2005年IMO中国国家集训队选拔考试

2005IMOOABCDACBDP,PBO1PAO2PQ,O1O2OEF.:PQCEDF.()n(n2),,:n,2,,.(.)()n,aj(j=1,2,,n),{1,2,,n}I,jI(1+aj)-112.:nj=1|aj|3.()a1,a2,,a6;b1,b2,,b6;c1,c2,,c61,2,,6.6i=1aibici.()n,x.:nk=1xkx-(x+1)kx+1n,[a]a.().fN+R,mk(+1)mkm,f(k+m)=f(k)+f(m).()1.1PJF=PAF=CAF=CDF,,PJCD.,IPCD.IPJ.EFD=180-ECD=180-EIJ,EFJI.,,IEFJO1O2IEPQJF,PQCEDF.1.:maxn2+1.G,G1.n2.n,nnGn2(1+2-1+2-2+)=2n22n2=n..,s,2-2-s2nn+1.n2+1n+12,2-2-s2nn+1nn2+1.,n2+1(1+2-1++2-s)n1.,1,1,,1,n2+12-1,2-1,,2-1n2+1,,2-s,2-s,,2-sn2+11,n.,maxn2+1.,max=n2+1.1+aj=rjeij,|j|,j=1,2,,n.jIrjeijIj-112..r,r0,||,|rei-1|12,12r32,||6,|rei-1||r-1|+||.2:2,,12r32,||6.,|rei-1|=|r(cos+isin)-1|=|(r-1)(cos+isin)+[(cos-1)+isin]||r-1|+(cos-1)2+sin2=|r-1|+2(1-cos)=|r-1|+2sin2|r-1|+||..,|I|,12jIrj32,jIj6,,|aj|=|rjeij-1||rj-1|+|j|,,nj=1|aj|nj=1|rj-1|+nj=1|j|=rj1|rj-1|+rj1|rj-1|+j0|j|+j0|j|.,rj1|rj-1|=rj1(rj-1)rj1(1+rj-1)-132-1=12,rj1|rj-1|=rj1(1-rj)rj1[1-(1-rj)]-1-12-1=1,nj=1|j|=j0j-j0j6-(-6)3.,nj=1|aj|12+1+33.S=6i=1aibici.,S666i=1aibici=66(6!)3=66!=725160.S161.a1b1c1,a2b2c2,,a6b6c66125,2612527,a1b1c1,a2b2c2,,a6b6c6127,126,261,2,3,4,5,63(),,125.a1b1c1=27,a2b2c2=25,,S=(a1b1c1-a2b2c2)2+2a1b1c1a2b2c2+(a3b3c3+a4b4c4)+(a5b5c5+a6b6c6)(27-25)2+2a1b1c1a2b2c2+2a3b3c3a4b4c4+2a5b5c5a6b6c6(33-5)2+2366i=1aibici=(33-5)2+725161,,S162.a1,a2,,a6;b1,b2,,b6;c1,c2,,c61,2,3,4,5,6;5,4,3,6,1,2;5,4,3,1,6,2,S=155+244+333+461+516+622=162.S162..,u,=u+,0,u.:u==-,0.,u(0),=u+,(u-u)=-.0,,u=u=...,k=1,2,,n,k=akx+bk=ck(x+1)+dk,ak=kx,ck=kx+1,0bkx,0dkx+1.S,S=nk=1[akx-ck(x+1)]=nk=1[(k-bk)-(k-dk)]=nk=1dk-nk=1bk.I={1kn|dk1}.f(k)=k-ck-1,kI,k=ck(x+1)+dkck+1,0f(k)n.,fI{1,2,,n}.f.,klI(kl)f(k)=f(l),k-l=ck-cl,,(ck-cl)x=dl-dk.,klI,dkdl(1,x+1),|dk-dl||x|.|ck-cl||x|=|k-l||x||x|,,..,k=ck(x+1)+dk,f(k)=ckx+(dk-1).kI,0dk-1x,,ck=af(k)dk-1=bf(k).,(kbk0)S=kIdk+kIdk-nk=1bkkIdk-kIbf(k)+kIdk=kI(dk-bf(k))+kIdk=kI1+kIdk|I|+(n-|I|)=n.f(n)=bn,b.:n2+3,f(n+1)-f(n)=f(n)-f(n-1).u,f(n+1)-f(n)=f(u+1)-f(u),f(n)-f(n-1)=f(u+1)-f(u).f(n+1)+f(u)=f(n)+f(u+1).,(n+1)u(+1)(n+1),nu+1(+1)n.(n+1)uu+1(+1)n.,nuu+1(+1)(n-1).,u,(n+1)uu+1(+1)(n-1).(+1)(n-1)-(n+1)=n-2-12,.,n0,n0-12+3n0,n03.f(n+1)-f(n)=f(n0)-f(n0-1),nn0-1.,f(n)=(n-n0+1)[f(n0)-f(n0-1)]+f(n0-1),nn0-1.km,mn0,mn0,mk(+1)m.f(k+m)=f(k)+f(m),(k+m-n0+1)[f(n0)-f(n0-1)]+f(n0-1)=(k-n0+1)[f(n0)-f(n0-1)]+f(n0-1)+(m-n0+1)[f(n0)-f(n0-1)]+f(n0-1).(n0-1)f(n0)=n0f(n0-1).f(n0-1)=b(n0-1),f(n0)=bn0.,f(n)=bn,nn0-1.:n,f(n)=bn.,f(n)bnn1.1,k,n1k(+1)n1,kn1,k+n1n1,f(k+n1)=f(k)+f(n1).,f(n1)=f(k+n1)-f(k)=(k+n1)b-kb=n1b,.1,n1n1(+1)n1.,b2n1=f(2n1)=f(n1)+f(n1).,f(n1)=n1b,.f(n)=bn,b,.,f(n)=bn,b.(2005)abca2+2b2+3c2=32.:3-a+9-b+27-c1.()DABCBC,PAD,DABPBME,ACPCFN.DE=DF,:DM=DN.()(1){an},n,a2n+12anan+2.(2){an},n,a2n+12anan+2.()2004n,1,2,,n2nn(nn),1.12004