TITLE: Can every simple polytope be inscribed in a sphere?
QUESTION [17 upvotes]: 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 $d$?

REPLY [21 votes]: 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.