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[几何] 来自人教群的正三角形外接圆三内切等圆切线长关系

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kuing posted 2016-5-29 21:52 |Read mode
豫F爱好者海阔天空(2154*****)  20:18:41
请教一题
QQ图片20160529214827.jpg


QQ截图20160529214937.png
证明,如图,设大圆和小圆半径分别为 $R$, $r$,由斯特瓦尔特定理,有
\[PD^2+r^2=PO_1^2=\frac{PA^2(R-r)+R^2r}R-r(R-r),\]
化简即得
\[\frac{PD^2}{PA^2}=\frac{R-r}R,\]
同理有另外两式,即得
\[PD:PE:PF=PA:PB:PC,\]
所以等价于证 $PB=PA+PC$,这是熟知的。

PS、中间那式子不知有没有更简单的证法

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original poster kuing posted 2016-5-29 23:23
PS2、如无意外,如果是外切的话,就是 $\frac{PD^2}{PA^2}=\frac{R+r}R$,所以原题改成外切也是成立的

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游客 posted 2016-5-30 11:18
建坐标系用三角函数做是代数法(易),几何要巧妙最好弄个旋转变化(难)。

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