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It is known that not every convex polytope (even polyhedron, e.g. this one) can be made inscribed, that is, we cannot always move its vertices so that

  • all vertices end up on a common sphere, and
  • the polytope has not changed its combinatorial type in the process.

Is there anything known about whether this is possible if we instead ask for simple polytopes, i.e., $d$-dimensional polytopes of vertex-degree $d$?

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  • $\begingroup$ Could you give a bit more detail about your reference to the triakis tetrahedron? The page you link to seems to include a picture of an inscription of the triakis tetrahedron on a sphere: en.wikipedia.org/wiki/Triakis_tetrahedron#/media/… . What am I missing? $\endgroup$
    – HJRW
    Commented Oct 2, 2020 at 10:42
  • $\begingroup$ @HJRW The picture shows a spherical polyhedron rather than a convex one (it is not the convex hull of finitely many points, has not only flat faces, etc.). The fact that the triakis tetrahedron is not inscribable can be found e.g. in "Six Topics on Inscribable Polytopes" by Padrol and Ziegler (p. 409 in "Advances in Discrete Differential Geometry"). $\endgroup$
    – M. Winter
    Commented Oct 2, 2020 at 10:46
  • $\begingroup$ I see it now! The straight line between two vertices of the underlying tetrahedron passes "below" the straight line between the vertices in the middle of the faces of the tetrahedron, so the convex hull contains additional edges and vertices. Thanks for the explanation. $\endgroup$
    – HJRW
    Commented Oct 2, 2020 at 10:50

1 Answer 1

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Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler

In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty completely. I think already a cube with corner cut, which is simple, will be a non-inscribable $3$-polytope. This can be checked with the following two lines of sage:

sage: C = polytopes.cube().intersection(Polyhedron(ieqs = [[15/8,1,1,1]])) 
....: C.graph().is_inscribable()                                                
False

sage: C.is_simple()                                                             
True

It's nice that Rivin's criterion is implemented in sage...

Here's an image of the graph of the "cube without one corner" 3-polytope, which is non-inscribable and simple:

I just checked that this is the smallest non-inscribable simple 3-polytope: all other simple 3-polytopes with up to 10 vertices are inscribable.

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