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Platonic Solids

In two dimensions, we have the regular polygons, which are polygons with equal angles and side lengths. In three dimensions, we have the platonic solids, which are polyhedra with congruent regular polygons as faces and the same number of faces meeting at each vertex.

Despite the fact that there are infinitely many regular polygons, there are only five platonic solids named after the number of faces they have, namely:

Tetrahedron

Hexahedron/Cube

Octahedron

Dodecahedron

Icosahedron

Proving the Nonexistence of Other Platonic Solids

Proving the existence of the platonic solids is nontrivial, but proving the nonexistence is not that hard. Consider a solid with faces of regular polygons. Then, at least three faces must intersect at each vertex. However, to make the polygon not just a flat surface, the three (or more) faces' angles must also sum to less than $360^\circ$! Then, simple casework on the polygon gives all the (possible) platonic solids:

Triangles:

3 triangles:

This gives the tetrahedron. Its dihedral angle is about $70.5$ degrees.

4 triangles:

This gives us the octahedron. Dihedral angle: $109.5$ degrees.

5 triangles:

This gives us the icosahedron. The dihedral angle is about $138.2$ degrees.

Squares: 3 squares:

This gives us the square. Dihedral angle: $90$ degrees.

Pentagons: 3 Pentagons:

This gives us the dodecahedron and since the sum of angles is $324^\circ$, it doesn't curve into $3D$ as much, so we have a quite rounded shape. Since we have pentagons instead of triangles, the dihedral angle is only about $116.56^\circ$.

Three hexagons have a sum of angles equal to $360^\circ$, so we are done. There are no other cases.

That's all there is for this explanation. Thanks for reading!