversiry of Nebraska, PO. Box 81606, Lincoln, NE 68501-1606Phone: 402-47 2-2257 ; Falir: 402-47 2-6087; email: [email protected] problzms and soluions for this AMC 12 were prEared by the MAAIs Committee on theAMC 10 and AMC 12 ander the direction ofAMC 12 Subcomrnittee Chair:

Profl Bernardo M. Abrego@ 2012 Mathematical Association of America

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2012AMC12B Solutions