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用向量变得简单了些，后面的计算也比刚才的好。
易证 $AH=2OD=2R\cos A$（不必涉及欧拉线），即得 $\vv{AH}=2\vv{OD}=\vv{OB}+\vv{OC}$，因此有
\begin{align*}
\vv{OH}&=\vv{OA}+\vv{AH}=\vv{OA}+\vv{OB}+\vv{OC}, \\
\vv{AD}&=\vv{OD}-\vv{OA}=-\vv{OA}+\frac{\vv{OB}+\vv{OC}}2,
\end{align*}
于是
\begin{align*}
OH\perp AD&\iff \vv{OH}\cdot2\vv{AD}=0 \\
& \iff \bigl(\vv{OA}+\vv{OB}+\vv{OC}\bigr)\cdot\bigl(-2\vv{OA}+\vv{OB}+\vv{OC}\bigr)=0 \\
& \iff -2OA^2+OB^2+OC^2-\vv{OA}\cdot\vv{OB}-\vv{OA}\cdot\vv{OC}+2\vv{OB}\cdot\vv{OC}=0 \\
& \iff 2\cos 2A=\cos 2B+\cos 2C.
\end{align*}
为了避免字母意义的混淆，这里就不用题目中的 $a$, $b$, $c$ 了，下面记 $ x=1/S_{\triangle HBC}$, $y=1/S_{\triangle HCA}$, $z=1/S_{\triangle HAB}$，设 $BC$ 边上的高为 $AE$，则
\[\frac yz=\frac{S_{\triangle HAB}}{S_{\triangle HCA}}=\frac{\frac12AH\cdot BE}{\frac12AH\cdot CE}=\frac{BE}{CE}=\frac{AE\cot B}{AE\cot C}=\frac{\tan C}{\tan B},\]
即
\[y\tan B=z\tan C,\]
同理有 $x\tan A=y\tan B$，于是可设
\[x\tan A=y\tan B=z\tan C=m,\]
由三角恒等式有
\begin{align*}
\tan A+\tan B+\tan C=\tan A\tan B\tan C &\iff\frac1x+\frac1y+\frac1z=\frac{m^2}{xyz} \\
&\iff m^2=xy+yz+zx,
\end{align*}
因此
\[\cos2A+1=\frac{1-\tan^2A}{1+\tan^2A}+1=\frac2{1+\tan^2A} =\frac2{1+\frac{m^2}{x^2}}=\frac2{1+\frac{xy+yz+zx}{x^2}} =\frac{2x^2}{(x+y)(x+z)},\]
同理对 $B$, $C$ 有类似的式子，代入上面得到的结论中有
\begin{align*}
2\cos 2A=\cos 2B+\cos 2C&\iff2(\cos 2A+1)=\cos 2B+1+\cos 2C+1 \\
&\iff \frac{4x^2}{(x+y)(x+z)}=\frac{2y^2}{(y+z)(y+x)}+\frac{2z^2}{(z+x)(z+y)} \\
&\iff 2x^2(y+z)=y^2(z+x)+z^2(x+y) \\
&\iff 2x(xy+zx+yz)=y(yz+xy+zx)+z(zx+yz+xy) \\
&\iff 2x=y+z.
\end{align*} |
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kuing
发表于 2014-1-7 16:30
isee
发表于 2014-1-7 21:11
法二哩？
