【数学A】3章：図形の性質（節末問題・章末問題）

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% 節末問題A3.1.1：（One More）★★
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$\triangle\mathrm{ABC}$の辺$\mathrm{BC},\mathrm{CA},\mathrm{AB}$の中点をそれぞれ$\mathrm{D},\mathrm{E},\mathrm{F}$とすると，$\triangle\mathrm{ABC}$の外心Oは，$\triangle\mathrm{DEF}$の垂心であることを証明せよ．

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% 節末問題A3.1.1の解答（One More）★★
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$\mathrm{O}$は$\triangle\mathrm{ABC}$の外心であるから，辺$\mathrm{BC}$の垂直二等分線上にある．

また，$\mathrm{D}$は辺$\mathrm{BC}$の中点であるから，この垂直二等分線は$\mathrm{D}$を通る．

中点連結定理より，

\[
\mathrm{FE}\parallel\mathrm{BC}
\]

であるから，辺$\mathrm{BC}$の垂直二等分線は$\mathrm{FE}$に垂直である．

したがって，$\mathrm{O}$は$\triangle\mathrm{DEF}$の頂点$\mathrm{D}$から辺$\mathrm{FE}$に引いた垂線上にある．

同様に，$\mathrm{O}$は，頂点$\mathrm{E}$から辺$\mathrm{FD}$に引いた垂線上，および頂点$\mathrm{F}$から辺$\mathrm{DE}$に引いた垂線上にある．

よって，$\mathrm{O}$は$\triangle\mathrm{DEF}$の垂心である．$\blacksquare$

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% 節末問題A3.1.2：（One More）★★
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右の図において，$\triangle\mathrm{ABC}$の重心をGとするとき，$\triangle\mathrm{ABC}$の面積と四角形ARGQの面積比を求めよ．

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% 節末問題A3.1.2の解答（One More）★★
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$(\text{四角形ARGQ})=\triangle\mathrm{ABC}-\triangle\mathrm{CBR}-\triangle\mathrm{CGQ}\cdots(\mathrm{i})$

また，Gは重心より，Rは辺ABの中点であるから，

\[
\triangle\mathrm{CBR}=\frac{1}{2}\triangle\mathrm{ABC}\cdots(\mathrm{ii})
\]

また，$\triangle\mathrm{CGQ}:\triangle\mathrm{CRA}=\mathrm{CG}\cdot\mathrm{CQ}:\mathrm{CR}\cdot\mathrm{CA}$

Gは重心より，$\mathrm{CG}:\mathrm{CR}=2:3,\mathrm{CQ}:\mathrm{CA}=1:2$であるから，

\[
\triangle\mathrm{CGQ}=\frac{2\cdot 1}{3\cdot 2}\triangle\mathrm{CRA}=\frac{1}{3}\triangle\mathrm{CRA}=\frac{1}{3}\cdot\frac{1}{2}\triangle\mathrm{ABC}=\frac{1}{6}\triangle\mathrm{ABC}\cdots(\mathrm{iii})
\]

(ii)，(iii)を(i)に代入すると，

\[
(\text{四角形ARGQ})=\triangle\mathrm{ABC}-\frac{1}{2}\triangle\mathrm{ABC}-\frac{1}{6}\triangle\mathrm{ABC}=\frac{1}{3}\triangle\mathrm{ABC}
\]

よって，$\triangle\mathrm{ABC}$の面積と四角形ARGQの面積比は，$3:1$

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% 節末問題A3.1.3：（One More）★★
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鋭角三角形である$\triangle\mathrm{ABC}$において，3つの頂点から対辺に下ろした垂線は1点で交わることを証明せよ．

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% 節末問題A3.1.3の解答（One More）★★
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3辺$\mathrm{BC},\mathrm{CA},\mathrm{AB}$の長さを，それぞれ$a,b,c$とする．また，3つの頂点$\mathrm{A,B,C}$から対辺へ下ろした垂線をそれぞれ$\mathrm{AD,BE,CF}$とする．

$\triangle\mathrm{ABE}$と$\triangle\mathrm{ACF}$において，

\[
\angle\mathrm{BEA}=\angle\mathrm{CFA}=90^{\circ}
\]

したがって，$\triangle\mathrm{ABE}\backsim\triangle\mathrm{ACF}$

これより，$\mathrm{AE}:\mathrm{AF}=\mathrm{AB}:\mathrm{AC}=c:b$

同様に考えると，$\triangle\mathrm{CBF}\backsim\triangle\mathrm{ABD}$より，$\mathrm{BF}:\mathrm{BD}=\mathrm{CB}:\mathrm{AB}=a:c$

また，$\triangle\mathrm{ACD}\backsim\triangle\mathrm{BCE}$より，$\mathrm{CD}:\mathrm{CE}=\mathrm{AC}:\mathrm{BC}=b:a$

ゆえに，

\[
\frac{\mathrm{BD}}{\mathrm{DC}}\cdot\frac{\mathrm{CE}}{\mathrm{EA}}\cdot\frac{\mathrm{AF}}{\mathrm{FB}}=\frac{\mathrm{BD}}{\mathrm{FB}}\cdot\frac{\mathrm{CE}}{\mathrm{DC}}\cdot\frac{\mathrm{AF}}{\mathrm{EA}}=\frac{c}{a}\cdot\frac{a}{b}\cdot\frac{b}{c}=1
\]

よって，チェバの定理の逆より，3つの頂点から対辺へ下ろした垂線は1点で交わる．$\blacksquare$

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% 節末問題A3.1.4：（One More）★★★
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$\triangle\mathrm{ABC}$の辺$\mathrm{BC,CA,AB}$を$2:1$に内分する点をそれぞれ$\mathrm{D,E,F}$とし，$\mathrm{AD}$と$\mathrm{BE}$，$\mathrm{BE}$と$\mathrm{CF}$，$\mathrm{CF}$と$\mathrm{AD}$の交点をそれぞれ$\mathrm{P,Q,R}$とする．このとき，次の問いに答えよ．

(1) $\mathrm{AP}:\mathrm{PR}:\mathrm{RD}=3:x:y$とするとき，$x,y$の値を求めよ．

(2) $\triangle\mathrm{ABC}$と$\triangle\mathrm{PQR}$の面積比を求めよ．

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% 節末問題A3.1.4の解答（One More）★★★
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(1) $\triangle\mathrm{ABD}$と直線CFについて，メネラウスの定理より，

\[
\frac{\mathrm{AF}}{\mathrm{FB}}\cdot\frac{\mathrm{BC}}{\mathrm{CD}}\cdot\frac{\mathrm{DR}}{\mathrm{RA}}=1
\]

$\frac{2}{1}\cdot\frac{3}{1}\cdot\frac{y}{x+3}=1$より，$x-6y=-3\cdots(\mathrm{i})$

また，$\triangle\mathrm{ACD}$と直線BEについて，メネラウスの定理より，

\[
\frac{\mathrm{AE}}{\mathrm{EC}}\cdot\frac{\mathrm{CB}}{\mathrm{BD}}\cdot\frac{\mathrm{DP}}{\mathrm{PA}}=1
\]

$\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{y+x}{3}=1$より，$x+y=4\cdots(\mathrm{ii})$

よって，(i)，(ii)より，$x=3,y=1$

(2) (1)より，$\mathrm{AP}:\mathrm{PR}:\mathrm{RD}=3:3:1$であるから，

\[
\triangle\mathrm{PBR}=\frac{\mathrm{PR}}{\mathrm{AD}}\triangle\mathrm{ABD}=\frac{3}{7}\triangle\mathrm{ABD}
\]

$\triangle\mathrm{ABD}=\frac{2}{3}\triangle\mathrm{ABC}$であるから，

\[
\triangle\mathrm{PBR}=\frac{3}{7}\triangle\mathrm{ABD}=\frac{3}{7}\cdot\frac{2}{3}\triangle\mathrm{ABC}=\frac{2}{7}\triangle\mathrm{ABC}
\]

また，(1)と同様に，$\mathrm{BQ}:\mathrm{QP}:\mathrm{PE}=3:3:1$であるから，$\triangle\mathrm{PQR}=\frac{\mathrm{QP}}{\mathrm{BP}}\triangle\mathrm{PBR}=\frac{3}{6}\triangle\mathrm{PBR}$

したがって，

\[
\triangle\mathrm{PQR}=\frac{3}{6}\triangle\mathrm{PBR}=\frac{3}{6}\cdot\frac{2}{7}\triangle\mathrm{ABC}=\frac{1}{7}\triangle\mathrm{ABC}
\]

よって，

\[
\triangle\mathrm{ABC}:\triangle\mathrm{PQR}=\triangle\mathrm{ABC}:\frac{1}{7}\triangle\mathrm{ABC}=7:1
\]

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% 節末問題A3.1.4の別解
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$\triangle\mathrm{ABP}=\frac{3}{7}\triangle\mathrm{ABD}=\frac{3}{7}\cdot\frac{2}{3}\triangle\mathrm{ABC}=\frac{2}{7}\triangle\mathrm{ABC}$

$\triangle\mathrm{BCQ},\triangle\mathrm{CAR}$も同様に，

\[
\triangle\mathrm{BCQ}=\triangle\mathrm{CAR}=\frac{2}{7}\triangle\mathrm{ABC}
\]

したがって，$\triangle\mathrm{PQR}=\triangle\mathrm{ABC}-(\triangle\mathrm{ABP}+\triangle\mathrm{BCQ}+\triangle\mathrm{CAR})=\frac{1}{7}\triangle\mathrm{ABC}$

よって，$\triangle\mathrm{ABC}$と$\triangle\mathrm{PQR}$の面積比は，

\[
\triangle\mathrm{ABC}:\triangle\mathrm{PQR}=\triangle\mathrm{ABC}:\frac{1}{7}\triangle\mathrm{ABC}=7:1
\]

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% 節末問題A3.2.1：（One More）★★★
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$\triangle\mathrm{ABC}$の内心を$\mathrm{I},\triangle\mathrm{BCI}$の外心を$\mathrm{O}$とする．4点$\mathrm{A},\mathrm{B},\mathrm{C},\mathrm{O}$は同一円周上にあることを示せ．

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% 節末問題A3.2.1の解答（One More）★★★
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点$\mathrm{I}$が$\triangle\mathrm{ABC}$の内心であるから，

\[
\angle\mathrm{IBC}=\angle\mathrm{IBA},\angle\mathrm{ICB}=\angle\mathrm{ICA}
\]

したがって，

\begin{align*}
\angle\mathrm{IBC}+\angle\mathrm{ICB}&=\frac{1}{2}\angle\mathrm{ABC}+\frac{1}{2}\angle\mathrm{ACB}\\
&=\frac{1}{2}(\angle\mathrm{ABC}+\angle\mathrm{ACB})\\
&=\frac{1}{2}(180^{\circ}-\angle\mathrm{BAC})\\
&=90^{\circ}-\frac{1}{2}\angle\mathrm{BAC}\cdots(\mathrm{i})
\end{align*}

また，点$\mathrm{O}$が$\triangle\mathrm{BCI}$の外心であるから$\mathrm{OB}=\mathrm{OI}=\mathrm{OC}$

ゆえに，$\angle\mathrm{BOI}=180^{\circ}-2\angle\mathrm{BIO},\angle\mathrm{COI}=180^{\circ}-2\angle\mathrm{CIO}$

したがって，

\begin{align*}
\angle\mathrm{BOC}&=\angle\mathrm{BOI}+\angle\mathrm{COI}\\
&=(180^{\circ}-2\angle\mathrm{BIO})+(180^{\circ}-2\angle\mathrm{CIO})\\
&=360^{\circ}-2(\angle\mathrm{BIO}+\angle\mathrm{CIO})\\
&=360^{\circ}-2\angle\mathrm{BIC}\cdots(\mathrm{ii})
\end{align*}

ここで，(i)より，

\begin{align*}
\angle\mathrm{BIC}&=180^{\circ}-(\angle\mathrm{IBC}+\angle\mathrm{ICB})\\
&=180^{\circ}-(90^{\circ}-\frac{1}{2}\angle\mathrm{BAC})\\
&=90^{\circ}+\frac{1}{2}\angle\mathrm{BAC}
\end{align*}

これを(ii)に代入すると，

\[
\angle\mathrm{BOC}=360^{\circ}-2(90^{\circ}+\frac{1}{2}\angle\mathrm{BAC})=180^{\circ}-\angle\mathrm{BAC}
\]

ゆえに，$\angle\mathrm{BOC}+\angle\mathrm{BAC}=180^{\circ}$

よって，四角形$\mathrm{ABCO}$は円に内接する，すなわち，4点$\mathrm{A},\mathrm{B},\mathrm{C},\mathrm{O}$は同一円周上にある．$\blacksquare$

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% 節末問題A3.2.2：（One More）★★
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右の図のように，半径$r,r^{\prime}(r>r^{\prime})$の円$\mathrm{O}$と$\mathrm{O}^{\prime}$が外接しており，さらに直線$l$にそれぞれ$\mathrm{A},\mathrm{B}$で接しているとする．このとき，次の問いに答えよ．

(1) 線分$\mathrm{AB}$の長さを$r,r^{\prime}$で表せ．

(2) 2つの円$\mathrm{O},\mathrm{O}^{\prime}$に外接し，さらに線分$\mathrm{AB}$に接する円$\mathrm{O}^{\prime\prime}$の半径$x$を求めよ．

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% 節末問題A3.2.2の解答（One More）★★
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(1) 円$\mathrm{O},\mathrm{O}^{\prime}$が外接しているから，$\mathrm{OO}^{\prime}=r+r^{\prime}$

$\mathrm{O}^{\prime}$から$\mathrm{OA}$に垂線$\mathrm{O}^{\prime}\mathrm{C}$を下ろすと，$\mathrm{OC}=r-r^{\prime}$

$\triangle\mathrm{O}\mathrm{O}^{\prime}\mathrm{C}$において，

\begin{align*}
\mathrm{O}^{\prime}\mathrm{C}&=\sqrt{\mathrm{OO}^{\prime 2}-\mathrm{OC}^2}\\
&=\sqrt{(r+r^{\prime})^2-(r-r^{\prime})^2}\\
&=2\sqrt{rr^{\prime}}
\end{align*}

よって，$\mathrm{AB}=\mathrm{O}^{\prime}\mathrm{C}={2}\sqrt{{r}{r}^{\prime}}$

(2) 中心$\mathrm{O}^{\prime\prime}$から$\mathrm{OA},\mathrm{O}^{\prime}\mathrm{B}$に垂線$\mathrm{O}^{\prime\prime}\mathrm{D},\mathrm{O}^{\prime\prime}\mathrm{E}$をそれぞれ下ろすと，

\begin{align*}
&\mathrm{DO}^{\prime\prime}=\sqrt{\mathrm{OO}^{\prime\prime 2}-\mathrm{OD}^2}=2\sqrt{rx},\\
&\mathrm{EO}^{\prime\prime}=\sqrt{\mathrm{O}^{\prime}\mathrm{O}^{\prime\prime 2}-\mathrm{O}^{\prime}\mathrm{E}^2}=2\sqrt{r^{\prime}x}
\end{align*}

$\mathrm{DE}=\mathrm{DO}^{\prime\prime}+\mathrm{EO}^{\prime\prime}$であるから，

\[
2\sqrt{rx}+2\sqrt{r^{\prime}x}=2\sqrt{rr^{\prime}}
\]

したがって，$(\sqrt{r}+\sqrt{r^{\prime}})\sqrt{x}=\sqrt{rr^{\prime}}$

ゆえに，$\sqrt{x}=\frac{\sqrt{rr^{\prime}}}{\sqrt{r}+\sqrt{r^{\prime}}}$

よって，$x=\frac{rr^{\prime}}{(\sqrt{r}+\sqrt{r^{\prime}})^2}$

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% 節末問題A3.2.3：（One More）★★
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右の図のように，半径が等しい2つの円$\mathrm{O},\mathrm{O}^{\prime}$と直線$l$がある．直線$l$上に中心があり，この2つの円$\mathrm{O},\mathrm{O}^{\prime}$に外接する円を作図せよ．

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% 節末問題A3.2.3の解答（One More）★★
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線分$\mathrm{OO}^{\prime}$の垂直二等分線を引き，直線$l$との交点を$\mathrm{A}$とする．

[2]線分$\mathrm{OA}$と円$\mathrm{O}$の交点を$\mathrm{B}$，線分$\mathrm{O}^{\prime}\mathrm{A}$と円$\mathrm{O}^{\prime}$の交点を$\mathrm{C}$とする．

[3]点$\mathrm{A}$を中心として，半径$\mathrm{AB}$の円をかく．この円が求める円である．

$\mathrm{A}$は$\mathrm{O}$と$\mathrm{O}^{\prime}$の垂直二等分線上にあるから，
\[
\mathrm{OA}=\mathrm{O}^{\prime}\mathrm{A}
\]

したがって，$\mathrm{OB}+\mathrm{AB}=\mathrm{O}^{\prime}\mathrm{C}+\mathrm{AC}$

また，円$\mathrm{O}$と円$\mathrm{O}^{\prime}$の半径が等しいから，$\mathrm{OB}=\mathrm{O}^{\prime}\mathrm{C}$

ゆえに，$\mathrm{AB}=\mathrm{AC}$

よって，円$\mathrm{A}$は2つの円$\mathrm{O},\mathrm{O}^{\prime}$に接する．

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% 節末問題A3.2.4：（One More）★★
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長さ1の線分が与えられたとき，連立方程式$x+y=3,xy=1$の解を長さとする線分を作図せよ．

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% 節末問題A3.2.4の解答（One More）★★
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与えられた線分を用いて長さ3の線分$\mathrm{AB}$を作り，$\mathrm{AB}$を直径とする半円をかく．

[2]線分$\mathrm{AB}$に平行で，線分$\mathrm{AB}$との距離が1である直線と，[1]の半円との交点の1つを$\mathrm{C}$とする．

[3]$\mathrm{C}$から線分$\mathrm{AB}$に垂線を引き，その交点を$\mathrm{D}$とすると，線分$\mathrm{AD},\mathrm{BD}$が求める線分である．

$\mathrm{AD}=x,\mathrm{BD}=y$とすると，$\mathrm{AD}+\mathrm{BD}=\mathrm{AB}$より，

\[
x+y=3
\]

また，方べきの定理より，$\mathrm{AD}\cdot\mathrm{DB}=\mathrm{CD}^2$であるから，$xy=1$

よって，線分$\mathrm{AD},\mathrm{BD}$は連立方程式$x+y=3,xy=1$の解を長さとする線分である．

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% 節末問題A3.3.1：（One More）★★
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正四面体$\mathrm{ABCD}$において，向かい合う2辺$\mathrm{AB}$と$\mathrm{CD}$は垂直であることを示せ．

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% 節末問題A3.3.1の解答（One More）★★
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辺$\mathrm{AB}$と辺$\mathrm{CD}$の中点をそれぞれ$\mathrm{M}$，$\mathrm{N}$とすると，$\triangle\mathrm{NAB}$は$\mathrm{NA}=\mathrm{NB}$の二等辺三角形であるから，

\[
\mathrm{MN}\perp\mathrm{AB}\cdots(\mathrm{i})
\]

また，$\triangle\mathrm{CAB}$は$\mathrm{CA}=\mathrm{CB}$の二等辺三角形であるから，

\[
\mathrm{CM}\perp\mathrm{AB}\cdots(\mathrm{ii})
\]

したがって，(i)，(ii)より，$\mathrm{AB}$は直線$\mathrm{MN}$，$\mathrm{CM}$を含む平面$\mathrm{MCD}$に垂直である．

$\mathrm{AB}$は平面$\mathrm{MCD}$上にある直線$\mathrm{CD}$に垂直である，すなわち，向かい合う2辺ABとCDは垂直である．$\blacksquare$

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% 節末問題A3.3.2：（One More）★★
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正四面体$\mathrm{OABC}$において，辺$\mathrm{AB}$の中点を$\mathrm{M}$とし，頂点$\mathrm{O}$から線分$\mathrm{CM}$に下ろした垂線を$\mathrm{OH}$とする．このとき，$\mathrm{OH}\perp$平面$\mathrm{ABC}$であることを示せ．

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% 節末問題A3.3.2の解答（One More）★★
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辺CMは正三角形である$\triangle\mathrm{CAB}$の中線であるから，

\[
\mathrm{CM}\perp\mathrm{AB}\cdots(\mathrm{i})
\]

辺OMは正三角形である$\triangle\mathrm{OAB}$の中線であるから，

\[
\mathrm{OM}\perp\mathrm{AB}\cdots(\mathrm{ii})
\]

また，与えられた条件より，$\mathrm{OH}\perp\mathrm{CM}$であるから，

\[
\mathrm{OH}\perp\mathrm{HM}\cdots(\mathrm{iii})
\]

(i)〜(iii)より，三垂線の定理を用いて，$\mathrm{OH}\perp\text{平面}\mathrm{ABC}$である．$\blacksquare$

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% 節末問題A3.3.3：（One More）★★★
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1辺の長さが3の正四面体がある．この正四面体を右の図のように，1つの頂点に集まる3つの辺においてそれぞれ3等分した点のうち，頂点に近い方の点を結んでできる正三角形を含む平面で切り，頂点を含む正四面体を取り除く．すべての頂点において同様に正四面体を取り除いたとき，残った立体の体積$V$を求めよ．

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% 節末問題A3.3.3の解答（One More）★★★
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右の図のように，正四面体$\mathrm{OABC}$の頂点$\mathrm{O}$から底面$\mathrm{ABC}$に垂線を下ろすと，その足$\mathrm{H}$は，正三角形$\mathrm{ABC}$の重心と一致する．

辺$\mathrm{BC}$の中点を$\mathrm{M}$とすると，

\[
\mathrm{AH}=\frac{2}{3}\mathrm{AM}=\frac{2}{3}\cdot\frac{\sqrt{3}}{2}\mathrm{AB}=\sqrt{3}
\]

したがって，

\[
\mathrm{OH}=\sqrt{\mathrm{OA}^2-\mathrm{AH}^2}=\sqrt{3^2-(\sqrt{3})^2}=\sqrt{6}
\]

正四面体$\mathrm{OABC}$の体積を$V_0$とすると，

\[
V_0=\frac{1}{3}\cdot\triangle\mathrm{ABC}\cdot\mathrm{OH}=\frac{1}{3}\cdot(\frac{1}{2}\cdot 3\cdot\frac{3\sqrt{3}}{2})\cdot\sqrt{6}=\frac{9\sqrt{2}}{4}
\]

また，取り除かれる正四面体の1辺の長さは1であるから，その体積を$V_1$とすると，

\[
V_1=\frac{1}{3}\cdot(\frac{1}{2}\cdot 1\cdot\frac{\sqrt{3}}{2})\cdot\sqrt{1^2-(\frac{\sqrt{3}}{3})^2}=\frac{\sqrt{2}}{12}
\]

取り除かれる正四面体の数は正四面体の頂点の数と同じであるから，4個取り除かれる．

よって，残った立体の体積$V$は，

\[
V=V_0-4V_1=\frac{9\sqrt{2}}{4}-4\cdot\frac{\sqrt{2}}{12}=\frac{23\sqrt{2}}{12}
\]

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% 節末問題A3.3.4：（One More）★★★
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1辺の長さが$a$の正四面体$\mathrm{ABCD}$の2辺$\mathrm{AB},\mathrm{CD}$の中点をそれぞれ$\mathrm{M},\mathrm{N}$とする．2直線CM，DMのなす角を$\alpha$，2直線$\mathrm{MN},\mathrm{AC}$のなす角を$\beta$とするとき，$\cos\alpha$と$\beta$を求めよ．

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% 節末問題A3.3.4の解答（One More）★★★
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CMとDMのなす角$\alpha$は$\angle\mathrm{CMD}$に等しい．

$\triangle\mathrm{ABC}$，$\triangle\mathrm{ABD}$はともに正三角形であり，$\mathrm{M}$は辺$\mathrm{AB}$の中点であるから，

\[
\mathrm{CM}\perp\mathrm{AB},\mathrm{DM}\perp\mathrm{AB}
\]

また，三平方の定理より，

\[
\mathrm{CM}=\mathrm{DM}=\sqrt{a^2-(\frac{a}{2})^2}=\frac{\sqrt{3}}{2}a
\]

よって，$\triangle\mathrm{CMD}$は$\mathrm{CM}=\mathrm{DM}$の二等辺三角形である．したがって，$\triangle\mathrm{CMD}$において余弦定理を用いると，

\[
\cos\alpha=\frac{\mathrm{CM}^2+\mathrm{DM}^2-\mathrm{CD}^2}{2\cdot\mathrm{CM}\cdot\mathrm{DM}}=\frac{\frac{3}{4}a^2+\frac{3}{4}a^2-a^2}{2\cdot(\frac{\sqrt{3}}{2}a)\cdot(\frac{\sqrt{3}}{2}a)}=\frac{1}{3}
\]

また，辺$\mathrm{BC}$の中点を$\mathrm{L}$とすると，$\mathrm{AC}\parallel\mathrm{ML}$であることから，$\mathrm{MN}$と$\mathrm{AC}$のなす角$\beta$は$\angle\mathrm{LMN}$に等しい．

$\mathrm{N}$は辺$\mathrm{CD}$の中点であるから，

\[
\mathrm{MN}\perp\mathrm{CD}
\]

したがって，$\angle\mathrm{CNM}=90^\circ$より，

\[
\mathrm{MN}=\sqrt{\mathrm{CM}^2-\mathrm{CN}^2}=\sqrt{(\frac{\sqrt{3}}{2}a)^2-(\frac{a}{2})^2}=\frac{a}{\sqrt{2}}
\]

$\mathrm{ML}=\mathrm{LN}=\frac{a}{2}$であるから，$\triangle\mathrm{LNM}$において余弦定理を用いると，

\[
\cos\beta=\frac{\mathrm{ML}^2+\mathrm{MN}^2-\mathrm{LN}^2}{2\cdot\mathrm{ML}\cdot\mathrm{MN}}=\frac{(\frac{a}{{2}})^2+(\frac{a}{\sqrt{2}})^2-(\frac{a}{{2}})^2}{2\cdot\frac{a}{2}\cdot\frac{a}{\sqrt{2}}}=\frac{1}{\sqrt{2}}
\]

よって，$\beta=45^\circ$

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% 章末問題A3.1：（One More）★★
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右の図のような四面体$\mathrm{ABCD}$において，$\mathrm{AP}:\mathrm{PB}=a_1:b_1$，$\mathrm{AQ}:\mathrm{QC}=a_2:b_2$，$\mathrm{AR}:\mathrm{RD}=a_3:b_3$とする．このとき，四面体$\mathrm{APQR}$と四面体$\mathrm{ABCD}$の体積比を求めよ．

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% 章末問題A3.1の解答（One More）★★
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\[
\triangle\mathrm{AQR}:\triangle\mathrm{ACD}=\mathrm{AQ}\cdot\mathrm{AR}:\mathrm{AC}\cdot\mathrm{AD}=a_2a_3:(a_2+b_2)(a_3+b_3)\cdots(\mathrm{i})
\]

$\mathrm{B}$から面$\mathrm{ACD}$に下ろした垂線の足を$\mathrm{M}$，$\mathrm{P}$から面$\mathrm{ACD}$に下ろした垂線の足を$\mathrm{N}$とすると，

\[
\mathrm{PN}:\mathrm{BM}=\mathrm{AP}:\mathrm{AB}=a_1:(a_1+b_1)\cdots(\mathrm{ii})
\]

よって，(i)，(ii)より，求める体積比は，

\[
a_2a_3\cdot a_1:(a_2+b_2)(a_3+b_3)\cdot(a_1+b_1)=a_1a_2a_3:({a}_1+b_1)({a}_2+b_2)({a}_3+b_3)
\]

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% 章末問題A3.2：（One More）★★
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1辺の長さが2の立方体がある．この立方体の各面（正方形）の対角線の交点を頂点とする正八面体について，次の問いに答えよ．

(1) 正八面体の1辺の長さを求めよ．

(2) 正八面体の体積を求めよ．

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% 章末問題A3.2の解答（One More）★★
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右の図のように，立方体を$\mathrm{ABCD}-\mathrm{EFGH}$とし，各面（正方形）の対角線の交点を頂点とする正八面体をPQRSTUとする．

(1) 辺$\mathrm{AB},\mathrm{BC}$の中点をそれぞれ$\mathrm{M},\mathrm{N}$とすると，

\[
\mathrm{QR}=\mathrm{MN}=\sqrt{1^2+1^2}=\sqrt{2}
\]

(2) PUは平面QRSTに垂直であり，四角錐$\mathrm{P}-\mathrm{QRST}$と四角錐$\mathrm{U}-\mathrm{QRST}$は合同であるから，求める正八面体の体積は，

\begin{align*}
2\times\text{（四角錐P}-\text{QRST）}&=2\cdot\{\frac{1}{3}\cdot(\text{四角形}\mathrm{QRST})\cdot\frac{1}{2}\mathrm{PU}\}\\
&=2\cdot\frac{1}{3}\cdot(\sqrt{2})^2\cdot 1\\
&=\frac{4}{3}
\end{align*}

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% 章末問題A3.3：（One More）★★★★
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各面が正三角形である正多面体は何種類あるか．すべて挙げよ．

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% 章末問題A3.3の解答（One More）★★★★
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正$m$面体（$m$は$4$以上の自然数）の頂点の数を$v$，辺の数を$e$，面の数を$f$とする．

正$m$面体の1つの頂点に集まる角の和は$360^\circ$より小さく，1つの頂点には少なくとも3つの面が集まる．

したがって，1つの頂点に集まる面の数を$n$（$n$は3以上の自然数）とすると，

\[
60^\circ\times n<360^\circ
\]

ゆえに，$n<6$であるから，$n=3,4,5$の場合について考える．

(i) $n=3$のとき

1つの頂点に3つの面が集まるから，

\[
v=\frac{3\times m}{3}=m,e=\frac{3\times m}{2}=\frac{3}{2}m,f=m
\]

これらをオイラーの定理に代入すると，

\[
m-\frac{3}{2}m+m=2
\]

したがって，$m=4$であるから，この正多面体は正四面体である．

(ii) $n=4$のとき

1つの頂点に4つの面が集まるから，

\[
v=\frac{3\times m}{4}=\frac{3}{4}m,e=\frac{3\times m}{2}=\frac{3}{2}m,f=m
\]

これらをオイラーの定理に代入すると，

\[
\frac{3}{4}m-\frac{3}{2}m+m=2
\]

したがって，$m=8$であるから，この正多面体は正八面体である．

(iii) $n=5$のとき

1つの頂点に5つの面が集まるから，

\[
v=\frac{3\times m}{5}=\frac{3}{5}m,e=\frac{3\times m}{2}=\frac{3}{2}m,f=m
\]

これらをオイラーの定理に代入すると，

\[
\frac{3}{5}m-\frac{3}{2}m+m=2
\]

したがって，$m=20$であるから，この正多面体は正二十面体である．

よって，(i)〜(iii)より，各面が正三角形である正多面体は，正四面体，正八面体，正二十面体の3種類ある．

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% 章末問題A3.4：（One More）★★★
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1辺の長さが8の立方体$\mathrm{ABCD}-\mathrm{EFGH}$において，辺$\mathrm{AB},\mathrm{AD}$の中点をそれぞれ$\mathrm{P},\mathrm{Q}$とする．このとき，次の問いに答えよ．

(1) 立体$\mathrm{APQ}-\mathrm{EFH}$の体積を求めよ．

(2) 台形$\mathrm{PQHF}$の面積を求めよ．

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% 章末問題A3.4の解答（One More）★★★
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(1) 線分$\mathrm{FP}$，線分$\mathrm{EA}$，線分$\mathrm{HQ}$の延長の交点をOとすると，$\mathrm{QA}:\mathrm{HE}=4:8=1:2$であるから，$\mathrm{OA}:\mathrm{OE}=1:2$

したがって，三角錐$\mathrm{OPQA}$と三角錐$\mathrm{OFHE}$の体積比は，

\[
1^3:2^3=1:8
\]

ここで，三角錐$\mathrm{OFHE}$の体積は，

\begin{align*}
\frac{1}{3}\cdot\triangle\mathrm{FHE}\cdot\mathrm{OE}&=\frac{1}{3}\cdot(\frac{1}{2}\cdot 8\cdot 8)\cdot 16\\
&=\frac{1}{3}\cdot 32\cdot 16=\frac{512}{3}
\end{align*}

よって，求める体積は，

\[
\frac{512}{3}\cdot(1-\frac{1}{8})=\frac{512}{3}\cdot\frac{7}{8}=\frac{3584}{24}=\frac{448}{3}
\]

(2) Oから$\mathrm{FH}$に垂線ORを下ろす．$\triangle\mathrm{OFH}$は二等辺三角形であるから，ORは$\mathrm{FH}$の垂直二等分線である．

$\mathrm{FP}=\sqrt{8^2+4^2}=4\sqrt{5},\mathrm{FH}=\sqrt{8^2+8^2}=8\sqrt{2}$より，$\mathrm{FO}=8\sqrt{5},\mathrm{RH}=4\sqrt{2}$

したがって，

\[
\mathrm{OR}=\sqrt{(8\sqrt{5})^2-(4\sqrt{2})^2}=12\sqrt{2}
\]

ゆえに，

\[
\triangle\mathrm{OFH}=\frac{1}{2}\cdot 8\sqrt{2}\cdot 12\sqrt{2}=\frac{1}{2}\cdot 192=96
\]

ここで，$\triangle\mathrm{OPQ}\backsim\triangle\mathrm{OFH}$であり，$\mathrm{OP}:\mathrm{OF}=1:2$より，$\triangle\mathrm{OPQ}$と$\triangle\mathrm{OFH}$の面積比は，

\[
1^2:2^2=1:4
\]

よって，台形$\mathrm{PQHF}$の面積は，$96\cdot(1-\frac{1}{4})=96\cdot\frac{3}{4}=72$

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% 節末問題A3.1.1の側注：（One More）★★
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% 側注1
OはBCの垂直二等分線上にある．

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% 節末問題A3.1.2の側注：（One More）★★
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% 側注1
$\triangle\mathrm{CBR},\triangle\mathrm{CGQ}$を$\triangle\mathrm{ABC}$で表すことを考える．

% 側注2
\begin{align*}
&\triangle\mathrm{CBR}:\triangle\mathrm{ABC}\\
&=\mathrm{BR}:\mathrm{BA}=1:2
\end{align*}

% 側注3
\begin{align*}
&\triangle\mathrm{CRA}:\triangle\mathrm{ABC}\\
&=\mathrm{AR}:\mathrm{AB}=1:2
\end{align*}

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% 節末問題A3.1.3の側注：（One More）★★
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% 側注1
$\angle\mathrm{A}$は共通の角であるから，2組の角がそれぞれ等しい．

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% 節末問題A3.1.4の側注：（One More）★★★
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% 側注1
$\frac{2}{1}\cdot\frac{3}{1}\cdot\frac{\mathrm{DR}}{\mathrm{RA}}=1$

% 側注2
$\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{\mathrm{DP}}{\mathrm{PA}}=1$

% 側注3
連立方程式を解く．

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% 節末問題A3.2.1の側注：（One More）★★★
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% 側注1
$\triangle\mathrm{BOI},\triangle\mathrm{COI}$は二等辺三角形であるから，$\angle\mathrm{BIO}=\angle\mathrm{IBO}$，$\angle\mathrm{CIO}=\angle\mathrm{ICO}$

% 側注2
(i)より，

\begin{align*}
&\angle\mathrm{IBC}+\angle\mathrm{ICB}\\
=&90^{\circ}-\frac{1}{2}\angle\mathrm{BAC}
\end{align*}

% 側注3
円に内接する四角形の対角の和は$180^{\circ}$である．

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% 節末問題A3.2.2の側注：（One More）★★
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% 側注1
三平方の定理を用いる．

% 側注2
三平方の定理を用いる．

% 側注3
$\sqrt{x}$をくくり出す．

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% 節末問題A3.2.3の側注：（One More）★★
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% 側注1
求める円の中心は，2つの円の中心O，$\mathrm{O}^\prime$からの距離が等しい．

% 側注2
$\mathrm{AC}$をかいてもよい．

% 側注3
$\mathrm{OA}-\mathrm{OB}=\mathrm{O}^{\prime}\mathrm{A}-\mathrm{O}^{\prime}C$

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% 節末問題A3.2.4の側注：（One More）★★
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% 側注1
$x+y=3$より，
\[
y=3-x
\]
これを$xy=1$に代入すると，$x(3-x)=1$となることから，方べきの定理の利用を考える．

% 側注2
作図によって得られた図形が条件を満たすことを確認する．

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% 節末問題A3.3.1の側注：（One More）★★
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% 側注1
直線$h$が平面$\alpha$上の交わる2直線に垂直$\Longrightarrow$直線$h\perp$平面$\alpha$

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% 節末問題A3.3.2の側注：（One More）★★
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% 側注1
点Mは辺ABの中点である．

% 側注2
$\mathrm{CM}\perp\mathrm{AB},\mathrm{OM}\perp\mathrm{AB},\mathrm{OH}\perp\mathrm{CM}\Longrightarrow\mathrm{OH}\perp\text{平面}\mathrm{ABC}$

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% 節末問題A3.3.3の側注：（One More）★★★
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% 側注1
なお，すべての頂点で正四面体を取り除いたときにできる残った多面体を{切頂四面体}ということがある．

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% 節末問題A3.3.4の側注：（One More）★★★
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% 側注1
2つの面ABC，ABDのなす角でもある．

% 側注2
三平方の定理を用いる．

% 側注3
三平方の定理を用いる．

% 側注4
$\triangle\mathrm{LNM}$は，$\angle\mathrm{L}=90^{\circ}$の直角二等辺三角形である．

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% 章末問題A3.1の側注：（One More）★★
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% 側注1
\begin{align*}
&\triangle\mathrm{AQR}\\
=&\frac{1}{2}\mathrm{AQ}\cdot\mathrm{AR}\sin\angle\mathrm{CAD},\\
&\triangle\mathrm{ACD}\\
=&\frac{1}{2}\mathrm{AC}\cdot\mathrm{AD}\sin\angle\mathrm{CAD}
\end{align*}

% 側注2
ABと面ACDのなす角を$\alpha$とすると，

\begin{align*}
&\mathrm{PN}:\mathrm{BM}\\
=&\mathrm{AP}\sin\alpha:\mathrm{AB}\sin\alpha\\
=&\mathrm{AP}:\mathrm{AB}
\end{align*}

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% 章末問題A3.2の側注：（One More）★★
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% 側注1
三平方の定理を用いる．

% 側注2
四角形QRSTは正方形である．また，$\mathrm{PU}=2$

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% 章末問題A3.3の側注：（One More）★★★★
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% 側注1
正三角形の1つの角は$60^\circ$であるから，$60^\circ\times n$

% 側注2
\begin{align*}
&v=\frac{\text{すべての面の頂点の数の和}}{1\text{つの頂点に集まる面の数}},\\
&e=\frac{\text{すべての面の辺の数の和}}{1\text{つの辺を共有する面の数}}
\end{align*}

% 側注3
オイラーの（多面体）定理
\[
v-e+f=2
\]

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% 章末問題A3.4の側注：（One More）★★★
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% 側注1
相似比が$m:n$のとき，体積比は$m^3:n^3$となる．

% 側注2
三角錐$\mathrm{OPQA}$と三角錐$\mathrm{OFHE}$の体積比が$1:8$であることから，立体$\mathrm{APQ}-\mathrm{EFH}$の体積は，三角錐OFHEの体積の$\frac{7}{8}$倍である．

% 側注3
三平方の定理を用いる．

% 側注4
$\triangle\mathrm{OAQ}\backsim\triangle\mathrm{OEH}$である．

% 側注5
相似比が$m:n$のとき，面積比は$m^2:n^2$である．