問題の解答
検索用コード（LaTeX）
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% 例題A3.1.10：メネラウスの定理と面積比（One More）★★★
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$\triangle\mathrm{ABC}$の辺$\mathrm{BC},\mathrm{CA},\mathrm{AB}$を$2:1$に内分する点をそれぞれL，M，Nとし，ALとCN，$\mathrm{AL}$とBM，$\mathrm{BM}$とCNの交点をそれぞれP，Q，Rとする．このとき，次の三角形の面積を$\triangle\mathrm{ABC}$の面積$S$を用いて表せ．
(1) $\triangle\mathrm{ABQ}$
(2) $\triangle\mathrm{PQR}$
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% 例題A3.1.10の解答（One More）★★★
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$\mathrm{CM}:\mathrm{AM}=2:1$より，$\mathrm{CA}:\mathrm{AM}=3:1$
また，$\mathrm{BL}:\mathrm{LC}=2:1$であるから，$\triangle\mathrm{BCM}$と直線ALについて，メネラウスの定理より，
\[
\frac{\mathrm{BL}}{\mathrm{LC}}\cdot\frac{\mathrm{CA}}{\mathrm{AM}}\cdot\frac{\mathrm{MQ}}{\mathrm{QB}}=1
\]
$\frac{2}{1}\cdot\frac{3}{1}\cdot\frac{\mathrm{MQ}}{\mathrm{QB}}=1$より，$\frac{\mathrm{MQ}}{\mathrm{QB}}=\frac{1}{6}$
したがって，$\mathrm{MQ}:\mathrm{QB}=1:6$
ゆえに，$\mathrm{MB}:\mathrm{QB}=7:6$
よって，
\[
\triangle\mathrm{ABQ}=\frac{6}{7}\triangle\mathrm{ABM}=\frac{6}{7}\cdot\frac{1}{3}\triangle\mathrm{ABC}=\frac{2}{7}S
\]
(2) (1)と同様に，$\triangle\mathrm{CAN}$と直線BM，$\triangle\mathrm{ABL}$と直線CNについて，メネラウスの定理より，
\[
\triangle\mathrm{BCR}=\triangle\mathrm{CAP}=\frac{2}{7}S
\]
よって，
\begin{align*}
\triangle\mathrm{PQR}&=\triangle\mathrm{ABC}-(\triangle\mathrm{ABQ}+\triangle\mathrm{BCR}+\triangle\mathrm{CAP})\\
&=S-3\cdot\frac{2}{7}S=\frac{1}{7}S
\end{align*}
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% 問題A3.1.10：メネラウスの定理と面積比（One More）★★★
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$\triangle\mathrm{ABC}$の辺$\mathrm{BC},\mathrm{CA},\mathrm{AB}$を$3:1$に内分する点をそれぞれL，M，Nとし，ALとCN，$\mathrm{AL}$とBM，$\mathrm{BM}$とCNの交点をそれぞれP，Q，Rとする．このとき，次の三角形の面積を$\triangle\mathrm{ABC}$の面積$S$を用いて表せ．
(1) $\triangle\mathrm{ABQ}$
(2) $\triangle\mathrm{PQR}$
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% 問題A3.1.10の解答（One More）★★★
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$\mathrm{CM}:\mathrm{AM}=3:1$より，$\mathrm{CA}:\mathrm{AM}=4:1$
また，$\mathrm{BL}:\mathrm{LC}=3:1$であるから，$\triangle\mathrm{BCM}$と直線ALについて，メネラウスの定理より，
\[
\frac{\mathrm{BL}}{\mathrm{LC}}\cdot\frac{\mathrm{CA}}{\mathrm{AM}}\cdot\frac{\mathrm{MQ}}{\mathrm{QB}}=1
\]
$\frac{3}{1}\cdot\frac{4}{1}\cdot\frac{\mathrm{MQ}}{\mathrm{QB}}=1$より，$\frac{\mathrm{MQ}}{\mathrm{QB}}=\frac{1}{12}$
したがって，$\mathrm{MQ}:\mathrm{QB}=1:12$
ゆえに，$\mathrm{MB}:\mathrm{QB}=13:12$
よって，
\[
\triangle\mathrm{ABQ}=\frac{12}{13}\triangle\mathrm{ABM}=\frac{12}{13}\cdot\frac{1}{4}\triangle\mathrm{ABC}=\frac{3}{13}S
\]
(2) (1)と同様に，$\triangle\mathrm{CAN}$と直線BM，$\triangle\mathrm{ABL}$と直線CNについて，メネラウスの定理より，
\[
\triangle\mathrm{BCR}=\triangle\mathrm{CAP}=\frac{3}{13}S
\]
よって，
\begin{align*}
\triangle\mathrm{PQR}&=\triangle\mathrm{ABC}-(\triangle\mathrm{ABQ}+\triangle\mathrm{BCR}+\triangle\mathrm{CAP})\\
&=S-3\cdot\frac{3}{13}S=\frac{4}{13}S
\end{align*}
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