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[几何] 来自网友的直角梯形过定点什么的

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kuing posted 2017-8-6 02:06 |Read mode
教学乡长 0:10:48
QQ截图20170806010755.jpg
(1)略;

(2)如图。
QQ截图20170806020339.png
由梅,有
\begin{align*}
\frac{NM}{ME}\cdot \frac{EP}{PA}\cdot \frac{AQ}{QN}&=1, \\
\frac{NM}{MF}\cdot \frac{FP}{PD}\cdot \frac{DQ}{QN}&=1,
\end{align*}
故由条件及 $ME=MF$,以上两式相除即得
\[\frac{AQ}{DQ}=\frac yx,\]
所以 $Q$ 为定点;

(3)设 $PM:PQ=k$,则
\[\frac1{xy}=\frac{\S{PAD}}{\S{PEF}}
=\frac{\S{PAQ}}{2\S{PEM}}+\frac{\S{PDQ}}{2\S{PFM}}
=\frac1{2kx}+\frac1{2ky},\]
所以
\[k=\frac{x+y}2,\]
下略。

其实和梯形关系不大,主要还是玩中间那些三角形……

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original poster kuing posted 2017-8-6 02:55
哦,其实(2)也可以用面积法,同样记 $PM:PQ=k$,我们有
\[
AQ:DQ=\S{PAQ}:\S{PDQ}=\frac{\S{PEM}}{kx}:\frac{\S{PFM}}{ky}=y:x,
\]
这样简单多了,根本用不着梅。

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original poster kuing posted 2017-8-6 13:42
第(3)问贴题者自解:
教学乡长 12:44:42
QQ图片20170806133700.png

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其妙 posted 2017-8-6 14:04
此题可以编拟一个向量模的题出来
妙不可言,不明其妙,不着一字,各释其妙!

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