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2006年北京市中学生数学竞赛()(5,25)1.PO,l,OP(y)1P(x)1.,P2().22.f(x)=sinx4.x1x2,x,f(x1)f(x)f(x2),|x1-x2|().(A)8(B)4(C)2(D)3.a1=12,a2=13+23,,an=1n+1+2n+1++nn+1,.Sn=1a1a2+1a2a3++1anan+1.S2006().(A)2(B)3(C)4(D)54.n,:n!=1!2!!n.1!!1+2!!2++59!!592006().(A)1(B)5(C)401(D)20055.xy(2x+9y)2006+(4x-y)2006=7777777,∀(x,y)∀.∀().(A)0(B)1(C)2(D)2006(7,35)1.|a|=1,|b|=2,c=a+b,c#a,ab.2.f(x)R,y=f(x)x=12.f(1)+f(2)++f(2006)=.3.∃ABC,a+ba=sinBsinB-sinA,cos(A-B)+cosC=1-cos2C.a+cb=.4.{an}a1=6,n%N+,n&2,an+an-1=9an-an-1+8.a70=.5.∃ABCAB=34,BC=510,CA=226.∃ABC.(10)∃ABC,r,BCACABrarbrc.:1ra+1rb+1rc=1r.注:,..(15)a1,a2,,an,b1,b2,,bnc1,c2,,cnaicib2i(i=1,2,,n).:(a1+a2++an)(c1+c2++cn)&(b1+b2++bn)2.31200710(15)∃ABCn,∃ABC(,,,3).A,B,C,n3.∋(.:.参考答案1.C.,(A)(B);(D)Px=l2,OP(POPP),,OP)OP);(C)y=lsinxl,1.2.B.,f(x1)f(x2)f(x),sinx14=-1,sinx24=1.x1=8k1-2,x2=8k2+2.,|x1-x2|&4.3.C.an=1n+1+2n+1++nn+1=n2,Sn=1a1a2+1a2a3++1anan+1=41!2+42!3++4n(n+1)=41-12+12-13++1n-1n+1=41-1n+1.,S2006=4(1-12007)4.4.D.2006=2!17!59,2006|60!.1!!1+2!!2++59!!59=1!!(2-1)+2!!(3-1)++59!!(60-1)=(2!-1!)+(3!-2!)++(60!-59!)=60!-1.,2005.5.A.xy∀.7|(2x+9y),∀7|(4x-y),,∀72006,∀7777.,7(2x+9y),7(4x-y).a%Z,7|[(a-3)(a-2)(a-1)a(a+1)(a+2)(a+3)]7|(a7-a)=a(a6-1).,7|[(2x+9y)6-1],7|[(4x-y)6-1].,7|{(2x+9y)2[(2x+9y)6!334-1]},7|{(4x-y)2[(4x-y)6!334-1]}.∀7|[(2x+9y)2+(4x-y)2]7|(x2+2y2).,70,1,2,4.,7|x,7|y,7|(x2+2y2).,7|(2x+9y),7|(4x-y),.,xy∀.1.120∗.c#ac+a=0(a+b)+a=0|a|2+|a|+|b|cos=0cos=-12.,ab=120∗.2.0.y=f(x)x=12,f(x)=f(1-x).f(x)R,f(1-x)=-f(x-1).,f(x)+f(x-1)=0.,f(1)+f(2)++f(2006)=0.3.5-12+5-12.cos(A-B)+cosC=1-cos2Ccos(A-B)-cos(A+B)=2sin2C2sinA+sinB=2sin2Cab=c2.a+ba=sinBsinB-sinAb2-a2=abab=5-12.32,cb=5-12.4.29.ak+ak-1=9ak-ak-1+8a2k-a2k-1=8(ak-ak-1)+9.k=2,3,,n,a2n-a21=8(an-a1)+9(n-1).a270-62=8(a70-6)+9!69,a270-8a70-609=0.a70=29().5.10.44,BC,∃ABCRt∃BCD,,BDC=90∗,BD=5,CD=15.DEA)F,DE=2,DF=5.BE=3,CF=10.,A)C=104=AC,A)B=34=AB.,AA).S∃ABC=S∃BCD-S∃ABE-S∃ACF-SDEAF=12!5!15-12!3!5-12!2!10-2!5=10.,∃ABC.55,AOaBOaCOa,∃AOaB=12rac,∃AOaC=12rab,∃BOaC=12raa.∃ABC=12r(a+b+c).∃ABC=∃AOaB+∃AOaC-∃BOaC,r(a+b+c)=ra(b+c-a),r!2p=ra(2p-2a).,rp=ra(p-a).,rp=rb(p-b),rp=rc(p-c).1ra+1rb+1rc=p-arp+p-brp+p-crp=1r+3p-2pp=1r.nfi(x)=aix2+2bix+ci(i=1,2,,n).i,ai0,aicib2i,i=(2bi)2-4aici0,,fi(x)&0,fi(x)=aix2+2bix+ci&0(i=1,2,,n).f(x)=f1(x)+f2(x)++fn(x)&0,f(x)=(a1+a2++an)x2+2(b1+b2++bn)x+(c1+c2++cn)&0.a1+a2++an0,,f(x)&0,,0.=[2(b1+b2++bn)]2-4(a1+a2++an)(c1+c2++cn)0.(a1+a2++an)(c1+c2++cn)&(b1+b2++bn)2.∃ABC(,)k,()p,q,(;;;;;;)r.,,p+2q.,,,∃ABCAB,.,2k+1.,p+2q=2k+1.,p=2(k-q)+1.,p,p1..(说明:.)(李延林)33200710