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发表于 2014-5-31 23:25:08
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显示全部楼层
根据楼上提供的链接,出于网友kuing 精彩而初等的分析特将其解答过程转载如下(图片引用原文):
引理1:锐角\(\triangle ABC\) 中,\(H\)为其垂心,则\(AH\*HD=BH\*HE=CH\*HF=k\),
\(\frac{k}{k+HA^2}+\frac{k}{k+HB^2}+\frac{k}{k+HC^2}=1\)
注:1.由\(AFH \sim CDH\) 可以得到\(\frac{AH}{FH}=\frac{CH}{DH}\) 即\(AH\*HD=CH\*HF\),同理易得\(AH\*HD=BH\*HE\)
2.\(\frac{S_{BHC}}{S_{ABC}}+\frac{S_{AHC}}{S_{ABC}}+\frac{S_{AHB}}{S_{ABC}}=1\)
\(\frac{HD}{AH+HD}+\frac{HE}{BH+HE}+\frac{HF}{CH+HF}=1\)
\(\frac{AH\*HD}{AH\*HD+HA^2}+\frac{BH\*HE}{BH\*HE+HB^2}+\frac{CH\*HF}{CH\*HF+HC^2}=1\)
在下面垂心四面体\(D-ABC\)的垂心\(H\)在四面体的内部,已知\(HA=a,HB=b,HC=c,HD=d\),体积为\(V\)
并记\(H\)在底面ABC的投影为\(E\),且记\(AE=x,BE=y,CE=z,HE=w\),由引理1,我们可以得到:
在\(\triangle ABC\) 中,设\(AE\*EF=k\),有 \(\frac{k}{k+x^2}+\frac{k}{k+y^2}+\frac{k}{k+z^2}=1\) ..........(1)
由于\(HE\perp \triangle ABC\) 易有\(x^2=a^2-w^2,y^2=b^2-w^2,z^2=c^2-w^2\)............................(2)
又\(\triangle AEH \sim \triangle DEF\), 有\(\frac{AE}{EH}=\frac{DE}{EF}\),即\(k=w(d+w)\).................(3)
将(2),(3)代入(1)并化简得到:\(3d^2w^4+2d(a^2+b^2+c^2+d^2)w^3+(a^2b^2+a^2+c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)w^2-a^2b^2c^2=0\).......(4)
设\(\triangle ABC\) 的面积为\(S\),其外接圆直径为D,则\(x=D\cos(A),y=D\cos(B),z=D\cos(C)\),代入\(\cos(A)^2+\cos(B)^2+\cos(C)^2+2\cos(A)\cos(B)\cos(C)=1\)
得到:\(D(x^2+y^2+z^2)+2xyz=D^3\).........................................................................................(5)
又\(a^2=(D\sin(A))^2=D^2-(D\cos(A))^2=D^2-x^2\),同理\(b^2=D^2-y^2\),\(c^2=D^2-z^2\).......(6)
代入\(16S^2=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4\).............................................................(7)
得到:\(16S^2=3D^4-2(x^2+y^2+z^2)D^2+2(x^2y^2+x^2z^2+y^2z^2)-(x^4+y^4+z^4)\)..........(8)
将(2)代入(5)和(8)并消除\(D\)得到:
\(256S^6+16(\sum_{a,b,c}^{}(a^4-10b^2c^2)+18w^2\sum_{a,b,c}^{}(a^2)-27w^4)S^4-8(\sum_{a,b,c}^{}(a^6b^2-4a^4b^4+a^2b^6+2a^2b^2c^4)+w^2\prod_{a,b,c}{}(a^2+b^2-2c^2))S^2+\prod_{a,b,c}^{}(a^2-b^2)^2=0\)…(9)
又\(S(d+w)=3V\).............................................................................................................................(10)
将(10)代入 (9) 消除\(S\)得到:
\(186624V^6+1296(d+w)^2(\sum_{a,b,c}^{}(a^4-10b^2c^2)+18w^2\sum_{a,b,c}^{}(a^2)-27w^4)V^4-72(d+w)^4(\sum_{a,b,c}^{}(a^6b^2-4a^4b^4+a^2b^6+2a^2b^2c^4)+w^2\prod_{a,b,c}^{}(a^2+b^2-2c^2))V^2+(d+w)^6\prod_{a,b,c}^{}(a^2-b^2)^2=0\)
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