A *convex prism* is a subset of $\mathbb{R}^3$ congruent to the Cartesian product of a convex polygon (the prism's base) with the interval $[0,1]$.



> 
> **Question.** If a family of congruent convex prisms tiles space (not necessarily in a face-to-face manner), must there exist a tiling of
>  the plane with polygons congruent to the prism's base?
> 
> 
> 



Space-tiling convex prisms
[The following is not quite an answer, but it refutes a natural
generalization suggested in the Comments, and is too long to be
a comment itself.]


Counterexample in ${\bf R}^N \times {\bf R}$ for some $N>2$:
any lattice hexagon $H$ with angles 
$90^\circ$, $90^\circ$, $135^\circ$, $135^\circ$, $135^\circ$, $135^\circ$ 
in that order. (That is, $H$ is obtained from a lattice rectangle by
truncating two adjacent vertices by two isosceles right lattice triangles, 
not necessarily congruent. Alternatively, remove two congruent
lattice triangles, not necessarily isoceles, related by a 90-degree rotation.)
Then $H$ does not tile the plane, but
four copies do tile a (non-simply-connected) polyomino. But it was
recently shown that *any* polyomino in some ${\bf Z}^n$
(which need not be simply connected, or even connected at all!) 
tiles ${\bf Z}^d$ for some $d$:



> 
> Vytautas Gruslys, Imre Leader, and Ta Sheng Tan: Tiling with arbitrary tiles.
>  *Proc. London Math. Soc.* (2016) **112** (6): 1019-1039.
>  <https://doi.org/10.1112/plms/pdw017> $\cong$ <http://arxiv.org/abs/1505.03697>
> 
> 
> 


(I learned about this from **Francisco Santos**'s accepted answer to 
[**Timothy Chow**'s 
 Mathoverflow question 49915](https://mathoverflow.net/questions/49915), which the MO algorithm helpfully put at 
the top of its list of questions "Related" to this one.)


