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a=cos2,b=cos2,c=cos2,0abc1,a+b+c=1.,0a4+b4+c4-2(a2+b2+c2)+1ab+bc+ca+abc.ab+bc+ca=u,abc=v,a2+b2+c2=1-2u,a4+b4+c4=2u2-4u+4v+1.,02u2+4vu+v.!u∀0,v∀0,,!,u=v=0,!20.,!u-2u2∀3v.u-2u2=u(1-2u)=(ab+bc+ca)(a2+b2+c2)∀(ab+bc+ca)#(a+b+c)23=13(ab+bc+ca)(a+b+c)∀13∃33a2b2c2#33abc=3abc=3v.,!,!.,..(刘康宁)2005年北京市中学生数学竞赛()(5,25)1.S={1,2,3,4,5},M={1,3,4},N={2,4,5},,(CSM)%(CSN)().(A)(B){1,3}(C){4}(D){2,5}2.ab.:a+b,ab;:a+b,ab.().(A),(B),(C),(D),3.cd,ab,().(A)a=-2(c+d),b=2(c+d)(B)a=c-d,b=-2c+2d(C)a=4c-25d,b=c-110d(D)a=c+d,b=2c-2d4.[a,b]f(x),c,x1&[a,b]x2&[a,b],f(x1)+f(x2)2=c,f(x)[a,b]∋(c.,f(x)=1gx[10,100]().(A)10(B)110(C)32(D)345.,3)4)5().(A)(B)(C)(D)(7,35)1.,,(-1,7).lA12,13B14,15,lA.2.∗ABC,AB=6+2,+ACB=2430,.AC+BC.3.2005x1,x2,−,x2005|x1-x2|+|x2-x3|+−+|x2004-x2005|+|x2005-x1|=1.|x1|+|x2|+−+|x2005|.4.x&R.f(x)=maxsinx,cosx,sinx+cosx2.5.6,,1:1123456411..(10)abc,f(x)=ax2+bx+cfa-b-c2a=0.:-11f(x)=0.(15)ABCDS,AB=a,BC=b,CD=c,DA=d.:(1)S=(p-a)(p-b)(p-c)(p-d),p=a+b+c+d2;(2)ABCD,S=abcd.(15)pa1,a2,−,apd(d0),a1p.:(1)p,p|d;(2)p=15,d30000.参考答案1.A.CSM={2,5},CSN={1,3},,(CSM)%(CSN)=.2.B.ab,a+b,a+b,ab.,∋ab(;∋ab(.3.D.a=-2(c+d)=-b,ab,(A);a=c-d=-12(-2c+2d)=-12b,ab,(B);a=4c-25d=4c-110d=4b,ab,(C).,cd.a=c+d.ab,∀,a=∀b,c+d=∀(2c-2d)(1-2∀)c=-(2∀+1)d.1-2∀=0,d,.1-2∀.0,c=2∀+12∀-1d,cd,.,ab.4.C.x1&[10,100],x2&[10,100],x1x2=103.,f(x1)+f(x2)2=1gx1+1gx22=1gx1x22=1g1032=32.5.B.456,3)4)5;,3)4)5,3k4k5k(k0),k.,,3)4)5.1.(-2,-1).ly=kx+b,252006113=12k+b,15=14k+b.k=815,b=115.ly=815x+115.,(-2,-1)l.x=-1,0,1,2,3y=815x+115,y.,lA(-2,-1).2.8+43.22,BCPPC=AC,AP,+APB=15,,PAB=6+215,.AC+BC,PB,PB,,+PAB=90,.AC+BCABsin15,=6+2sin15,.,∗MKN,+MKN=90,,NK=1,MN=2.,+NMK=30,,KM=3.33,KMP,MP=MN=2,PN,+NPK=15,.tan15,=12+3=2-3.PN=6+2.,sin15,=NKPN=16+2=6-24.AC+BCABsin15,=6+26-24=4(6+2)24=8+43.3.12.ij,|xi-xj||xi|+|xj|.1=|x1-x2|+|x2-x3|+−+|x2004-x2005|+|x2005-x1||x1|+|x2|+|x2|+|x3|+−+|x2004|+|x2005|+|x2005|+|x1|=2(|x1|+|x2|+−+|x2004|+|x2005|).,|x1|+|x2|+−+|x2005|∀12.,x1=12,x2=x3=−=x2005=0,|x1|+|x2|+−+|x2005|=12.,|x1|+|x2|+−+|x2005|12.4.1-22.f(x)=maxsinx,cosx,sinx+cosx2=maxsinx,cosx,sinx+!4,,f(x)1.y=sinx,y=cosxmax{sinx,cosx}-22,x=5!4,-22.x=5!4,sinx+!4-1.,f(x)-22.5.17.4,:4,,.4,,.,,:()()5;()()2;()()6;()()4.17.(1)fa-b-c2a=0,f(x)=ax2+bx+c.f(x)=ax2+bx+c26,.,a-b-c2a=-b2a,a=c.#=0,b2-4a2=0,b=/2a.x=-b2a=2a2a=1.,f(x)=ax2/2ax+a=a(x/1)2,-11f(x)=0.f(x)=ax2+bx+c,fa-b-c2a=a(a-b-c)24a2+b(a-b-c)2a+c=(a-b-c)2+2b(a-b-c)+4ac4a=(a-b-c)(a-b-c+2b)+4ac4a=(a-c)2-b2+4ac4a=(a+c)2-b24a=(a-b+c)(a+b+c)4a=f(-1)f(1)4a,fa-b-c2a=0,f(-1)f(1)4a=0f(-1)=0f(1)=0.-11f(x)=0.(1)AC.ABCD,,+B++D=180,.cosB=-cosD,sinB=sinD.,S=S∗ABC+S∗ADC=12(absinB+cdsinD),4S=2(ab+cd)sinB.,16S2=4(ab+cd)2sin2B.a2+b2-2abcosB=AC2=c2+d2-2cdcosD,a2+b2-2abcosB=c2+d2+2cdcosB.a2+b2-c2-d2=2(ab+cd)cosB.(a2+b2-c2-d2)2=4(ab+cd)2cos2B.!+!16S2+(a2+b2-c2-d2)2=4(ab+cd)2.16S2=(2ab+2cd)2-(a2+b2-c2-d2)2=(2ab+2cd+a2+b2-c2-d2)#(2ab+2cd-a2-b2+c2+d2)=[(a+b)2-(c-d)2][(c+d)2-(a-b)2]=(a+b+c-d)(a+b-c+d)#(c+d+a-b)(c+d-a+b)=16(p-a)(p-b)(p-c)(p-d).,S=(p-a)(p-b)(p-c)(p-d).(2)ABCD,(1)S=(p-a)(p-b)(p-c)(p-d).ABCD,,a+c=b+d=p.p-a=c,p-b=d,p-c=a,p-d=b.(1),S=abcd.(1)a1p,d0,,a1,a2,−,app.,ai(i=1,2,−,p)p.a1,a2,−,appp-1,,p.aman(nm).,an-am=(n-m)dp.0n-mp,p,,p!(n-m).,p|d.(2)p1,p2,−,p1515d(d0).15,,d,2|d.p2,p3,p43,p23,(1),3|d.p3,p4,p5,p6,p75,p35,(1),5|d.2|d,3|d5|d(2,3,5)=1,30|d.,p1∀3p2∀33.p2,,p2∀37.,p2,p3,−,p87,p27,(1),7|d.,p2,p3,−,p1211,p211,(1),11|d.p2,p3,−,p1413,p213,(1),13|d.(30,7,11,13)=1,,30∃7∃11∃13|d,30030|d.d∀3003030000.(周春荔)2720061