初中数学竞赛题

设△ABC的三边长a,b,c上的高与内角平分线分别为ha，hb，hc，wa，wb，wc，外接与内切圆半径分别为R,r. 求证:ha/(wa)^2+hb/(wb)^2+hc/(wc)^2=1/R+1/2r.证明 记△ABC的面积为△, 周长为s。则有(wa)^2=4bcs(s-a)/(b+c)^2,△=a*ha/2=abc/(4R)=s*r=√[(s-a)(s-b)(s-c)s]Σha/(wa)^2=Σ△(b+c)^2/[2abcs(s-a)]=[△/(abcs)]* Σ(b+c)^2/(b+c-a)={△/[abcs(b+c-a)(c+a-b)(a+b-c)]}*Σ(b+c)^2(c+a-b)(a+b-c)注意恒等式: △/[abcs(b+c-a)(c+a-b)(a+b-c)]=1/(32R*s^2*r^2)而Σ(b+c)^2(c+a-b)(a+b-c)=Σ[a^2(b+c)^2-(b^2-c^2)^2]=Σ[a^2*b^2+a^2*c^2+2a^2bc-b^4-c^4+2b^2*c^2]=4Σb^2*c^2-2Σa^4+2abc(a+b+c)=32△^2+16R*s^2*r=16s^2*r*(R+2r)所以 Σha/(wa)^2=16s^2*r*(R+2r)/(32R*s^2*r^2)=1/R+1/2r. 收起