# A polytope with congruent facets and an insphere that is not facet-transitive?

Is there a $$d$$-dimensional convex polytope (convex hull of finitely many points, not contained in a proper subspace), with $$d\ge 4$$ and the following properties?

• All facets are congruent,
• it has an insphere (a sphere to which each facet is tangent to), and
• it is not facet-transitive.

In 3-dimensional space there is an example with the "memorable" name Pseudo-deltoidal icositetrahedron, depicted below. I believe its the only such polyhedron. I am not aware of any higher dimensional examples.

• This seems possibly related to a previous MO question about irregular, but fair dice. Is your 3D example the same as the example described in this answer? mathoverflow.net/questions/46684/… Jun 3 '20 at 17:54
• @YoavKallus Interesting link! Yes that's exactly the same polyhedron. Jun 3 '20 at 18:14

## 1 Answer

First, a simple remark: If a polytope with congruent facets is inscribed in a sphere, then it is circumscribed about a sphere as well, and the two spheres are concentric.

Next, there is a series of examples described and pictured in my old question Can the sphere be partitioned into small congruent cells? . Each of these examples is what you want in $$R^3$$. If you begin with any one such example and place it on a great 2-sphere of the 3-sphere in $$R^4$$ (say, the "equator"), then suspend it from the poles, you will get an example answering your question. The construction generalizes inductively to all higher dimensions.

• Thank you for your answer! Very interesting linked question of yours. However, I object to the claim in the first paragraph: the rhombic dodecahedron has congruent faces, an insphere that touches all faces, but no circumsphere that contains all vertices. Jun 3 '20 at 22:10
• You are right, an additional assumption is needed, namely the facets should be inscribed in a sphere of one dimension lower. Thank you, I am correcting my mistake. Jun 3 '20 at 23:27
• I know it is not relevant for the answer (which I am going to accept soon), but how can I see that your modified first paragraph indeed holds? Jun 4 '20 at 8:52
• Proof. Let P be a convex polytope with congruent facets and inscribed in a sphere S centered at the origin. Obviously, the origin is an interior point of P. Now, let S_0 be the largest sphere centered at the origin and contained in P. Then S_0 touches the boundary of P at at least one point, and the point must lie on some facet F of P. Since P is inscribed in a sphere, the facet is inscribed in a circle lying on S, hence S_0 touches F at the center of the circle. Since all facets are congruent, the distance from the origin to the center of each circle circumscribed about any facet is the same. Jun 4 '20 at 12:49
• Thank you. This direction now seems plausible to me for faces of all dimensions, as long as they are congruent. Jun 4 '20 at 16:12