【数学A】例題3.3.3：多面体の面・辺・頂点の数（One More）★★★

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% 例題A3.3.3：多面体の面・辺・頂点の数（One More）★★★
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正二十面体において，各辺の中点を通る平面ですべてのかどを切り取ったときにできる多面体の面の数$f$，辺の数$e$，頂点の数$v$をそれぞれ求めよ．

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% 例題A3.3.3の解答（One More）★★★
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正二十面体は，各面が正三角形であり，1つの頂点に集まる面の数は5個である．

したがって，正二十面体の辺の数は$3\times 20\div 2=30$

頂点の数は，$3\times 20\div 5=12\cdots(\mathrm{i})$

ここで，正二十面体の1つのかどを切り取ると，新しい面として正五角形の面が$1$つできる．

(i) より，新しく増える面として正五角形が12個できる．

ゆえに，面の数は，$f=20+12=32$

辺の数は，正五角形が12個あるから，$e=5\times 12=60$

頂点の数は，オイラーの（多面体）定理から，$v=60-32+2=30$

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% 問題A3.3.3：多面体の面・辺・頂点の数（One More）★★★
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正十二面体において，各辺の中点を通る平面ですべてのかどを切り取ったときにできる多面体の面の数$f$，辺の数$e$，頂点の数$v$をそれぞれ求めよ．

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% 問題A3.3.3の解答（One More）★★★
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正十二面体は，各面が正五角形であり，1つの頂点に集まる面の数は3個である．

したがって，正十二面体の辺の数は$5\times 12\div 2=30$

頂点の数は，$5\times 12\div 3=20\cdots(\mathrm{i})$

ここで，正十二面体の1つのかどを切り取ると，新しい面として正三角形の面が$1$つできる．

(i) より，新しく増える面として正三角形が20個できる．

ゆえに，面の数は，$f=12+20=32$

辺の数は，正三角形が20個あるから，

\[
e=3\times 20=60
\]

頂点の数は，オイラーの（多面体）定理から，$v=60-32+2=30$

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% 例題A3.3.3の側注：多面体の面・辺・頂点の数（One More）★★★
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% 側注1
先に，正二十面体の辺の数と頂点の数を求める．

% 側注2
新しく増える面を足し合わせる．

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% 問題A3.3.3の側注：多面体の面・辺・頂点の数（One More）★★★
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% 側注1
先に，正十二面体の辺の数と頂点の数を求める．

% 側注2
なお，正二十面体の各辺の中点でかどを切り取ったときにできる多面体と，同じ多面体となる．

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% 例題A3.3.3の考え方・補足など：多面体の面・辺・頂点の数（One More）★★★
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% 考え方
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先に，正二十面体の辺の数と頂点の数を求め，次にすべてのかどを切り取ったときにできる多面体について考えるとよい．正多面体の辺の数は，

\[
(1\text{つの面の辺の数})\times(\text{面の数})\div 2
\]

で求まり，正多面体の頂点の数は，

\[
(1\text{つの面の頂点の数})\times(\text{面の数})\div(\text{1つの頂点に集まる面の数})
\]

で求めることができる．また，多面体の頂点の数$v$，辺の数$e$，面の数$f$について，オイラーの（多面体）定理$v-e+f=2$を利用するとよい．なお，すべてのかどを切り取ったときにできる多面体は右の図のようになり，二十面十二面体ということがある．

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% オイラーの（多面体）定理
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凸多面体で，頂点，辺，面の数をそれぞれ$v,e,f$とすると，

\[
v-e+f=2
\]