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I hope this is not too trivial for this forum. I was wondering if someone has come across this polytope.

You start with the rhombic dodecahedron, subdivide it into four parallellepipeds, and then fill the space between the four parallellepipeds with a tetrahedron, six parallellepipeds and four prisms (hopefully i counted correctly), so as to obtain a convex polytope.

Does this have a name? could someone provide a link to a picture?

edited Feb 14 at 13:30

asked Feb 14 at 10:00

Camilo Sarmiento
381●2●10

	No real answer, but I think you want six parallelepipeds, not four. – Klaus Draeger Feb 14 at 13:05
	fixed, thanks... – Camilo Sarmiento Feb 14 at 13:30

Joseph O'Rourke · Accepted Answer · 2013-02-17 02:32:57Z UTC

0

Is this it?
Rhombic Dodecahedron + Tetrahedron

answered Feb 17 at 2:32

Joseph O'Rourke
43.1k●1●49●214

	wow thanks. i think so, although it would help a lot if it could be seen from other perspectives, or even exploded according to the above subdivision. – Camilo Sarmiento Feb 18 at 8:47
	This doesn't look right to me. I would expect faces that are adjacent to equilateral triangles to all be rectangles. In your left picture, they look more like rhombi... – André Henriques Feb 24 at 21:45
	@André: I don't doubt you. I formed this as the Minkowski sum of a rhombic dodecahedron and a regular tetrahedron, but perhaps I did not scale one with respect to the other appropriately. Cannot check now... – Joseph O'Rourke Feb 25 at 0:44

André Henriques · Answer 2 · 2013-02-14 22:26:43Z UTC

2

It's the Minkowski sum of the rhombic dodecahedron with a regular tetrahedron. (The rhombic dodecahedron is it self the Minkowski sum of four segments).

answered Feb 14 at 22:26

André Henriques
14.8k●2●30●83

indeed. have you casually seen a picture of it with the decomposition written in the question? – Camilo Sarmiento Feb 16 at 16:13

Who knows this convex polytope?

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