【数学I】例題4.3.7：中線定理（One More）★★

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% 例題I4.3.7：中線定理（One More）★★
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(1) $\triangle\mathrm{ABC}$において，辺BCの中点をMとする．このとき，

\[
\mathrm{AB}^2+\mathrm{AC}^2=2(\mathrm{AM}^2+\mathrm{BM}^2)
\]

が成り立つことを証明せよ．

(2) $\mathrm{AB}=4,\mathrm{BC}=8,\mathrm{CA}=5$である$\triangle\mathrm{ABC}$において，辺BCの中点をMとするとき，線分AMの長さを求めよ．

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% 例題I4.3.7の解答（One More）★★
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(1) $\angle\mathrm{AMB}=\theta$とすると，$\angle\mathrm{AMC}=180^{\circ}-\theta$

$\triangle\mathrm{ABM}$において，余弦定理より，

\[
\mathrm{AB}^2=\mathrm{AM}^2+\mathrm{BM}^2-2\cdot\mathrm{AM}\cdot\mathrm{BM}\cdot\cos\theta\cdots(\mathrm{i})
\]

$\triangle\mathrm{ACM}$において，余弦定理より，

\[
\mathrm{AC}^2=\mathrm{AM}^2+\mathrm{CM}^2-2\cdot\mathrm{AM}\cdot\mathrm{CM}\cdot\cos(180^\circ-\theta)\cdots(\mathrm{ii})
\]

(i)と(ii)の辺々を足し合わせると，$\mathrm{BM}=\mathrm{CM}$より，

\begin{align*}
\mathrm{AB}^2+\mathrm{AC}^2&=2(\mathrm{AM}^2+\mathrm{BM}^2)-2\mathrm{AM}\cdot\mathrm{BM}\{\cos\theta+\cos(180^{\circ}-\theta)\}\\
&=2(\mathrm{AM}^2+\mathrm{BM}^2)-2\mathrm{AM}\cdot\mathrm{BM}(\cos\theta-\cos\theta)\\
&=2(\mathrm{AM}^2+\mathrm{BM}^2)\blacksquare
\end{align*}

(2) $\mathrm{AB}=4,\mathrm{BM}=\frac{1}{2}\mathrm{BC}=4,\mathrm{AC}=5$とすると，(1)より，

\[
4^2+5^2=2(\mathrm{AM}^2+4^2)
\]

したがって，$\mathrm{AM}^2=\frac{9}{2}$

よって，$\mathrm{AM}>0$より，$\mathrm{AM}=\frac{3}{\sqrt{2}}$

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% 問題I4.3.7：中線定理（One More）★★
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$\mathrm{AB}=5,\mathrm{BC}=10,\mathrm{CA}=6$である$\triangle\mathrm{ABC}$において，辺BCの中点をMとするとき，線分AMの長さを求めよ．

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% 問題I4.3.7の解答（One More）★★
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中線定理より，
\[
\mathrm{AB}^2+\mathrm{AC}^2=2(\mathrm{AM}^2+\mathrm{BM}^2)
\]

これに，$\mathrm{AB}=5,\mathrm{BM}=\frac{1}{2}\mathrm{BC}=5,\mathrm{AC}=6$を代入すると，

\[
5^2+6^2=2(\mathrm{AM}^2+5^2)
\]

したがって，$\mathrm{AM}^2=\frac{11}{2}$

よって，$\mathrm{AM}>0$より，$\mathrm{AM}=\frac{\sqrt{22}}{2}$

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% 例題I4.3.7の側注：中線定理（One More）★★
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% 側注1
$\cos(180^{\circ}-\theta)=-\cos\theta$

% 側注2
中線定理を用いると，楽に中線の長さを求めることができる．中線定理を用いずに，$\triangle\mathrm{ABC},\triangle\mathrm{ABM}$でそれぞれ余弦定理を用いたり，三平方の定理を用いたりしてもよいが，計算に手間が掛かる．

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% 問題I4.3.7の側注：中線定理（One More）★★
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% 側注1
中線定理を用いると，楽に中線の長さを求めることができる．

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% 例題I4.3.7の考え方・補足など：中線定理（One More）★★
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% 考え方
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$\triangle\mathrm{ABC}$において，辺BCの中点をMとするとき，AMを中線といい，$\mathrm{AB}^2+\mathrm{AC}^2=2(\mathrm{AM}^2+\mathrm{BM}^2)$を中線定理という（パップスの中線定理ともいう）．$\angle\mathrm{AMB}=\theta,\angle\mathrm{AMC}=180^{\circ}-\theta$とすると，$\cos(180^{\circ}-\theta)=-\cos\theta$を利用して証明することができる．