Find, with a proof, the variety of vertices, edges, and faces that a dodecahedron.

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Its is clear that their are $20$ vertices, $30$ edges, and $12$ faces. I am no sure how to prove this though.


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For vertices, there are $12$ encounters times $5$ vertices per face but since each confront is associated to $3$ vertices the is counted 3 times. Therefore, $V = 12 imes 5 div 3 = 20$.

For edges, there space $12$ faces times $5$ edge per confront but since each edge joins $2$ encounters it is counted twice. Therefore, $E = 12 imes 5 div 2 = 30$.

Do you watch the pattern? try the same exercise with a tetrahedron and an octahedron to check out if you obtain it.


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