Is there a neat way to show (or a reference that already proves) that

  • the 4-cube is the only convex 4-polytope in which all facets are regular 3-cubes?
  • the 24-cell is the only convex 4-polytope in which all facets are regular octahedra?
  • the 120-cell is the only convex 4-polytope in which all facets are regular dodecahedra?

Note that I do require that the facets are regular, so e.g. just any cubical polytope does not fit.

I am aware of Gosset's semiregular polytopes (vertex-transitive and all facets are regular polytopes), for which this statement is true, but I do not require vertex-transitivitiy here.

However, if my statement turns out to be wrong in this general form, then I wonder whether it holds when I require that all vertices are on a common sphere.


2 Answers 2


This is Theorem 1 (actually, Satz 1) of Roswitha Blind, Konvexe Polytope mit kongruenten regulären $(n- 1)$-Seiten im $\Bbb{R}^n$ $(n \ge 4)$, Comment. Math. Helvetici 54 (1979) 304--308. The short proof follows from two short lemmas, one of which cites Coxeter's Regular Polytopes and an article by Shephard.

You might also be interested in a related question from last year with helpful links, which references a second 1979 paper of Blind in her program to explore the analogue of Johnson solids for 4-polytopes.


Within 4D we also have the bipyramid on a tetrahedral base, as well as the bipyramid on an icosahedral base, which both use 8 or 40 tetrahedra solely.

In the papers of the Blind couple there are further polytopes, with various regular facets listed too. But your restriction to just a single type of facets and your restriction to 4D comes down to just these 2, except of the six 4D regular polytopes - at least if you'd further restrict to convexity.

--- rk


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