【数学A】例題2.2.17：期待値（図形）（One More）★★★★

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1辺の長さが1の正六角形ABCDEFの6つの頂点から異なる3点を無作為に選び，それらの3点を頂点とする三角形をつくる．このとき，三角形の面積の期待値を求めよ．

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% 例題A2.2.17の解答（One More）★★★★
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3つの頂点の選び方の総数は${}_6\mathrm{C}_3=20(\text{通り})$

三角形の形は次の(i)～(iii)の3種類がある．

(i) 正六角形と2辺を共有するとき

3辺が$1,1,\sqrt{3}$の二等辺三角形となり，その面積は，$\frac{1}{2}\cdot 1\cdot 1\cdot\sin 120^{\circ}=\frac{\sqrt{3}}{4}$

このような三角形は，正六角形の各頂点に対して1つずつできるから，6通り

(ii) 正六角形と1辺だけを共有するとき

3辺が$1,\sqrt{3},2$の直角三角形となり，その面積は，$\frac{1}{2}\cdot 1\cdot\sqrt{3}=\frac{\sqrt{3}}{2}$

このような三角形は，$\mathrm{AD},\mathrm{BE},\mathrm{CF}$を斜辺としたときに，それぞれ4通りずつできるから，$3\times 4=12(\text{通り})$

(iii) 正六角形と辺を共有しないとき

1辺が$\sqrt{3}$の正三角形となり，その面積は，$\frac{1}{2}\cdot\sqrt{3}\cdot\sqrt{3}\cdot\sin 60^{\circ}=\frac{3\sqrt{3}}{4}$

このような三角形は，$\triangle\mathrm{ACE},\triangle\mathrm{BDF}$の2通り

よって，(i)〜(iii)より，求める期待値は，

\[
\frac{\sqrt{3}}{4}\times\frac{6}{20}+\frac{\sqrt{3}}{2}\times\frac{12}{20}+\frac{3\sqrt{3}}{4}\times\frac{2}{20}=\frac{9\sqrt{3}}{20}
\]

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% 問題A2.2.17：期待値（図形）（One More）★★★★
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1辺の長さが1の正六角形ABCDEFの6つの頂点から異なる3点を無作為に選び，それらの3点を頂点とする三角形をつくる．このとき，三角形の周の長さの期待値を求めよ．

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% 問題A2.2.17の解答（One More）★★★★
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3つの頂点の選び方の総数は${}_6\mathrm{C}_3=20(\text{通り})$

三角形の形は次の(i)～(iii)の3種類がある．

(i) 正六角形と2辺を共有するとき

3辺が$1,1,\sqrt{3}$の二等辺三角形となり，その周の長さは，$1+1+\sqrt{3}=2+\sqrt{3}$

このような三角形は，正六角形の各頂点に対して1つずつできるから，6通り

(ii) 正六角形と1辺だけを共有するとき

3辺が$1,\sqrt{3},2$の直角三角形となり，その周の長さは，$1+\sqrt{3}+2=3+\sqrt{3}$

このような三角形は，$\mathrm{AD},\mathrm{BE},\mathrm{CF}$を斜辺としたときに，それぞれ4通りずつできるから，$3\times 4=12(\text{通り})$

(iii) 正六角形と辺を共有しないとき

1辺が$\sqrt{3}$の正三角形となり，その周の長さは，$3\cdot\sqrt{3}=3\sqrt{3}$

このような三角形は，$\triangle\mathrm{ACE},\triangle\mathrm{BDF}$の2通り

よって，(i)〜(iii)より，求める期待値は，

\[
(2+\sqrt{3})\times\frac{6}{20}+(3+\sqrt{3})\times\frac{12}{20}+3\sqrt{3}\times\frac{2}{20}=\frac{12+6\sqrt{3}}{5}
\]

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% 例題A2.2.17の考え方・補足など：期待値（図形）（One More）★★★★
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三角形の形は，(i)正六角形と2辺を共有するとき，(ii)正六角形と1辺だけを共有するとき，(iii)正六角形と辺を共有しないときの3つの場合がある（3種類）．それぞれの場合分けをして，面積を求める．(i)，(iii)は三角形の面積の公式を用いるとよい（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，$S=\frac{1}{2}ab\sin\theta$）．