鋭角三角形 ABC の \(\angle \text{A} , \angle \text{B} , \angle \text{C}\) の大きさをそれぞれ \(\alpha , \beta , \gamma\) で表す. 点 D , E , F はそれぞれ辺 CA , AB , BC 上にあり, \(\text{DE} \perp \text{AB}\) , \(\text{EF} \perp \text{BC}\) , \(\text{FD} \perp \text{CA}\) を満たす. 次の問いに答えよ.
(1) △ABC と △DEF は相似であることを示せ.
(2) \(\dfrac{\text{BC}}{\text{EF}} = \dfrac{1}{\tan \alpha} +\dfrac{1}{\tan \beta} +\dfrac{1}{\tan \gamma}\) を示せ.
(3) \(\alpha\) が一定のとき, \(\dfrac{\text{BC}}{\text{EF}}\) を最小にするような \(\beta , \gamma\) を \(\alpha\) で表せ.
【 解 答 】
(1)
\[ \angle \text{FDE} = \dfrac{\pi}{2} -\angle \text{ADE} =\dfrac{\pi}{2} -\left( \dfrac{\pi}{2} -\alpha \right) =\alpha \] 同様にすれば \[ \angle \text{DEF} = \dfrac{\pi}{2} -\angle \text{BEF} = \dfrac{\pi}{2} -\left( \dfrac{\pi}{2} -\beta \right) = \beta \] よって, \(2\) 角がそれぞれ等しいので \[ \triangle \text{ABC} \sim \triangle \text{DEF} \]
(2)
\(\text{BC} = a\) , \(\text{AC} = b\) とおく.
正弦定理より
\[\begin{align}
\dfrac{a}{\sin \alpha} & = \dfrac{b}{\sin \beta} \quad ... [1] \\
\text{∴} \quad b & = \dfrac{a \sin \beta}{\sin \alpha}
\end{align}\]
(1) の結果より, \(\dfrac{\text{BC}}{\text{EF}} = \dfrac{\text{CA}}{\text{FD}} = k\) とおけるので
\[\begin{align}
\text{BF} & = \dfrac{\text{EF}}{\tan \beta} = \dfrac{a}{k \tan \beta} \quad , \\
\text{FC} & = \dfrac{\text{FD}}{\sin \gamma} = \dfrac{a \sin \beta}{k \sin \alpha \sin \gamma} \quad ( \ \text{∵} \ [1] \ )
\end{align}\]
\(\text{BC} = \text{BF} +\text{FC}\) なので
\[\begin{align}
a & = \dfrac{a}{k \tan \beta} +\dfrac{a \sin \beta}{k \sin \alpha \sin \gamma} \\
\text{∴} \quad k & = \dfrac{1}{\tan \beta} +\dfrac{\sin \beta}{\sin \alpha \sin \gamma}
\end{align}\]
ここで
\[\begin{align}
\sin \beta & = \sin \left\{ \pi -( \alpha +\gamma ) \right\} =\sin ( \alpha +\gamma ) \\
& = \sin \alpha \cos \gamma +\cos \alpha \sin \gamma
\end{align}\]
なので
\[\begin{align}
k & = \dfrac{1}{\tan \beta} +\dfrac{\sin \alpha \cos \gamma +\cos \alpha \sin \gamma}{\sin \alpha \sin \gamma} \\
& = \dfrac{1}{\tan \beta} +\dfrac{\cos \alpha}{\sin \alpha} +\dfrac{\cos \gamma}{\sin \gamma} \\
& = \dfrac{1}{\tan \alpha} +\dfrac{1}{\tan \beta} +\dfrac{1}{\tan \gamma}
\end{align}\]
(3)
(2) の結果を用いれば \[ k = \dfrac{1}{\tan \alpha} +\dfrac{\sin \alpha}{\sin \beta \sin \gamma} \] ここで \[\begin{align} \sin \beta \sin \gamma & = -\dfrac{1}{2} \left\{ \cos ( \beta +\gamma ) -\cos ( \beta -\gamma ) \right\} \\ & = -\dfrac{1}{2} \left\{ \cos ( \pi -\alpha ) -\cos ( \beta -\gamma ) \right\} \\ & = \dfrac{1}{2} \left\{ \cos \alpha +\cos ( \beta -\gamma ) \right\} \end{align}\] なので \[ k =\dfrac{1}{\tan \alpha} +\dfrac{2 \sin \alpha}{\cos \alpha +\cos ( \beta -\gamma )} \] \(\alpha\) は定数なので, \(k\) が最小になるのは, \(\cos ( \beta -\gamma )\) が最大になるときで \[ \beta = \gamma = \underline{\dfrac{\pi -\alpha}{2}} \]