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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.

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    $\begingroup$ 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/… $\endgroup$
    – Yoav Kallus
    Commented Jun 3, 2020 at 17:54
  • $\begingroup$ @YoavKallus Interesting link! Yes that's exactly the same polyhedron. $\endgroup$
    – M. Winter
    Commented Jun 3, 2020 at 18:14

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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.

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  • $\begingroup$ 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. $\endgroup$
    – M. Winter
    Commented Jun 3, 2020 at 22:10
  • $\begingroup$ 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. $\endgroup$
    – Wlodek Kuperberg
    Commented Jun 3, 2020 at 23:27
  • $\begingroup$ 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? $\endgroup$
    – M. Winter
    Commented Jun 4, 2020 at 8:52
  • $\begingroup$ 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. $\endgroup$
    – Wlodek Kuperberg
    Commented Jun 4, 2020 at 12:49
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    $\begingroup$ Thank you. This direction now seems plausible to me for faces of all dimensions, as long as they are congruent. $\endgroup$
    – M. Winter
    Commented Jun 4, 2020 at 16:12

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