Do non-convex platonic solids exist? Consider a solid with the following properties - 1. It is composed of congruent, regular polygons. 2. At each vertex, the same number of edges and faces meet. This is the same as the requirement for the Platonic solids, but the solid need not be convex. Of course, the five Platonic solids will satisfy these conditions, but are there any others? EDIT: consider the topological proof given in the Wikipedia article on Platonic solids - < If we require the Euler characteristic to be 1 instead of 2 (as in the proof) we get - $$\frac{1}{p} + \frac{1}{q} = \frac{1}{2} + \frac{1}{2E}$$ This still keeps open the possibility (using the same argument as for the Platonic solids given in that proof) of five such solids with Euler characteristic 1 (so they won't be convex). Question is, do these solids exist?

The central question here was if there exist solids that satisfy all requirements of the Platonic solids, but aren't convex. On thinking about this, at least one such solid exists (which proves they are possible). When constructing an Icosahedron, we take five triangles and form a bowl out of them. Then, we put an intermediate "ring" of triangles on top of this and finally, another bowl composed of five triangles is put on the very top. Now, imagine putting the top bowl upside down. This leads to a concave version of the Icosahedron. Note that the Euler characteristic of this non-convex Icosahedron is still 2. And based on the edit to my question, there is the possibility of a solid with the same vertices, edges and faces as an Icosahedron (or any other Platonic solid) but having Euler characteristic 1. I'm interested in how we might find these solids as well, but I'll save that for another question.

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