I will use the following definitions

Platonic graph: A 3-connected planar graph with faces bounded by the same number of edges and vertices having the same number of incident edges.

(remark: the faces of a 3-connected planar graph are well-defined due to Whitney's theorem)

Combinatorially regular polyhedron: A polyhedron with Platonic vertex-edge graph.

Platonic solid: A combinatorially regular convex polyhedron with congruent faces of regular polygons.

Suppose we know the existence of the 5 Platonic graphs, but we don't know the existence of Platonic solids. How can we prove that they exist?

The existence of combinatorially regular convex polyhedra follows from the existence of the Platonic graphs by Steinitz's theorem.

But how can we know, that every combinatorially regular convex polyhedron can be continuosly deformed so that their faces become congruent regular polygons?


1 Answer 1


Consider the docecahedron as an example: Between a minuscule regular pentagon on $S^2$ having angles slightly over $108^\circ$ at the vertices and a hemispherical regular pentagon with angles $180^\circ$ at the vertices there is a regular pentagon on $S^2$ with angles $120^\circ$ at the vertices. Twelve such pentagons will tile $S^2$ in the desired way: Just place one such pentagon anywhere on $S^2$. Five more will precisely fit around it. Put another five into the five indentations on the outside of this configuration, and a hole will remain for the twelfth pentagon.

  • 1
    $\begingroup$ >Twelve such pentagons will tile S2 in the desired way.< Why? How do we know that they will cover the sphere without overlapping? $\endgroup$
    – mma
    Apr 28, 2015 at 4:59
  • $\begingroup$ One way would be to construct the symmetries of the graph as a spherical kaleidoscope and then to derive the two associated regular solids (corresponding to the graph and its dual) from that kaleidoscope. But is that a sufficiently direct "how" for you? $\endgroup$ Mar 9, 2020 at 7:29

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