2003年第三届中国西部数学奥林匹克试题及解答

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2003(2003-09-2709-28,)1.1,2,3,4,5,6,7,8,10..2.2na1,a2,,a2n2n-1i=1(ai+1-ai)2=1.(an+1+an+2++a2n)-(a1+a2++an).3.n.un,:d,und1,3,5,,2n-1d.4.:ABCDPABBCCDDA,ABCD.5.{an}:a0=0,an+1=kan+(k2-1)a2n+1,n=0,1,2,,k.:{an},2k|a2n,n=0,1,2,.6.ABCD,ABBCCDDAA1B1C1D1,A1B1B1C1C1D1D1A1,EFGHA1B1B1C1C1D1D1A1.:EFGHABCD.7.x1x2x3x4x55i=111+xi=1.:5i=1xi4+x2i1.8.16502275.,11.:928.1.a1a2a3a4,a1a2a3a4.59,a46.,a1+a2+a3+a416.1116.2.n=1,(a2-a1)2=1,a2-a1=1.1.n2,x1=a1,xi+1=ai+1-ai,i=1,2,,2n-1.2ni=2x2i=1,ak=x1+x2++xk,k=1,2,,2n.(an+1+an+2++a2n)-(a1+a2++an)=n(x1+x2++xn)+nxn+1+(n-1)xn+2++x2n-[nx1+(n-1)x2++xn]=x2+2x3++(n-1)xn+nxn+1+(n-1)xn+2++x2n[12+22++(n-1)2+n2+(n-1)2++12]12(x22+x23++x22n)12=n2+216(n-1)n(2(n-1)+1)12=n(2n2+1)3.ak=a1+3k(k-1)2n(2n2+1),k=1,2,,n+1,an+k+1=a1+3[2n2-(n-k)(n-k-1)]2n(2n2+1),k=1,2,,n-1,.,(an+1+an+2++a2n)-(a1+a2++an)n(2n2+1)3.3.un=2n-1.(1)un2n-1.u11,n2.1,3,,2n-12n-11,2(n+1)-1,2(n+2)-1,,2(n+2n-2)-12n-10,,un2n-1.(2)un2n-1.9220042d1d2n-1.2n-1:2(a+1)-1,2(a+2)-1,,2(a+2n-1)-1.st,(2s-1)d2n-1(2s+1)d,(2t-1)d2(a+1)-1(2t+1)d.,1,3,,2n-1ds.[2(t+s)-1]d2(a+2n-1)-1.,[2(t+s)-1]d=(2t-1)d+(2s-1)d+d2(a+1)-3+2n-1+2n-1=2(a+2n-1)-1.,un2n-1.,un=2n-1.4.d(P,l)Pl..SAT=,SATPASATmBC,AB=AC=msin.PABC,PASATm;PABC,PASATm.2,SPAB+SPAC=SABC,d(P,AB)+d(P,AC)=m.2,QABC,SQAB+SQACSABC,d(Q,AB)+d(Q,AC)m;QABC,SQAB+SQACSABC,d(Q,AB)+d(Q,AC)m.3(1)ABCD,BCADF,BACDE.Pl1l2,l1ABCD,l2BCAD,3.SQ,d(P,AB)+d(P,BC)+d(P,CD)+d(P,DA)=d(Q,AB)+d(Q,BC)+d(Q,CD)+d(Q,DA)=[d(Q,AB)+d(Q,CD)]+[d(Q,BC)+d(Q,DA)][d(P,AB)+d(P,CD)]+[d(P,BC)+d(P,DA)]..(2)ABCD,.5.a2n+1-2kanan+1+a2n-1=0.,a2n+2-2kan+1an+2+a2n+1-1=0.,a2n+2-a2n-2kan+1an+2+2kanan+1=0,(an+2-an)(an+2+an-2kan+1)=0.,{an},,an+2=2kan+1-an.a0=0,a1=1,{an}.{an},2k|(an+2-an).,2k|a0,2k|a2n,n=0,1,2,.46.4,IABCD.HD1A1,AA1AD1AI,HAI,AIA1D1.ID1AD1,IHIA=ID21=r2,r.,EBI,IEIB=r2.,IEIB=IHIA,AHEB.,EHI=ABE.,IHG=ADG,IFE=CBE,IFG=CDG.EHG+EFG=ABC+ADC.,ABCDEFGH.03,.,EFGHABCD.7.yi=11+xi,i=1,2,,5,xi=1-yiyi,i=1,2,,5,5i=1yi=1.,5i=1xi4+x2i1Z5i=1-y2i+yi5y2i-2yi+11Z5i=1-5y2i+5yi5y2i-2yi+15Z5i=1-1+3yi+15y2i-2yi+15Z5i=13yi+15yi-152+4510.5i=13yi+15yi-152+455i=13yi+145=545i=1(3yi+1)=54(3+5)=10,.8.iai,75-ai.22i=1(C2ai+C275-ai)11C275.,11,11C275.22i=1(a2i-75ai)-30525,22i=1(2ai-75)21650.,22i=1(2ai-75)22222i=1(2ai-75)236300.,22i=1(2ai-75)191.,22i=1ai191+16502921.,928.(2003)2003(6,36)1.f(x)=logax(a0,a1).f(x1x2x2003)=8,f(x21)+f(x22)++f(x22003)().(A)4(B)8(C)16(D)2loga812.1,S-ABC,OABC.OSA=,OSB=,OSC=,,tantantan().(A)[22,+)(B)(0,22)(C)[1,22](D)(1,22)3.1,2,,99,,.1.9,().(A)1(B)2(C)3(D)411,22,33,44,55,66,77,88,99,1()2481631621242484964.,().(A)(B)(C)(D)5.10,253.15,11,12,5,1015().(A)90(B)100(C)110(D)1201320042

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