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This is again a natural category of polyhedra without having an own name. Is it possible, that their graphs are the same as the graphs of polyhedra with faces of regular polygons?

My question is explicitely: Is it true, that every such polihedron can be distorted by keeping the length of its edges to a polyhedron having regular polygon faces? Or in other words: is it true, that if a 3-connected planar graph $G$ can be embedded into the metric $S^2$ so, that all the edges of the embedded graph have the same arc length, then $G$ is the graph of some polyhedron with regular polygon faces?

For example, a twsted cube has the same graph as the cube, and I haven't found a counterexample so far.

mma
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    Here's how I would try to construct an example. Take a rhombus. In fact, take 4 of them, identical, and have them meet at one point where they each have an acute angle. Then try to fill out the rest of a polyhedron in such a way that all the edge lengths are equal. If you can do that, you win: you can't make all the rhombi squares, since four squares can't meet at a vertex of a polyhedron. – Gerry Myerson Apr 21 '15 at 09:23
  • Wow! I think I've found one: the Rhombic dodecahedron, thanks! – mma Apr 21 '15 at 10:47
  • However, I don't know whether its vertexes lie on a sphere, or not. If not, then the two kind of formulations of my question aren't equivalent. So, in this case the second formulation can be true in spite of this example. – mma Apr 21 '15 at 10:53
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    There's a drawing of something called a spherical rhombic dodecahedron on that Wikipedia page, though it doesn't say whether all the edge lengths are equal. – Gerry Myerson Apr 21 '15 at 12:45

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