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[几何] 其外接圆半径是内切圆直径证等边三角形

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isee posted 2019-7-17 16:30 |Read mode
若三角形外接圆的半径等于内切圆的直径,求证:此三角形为等边三角形。

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kuing posted 2019-7-17 16:45
\begin{align*}
R-2r&=\frac{abc}{4S}-\frac{4S}{a+b+c}\\
&=\frac1{4S}\left( abc-\frac{16S^2}{a+b+c} \right)\\
&=\frac1{4S}\bigl( abc-(a+b-c)(b+c-a)(c+a-b) \bigr),
\end{align*}最后这条式子你认识的吧?

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original poster isee posted 2019-7-17 16:54
回复 2# kuing


舒尔不等式(Schur's inequality) 最特殊情况,(这个角度)这么一看,此题(对新手)难度相当大啊。。。。。

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kuing posted 2019-7-17 17:00
回复 3# isee

对新手,直接让他证明 schur 不等式反而难,现在这个形式才好证,况且已经是三边,都不用讨论括号的正负问题了,直接均值就行。

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