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72013AIMEI Solutions 21. (Answer: 150)Letr betheratethatTombicyclesinminutespermile .ThenTomrunsat therateof2r andswimsat therateoflOr. Becausefouranda quarterhoursis255minutes,

1- . lOr +30r +8.2r = 255.2 .This simplifiest o51r= 255, so r =5. Tom bicycles at the rateof5minutespermile,so thenumberofminuteshespendsbicyclingis30 .5= 150.

2. (Answer: 200)f anumbersatisfiesthegiven conditions, thenbecause itsfirstand lastdigitsareequal,andthenumberis amultipleof5,itsfirstand lastdigitsmust be5.Thesumofthemiddlethreedigitsmustalsobeamultipleof5. f onechoosesanytwo valuesfor thefirsttwo ofthose threedigits , therewillbe two possiblechoicesforthethirddigit thatwill makethesumequaltoamultipleof5. Thusthereare10 102=200choicesforsuchanumber .

3. (Answer: 018)D, ,e

'" Y IF

A E B

Letthesidesofthetwosmallersquareshave lengthsx andy so that thesquareABCD hasside lengthx + y. It is given that x 2 + y2 = i'o(x + y 2. Then1O(x2 + y2 =9(x2 +y2 +18xy , andx 2 + y2 = 18xy. The requested sum is;: + 11 = +I ' =18.I! ' "'II

4. (Answer: 429)Thecolorsof the 13 cellsin thearraycanbechosen in =1287ways. Foranarrayto satisfy the givensymmetry condition, the centersquare mustbeblue,andeachofthethree-square"L" shapesmustcontain2redsquaresand1

2013AIMEI Solutions14. (Answer: 036)

NoticethatQ+iP = 1+ (- sinB+ i cos0) + cos2B-i sin20 + (s in30-icos38)+ ..

= 1+ + e e i 9 + i 3 e 3 i 9 + .2 4 82 2(2 + sin8+ icos8)ie i92- 5+ 4sinB

ThusP = o ~ 9 andQ =42s!n9. So~ n 9 5 4 ~ n 92V2 P cos8 - ,7 Q 2+sin8

32+ 32sinO+ 8sin2 8= 49 - 49si n2 0,57sin2B+32sinB- 17= (3sinO- 1)(19sinO+17)= 0,

and 17 1inu = or 19 3 '

The requested sumfor thenegativesolutionis 17+19 = 36.15. (Answer: 272)

Let d = a - b = c - a. Then thereareintegersm and n so that np _ d =B - A =C - B =mp 2d, implying that3d = (n - m)p. Notethat2d < p, so3cannotdividen - m. It follows thatp = 3,and (b, a, c) = (0,1,2).Thereforethe ordered triples thatmeet the conditions areprecisely those of the form1 +3j ,3+3j+3k ,5+3j+6k , wherej20, k 2 0, and j 2k 31. Thus

15 15.16N = ~ ) 3 2 - 2k = 1632- 2 = 272.k=O

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2 ( J J ~ A AlC 12B S ( ) l l j o n ~ s.... "/ i ''- - --"() ..-- ~ ~ - ' .( , 1(. .[ r \ .... i \\, \ f J\. IJ d \' ..I,

hilS there are C ~ O ) pflir:-; of panlhoJas and tiJ(' pa i r:; tlia t do nut intersecl arcexactly Ihose \ \ I t ( ) ~ ( Ji rccl riees have t he saJ1lc s lop(" aud wl lo::,e }-iull'l'Ccpb ] Iave111(' same sign . There ,WI' G ilillert'llt slopes ane! 2 G) = G pairs of .1J-u lkrcl'ptswitlr l ir e :;W' I jeltl,,!'. P(:.) ",,11Ill' fa!'! ol'('d illto t Ire product uf pai rwise dUfon'lll lilll',lI ' p()IYlttlllJi;d:-; 01' t ltv fill'1ll( ;; - I ') wit Ir ( E Z /Juel (]lI

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2(J) J l lM C J2B Sollltions {J/\s a quadratic ccplalion i l t log .r. Lh(' --11111 oi' tite two :;ulu l iollS log .!1 Hlld log .":,is equal 10 1I1(, l]('galiy(' of thl' lil1C'nJ ('odTicic'lIl. I t follow:; I l l ' ll

1 (( - I;) 11K)log(.I'I.l':.d + log .1'1 = 8(71og I I I + Glog 1/) = l o ~ /11'11 .Lc,t N = I'I.I":.! b!' the product of tlt(' sO l lltiullS. Suppos(' p i::; 11 pr ime (lividingm, Lpl p il.nd Ii ' b (' Il l(' largest PO\\'C'I"S of Ii I IIHI di \ ;ck II/ n.lld II r('sIH'C'1 iV0ly.Th en ]1,,,"'61 is Lll l ' largesl pow('r of /, tha i di " idf's II I ' /I I; = Xli. I fo llows I hal( II+Gb == 0 (mod 8) . IJ II is odd. t 1 (11 I hcJ'f' is no solu l iun to 7o + 0/; == f) (mod x)h r(,H lISl' 7(1 is not diyi;;ihlc' by gc cl (6.8) = 2. If 0 == () ( Illod 1.-:). 1111'11 11('('n 11;;((/ > (J . il follows t hai - /11' 1/ (; ::::: (l )T jlSO ? i ' iG SO X 2: '27 = 128 . If(J == :2 (1110

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7013 .1 C12 B Soiutiw is ( i?rOl ll tlw eq llat ions , I + b = 2 - c am J (12 + 1> = 12 - (.2. Let:r b e an ar b itraryrl',u uum uer. t bcn ( : r - I J . ) 2 + ( a - I ) ~ 2: 0; th a t. is, 2. ,.2_2 u +b).r + u 2 +b 2 ) 2: O.T i l l iS

23.2 - 2(2 - e)l + (12 - c2) 2: ()ror tI ll nml vI-d ues .1 . TIUlt IU('flm; t.he- discr im imlI 1 ~ ( 2 - (-)2 - .1 2( 12 - (.2) S'. O.Simplifying and factoring givcb (3c - }U)(c+2) :s O So the range of values of c i::;-2 < e -$ B oth maximum and [ll in imum are ultn inablc by letti ng (a, b, c) =(2 . 2.-2) tllH l ((I .b,c) = ( - ~ . l J : ) . Therefo re I]w differellce between tlterumdlllulll awl mjn.imulll po::>siblc value::> of r: is - (- 2) = -W .

18. Answer (B):Lf 111(' g the 1.

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2 ) J:3 ltMCJ2B Solu tiolJl'> 1After Lhis II I omen!. h('(' A will tr ave l Ntsl 10 the poinl (3 ,3 . 2) Hl ld hee B willtl 'av('1 w('st 10111(' point ( - 1. - -1.0) . Tl l( ir ri ist.Hn('( ai'tpr travdi ug am' rool wi llhe

)(3 - ( _ 4))2 + (:3 - ( - I)j2 = v'102 > I D.!l enc c ))('(' A is I rave lin g eH ,)l Ilnd b('(' B is IT1Jvd ing Wl'st whe n t hey are exacl ly10 /('('1 Hway fro m ene l! ot her .

12. Answer (D ): Ci t ies C RIllI E Hn d t ltP. roa ds lead ing in and o ll l of t hem ca n 1('rcplaced by a sC'cond A D ran d find a SCCUI l J B D road, respcc ti vely. If' r()IllpsRl't' cI (os ignat ed by Ill ' lisl or due s t hey visil in order, the ll there nre I I y p ( ~or rOll tes : AB D ! I DF3 . AD.IlBDB . ADBADB, and A D BDAB_ Eael! type ofrout e repn'scnls -I nct ual 1"011 1('s_becansc the I rip hl'l Wf'f' lI !1 and D calt iI II'i 11(1'Il lc dcl.otll' tltwnp; iJ E ei t\lI't' t he first or the "C'colld t imc _ and a m i l t l choicer\pp lies 101' tIl e t rip h ('twcen B a u d D . Tlw rC'fore th ere arc ..) 4 = JG d iffcH'liIroll les .

1:3. Answer (D): Lei the dCgT(,1' 1nC'llsur cs o f 1h(' Bllgle" he ;l S o \ \ ill 1/11'ri g1Ire . T ll (, ang l('s or a I t'i< lllgic: fur lll nJl ari ll ull('1ic progressio ll i f Hn el ollly irt lie media n ang le- is (iO o. so o ll e of :1 . .'/. or ISO - .,. - J 1I1ll ' l t be equal 10 GO .13)' " ), 111111 ('1r\' of t hc' ro l(' o r I h(' Ir iang les A B D and DC B . aSHUllU' tltHt. . . ::; .1/.13e('Htlsc .r ::; .II < 180 -- .1' and :,; < 180 - II ::; 180 - J:. i L follo\\'s that thf'nrit b11lNic r o g r e ~ of t he allgl f's in AB eD frolll smalle::.t to larg(>sl 111 111 1)('c: itl lPr .1: . !/ _1 80 .IJ_180 - . / ,o r.J', 18 ll - Y 1I, l80 - .J' . T huseither ;c + 180 - y = '2 J.in whic" case 3.11 = :r + ]80 : or .r + 11 = 2(180 - !I ), in whicli r a ~ t 3" = 360 - .r.Ncit\wr of t hese is compatible -with /I GO (11 11 ronnel' rOt'(TR I = n ;lIld 111('laLter forcC's :1' = ISO), so ('illH'l' ;r = 60 or .r + I = l20.

n/>'C ' 0" .1

F irst sup pose thaI :r = 60 . If 3 J = .1+ 180 . thel1 IJ = 80. and the scqUC'lJ('(' ofanglps ill ABeD is ( /" .1} 180 - .1}- H:;O - .1 = (GO. 80.100 . 120). IT 3,1} = :J(iO :)'.Ll ICll ,lj = j 00. and i Ite S(,C]l1PDCe of a ngles in ABCD is (.1'. 180 - .fl . 11_ U)l) - :r) =((iO. SO. l OU . 120). Fi lHtl] \'. suppose that.J' 1I - 120. 1I :3 J = -r-! 180, th en II = 75.Rnd s ( q l l c l l ( e o i l l ABCDis (.1' . /) . 180- } . 180 .7') = ( Ifi, 75. 10:>_ 11;) .I f 3!J = alio - .r. Il w ll IJ = 120 an d .1. = 0_ which is p o s ~ i b l ( .The reforC', th e Hllm in d c g n e ~ of lite t \\'o lm-ge.'or possible I.lngles is 10;>+ 1 :3,. =24 0.

U. Answer (C): L( 'I' the two Sf'quences b e (a ,,) Hnd (b,,), fut'(, li t(' t'('qltin'dperi1tld (' r equ ais

.c\'C' + A U f H D 13'R -t (" II t (";;; + 1) ;\ -+ D'I3 + ' f3 1 ( '2('(CD + DE + E A + ..IB + fW ) = 2,'.

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:l{) i ;1 A M e 12D SOil ltiollS ()Fro m IL(' equations, l + h = 2 - (. il1l d u,l + b2 = 12 - r2 . Lpi .1: 1> an ar bitraryrpal number. IhpTI (.r - af + (.1 _ /i 2 2. 0; i hat is . 2.,.1 2(0 +/i ).r T ( ( I ~ + Il 2: O.Th lls

?2,.- - 2(2 - (-)'1+ 12 - (.-) 2.. 0for all rea l l J ( s .r . T hat l l i e ~ w . , > 11lC' discriminan t 4(2 - ( )'.1 - -1 2 12 - (,2) {J.Sim pu lyi ng ilnd c t o r i n g givf' -i (3(:- ]O)(r'+ 2) OSo til e J"illlgC or value;; of t is-2 (: :;; Jf . Bol Ii I l lHx i l l ll l l l1 an d minimulll arc attainaulc by let t ing (0 . /i, r: =(2 , 2. -2) a ile (J ./J,c) = 1 j d ) T he refore th e d ifferllC( , i>IW(,f'1l 1ilpmax i mum cwcl minimum p ossi ble val ues of c is - (- 2) =

18. Answer (B ):II" till' gctlll(" stiuts wit h 2013 coills Hw l .kllnH starts. 1\1 J l \ ( ; { ' ~ til(' ga llw 10 Il l(' pr r \'iollli rase of 201:3 ("OUL" wllcn' s lJ(' wins. U13m-J-mra ,.,t ar1 s. she s