Is the rhombic dodecahedron the only isohedral polyhedron that tiles 3-space (other than the cube)?

Question

Is the rhombic dodecahedron

the only face-transitive (or isohedral, i.e. all faces are the same) polyhedron that seamlessly tiles 3-dimensional Euclidean space (other than the cube)?

I'm looking for an answer to this question and although Wikipedia provides a lot of lists for 3-tesselations I cannot find a definite closure. In particular, if the above statement were true I'd expect it to be listed on the tesselation's Wiki page, but no such statement exists, which leaves some doubt in me whether it is actually the case.

Answer 1

Does the (irregular) space-filling octahedron meet your criteria? (The tiles are not all obtained from a single tile by translation, i.e., this is not a lattice tiling. Instead, there are three families of mutually-orthogonal tiles in a tessellation.)

Answer 2

The rhombohedron works, though you may consider it too closely related to the cube to be of any interest:

Answer 3

No. The body-centered cubic tetrahedron tiles it as well.

Please note that an isohedral polyhedron is not just a polyhedron in which all faces are congruent, but one in which all the faces also lie in the same symmetry orbit (which unfortunately is not the case for the irregular space-filling octahedron).

Answer 4

The rhombohedron provided by Kaj Hansen does provide additional translational symmetries beyond those of a cubic tesselation — if the polyhedron is shaped properly.

Below is a reproduction of Kaj Hansen's picture, with additional lines indicating three pairs of parallel face diagonals.

In a rhomohedron the diagonals shown are congruent to each other. But if we shape the polyhedron so that these diaginals are also cobgruent to the edges, then the translations along these diaginals become symmetrically equivalent to those along the edges of the rhombohedron and the lattice becomes face-centered cubic or cubic close-packed.

The condition for this extra congruency is as follows: Body diagonal AB in the above figure measures $\sqrt6$ times the edges of the rhombohedron. Then the other body diagonals measure $\sqrt2$ times the edges and are also perpendicular to each other, making the six vertices besides A and B those of a regular octahedron. This regular octahedron corresponds the the face centers of the cubic unit cells with center of the octahedron corresponding to the empty body center of the cube. AB itself is a body diagonal of tge face-centered cube.

Answer 5

The rhombic dodecahedron could be split-up into six isohedral octahedrons mentioned above. But each octahedron could be split-up further into four isohedral tetrahedrons (so the rhombic dodecahedron consists of 24 isohedral tetrahedrons).

Below is the view of this isohedral tetrahedron inside the half of the cube:

Answer 6

Truncated octahedron also tesselates 3-dimensional space.