# Convex polyhedra jammed in $k$ disjoint holes

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...

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(1) Why is the question restricted to convex polyhedra instead of looking at all convex bodies? The "best" among polyhedra may not exist, but among all bodies existence is more likely. (2) If $P$ were a ball, then the question would be about packing $k$ congruent spherical caps of maximum size on the sphere (a hard problem already), so this is a nice generalization. – Wlodek Kuperberg Jan 18 '14 at 3:57
One should be able to lower bound d by looking at packing a ball with spherical caps, as Neil Sloane has done elsewhere. – The Masked Avenger Jan 18 '14 at 4:05
In fact, capping a polyhedron makes more sense than placing holes around a polyhedron. – The Masked Avenger Jan 18 '14 at 4:07
@JosephO'Rourke Correct me if I'm wrong: It seems that if $P$ is a polyhedron with $v$ vertices and $k>v$, then $d(P,k)=0$, but if $K$ is strictly convex, then $d(K,k)\neq 0$ for all $k$. – Wlodek Kuperberg Jan 18 '14 at 4:41
@JosephO'Rourke: A suggestion. Instead of measuring the depth of the hole (cap) with respect to the circumscribed sphere, measure it with respect to the width of the body relative to the two support planes parallel to the cutting plane. This way the depth and the quantity $d(K,m)$ become affine invariant. Then, by compactness of the space of affine equivalence classes of convex bodies [Macbeath 1951, MR0045406] there exist both a best and a worst body for the values of $d(K,m)$. I do not dare to conjecture anything, but the ball (ellipsoid) and for $m=4$ the tetrahedron come to mind naturally. – Wlodek Kuperberg Jan 18 '14 at 15:03