For a given convex polyhedron $P \subset \mathbb{R}^3$, I was imagining finding the optimal "fixing" of $P$ in holes (or jamming them in "mud"), which led to the following question.

First, scale $P$ to have a minimal circumscribed sphere of radius $1$.
Let a *truncation of $P$* be an intersection of $P$ with a halfspace $H$,
which naturally defines a topological circle on the surface of $P$
where $H$ cuts through that surface, and a maximal *depth* $d$
from $H$ to points of $P$ cut off by the truncation.
I am thinking of $d$ as a measure of stability when $P$ is jammed
into a hole of depth $d$.

Now, suppose one desires that $P$ be jammed into $k$ distinct holes, none of which overlap in the sense that the open disks on the surface defined by the truncation circles determined by the planes in which the holes are sunk, are all disjoint. The question is,

Q. Given $P$ and $k$, what is the optimal $d=d(P,k)$ achievable, that is, what is the largest $d$ over the minimum of the $k$ truncations, i.e., so that every hole is at least $d$ deep.

A natural example is that $P$ is the cube $C$ inscribed in a unit-radius sphere,
and so with corner coordinates $\pm \frac{1}{\sqrt{3}} \approx 0.58$. For $k=2$,
one can achieve $d=1$:

For $k=8$, it is natural to truncate each vertex of the cube maximally, carving out the cuboctahedron. If I've computed correctly, then $d(C,8)=2/(3\sqrt{3}) \approx 0.38$
(meaning it could be stuck into $8$ distinct holes, each $38$% of the circumradius deep):

What seems more difficult to analyze are the situations when $k$ and the symmetries of $P$ are unrelated, e.g., $k=3$ or $k=5$ for the cube $C$.

Ultimately I am seeking to identify the polyhedra that are the "best"—most stable—for $k$ holes. Likely this is a well-studied problem from another perspective? It is, in a sense, a particular partition of the surface into nonoverlapping (topological) disks, and so in that sense a type of packing...