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[几何] 正三角形的三等分角平分線

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Zach Posted 2024-11-4 14:05 |Read mode
(1)(2) ok
(3) 怎麼作呢?(可以應用(1)(2)?)
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Aluminiumor Posted 2024-11-4 14:50
$$\frac{a}{\sin20^\circ}=\frac{b}{\sin60^\circ}=\frac{1}{\sin100^\circ}$$
令 $\theta=10^\circ$ 则
$$a=\frac{\sin2\theta}{\cos\theta},b=\frac{\sqrt{3}}{2\cos\theta}$$
$$\begin{align*}
\frac1a-b
&=\frac{\cos\theta}{\sin2\theta}-\frac{\sqrt{3}}{2\cos\theta}\\
&=\frac{\cos\theta-\sqrt{3}\sin\theta}{\sin2\theta}\\
&=\frac{2\cos(\theta+60^\circ)}{\sin2\theta}\\
&=\frac{2\cos70^\circ}{\sin20^\circ}\\
&=2
\end{align*}$$
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