Because of Euler we know that for polyhedra without holes we have
V+F=E+2. As a result we can't have a polyhedron made entirely of
hexagons, and need at least 12 pentagons or an equivalent set of
shapes. Etc.

But things change if we allow intersecting faces. Clearly we have
a lot of scope for variations, but we can always retain limits to
make things reasonable.

But ...

Can we make a "polyhedron" consisting only of hexagonal "faces"?

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@ColinTheMathmo I don't know the answer, but I remember that Branko Gruenbaum was very interested in polyhedra where regularity conditions are maintained at the potential expense of allowing faces to self intersect.

Maybe you know all this...

@bremner I didn't know that, thanks. Someone has also pointed me at this:

en.wikipedia.org/wiki/Szilassi

@ColinTheMathmo Oh right. I was guessing you had convex faces in mind. In fact even self intersecting faces can make some sense. Consider e.g. a 5-cycle 13524.

@bremner Yeah, we know that with flat, convex, non-interesting faces, and something without holes, you can't have all hexagons. So the question is, what conditions do we relax? What is the *minimum* amount of relaxation we can do?

It's an odd question, but there's something there.

@ColinTheMathmo @bremner You can even make all faces be right-angled L-shaped hexagons: 11011110.github.io/blog/2009/0

@ColinTheMathmo @bremner Or for another interesting example, Grünbaum in digital.lib.washington.edu/res suggests the small triambic icosahedron en.wikipedia.org/wiki/Small_tr, a self-intersecting polyhedron with congruent convex-hexagon faces

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