【数学A】例題3.3.4：多面体の切断・体積（One More）★★★

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% 例題A3.3.4：多面体の切断・体積（One More）★★★
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1辺の長さが4の立方体$\mathrm{ABCD}-\mathrm{EFGH}$において，辺$\mathrm{AB},\mathrm{AD},\mathrm{CG}$の中点をそれぞれ$\mathrm{P},\mathrm{Q},\mathrm{R}$とする．この3点$\mathrm{P},\mathrm{Q},\mathrm{R}$を通る平面で立方体を切断したとき，点$\mathrm{C}$を含む部分の立体の体積を求めよ．

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% 例題A3.3.4の解答（One More）★★★
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直線$\mathrm{PQ}$と直線$\mathrm{BC}$との交点を$\mathrm{S}$，直線$\mathrm{PQ}$と直線$\mathrm{CD}$との交点を$\mathrm{T}$とし，線分$\mathrm{RS}$と辺$\mathrm{BF}$の交点を$\mathrm{U}$，線分$\mathrm{RT}$と辺$\mathrm{DH}$の交点を$\mathrm{V}$とすると，切り口は五角形$\mathrm{PQVRU}$となる．

ここで，求める立体は，三角錐$\mathrm{RCST}$から三角錐$\mathrm{SPBU}$と三角錐$\mathrm{TDQV}$を除いたものである．

$\triangle\mathrm{APQ}\equiv\triangle\mathrm{BPS}\equiv\triangle\mathrm{DQT}$より，

\[
\mathrm{BP}=\mathrm{BS}=2,\mathrm{DQ}=\mathrm{DT}=2
\]

$\triangle\mathrm{CRS}\backsim\triangle\mathrm{BUS},\triangle\mathrm{CRT}\backsim\triangle\mathrm{DVT}$であり，相似比はともに$3:1$であるから，

\[
\mathrm{BU}=\frac{2}{3},\mathrm{DV}=\frac{2}{3}
\]

三角錐$\mathrm{RCST}$の体積は，$\frac{1}{3}\cdot\triangle\mathrm{CST}\cdot\mathrm{CR}=\frac{1}{3}\cdot(\frac{1}{2}\cdot 6\cdot 6)\cdot 2=12$

三角錐$\mathrm{SPBU}$の体積は，$\frac{1}{3}\cdot\triangle\mathrm{PBU}\cdot\mathrm{BS}=\frac{1}{3}\cdot(\frac{1}{2}\cdot 2\cdot\frac{2}{3})\cdot 2=\frac{4}{9}$

三角錐$\mathrm{TDQV}$の体積は，$\frac{1}{3}\cdot\triangle\mathrm{DQV}\cdot\mathrm{DT}=\frac{1}{3}\cdot(\frac{1}{2}\cdot 2\cdot\frac{2}{3})\cdot 2=\frac{4}{9}$

よって，求める体積は，$12-(\frac{4}{9}+\frac{4}{9})=\frac{100}{9}$

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% 問題A3.3.4：多面体の切断・体積（One More）★★★
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右の図のような三角錐$\mathrm{P-EQR}$と，1辺の長さが2の立方体$\mathrm{ABCD-EFGH}$における共通部分の立体の体積を求めよ．ただし，$\mathrm{AP},\mathrm{FQ},\mathrm{HR}$の長さを1とする．

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% 問題A3.3.4の解答（One More）★★★
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求める立体は，三角錐$\mathrm{P-EQR}$から立方体に重なっていない部分の3つの三角錐の体積を除いたものである．

三角錐$\mathrm{P-EQR}$の体積を$V$とすると，

\[
V=\frac{1}{3}\cdot\triangle\mathrm{EQR}\cdot\mathrm{EP}=\frac{1}{3}\times\{\frac{1}{2}\times(2+1)\times(2+1)\}\times(2+1)=\frac{9}{2}
\]

三角錐$\mathrm{P-EQR}$において，立方体に重なっていない部分の3つの三角錐の体積は，

\[
\mathrm{AP}:\mathrm{EP}=\mathrm{FQ}:\mathrm{EQ}=\mathrm{HR}:\mathrm{ER}=1:3
\]

より，それぞれ$\frac{1}{27}V$である．

よって，求める体積は，$V-3\times\frac{1}{27}V=\frac{8}{9}V=4$

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% 例題A3.3.4の側注：多面体の切断・体積（One More）★★★
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% 側注1
$\mathrm{P},\mathrm{Q},\mathrm{R}$は同じ切断面上にある．ただし，切り口の周上では，$\mathrm{P}$と$\mathrm{R}$の間に$\mathrm{U}$，$\mathrm{Q}$と$\mathrm{R}$の間に$\mathrm{V}$があるため，$\mathrm{P}$と$\mathrm{R}$，$\mathrm{Q}$と$\mathrm{R}$は直接結ぶことはできないので注意すること．

% 側注2
$\mathrm{CR}:\mathrm{BU}=\mathrm{CS}:\mathrm{BS}$より，$2:\mathrm{BU}=6:2$であるから，$\mathrm{BU}=\frac{2}{3}$同様にして，$\mathrm{DV}=\frac{2}{3}$

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% 問題A3.3.4の側注：多面体の切断・体積（One More）★★★
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% 側注1
3つの三角錐は，三角錐$\mathrm{P-EQR}$と相似である．

% 側注2
相似比が$1:3$であるから，体積比は$1^3:3^3$となることを利用する．

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% 例題A3.3.4の考え方・補足など：多面体の切断・体積（One More）★★★
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% 考え方
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切断する面と立体の面との共有点に注意すること．また，求める立体は，右の図のように交点をおくと，三角錐RCSTから三角錐SPBUと三角錐TDQVを除いたものであると考えるとよい．