Archimedean Solids Definition: Is Uniformity Necessary?

Question

Is every convex and isogonal polyhedron uniform? That is, are the faces of a convex isogonal polyhedron all regular polygons?

If so, then the Archimedean solids could be defined as convex and isogonal, ditching uniformity. If not, then any counterexample would be helpful.

The reason I think this would be interesting is because the Catalan solids need not be uniform, since the dual of a uniform polyhedron need not be uniform. However, it would make the definitions symmetrical to define the Catalan solids as convex isohedral polyhedra, setting aside the Platonic solids and infinite families.

Answer 1

A rectangular box with different side lengths, e.g. side lengths $2$-by-$2$-by-$4$, has all angles $90^\circ$. So it is convex and isogonal but not regular.

Answer 2

uniformity for polyhedra requires 2 things:

1. vertex transitivity (i.e. isogonality)
2. equal edge size throughout (say unity)

wrt. higher dimensional polytopes you even would have to add

3. hierarchicality (i.e. all facial elements are uniform on their own, where - for induction start - in 2D uniformity = regularity)

--- rk

Answer 3

In principle, there are infinitely many polyhedra where all edges are equal and all faces are regular polygons. Just pick any regular polygon as a base for a prism and have its height equal the edge length of the base. Antiprisms with equilateral triangular lateral faces work, too.

Archimedean solids are typically defined to require more than one axis having the maximum rotational symmetry for that polyhedron, a restriction that admits only the cube among prisms and only the regular octahedron and regular tetrahedron (think of two opposing edges as di-gon "bases") among antiprisms.