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源自知乎提问
FAQ了
题：等轴双曲线内接三角形的垂心在双曲线上.
不妨设等轴双曲线为 $xy=1$ ， $A(a,1/a)$ , $B(b,1/b)$ , $C(c,1/c)$ ，三角形 ABC 垂心为 $H(x,y)$ ，下面证明垂心坐标满足 $xy=1.$
\begin{gather*}
\left\{\begin{aligned}\overrightarrow {AH}\cdot \overrightarrow{CB}&=0,\\[1ex]
\overrightarrow{BH}\cdot \overrightarrow{AC}&=0,\end{aligned}\\[1ex]
\right.\\[1ex]
\left\{\begin{aligned} (x-a)(b-c)+(y-\frac1a)(\frac1b-\frac1c)&=0,\\[1ex]
(x-b)(c-a)+(y-\frac1b)(\frac1c-\frac1a)&=0,\end{aligned}\right.\\[1ex]
\left\{\begin{aligned} bc(x-a)&=y-\frac1a,\\[1ex]
ac(x-b)&=y-\frac1b,\end{aligned}\right.\\[1ex]
\frac{x-a}{x-b}=\frac{ay-1}{by-1},\\[1ex]
\frac{x-a}{a-b}=\frac{ay-1}{by-ay},\\[1ex]
x-a=\frac{ay-1}{-y},\\[1ex]
xy=1.
\end{gather*} 得证. |
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hbghlyj
posted 2023-3-5 19:21
Ly-lie
posted 2023-3-5 21:05
