TITLE: Is there a midsphere theorem for 4-polytopes?
QUESTION [21 upvotes]: The (remarkable) midsphere theorem says that each combinatorial
type of convex polyhedron may be realized by one all of whose edges are
tangent to a sphere
(and the realization is unique if the center of gravity is specified).
          

Q1. Is there an analogous theorem for 4-polytopes,
  that each combinatorial type may be realized by a polytope
  with ridges (or edges?) tangent to a 3-sphere?

Because the proofs of the midsphere theorem rely on the
Koebe–Andreev–Thurston circle-packing theorem,
a related query is:

Q2. Is there a generalization of the circle-packing
  theorem to sphere-packing?

Both questions may be generalized to arbitrary dimension.
I suspect the answer to both questions may be No,
in which case a pointer would suffice.  Thanks!

REPLY [5 votes]: In a recent paper of Padrol and me, we studied several generalizations of this problem.  http://arxiv.org/pdf/1508.03537v1.pdf
Regarding Q1, Yoav already mentioned Schulte's work, and Gil mentioned that stacked polytopes won't work.  In fact,

For any 0 ≤ k ≤ d − 3, there are stacked d-polytopes that are not k-scribable.

But the other side of the story is much more joyful:

Every stacked polytope is (d − 1)-scribable (i.e. circumscribable) and (d − 2)-scribable (i.e. ridge-scribable).  Dually, it means that every truncated polytope is edge-scribable and inscribable.  In particular, the $1$-skeleton of every truncated polytope is the tangency graph of a sphere packing.

We also studied the weak scribabilities, also mentioned in Schulte's work.
The most interesting generalization (which I would like to make some advertisement here) is the $(i,j)$-scribability problem: 

Can we realize every $d$-polytope with all $i$-faces "outside" the sphere and all $j$-faces "intersecting" the sphere, $0\le i \le j \le d-1$?

The case of $i=j$ is just the classical scribability problem.  For $i<j$, we constructed examples with no such realization for all but two cases: $j-i=d-2$ for odd $d$, or $j-i=d-1$ for any $d$.