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[几何] 外角三等分线何时存在

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hbghlyj posted 2024-12-7 17:44 |Read mode
从图中看出,$B$ 相对于 $A$ 的外角三等分线当且仅当 $(π−φ)/3<θ$ 时存在。但是,我们希望用边长 $a$、$b$ 和 $c$ 而不是角度来表示。如何证明 $(π − φ)/3 < θ$ 等价于 $c^2 + bc - a^2 >0$
Screenshot 2024-12-07 094005.png

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original poster hbghlyj posted 2024-12-7 17:54
$\angle A+θ<3θ$
$\iff\angle A<2θ$
$\iff\cos\frac{\angle A}2>\cosθ$
$\iff\sqrt{s(s-a)\over bc}>\frac{a^2+b^2-c^2}{2ab}$
$\iff\sqrt{(a+b+c)(-a+b+c)\over bc}>\frac{a^2+b^2-c^2}{ab}$ WolframAlpha
$\iff c^2+bc-a^2>0$

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original poster hbghlyj posted 2024-12-7 18:15
求外角三等分线的长希望用边长 $a$、$b$ 和 $c$ 来表示。

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