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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?

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No real answer, but I think you want six parallelepipeds, not four. –  Klaus Draeger Feb 14 '13 at 13:05
    
fixed, thanks... –  Camilo Sarmiento Feb 14 '13 at 13:30
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Is this it?
           Rhombic Dodecahedron + Tetrahedron

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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 '13 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 '13 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 '13 at 0:44
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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).

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