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.

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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