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I recently posted Engel-38, a 38-sided plesiohedron found by P. Engel in 1980. The set of vertices was a question here. A plesiohedron is a space-filling polyhedra that is also a Voronoi cell.

The plesiohedrons are a subset of stereohedrons, which do not need to be Voronoi cells.

For this question, I'm looking for a simple convex space-filling polyhedron that meets a few requirements.

  1. Fills space in a unique, obvious way. Cubes would fail, since columns of cubes can be skew to each other. The Engel-38 variations would fail because they are not obvious.
  2. Let the spacefiller $S$ have arbitrary point $a$ inside, then surround $S$ with copies of itself using the method in 1. Connect $a$ to all corresponding $a'$ for a set of edges equal to the faces. For any $a$, there must be two or more edge-face pairs that are not perpendicular.

Is there a stereohedron that is somewhat obviously not a plesiohedron? (I know several complicated, non-obvious examples.)

The next step up would be a space-filling polyhedron which is not a stereohedron. A good example of this is the Schmitt–Conway–Danzer biprism. Are other examples of this type known?

Ed Pegg
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  • I don't understand "Connect $a$ to all corresponding $a'$ for a set of edges equal to the faces" - are you assuming the polyhedron is convex? – RavenclawPrefect Sep 16 '22 at 22:42
  • For example, in a unit cube, the point (a,b,c) would connect to (a+1,b,c), (a-1,b,c), (a,b+1,c), (a,b-1,c), (a,b,c+1) and (a,b,c-1). But all of these lines are perpendicular to the faces. Added the word convex. – Ed Pegg Sep 16 '22 at 22:51
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    Take any reasonable unique space-filler (a truncated octahedron will do) and stretch it 10-fold along some arbitrary direction. How about that? – Ivan Neretin Sep 24 '22 at 11:25
  • Ivan, I think that might work. If there are two separate points that are centers for half the faces, then there isn't a single point, and the shape then is obviously not Voronoi. – Ed Pegg Sep 24 '22 at 12:18

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