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[几何] 三角形四线交四边形面积

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kuing posted 2019-3-4 16:28 |Read mode
某大大问道:
QQ截图20190304161852.jpg
是不是有个梗
不知以前有没有写过,懒得找了,反正是简单东西,直接写个一般式,以后代数字好了。

QQ截图20190304162058.jpg
记 `AG:AC=a`, `AF:AC=b`, `BD:BC=x`, `BE:BC=y`,如图所示,则 $\S{ADE}=y-x$,由梅氏定理有
\[
\frac{AI}{ID}\cdot x\cdot\frac{1-b}b=1\riff\frac{AI}{AD}=\frac b{x(1-b)+b},
\]同理有
\begin{align*}
\frac{AK}{AE}&=\frac b{y(1-b)+b},\\
\frac{AH}{AD}&=\frac a{x(1-a)+a},\\
\frac{AJ}{AE}&=\frac a{y(1-a)+a},
\end{align*}而
\[
\frac{S_{HIKJ}}{\S{ADE}}=\frac{\S{AIK}-\S{AHJ}}{\S{ADE}}=\frac{AI}{AD}\cdot\frac{AK}{AE}-\frac{AH}{AD}\cdot\frac{AJ}{AE},
\]故
\[S_{HIKJ}=(y-x)\left( \frac{b^2}{\bigl(x(1-b)+b\bigr)\bigl(y(1-b)+b\bigr)}-\frac{a^2}{\bigl(x(1-a)+a\bigr)\bigl(y(1-a)+a\bigr)} \right),\]化简即得
\[S_{HIKJ}=\frac{(b-a)(y-x)\bigl(ab(x+y)+xy(a+b)-2abxy\bigr)}{(a+x-ax)(a+y-ay)(b+x-bx)(b+y-by)}.\]
对于原题,代 `a=x=1/4`, `b=y=3/4` 得 `128/455`。

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realnumber posted 2019-3-4 22:00
小奥六的内容,图中所画的(A为顶点的)三角形面积都可以算出来.

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original poster kuing posted 2019-3-4 22:02
回复 2# realnumber

小奥方法怎么算?

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realnumber posted 2019-3-4 22:17
QQ截图20190304221209aaa.png
也是算出F分AE,CD的比例.他们好象用利用面积算的,不确定小奥有没有,添平行线三角形相似.

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isee posted 2019-3-5 09:46
回复 3# kuing

虽然没见到标答,但实质(面积比等价于线段比)不会发生变化

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游客 posted 2019-3-5 14:03
是不是小学现在都可以“梅涅”了?

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