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
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.
Can we make a "polyhedron" consisting only of hexagonal "faces"?
@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...
@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: https://11011110.github.io/blog/2009/09/18/not-nauru-graph.html
@ColinTheMathmo @bremner Or for another interesting example, Grünbaum in https://digital.lib.washington.edu/researchworks/handle/1773/4593 suggests the small triambic icosahedron https://en.wikipedia.org/wiki/Small_triambic_icosahedron, a self-intersecting polyhedron with congruent convex-hexagon faces
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