(Formerly on MSE.)
Here is a fairly natural question: Can three-dimensional space be filled with convex polyhedra of the same incidence structure (if not the same geometry) as the regular dodecahedron, such that four pieces meet at every vertex?
We don't require that the cells be congruent to one another, but we'll say that each dihedral angle must be strictly less than 180 degrees so as to forbid degenerate cases that actually form polyhedra with fewer faces. We'll also require that the tiling be "face-to-face", such that every point in space is in one of four categories:
- Strictly within the interior of a dodecahedron.
- On the boundary of exactly two (closed) dodecahedra, and strictly within the interior of a face of each.
- On the boundary of exactly three dodecahedra, and strictly within the interior of an edge of each.
- On the boundary of exactly four dodecahedra, and a vertex of each.
That is, every face meets exactly one congruent face on the opposite side, every edge is shared between all its members without any sliding or offsets, etc. This makes the combinatorial incidence structure especially straightforward.
I strongly suspect the answer is "no", and I'd like to give the following argument for it:
As we build up any attempt at such a construction, we will be "filling out" the incidence structure of the cells of the 120-cell, with each decision forced by the way in which the dodecahedra must meet one another.
So we can place an initial cell, and lay down its 12 neighbors; this will force 20 more to fill out the vertices one edge away from a vertex of the original cell in a second layer. After placing these, we'll lay down another 12 touching our first layer on the face opposite that which they touch the central cell. After that, we'll place another 30, then another 12, then another 20, then another 12, and then we'll find that the resulting union is a single dodecahedron, requiring exactly one more piece at each vertex, which is now forced to be concave and in fact to "wrap around" the entire arrangement in one go, its interior equal to the full rest of space.
However, this argument doesn't seem incredibly rigorous, and it rests heavily on a particular four-dimensional object in a way that doesn't clearly generalize to similar questions. I'd like something more systematic.
This argument is perhaps easier to feel the force of for the analogous two-dimensional question of tiling the plane with convex pentagons three to a vertex. When one tries, they cannot help but find themselves drawing a skeleton of the dodecahedron, and after completing two rings around a central pentagon there is nowhere to go:
In the two-dimensional case, this less-than-perfectly-rigorous argument can be replaced with a more precise appeal to some invariants using Euler's formula:
Consider a finite patch of pentagons in a tiling. Assign to every pentagon in this tiling a score as follows: if $a$ vertices are shared by one other pentagon in the patch, and $b$ vertices are shared by two other pentagons, the score is $1-a/4-b/6$. Using Euler's formula $V-E+F=2$, it's easy enough to see that the total score in a finite simply connected patch is always equal to $1$.
Now, suppose we had a tiling of the plane. Take a simply-connected patch $P$ of at least 7 pentagons, and add all pentagons sharing an edge with any of those 7 to create a larger simply connected patch $Q$. Each of our starting pentagons now has a score of $1/6$ (since $b=5$ for a fully surrounded pentagon), so they contribute more than $1$ to the total in $Q$. This means that at least one of the outer pentagons in $Q$ must have a negative score. But there is only one local arrangement that assigns a negative score to a pentagon, and it requires that none of that pentagon's neighbors be fully surrounded; since all of these outer pentagons border an inner pentagon, this is a contradiction. So it is impossible to arrange pentagons in the plane in a degree-3 way, even far enough out to surround a group of 7 central pentagons.
(Note that this argument doesn't actually use convexity, only the graph-theoretic incidence structure.)
However, I don't see how to perform an analogous argument in the polyhedral case, because the Euler characteristic for polychora is $V-E+F-C=0$, which is invariant to e.g. having twice as many of everything so doesn't naturally lead to any upper bounds.
I've spent some time playing with different possible invariants, and despite my intuitive feeling that some kind of resource is being used up as more dodecahedra are added, I haven't found a metric that describes this in a satisfactory way; most angles of attack on the problem end up petering out in a way that relates to the zero Euler characteristic.
What sorts of machinery can be used to settle questions like these? Are there existing methods in the literature I should examine?
