問題の解答
検索用コード（LaTeX）
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底面の中心がOで半径が3，高さが$6\sqrt{2}$の直円錐がある．直円錐の頂点をA，底面の直径の両端をB，Cとし，線分AC上に$\mathrm{AP}=3$となる点Pをとる．側面上において，点Bから点Pまでに至る最短距離を求めよ．
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% 例題I4.3.11の解答（One More）★★★
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側面を直線ABに沿って切り開いた展開図は，右の図のように，中心A，半径ABの扇形となる．
三角形ABOにおいて，$\mathrm{BO}=3,\mathrm{AO}=6\sqrt{2}$より，
\[
\mathrm{AB}=\sqrt{3^2+(6\sqrt{2})^2}=9
\]
求める最短距離の長さは，展開図において，線分BPの長さである．
弧$\mathrm{BCB}^{\prime}$の長さは，$2\pi\cdot 3=6\pi$
扇形の半径は9であるから，中心角$\angle\mathrm{BAB}^\prime$は，
\[
\angle\mathrm{BAB}^{\prime}=360^{\circ}\times\frac{6\pi}{2\pi\cdot 9}=120^{\circ}
\]
したがって，$\angle\mathrm{BAC}=60^{\circ}$
$\triangle\mathrm{ABP}$において，余弦定理より，
\[
\mathrm{BP}^2=9^2+3^2-2\cdot 9\cdot 3\cdot\cos 60^{\circ}=90-54\cdot\frac{1}{2}=63
\]
よって，$\mathrm{BP}>0$より，$\mathrm{BP}=3\sqrt{7}$
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% 問題I4.3.11：空間図形における最短距離（One More）★★★
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底面の中心がOで半径が2，高さが$2\sqrt{15}$の直円錐がある．直円錐の頂点をA，底面の直径の両端をB，Cとし，線分AC上に$\mathrm{AP}=\sqrt{2}$となる点Pをとる．側面上において，点Bから点Pまでに至る最短距離を求めよ．
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% 問題I4.3.11の解答（One More）★★★
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側面を直線ABに沿って切り開いた展開図は，右の図のように，中心A，半径ABの扇形となる．
{三角形ABO}において，$\mathrm{BO}=2,\mathrm{AO}=2\sqrt{15}$より，
\[
{\mathrm{AB}=\sqrt{2^2+(2\sqrt{15})^2}=8}
\]
求める最短距離の長さは，展開図において，線分BPの長さである．
弧$\mathrm{BCB}^{\prime}$の長さは，$2\pi\cdot 2=4\pi$
扇形の半径は$8$であるから，中心角$\angle\mathrm{BAB}^\prime$は，
\[
\angle\mathrm{BAB}^{\prime}=360^{\circ}\times\frac{4\pi}{2\pi\cdot 8}=90^{\circ}
\]
したがって，$\angle\mathrm{BAC}=45^{\circ}$
$\triangle\mathrm{ABP}$において，余弦定理より，
\[
\mathrm{BP}^2=8^2+(\sqrt{2})^2-2\cdot 8\cdot\sqrt{2}\cdot\cos 45^{\circ}=66-16\sqrt{2}\cdot\frac{\sqrt{2}}{2}=50
\]
よって，$\mathrm{BP}=\sqrt{50}=5\sqrt{2}$
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