【数学A】例題3.1.8：チェバの定理・メネラウスの定理（One More）★

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% 例題A3.1.8：チェバの定理・メネラウスの定理（One More）★
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右の図のような$\triangle\mathrm{ABC}$において，$x:y$を求めよ．

(1)

(2)

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% 例題A3.1.8の解答（One More）★
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(1) $\triangle\mathrm{ABC}$において，チェバの定理より，

\[
\frac{\mathrm{AP}}{\mathrm{PB}}\cdot\frac{\mathrm{BQ}}{\mathrm{QC}}\cdot\frac{\mathrm{CR}}{\mathrm{RA}}=1
\]

したがって，

\[
\frac{x}{y}\cdot\frac{3}{5}\cdot\frac{6}{2}=1
\]

ゆえに，$\frac{x}{y}=\frac{5}{9}$

よって，$x:y=5:9$

(2) $\triangle\mathrm{ABC}$と直線PQについて，メネラウスの定理より，

\[
\frac{\mathrm{AP}}{\mathrm{PB}}\cdot\frac{\mathrm{BQ}}{\mathrm{QC}}\cdot\frac{\mathrm{CR}}{\mathrm{RA}}=1
\]

したがって，

\[
\frac{2}{5}\cdot\frac{7}{1}\cdot\frac{x}{y}=1
\]

ゆえに，$\frac{x}{y}=\frac{5}{14}$

よって，$x:y=5:14$

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% 問題A3.1.8：チェバの定理・メネラウスの定理（One More）★
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右の図のような$\triangle\mathrm{ABC}$において，$x:y$を求めよ．

(1)

(2)

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% 問題A3.1.8の解答（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) $\triangle\mathrm{ABC}$において，チェバの定理より，$\frac{\mathrm{AP}}{\mathrm{PB}}\cdot\frac{\mathrm{BQ}}{\mathrm{QC}}\cdot\frac{\mathrm{CR}}{\mathrm{RA}}=1$

したがって，$\frac{x}{y}\cdot\frac{5}{3}\cdot\frac{5}{3}=1$

ゆえに，$\frac{x}{y}=\frac{9}{25}$

よって，$x:y=9:25$

(2) $\triangle\mathrm{ABC}$と直線PQについて，メネラウスの定理より，$\frac{\mathrm{AP}}{\mathrm{PB}}\cdot\frac{\mathrm{BQ}}{\mathrm{QC}}\cdot\frac{\mathrm{CR}}{\mathrm{RA}}=1$

したがって，$\frac{3}{4}\cdot\frac{7}{2}\cdot\frac{x}{y}=1$

ゆえに，$\frac{x}{y}=\frac{8}{21}$

よって，$x:y=8:21$

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% 例題A3.1.8の側注：チェバの定理・メネラウスの定理（One More）★
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% 側注1
$\frac{[1]}{[2]}\cdot\frac{[3]}{[4]}\cdot\frac{[5]}{[6]}=1$

% 側注2
$\frac{[1]}{[2]}\cdot\frac{[3]}{[4]}\cdot\frac{[5]}{[6]}=1$

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% 問題A3.1.8の側注：チェバの定理・メネラウスの定理（One More）★
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% 側注1
$\frac{[1]}{[2]}\cdot\frac{[3]}{[4]}\cdot\frac{[5]}{[6]}=1$

% 側注2
$\frac{[1]}{[2]}\cdot\frac{[3]}{[4]}\cdot\frac{[5]}{[6]}=1$

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% 例題A3.1.8の考え方・補足など：チェバの定理・メネラウスの定理（One More）★
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% 考え方
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(1) 3頂点からの直線が1点で交わるとき，チェバの定理を利用する．

\[
{\frac{\mathrm{AP}}{\mathrm{PB}}\cdot\frac{\mathrm{BQ}}{\mathrm{QC}}\cdot\frac{\mathrm{CR}}{\mathrm{RA}}=1}
\]

(2) 直線が3辺またはその延長と交わるとき，メネラウスの定理を利用する．

\[
{\frac{\mathrm{AP}}{\mathrm{PB}}\cdot\frac{\mathrm{BQ}}{\mathrm{QC}}\cdot\frac{\mathrm{CR}}{\mathrm{RA}}=1}
\]

(1)

(2)