Definitions of Platonic and Archimedean Solids using Symmetry Groups? A Platonic Solid is defined to be a convex polyhedron where all the faces are congruent and regular, and the same number of faces meet at each vertex. An Archimedean Solid drops the requirement that all the faces have to be the same, but they must still all be regular, and each vertex must have the same arrangement of faces. 
However, for Archimedean Solids, the pseudorhombicuboctahedron fits this definition, despite not being vertex-transitive (meaning that the rotation group of the solid does not act transitively on the vertices). 
I was wondering: For Platonic Solids, is it equivalent to define them as convex polyhedra that are face, vertex, and edge-transitive (where for Archimedean solids, we drop the face-transitivty condition)? Face-transitivity forces all the faces to be congruent, edge-transitivity forces all the faces to be equilateral, and vertex-transitivity forces the same number (or arrangement, in terms of Archimedean solids). It's not immediately obvious to me that these conditions force the faces to be equiangular as well as equilateral...does it indeed follow, or is there a counterexample?

The main observation here is the following:
Being vertex transitive (and considering bounded solids only) requires the vertices to lie on sphere. Faces on the other hand are bound to be planar by definition, thence those are contained within a supporting plane. The intersection of the plane and the sphere (ball) simply provides a circle (disc). Now, using equisized edges only (being an Archimedean solid), it becomes evident that the edge circuit of each polygonal face has all its vertices on that very circle, they will be evenly spaced appart, so in effect those each happen to be regular polygons only.
In fact this even would apply for their non-convex counterparts as well, just that the faces then might become non-convex regular polygrams as well. But still all corner angles thereof would be forced to the same size.
Note that the same argument also applies to the other non-flat geometry, i.e. for the hyperbolic tilings. Again the cross-section of a planar face plane and the there supporting hyperboloid defines a circle. - In contrast this argument would break down within any flat geometry (euclidean space of arbitrary dimension). So for instance there is a vertex- edge-, and face-transitive tiling, which still isn't a regular tiling of the plane: the rhombic tiling (of non-square rhombs).
--- rk
