△ABC において, \(\angle \text{BAC} = 90^{\circ}\) , \(\left| \overrightarrow{\text{AB}} \right| = 1\) , \(\left| \overrightarrow{\text{AC}} \right| = \sqrt{3}\) とする. △ABC の内部の点 P が \[ \dfrac{\overrightarrow{\text{PA}}}{\left| \overrightarrow{\text{PA}} \right|} +\dfrac{\overrightarrow{\text{PB}}}{\left| \overrightarrow{\text{PB}} \right|} +\dfrac{\overrightarrow{\text{PC}}}{\left| \overrightarrow{\text{PC}} \right|} = \overrightarrow{0} \] を満たすとする.
(1) \(\angle \text{APB}\) , \(\angle \text{APC}\) を求めよ.
(2) \(\left| \overrightarrow{\text{PA}} \right|\) , \(\left| \overrightarrow{\text{PB}} \right|\) , \(\left| \overrightarrow{\text{PC}} \right|\) を求めよ.
【 解 答 】
(1)
\[ \dfrac{\overrightarrow{\text{PA}}}{\left| \overrightarrow{\text{PA}} \right|} +\dfrac{\overrightarrow{\text{PB}}}{\left| \overrightarrow{\text{PB}} \right|} +\dfrac{\overrightarrow{\text{PC}}}{\left| \overrightarrow{\text{PC}} \right|} = \overrightarrow{0} \quad ...\text{[A]} \] [A] の両辺について, \(\overrightarrow{\text{PA}}\) との内積を考えると \[\begin{align} \overrightarrow{\text{PA}} \cdot \left( \dfrac{\overrightarrow{\text{PA}}}{\left| \overrightarrow{\text{PA}} \right|} +\dfrac{\overrightarrow{\text{PB}}}{\left| \overrightarrow{\text{PB}} \right|} +\dfrac{\overrightarrow{\text{PC}}}{\left| \overrightarrow{\text{PC}} \right|} \right) = \overrightarrow{\text{PA}} & \cdot \overrightarrow{0} \\ \left| \overrightarrow{\text{PA}} \right| + \left| \overrightarrow{\text{PA}} \right| \cos \angle \text{APB} + \left| \overrightarrow{\text{PA}} \right| \cos \angle \text{APC} & = 0 \end{align}\] \(\left| \overrightarrow{\text{PA}} \right| \neq 0\) なので \[ \cos \angle \text{APB} + \cos \angle \text{APC} = -1 \quad ... [1] \] 同様に, [A] の両辺について, \(\overrightarrow{\text{PB}} , \overrightarrow{\text{PC}}\) との内積を考えれば \[\begin{align} \cos \angle \text{BPC} + \cos \angle \text{APC} = -1 \quad ... [2] \\ \cos \angle \text{APB} + \cos \angle \text{BPC} = -1 \quad ... [3] \end{align}\] [1] ～ [3] をとくと \[ \cos \angle \text{APB} = \cos \angle \text{APC} = \cos \angle \text{APB} ＝-\dfrac{1}{2} \] よって \[ \text{APB} = \angle \text{APC} = \angle \text{APB} = \underline{120^{\circ}} \]
(2)
\(x = \left| \overrightarrow{\text{PA}} \right|\) , \(y = \left| \overrightarrow{\text{PB}} \right|\) , \(z = \left| \overrightarrow{\text{PC}} \right|\) とおく.
また, \(\alpha = \angle \text{PAB}\) とおくと
\[\begin{align}
\angle \text{PAC} & = 90^{\circ} -\alpha \\
\angle \text{PBA} & = 180^{\circ} -120^{\circ} -\alpha = 60^{\circ} -\alpha
\end{align}\]
また, 条件より \(\angle \text{ABC} = 60^{\circ}\) .
\[
\angle \text{PBC} = 60^{\circ} -( 60^{\circ} -\alpha ) = \alpha
\]
したがって
\[
\angle \text{PAB} = \angle \text{PBC}
\]
これと (1) の結果から, \(2\) 角が等しいので
\[
\triangle \text{ABP} \sim \triangle \text{BCP}
\]
さらに \(\text{AB} = 1\) , \(\text{BC} = 2\) なので, 相似比は \(1 : 2\) だから
\[\begin{align}
x : y & = y : z = 1 : 2 \\
\text{∴} \quad y & = 2x , \quad z = 4x \quad ... [4]
\end{align}\]
ここで, \(\triangle \text{ABP}\) について, 余弦定理から
\[\begin{align}
x^2 +(2x)^2 -2 x & \cdot 2x \cdot \cos 120^{\circ} = 1 \\
7x^2 & = 1 \\
\text{∴} \quad x & = \dfrac{\sqrt{7}}{7}
\end{align}\]
よって, [4] に代入すれば
\[
\left| \overrightarrow{\text{PA}} \right| = \underline{\dfrac{\sqrt{7}}{7}} , \quad \left| \overrightarrow{\text{PB}} \right| = \underline{\dfrac{2 \sqrt{7}}{7}} , \quad \left| \overrightarrow{\text{PC}} \right| = \underline{\dfrac{4 \sqrt{7}}{7}}
\]