# 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 - https://en.wikipedia.org/wiki/Platonic_solid

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?

• If you drop the regularity requirement, you get Kepler-Poinsot polyhedra. – Noah Schweber Dec 11 '17 at 2:49
• @NoahSchweber - these are interesting, but I don't think they satisfy the requirement that each vertex should have the same number of faces and edges intersecting at it. – Rohit Pandey Dec 11 '17 at 2:53
• They do, actually… the faces or vertex arrangements are regular star polyhedra. – Parcly Taxel Dec 11 '17 at 2:55
• @ParclyTaxel can you provide an example of such a polyhedron? – Rohit Pandey Dec 11 '17 at 3:01
• The Kepler-Poinsot polyhedra do satisfy the requirements because the non-convex "vertices" are not considered vertices of the polyhedra, they merely happen to be points where three faces intersect. For example, the great dodecahedron is considered to have only 12 vertices and 12 pentagonal faces. However, since the faces pass through each other, I suppose you cannot really call it a "solid". It does not have a clearly defined interior, it is merely an arrangement of 12 pentagons with nice regularity properties. – Rahul Dec 11 '17 at 3:16