# Polyhedra that can pack 3-space only in a non-vertex-to-vertex fashion

Question: Are there polyhedral units (convex or otherwise) that can pack 3D space without gaps only such that the arrangement is not vertex-to-vertex?

Same question can be asked with 'edge to edge'.

Note 1: In a vertex to vertex arrangement, any vertex of any unit touches any other unit only at a vertex - and not at an intermediate point of an edge or a face. In an edge to edge arrangement, any edge of any unit cannot touch another unit at an interior point of a face.

It appears unlikely that there are convex polyhedral units that pack 3-space only such that some vertices touch another unit at an interior point of a face.

Note 2: In 2D, at least some of the pentagons that tile the plane (discovered by Marjorie Rice) appear to tile only in non-vertex-to-vertex manner.

• Do rectangular prisms of the tilings alluded to in Note 2 work? Feb 2, 2020 at 15:06
• Thank you very much. Yes, they should work as examples of polyhedral which pack perfectly only when an edge of a unit touches the interior of faces of another unit. But are there non-prismatic polyhedrons that pack perfectly and have this property (Prisms are 2D in some sense, aren't they?)? And one still suspects that there probably are no convex polyhedrons that are perfect packers only when some vertices touch an interior point of a face of another unit. Feb 4, 2020 at 13:43

It seems that there are convex polyhedra which tile 3D space, for which any such tiling must have a vertex of some tile lying on the interior of a face of another tile.

Consider the polyhedron which is the convex hull of the following points:

$$(0,0,4), (2,0,2), (2,0,-2), (0,0,-4), (-2,0,-2), (-2,0,2), (3,\sqrt{3},0), (-3,\sqrt{3},0), (0,2\sqrt{3},2), (0,2\sqrt{3},-2)$$

This polyhedron can tile an unbounded hexagonal prism: Place three copies of the tile by rotating the shape defined above by 0, 120, and 240 degrees around the edge through $$(0,2\sqrt{3},2)$$ and $$(0,2\sqrt{3},-2)$$. Then take a copy of those three tiles, rotate them by 60 degrees around the same line and shift by 4 in the $$z$$ direction. These six tiles can then be repeated indefinitely by shifting by 8 in the $$z$$ direction. So far, everything is vertex-to-vertex. But notice that there are regularly spaced vertices lying in the middle of each side of each hexagonal prism, and the location of these are offset between adjacent sides of the prism. We can put two hexagonal prisms together aligning these points vertex-to-vertex, but then the prism which touches those two cannot be placed to have all the vertices line up.

Of course, I have not proven that there is not some other way to tile 3D space with this tile. But I suspect that the only tilings with this tile involve forming hexagonal prisms, in which case the result holds. Note also that the polyhedron I described can be stretched by an arbitrary factor in the $$z$$ direction to form a family of examples of this form, in case the parameters I chose above happen to admit some surprising alternate tiling.

Assume "polyhedron" is being understood simply as a (numerically) finite intersection of halfspaces.

Then you well could tile 3D space by the "polyhedron" 1, which is just the upper halfspace $$(0,0,1)\cdot (x,y,z)\ge 0$$, and the "polyhedron" 2, which is just the lower halfspace $$(0,0,-1)\cdot (x,y,z)\ge 0$$.

This tesselation then obviously isn't vertex to vertex, as neither of those "polyhedra" have any vertices at all. Moreover both used "polyhedra" are clearly convex each. (It's just that those are not compact.)

Btw., by the same argument that very tesselation is not edge to edge either, as neither of those "polyhedra" have any edges at all.

--- rk

• Are not these strictly speaking vertex-to-vertex and edge-to-edge? I mean, since there are no instances of the condition to check, the condition is trivially satisfied. Feb 24, 2020 at 7:47