

 Are there any regular solids other than cubes that can fill 3D space entirely?

 How about nonregularsolids, not based on cubes? Like, uh, those uh, math tiles, the kite and delta, except not flat? Although, I suppose if they were a quarterinch thick they would have to count, but besides them
 MC
I think dodecahedrons can…
Well, there would have to be nonregular solids that could fill space completely. There are several ways to arrange spheres to fill a space with each one touching others in the same way. For example, hexagonal prisms could do it, if you stacked them one layer at a time.
As for the trickier issue, regular polyhedra, I’ve long wondered about that. There are only five, and we know the cube can, so that only leaves four to try. I’m pretty sure the icosohedron (20 sides, all of which are triangles) can be easily eliminated. That just leaves the tetrahedron (a pyramid with a triangular base), the octahedron (two regular pyramids joined at the base), and the dodecahedron (12 pentagonal sides).
The cube is the only regular polyhedron that can tile space. In order for a polyhedron to tile space it must have a dihedral angle (angle between two faces that share an edge) that is an integral divisor of 360 degrees.
The dihedral angles for the five regular polyhedra are as follows:
Tetrahedron: about 70.53 deg
Cube: 90 deg
Octahedron: about 109.47 deg
Dodecahedron: about 116.57 deg
Icosahedron: about 138.19 deg
You can see that the cube is the only one that has a dihedral angle that works.
For nonregular solids or when you can use more than one solid it gets more complicated.
P. Schmitt discovered a nonconvex aperiodic polyhedral spacefiller around 1990, and a convex polyhedron known as the SchmittConway Biprism which fills space only aperiodically was found by J. H. Conway in 1993, but I can’t find a picture or any other info other than the bare mention. Conway is the kites and darts guy.
Cabbage’s post is both earlier and better.
Here are some more notquiteregular solids which tile space…
The rhombic dodecahedron: 12 rhombic faces; if you glue a foursided pyramid to each face of a cube you get this, provided the pyramids have just the right height (you want the sides of the pyramids to match up in pairs to give the twelve rhombi which make up the solid).
The truncated octahedron: lop the points off of an octahedron, to get a shape with 8 hexagonal and 6 square faces.
These are both periodic tilings, unlike the SchmittConway biprism, and hence aren’t quite as interesting, but they’re both still neat.
I fou mean regular polyhedra, I ges the choices are the cube, tetrahedron (solid with four regular triangles),
and dodecahedron (solid with 12 regular pentagons)
Math Geek, are you sure about the truncated octahedron? I can’t for the life of me figure out how to tile them.
I’m sure, although unfortunately I can’t remember where I first saw the tiling.
There are two kinds of dihedral angles in the truncated octahedron (assuming that all edge lengths are equal and all faces are regular polyhedrons): the dihedral angle where two hexagons meet is acos(1/3)=109.47… degrees, while the dihedral angle where a hexagon and a square meet is acos(1/sqrt(3))=125.26… degrees. Since 109.47+2(125.26)=359.99 (roundoff error), you have to arrange the tiles so that two “hexagonsquare” edges always meet one “hexagonhexagon” edge.
So…start with a single “truncocta”. Glue a new truncocta onto every hexagonal face, but twisted so that hexagons on the new truncoctas are next to squares on the original truncocta, and vice versa. Then another set of new truncoctas should fit perfectly onto the square faces of the original truncocta, covering it completely. Add additional layers in the same fashion.
(Having just reread that last paragraph, I really really wish I had diagrams to show you…oh well.)
I mean regular polygons, of course…