几何恒等式

证明 记△ABC的面积为△, 半周长为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