問題の解答
検索用コード（LaTeX）
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% 例題A3.2.8：互いに接する円（One More）★★
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半径$6$の$2$つの円$\mathrm{A}$，$\mathrm{B}$が外接しており，それぞれが半径$16$の円$\mathrm{O}$に内接している．このとき，右の図のように，円$\mathrm{A}$，$\mathrm{B}$の両方に外接し，さらに円$\mathrm{O}$に内接する円$\mathrm{C}$の半径$r$を求めよ．
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% 例題A3.2.8の解答（One More）★★
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$\triangle\mathrm{CAB}$は$\mathrm{CA}=\mathrm{CB}=r+6$の二等辺三角形，$\triangle\mathrm{OAB}$は$\mathrm{OA}=\mathrm{OB}=16-6=10$の二等辺三角形である．
したがって，ABの中点をMとすると，2つの円A，BはMで接し，$\mathrm{CM}\perp\mathrm{AB},\mathrm{OM}\perp\mathrm{AB}$であるから，C，O，Mは一直線上にある．
$\triangle\mathrm{OAM}$において，
\begin{align*}
\mathrm{OM}&=\sqrt{\mathrm{OA}^2-\mathrm{AM}^2}\\
&=\sqrt{10^2-6^2}\\
&=8
\end{align*}
また，$\mathrm{CO}=16-r$であり，$\triangle\mathrm{CAM}$において，$\mathrm{CA}^2=\mathrm{AM}^2+\mathrm{CM}^2$が成り立つから，
\[
(r+6)^2=6^2+\{(16-r)+8\}^2
\]
よって，これを解いて，$r=\frac{48}{5}$
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% 問題A3.2.8：互いに接する円（One More）★★
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右の図のように，半径が等しい3つの円A，B，Cが互いに外接し，さらに，それぞれ半径1の円Oに内接している．このとき，円Aの面積$S$を求めよ．
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% 問題A3.2.8の解答（One More）★★
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右の図のように，内接円の半径を$r$，線分ABと内接円の交点をDとする．
$\angle\mathrm{OAD}=30^{\circ},\angle\mathrm{ODA}=90^{\circ}$より，
\[
\mathrm{OA}:\mathrm{AD}=2:\sqrt{3}
\]
$(1-r):r=2:\sqrt{3}$より，
\[
2r=\sqrt{3}(1-r)
\]
したがって，$r=2\sqrt{3}-3$
よって，求める面積$S$は，
\[
S=\pi r^2=\pi(2\sqrt{3}-3)^2=(21-12\sqrt{3})\pi
\]
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