V+F-E=2

Euler's celebrated polyhedron formula V+F-E=2 is part of the MT365 course. Worked on question 3 of MT365 TMA01 today ( close to the cut-off date by the way ). It was a proof which required V+F-E=2 and two Handshaking Lemma's: one for graphs and for one polyhedra.

A few years ago I read that Euler proved that the only possible -regular- polyhedra are:
- the Tetrahedron with triangle faces and V4-F4-E6,
- the Cube with square faces and V8-F6-E12,
- the Icosahedron with triangle faces and V6-F8-E12,
- the Dodecahedron with pentagon faces and V20-F12-E30 and finally,
-  the Icosahedraon with triangle faces and V12-F20-E30.

I think that I can prove this now too, if I can start from V+F-E=2 that is. I can at least reproduce the regular polyhedra. Part of question 3 was giving the names of at least three semi regular polyhedra with both triangle faces and pentagon faces.