【数学A】例題3.2.2：接線の長さ（One More）★★

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% 例題A3.2.2：接線の長さ（One More）★★
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右の図のように，$\triangle\mathrm{ABC}$において，$\mathrm{AB}=5$，$\mathrm{BC}=7$，$\mathrm{CA}=6$とする．また，$\triangle\mathrm{ABC}$の内接円と辺AB，BC，CAの接点を，それぞれ点$\mathrm{D}$，$\mathrm{E}$，$\mathrm{F}$とするとき，$\mathrm{AD}$の長さを求めよ．

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% 例題A3.2.2の解答（One More）★★
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$\mathrm{AD}=x$とすると，$\mathrm{BD}=\mathrm{BE}=5-x\cdots(\mathrm{i})$

また，$\mathrm{AF}=x$であるから，

\[
\mathrm{FC}=\mathrm{EC}=6-x\cdots(\mathrm{ii})
\]

(i)，(ii)より，

\[
\mathrm{BC}=\mathrm{BE}+\mathrm{EC}=(5-x)+(6-x)=7
\]

したがって，$x=2$

よって，$\mathrm{AD}=2$

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% 例題A3.2.2の別解
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$\mathrm{AD}=x$，$\mathrm{BE}=y$，$\mathrm{CF}=z$とすると，$\mathrm{AD}=\mathrm{AF}$，$\mathrm{BE}=\mathrm{BD}$，$\mathrm{CF}=\mathrm{CE}$であるから，

\[
x+y=5,y+z=7,z+x=6
\]

辺々を足し合わせると，$2(x+y+z)=18$

したがって，$x+y+z=9$

ゆえに，$y+z=7$より，$x=2$

よって，$\mathrm{AD}=2$

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% 問題A3.2.2：接線の長さ（One More）★★
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$\triangle\mathrm{ABC}$において，$\mathrm{AB}=7$，$\mathrm{BC}=9$，$\mathrm{CA}=8$とする．また，$\triangle\mathrm{ABC}$の内接円と辺AB，BC，CAの接点を，それぞれ点$\mathrm{D}$，$\mathrm{E}$，$\mathrm{F}$とするとき，$\mathrm{AD}$の長さを求めよ．

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% 問題A3.2.2の解答（One More）★★
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$\mathrm{AD}=x$とすると，$\mathrm{BD}=\mathrm{BE}=7-x\cdots(\mathrm{i})$

また，$\mathrm{AF}=x$であるから，$\mathrm{FC}=\mathrm{EC}=8-x\cdots(\mathrm{ii})$

(i)，(ii)より，

\[
\mathrm{BC}=\mathrm{BE}+\mathrm{EC}=(7-x)+(8-x)=9
\]

したがって，$x=3$

よって，$\mathrm{AD}=3$

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% 問題A3.2.2の別解
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$\mathrm{AD}=x$，$\mathrm{BE}=y$，$\mathrm{CF}=z$とすると，$\mathrm{AD}=\mathrm{AF}$，$\mathrm{BE}=\mathrm{BD}$，$\mathrm{CF}=\mathrm{CE}$であるから，

\[
x+y=7,y+z=9,z+x=8
\]

辺々を足し合わせると，$2(x+y+z)=24$

したがって，$x+y+z=12$

ゆえに，$y+z=9$より，$x=3$

よって，$\mathrm{AD}=3$

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% 例題A3.2.2の側注：接線の長さ（One More）★★
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% 側注1
\begin{align*}
&\mathrm{AD}+\mathrm{DB}=\mathrm{AB},\\
&\mathrm{BE}+\mathrm{EC}=\mathrm{BC},\\
&\mathrm{CF}+\mathrm{FA}=\mathrm{CA}
\end{align*}

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% 問題A3.2.2の側注：接線の長さ（One More）★★
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% 側注1
\begin{align*}
&\mathrm{AD}+\mathrm{DB}=\mathrm{AB},\\
&\mathrm{BE}+\mathrm{EC}=\mathrm{BC},\\
&\mathrm{CF}+\mathrm{FA}=\mathrm{CA}
\end{align*}

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% 例題A3.2.2の考え方・補足など：接線の長さ（One More）★★
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% 考え方
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円の外部の点から円に引いた2本の接線の長さは等しいことを利用する．求める長さを$x$などとおいて，方程式を作るとよい．

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% 接線の長さ
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円Oの外部の点Aから，その円Oに引いた2本の接線について，2つの接線の長さは等しい．

\[
\mathrm{AP}=\mathrm{AQ}
\]