Note that a Platonic solid is a regular polyhedron, so all of its faces are regular and congruent. Briefly, the Euler characteristic says that vertices minus edges plus faces equals two for all polyhedra (this can be proven by puncturing one face, mapping the polyhedra to a planar graph and triangulating it. With this relationship, the fact that regularity implies if the polyhedron has faces with n edges each and m faces meet at each vertex then the following equations hold:
V - E + F = 2
2E = nF
mE = 2V
The only solutions to these equations amongst the integers give us the tetrahedron, cube, octahedron, dodecahedron and icosahedron.
Yeah, I know that (although I don't have the equations committed to memory!). My point was that if this topological
relationship on the net works, then the geometric
condition of congruency and regularity is unnecessary and the proof is more general. That is, the proof using the Euler formula demonstrates that one could not make a polyhedron of (say) heptagons (regular or otherwise, not necessarily congruent) where some constant number N
≥3) heptagon vertices meet at each vertex of the polyhedron. My loose geometric reasoning only works for regular congruent polygons.
But I think it doesn't demonstrate (nor does it attempt to) that one cannot make a polyhedron using regular congruent polygons but with varying
numbers of polygons meeting at each vertex - imagine two tetrahedra pushed together and the joining faces removed. But triangles are always the most potent geometers! Squares won't work (unless one accepts coplanar faces), so that probably only applies to triangles.
Can one pave a polyhedron with congruent heptagons, or congruent hexagons (relaxing the condition on constant number of faces at each vertex)? You can have three faces meeting if the fatter corners are kept apart from each other, but do those fat corners eventually bite you on the backside? This is left as an exercise for the reader, or for Lord Monckton, as he's such a friggin' mathematical genius, in his own mind.