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--- $\small\exists P\,\forall V\in\{A,B,C\}:\overline{VP}\perp\overline{PV'}\,\small(V'$是$\,\triangle\,$内心关于$\,\small V\,$对边的镜像$).$ (http://mathchina.elinkage.net/cgi-bin/topic.cgi?forum=1&topic=1042)


-- 作者: HOFFMAN
-- 发布时间: 2020/06/07 03:31pm

[b]题:[/b]试证$\;\small\exists P:\overline{VP}\perp\overline{PV'}\,(\forall V\in\{A,B,C\})$
$\qquad\small(V'$为$\,\triangle ABC\,$的内心关于$\,\small V\,$的对边的镜像$).$


-- 作者: elim
-- 发布时间: 2020/06/10 09:34am

下面是主贴问题的复数处理. 纯几何的证明很漂亮. 待续...
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设内切圆是${\small\odot(O,1),\;T_{\theta}=}e^{i\theta}\small\;(\theta=\alpha,\beta,\gamma=-\alpha,\,0<\alpha<\frac{\pi}{2}<\beta\le\pi)\,$为切
点. 于是${\small\,A=}\sec\alpha,\;{\small B,\,C=}\sec\frac{\beta\mp\alpha}{2}\text{cis}\big(\frac{\beta\pm\alpha}{2}\big),\,$记$\small\,V'$为$\triangle ABC$内心
关于顶点$\small\,V\,$之对边的镜像,则$\small\,A'=2e^{i\beta},\,B\ '=2e^{-i\alpha},\,C\ '=2e^{i\alpha}$
$\because\;z\,$关于射线$\,tw\small(\ne 0)$的镜像是$\,\bar{z}\big(\frac{w}{|w|}\big)^2{\small=\,}\bar{z}\frac{w}{\overline{w}},\small\;V'$对$\small\,V'+V\,$的镜像
${\small P_V=\overline{V'}}{\frac{(V'+V)^2}{|V'+V|^2}}\in\odot\big({\scriptsize\dfrac{V'+V}{2},\dfrac{|V'-V|}{2}}\big).\;$故$\scriptsize\,\overline{VP_V}\perp\overline{P_VV'}.$下证$\scriptsize\,P_A=P_B=P_C:$
$\frac{P_A}{2}{\small=e^{-i\beta}}{\large\frac{2e^{i\beta}+\sec\alpha}{2e^{-i\beta}+\sec\alpha}}{\small=e^{-i\beta}}{\large\frac{(e^{i\alpha}+e^{-i\alpha})e^{i\beta}+1}{(e^{i\alpha}+e^{-i\alpha})e^{-i\beta}+1}}{\small=e^{i\beta}}\large\frac{e^{i\alpha}+e^{-i\alpha}+e^{-i\beta}}{e^{i\alpha}+e^{-i\alpha}+e^{i\beta}}$
$\frac{P_B}{2}{\small=e^{i\alpha}}\frac{2e^{-i\alpha}+\sec\frac{\beta-\alpha}{2}e^{i\frac{\beta+\alpha}{2}}}{2e^{i\alpha}+\sec\frac{\beta-\alpha}{2}e^{-i\frac{\beta+\alpha}{2}}}{\small=e^{i\alpha}}\frac{(e^{i\frac{\beta-\alpha}{2}}+e^{-i\frac{\beta-\alpha}{2}})e^{-i\alpha}+e^{i\frac{\beta+\alpha}{2}}}{(e^{i\frac{\beta-\alpha}{2}}+e^{-i\frac{\beta-\alpha}{2}})e^{i\alpha}+e^{-i\frac{\beta+\alpha}{2}}}$
$\qquad{\small=e^{i\alpha}}\frac{e^{i\frac{\beta-\alpha}{2}}((1+e^{-i(\beta-\alpha)})e^{-i\alpha}+e^{i\alpha})}{e^{-i\frac{\beta-\alpha}{2}}((1+e^{i(\beta-\alpha)})e^{i\alpha}+e^{-i\alpha})}{\small=e^{i\beta}}\large\frac{e^{i\alpha}+e^{-i\alpha}+e^{-i\beta}}{e^{i\alpha}+e^{-i\alpha}+e^{i\beta}}$
$\frac{P_C}{2}{\small=e^{-i\alpha}}\frac{(e^{i\frac{\beta+\alpha}{2}}+e^{-i\frac{\beta+\alpha}{2}})e^{i\alpha}+e^{i\frac{\beta-\alpha}{2}}}{(e^{i\frac{\beta+\alpha}{2}}+e^{-i\frac{\beta+\alpha}{2}})e^{-i\alpha}+e^{-i\frac{\beta-\alpha}{2}}}{\small=e^{-i\alpha}}\frac{e^{i\frac{\beta+\alpha}{2}}((1+e^{-i(\beta+\alpha)})e^{i\alpha}+e^{-i\alpha})}{e^{-i\frac{\beta+\alpha}{2}}((1+e^{i(\beta+\alpha)})e^{-i\alpha}+e^{-i\alpha})}\small\;\;\;\square$


-- 作者: elim
-- 发布时间: 2020/06/30 07:27am

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