KakomonRikei_MatsuyamaSota.tex

\documentclass[a4j, 11pt]{jarticle}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

\begin{document}
\section*{2}
% (2)
\subsection*{(2)}
$\frac{\pi}{2} < \angle \mathrm{B'AC'} < \pi$のとき, $\cos \angle \mathrm{B'AC} < 0$なので,
$\overrightarrow{\mathrm{AB'}} \cdot \overrightarrow{\mathrm{AC'}} < 0$
であることを示せば良い.
\begin{align*}
	&\quad \overrightarrow{\mathrm{AB'}} \cdot \overrightarrow{\mathrm{AC'}} \\
	%
	&= (\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BB'}}) \cdot
	(\overrightarrow{\mathrm{AC}} + \overrightarrow{\mathrm{CC'}}) \\
	%
	&= \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{AC}}
	+ \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CC'}}
	+ \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{AC}}
	+ \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{CC'}} \\
	%
	&= 0 + (\overrightarrow{\mathrm{AB'}} + \overrightarrow{\mathrm{B'B}}) \cdot
	\overrightarrow{\mathrm{CC'}} + \overrightarrow{\mathrm{BB'}} \cdot
	(\overrightarrow{\mathrm{AC'}} + \overrightarrow{\mathrm{C'C}})
	+ \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{CC'}} \\
	%
	&= \overrightarrow{\mathrm{AB'}} \cdot \overrightarrow{\mathrm{CC'}}
	+ \overrightarrow{\mathrm{B'B}} \cdot \overrightarrow{\mathrm{CC'}}
	+ \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{AC'}}
	+ \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{C'C}}
	+ \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{CC'}} \\
	%
	&= - \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{CC'}}
	- \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{CC'}}
	+ \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{CC'}} \\
	%
	&= - \overrightarrow{\mathrm{BB'}} \cdot \overrightarrow{\mathrm{CC'}} \\
	%
	&= - |\mathrm{BB'}| |\mathrm{CC'}| \cos 0 < 0
	\quad (\because \mathrm{BB'} \parallel \mathrm{CC'}, 
	\quad |\mathrm{BB'}| > 0 \text{かつ} |\mathrm{CC'}| > 0)
	%
\end{align*}
したがって, 
$\overrightarrow{\mathrm{AB'}} \cdot \overrightarrow{\mathrm{AC'}} < 0$が成り立つので, 
$\angle \mathrm{B'AC'} > \frac{\pi}{2}$が成立する.
\end{document}