【数学I】例題4.3.4：円に内接する四角形1（One More）★★

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% 例題I4.3.4：円に内接する四角形1（One More）★★
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円に内接する四角形ABCDにおいて,$\mathrm{AB}=4,\mathrm{BC}=3,\mathrm{CD}=1,\angle\mathrm{ABC}=60^{\circ}$とする．このとき，次の値を求めよ．

(1) 対角線ACの長さ

(2) ADの長さ

(3) 四角形ABCDの面積

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(1) $\triangle\mathrm{ABC}$において，余弦定理より，$\mathrm{AC}^2=4^2+3^2-2\cdot 4\cdot 3\cdot\cos 60^{\circ}=13$

よって，$\mathrm{AC}>0$より，$\mathrm{AC}=\sqrt{13}$

(2) 四角形ABCDは円に内接するから，

\[
D=180^{\circ}-B=180^{\circ}-60^{\circ}=120^{\circ}
\]

したがって，$\triangle\mathrm{ACD}$において，余弦定理より，

\[
(\sqrt{13})^2=1^2+\mathrm{AD}^2-2\cdot 1\cdot\mathrm{AD}\cdot\cos 120^{\circ}
\]

整理すると，$\mathrm{AD}^2+\mathrm{AD}-12=0$より，$(\mathrm{AD}+4)(\mathrm{AD}-3)=0$

よって，$\mathrm{AD}>0$より，$\mathrm{AD}=3$

(3) 四角形ABCDの面積を$S$とすると，

\begin{align*}
S&=\triangle\mathrm{ABC}+\triangle\mathrm{ACD}\\
&=\frac{1}{2}\cdot 4\cdot 3\cdot\sin 60^{\circ}+\frac{1}{2}\cdot 1\cdot 3\cdot\sin 120^{\circ}\\
&=\frac{1}{2}\cdot 4\cdot 3\cdot\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot 1\cdot 3\cdot\frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{4}
\end{align*}

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% 問題I4.3.4：円に内接する四角形1（One More）★★
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円に内接する四角形ABCDにおいて，$\mathrm{AB}=8,\mathrm{BC}=4,\mathrm{CD}=4,\angle\mathrm{BCD}=120^{\circ}$とする．このとき，次の値を求めよ．

(1) 対角線BDの長さ

(2) ADの長さ

(3) 四角形ABCDの面積

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% 問題I4.3.4の解答（One More）★★
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(1) $\triangle\mathrm{BCD}$において，余弦定理より，$\mathrm{BD}^2=4^2+4^2-2\cdot 4\cdot 4\cdot\cos 120^{\circ}=48$

よって，$\mathrm{BD}>0$より，$\mathrm{BD}=4\sqrt{3}$

(2) 四角形ABCDは円に内接するから，
\[
\angle\mathrm{BAD}=180^\circ-\angle\mathrm{BCD}=60^\circ
\]

したがって，$\triangle\mathrm{ABD}$において，余弦定理より，

\[
(4\sqrt{3})^2=\mathrm{AD}^2+8^2-2\cdot\mathrm{AD}\cdot 8\cos 60^{\circ}
\]

整理すると，$\mathrm{AD}^2-8\mathrm{AD}+16=0$より，$(\mathrm{AD}-4)^2=0$

よって，$\mathrm{AD}>0$より，$\mathrm{AD}=4$

(3) 四角形ABCDの面積を$S$とすると，

\begin{align*}
S&=\triangle\mathrm{ABD}+\triangle\mathrm{BCD}\\
&=\frac{1}{2}\cdot 8\cdot 4\cdot\sin 60^{\circ}+\frac{1}{2}\cdot 4\cdot 4\cdot\sin 120^{\circ}\\
&=\frac{1}{2}\cdot 8\cdot 4\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot 4\cdot 4\frac{\sqrt{3}}{2}=12\sqrt{3}
\end{align*}

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% 例題I4.3.4の側注：円に内接する四角形1（One More）★★
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% 側注1
四角形の面積は，2つの三角形の面積の和と考えられる．

% 側注2
2辺の長さが$a,b$であり，その間の角が$\theta$のとき，三角形の面積$S$は，$S=\frac{1}{2}ab\sin\theta$

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% 問題I4.3.4の側注：円に内接する四角形1（One More）★★
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% 側注1
四角形の面積は，2つの三角形の面積の和と考えられる．

% 側注2
2辺の長さが$a,b$であり，その間の角が$\theta$のとき，三角形の面積$S$は，$S=\frac{1}{2}ab\sin\theta$

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% 例題I4.3.4の考え方・補足など：円に内接する四角形1（One More）★★
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% 考え方
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(1) 2辺とその間の角が与えられているので，余弦定理を用いる．

(2) 円に内接する四角形の対角の和は$180^\circ$であることを用いる．$B+D=180^{\circ}$より，$D$を求めて，$\triangle\mathrm{ACD}$において余弦定理を用いる．

(3) 2つの三角形$\triangle\mathrm{ABC},\triangle\mathrm{ACD}$に分けて，それぞれに対して三角形の面積の公式を用いる．

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% 余談
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覚えて利用するようなものではないが，円に内接する四角形の面積を求める公式として，次のようなブラーマグプタの公式がある．ブラーマグプタの公式に$d=0$を代入すると，ヘロンの公式が得られる．

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% ブラーマグプタの公式
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円に内接する四角形の4辺の長さを$a,b,c,d$とし，$2s=a+b+c+d$とすると，四角形の面積$S$は
\[
S=\sqrt{(s-a)(s-b)(s-c)(s-d)}
\]