【数学A】例題3.1.7：三角形の面積比（One More）★★

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% 例題A3.1.7：三角形の面積比（One More）★★
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$\triangle\mathrm{ABC}$において，線分$\mathrm{AB}$を$2:1$に内分する点を$\mathrm{D}$，線分BCの中点をE，線分$\mathrm{CA}$を$1:3$に内分する点を$\mathrm{F}$とする．このとき，$\triangle\mathrm{ABC}$と$\triangle\mathrm{DEF}$の面積比を求めよ．

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% 例題A3.1.7の解答（One More）★★
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$\triangle\mathrm{DEF}=\triangle\mathrm{ABC}-(\triangle\mathrm{ADF}+\triangle\mathrm{BED}+\triangle\mathrm{CFE})\cdots(\mathrm{i})$

また，$\triangle\mathrm{ADF}:\triangle\mathrm{ABC}=\mathrm{AD}\cdot\mathrm{AF}:\mathrm{AB}\cdot\mathrm{AC}$より，

\[
\triangle\mathrm{ADF}=\frac{2\cdot 3}{3\cdot 4}\triangle\mathrm{ABC}=\frac{1}{2}\triangle\mathrm{ABC}
\]

同様に，

\[
\triangle\mathrm{BED}=\frac{1\cdot 1}{2\cdot 3}\triangle\mathrm{ABC}=\frac{1}{6}\triangle\mathrm{ABC},\triangle\mathrm{CFE}=\frac{1\cdot 1}{4\cdot 2}\triangle\mathrm{ABC}=\frac{1}{8}\triangle\mathrm{ABC}
\]

したがって，これらを(i)に代入すると，

\[
\triangle\mathrm{DEF}=\triangle\mathrm{ABC}-(\frac{1}{2}+\frac{1}{6}+\frac{1}{8})\triangle\mathrm{ABC}=\frac{5}{24}\triangle\mathrm{ABC}
\]

よって，

\[
\triangle\mathrm{ABC}:\triangle\mathrm{DEF}=\triangle\mathrm{ABC}:\frac{5}{24}\triangle\mathrm{ABC}=24:5
\]

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% 問題A3.1.7：三角形の面積比（One More）★★
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$\triangle\mathrm{ABC}$において，線分$\mathrm{AB}$を$4:1$に内分する点を$\mathrm{D}$，線分BCを$3:1$に内分する点を$\mathrm{E}$，線分$\mathrm{CA}$を$1:3$に内分する点を$\mathrm{F}$とする．このとき，$\triangle\mathrm{ABC}$と$\triangle\mathrm{DEF}$の面積比を求めよ．

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% 問題A3.1.7の解答（One More）★★
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$\triangle\mathrm{DEF}=\triangle\mathrm{ABC}-(\triangle\mathrm{ADF}+\triangle\mathrm{BED}+\triangle\mathrm{CFE})\cdots(\mathrm{i})$

また，$\triangle\mathrm{ADF}:\triangle\mathrm{ABC}=\mathrm{AD}\cdot\mathrm{AF}:\mathrm{AB}\cdot\mathrm{AC}$より，

\[
\triangle\mathrm{ADF}=\frac{4\cdot 3}{5\cdot 4}\triangle\mathrm{ABC}=\frac{3}{5}\triangle\mathrm{ABC}
\]

同様に，

\[
\triangle\mathrm{BED}=\frac{3\cdot 1}{4\cdot 5}\triangle\mathrm{ABC}=\frac{3}{20}\triangle\mathrm{ABC},\triangle\mathrm{CFE}=\frac{1\cdot 1}{4\cdot 4}\triangle\mathrm{ABC}=\frac{1}{16}\triangle\mathrm{ABC}
\]

したがって，これらを(i)に代入すると，

\[
\triangle\mathrm{DEF}=\triangle\mathrm{ABC}-(\frac{3}{5}+\frac{3}{20}+\frac{1}{16})\triangle\mathrm{ABC}=\frac{3}{16}\triangle\mathrm{ABC}
\]

よって，

\[
\triangle\mathrm{ABC}:\triangle\mathrm{DEF}=\triangle\mathrm{ABC}:\frac{3}{16}\triangle\mathrm{ABC}=16:3
\]

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% 例題A3.1.7の側注：三角形の面積比（One More）★★
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% 側注1
$\triangle\mathrm{ABC}$の面積から，$\triangle\mathrm{DEF}$のまわりの三角形の面積を引く．

% 側注2
与えられた条件より，

\begin{align*}
&\mathrm{AD}:\mathrm{DB}=2:1,\\
&\mathrm{BE}:\mathrm{EC}=1:1,\\
&\mathrm{CF}:\mathrm{FA}=1:3
\end{align*}

である．

% 側注3
$\triangle\mathrm{DEF}$を$\triangle\mathrm{ABC}$で表す．

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% 問題A3.1.7の側注：三角形の面積比（One More）★★
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% 側注1
$\triangle\mathrm{ABC}$の面積から，$\triangle\mathrm{DEF}$のまわりの三角形の面積を引く．

% 側注2
与えられた条件より，

\begin{align*}
&\mathrm{AD}:\mathrm{DB}=4:1,\\
&\mathrm{BE}:\mathrm{EC}=3:1,\\
&\mathrm{CF}:\mathrm{FA}=1:3
\end{align*}

である．

% 側注3
$\triangle\mathrm{DEF}$を$\triangle\mathrm{ABC}$で表す．

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% 例題A3.1.7の考え方・補足など：三角形の面積比（One More）★★
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% 考え方
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例えば，$\triangle\mathrm{ABC}$と$\triangle\mathrm{ADF}$の面積比は，$\angle\mathrm{A}$が共通であるから，

\[
\triangle\mathrm{ADF}:\triangle\mathrm{ABC}=\frac{1}{2}\cdot\mathrm{AD}\cdot\mathrm{AF}\cdot\sin A:\frac{1}{2}\cdot\mathrm{AB}\cdot\mathrm{AC}\cdot\sin A
\]

より，$\triangle\mathrm{ADF}:\triangle\mathrm{ABC}=\mathrm{AD}\cdot\mathrm{AF}:\mathrm{AB}\cdot\mathrm{AC}$となることを利用する．