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

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Its is clear that their space \$20\$ vertices, \$30\$ edges, and \$12\$ faces. Ns am not sure exactly how to prove this though.

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

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

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

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What space the formulas because that the variety of vertices, edges, faces, cells, 4-faces, ..., \$n\$-faces, of convex continual polytopes in \$n geq 5\$ dimensions?
Probability the connecting 2 vertices in a dodecahedron will result in a segment inside the dodecahedron

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