正三角形

\[
\begin{aligned}
&\text{四邊形 } APQR \text{ 為平行四邊形} \\
&\Rightarrow \frac{AP}{AB} = \frac{CQ}{CB} = \frac{CR}{AC} \\
&\Rightarrow AP = CR \quad (\because AB = AC) \\
\\
&\text{設 } AP = CR = x,\quad AR = 13 - x \\
\\
&\text{平行四邊形面積：} \\
&x(13 - x) \sin 60^\circ = 20\sqrt{3} \\
&\Rightarrow \frac{\sqrt{3}}{2} x(13 - x) = 20\sqrt{3} \\
&\Rightarrow x^2 - 13x + 40 = 0 \\
&\Rightarrow x = 5 \text{ 或 } 8 \\
\\
&\text{取 } AP=5,\ AR=8 \ (\text{或反之})，\text{餘弦定理求 } PR： \\
&PR = \sqrt{AP^2 + AR^2 - 2 \cdot AP \cdot AR \cdot \cos 60^\circ} \\
&\quad = \sqrt{25 + 64 - 2 \times 5 \times 8 \times \frac{1}{2}} \\
&\quad = \sqrt{89 - 40} = \sqrt{49} = 7
\end{aligned}
\]

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在 \(\triangle ABE\) 中，

\[
\angle ABE = 60^\circ - 15^\circ = 45^\circ,\quad
\angle AEB = 180^\circ - 15^\circ - 45^\circ = 120^\circ.
\]

由正弦定理：
\[
\frac{AE}{\sin 45^\circ} = \frac{BE}{\sin 15^\circ} = \frac{AB}{\sin 120^\circ}
\]
其中 \(AB = 1\)，因此
\[
\frac{1}{\sin 120^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}.
\]

於是
\[
BE = \frac{\sin 15^\circ}{\sin 120^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{3}}{2}}
= \frac{\sqrt{6} - \sqrt{2}}{2\sqrt{3}},
\]
\[
AE = \frac{\sin 45^\circ}{\sin 120^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}
= \frac{\sqrt{2}}{\sqrt{3}}.
\]

由對稱性（\(\triangle ABE \cong \triangle CAD\)）得 \(AD = BE\)。
正三角形 \(DEF\) 的邊長為
\[
DE = AE - AD = \frac{\sqrt{2}}{\sqrt{3}} - \frac{\sqrt{6} - \sqrt{2}}{2\sqrt{3}}
= \frac{2\sqrt{2} - (\sqrt{6} - \sqrt{2})}{2\sqrt{3}}
= \frac{3\sqrt{2} - \sqrt{6}}{2\sqrt{3}}.
\]

有理化：
\[
\frac{3\sqrt{2} - \sqrt{6}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}
= \frac{3\sqrt{6} - 3\sqrt{2}}{6}
= \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}.
\]

故正三角形 \(DEF\) 的邊長為 \(\displaystyle \frac{\sqrt{6} - \sqrt{2}}{2}\)。

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