What regular polyhedra can be tightly packed?

Question

By "tightly packed", I mean that 3-dimensional space can be occupied solely by a collection of these same-sized regular polyhedra with no air gaps in between.

I can think of three: Tetrahedra, cubes, and dodecahedra. Are there others?

Answer 1

The only regular polyhedron that tiles three-dimensional space is the cube.

The tetrahedron does not, although this is a very old misconception dating back to Aristotle!

The regular dodecahedron also does not tile space. What you describe in a comment (taking the Voronoi cells of spheres packed in an FCC lattice) does produce a face-transitive polyhedron with 12 faces that tiles three-dimensional space, but its faces aren't pentagons, they're rhombi! This results in the rhombic dodecahedron, shown tiling 3D space below:

One easy way to see that none of the Platonic solids besides the cube fill space is to look at their dihedral angles (the angle between two faces where they meet at an edge) - in no case besides the cube do these angles divide 360 degrees, so you can't fit them around an edge without a gap.

If you relax regularity and request only that the faces be regular polygons, then we can expand our domain to include the Archimedean and Johnson solids, which gives us the triangular prism, the hexagonal prism, the truncated octahedron, and the gyrobifastigium.

Answer 2

As described by RavenclawPrefect, the only Platonic sold capable of packing all of space is the cube. However, some less regular polyhedra work:

• Trivially, you can make a prism (right or oblique) using any base that tiles the plane. There are infinitely many choices, including parallepipeds.

• The truncated octahedron. If you cut the corners of a regular octahedron the right way, you get an Archimedean solid consisting of square and regular hexagonal faces, and this will tile space. Conway and Guy's Book of Numbers uses packings of this figure to represent three-dimensional figurate number patters. Centers of these polyhedra form a face-centered cubic lattice.

• The squashed octahedron. Take a regular octahedron with edge length $s$ and push two opposing vertices together until the distance between them is also $s$ (making eight of the edges shorter than $s$). To fill space with this figure, fit the copies together using three mutually perpendicular orientations of the shortened axis. The vertices of the polyhedra form a body-centered cubic lattice.

Answer 3

And then there is a further honeycomb of regular polyhedra, all (each) of equal size: the "alternated cubic honeycomb" built from octahedra and tetrahedra in the same alternating manor, as in the cuboctahedron do alternate triangles and squares.

Sure, this polytopal honeycomb has not a single building block, instead there are two, but it still packs the 3-dimensional space tightly, i.e. it can be occupied solely by a collection of these same-sized regular polyhedra with no air gaps in between. Thus it still conforms to the original quest in some way.

--- rk