Does anyone recognize / know the name of the convex polyhedron depicted below as the intersection of a Cuboctahedron and a Rhombicdodecahedron? Please note you have to interpret this picture and extract the intersection. Its faces are squares, equilateral triangles, and rectangles. The presence of rectangles takes it out of the realm of the Archimedean and Johnson solids.

Thanks for any pointers!

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    $\begingroup$ It looks like a rhombicuboctahedron whose triangular faces have been shrunk. $\endgroup$ – Rahul Jun 19 '11 at 2:19
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    $\begingroup$ This particular one is named Fred. $\endgroup$ – Will Jagy Jun 19 '11 at 2:24
  • $\begingroup$ At least in my view, it does not appear convex. A line segment from each point to a neighboring one looks outside the volume. $\endgroup$ – Ross Millikan Jun 19 '11 at 2:32
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    $\begingroup$ @Ross: "Please note you have to interpret this picture and extract the intersection." $\endgroup$ – Rahul Jun 19 '11 at 2:35
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    $\begingroup$ Thanks for everyone's help! @amWhy has nailed it in her/his answer. The cuboctahedron and rhombicdodecahedron are both arranged around a common midsphere, which I was exploring. (See my MO question on midspheres.) $\endgroup$ – Joseph O'Rourke Jun 19 '11 at 13:04

I believe this is what's referred to as a Cuboctahedron-Rhombic Dodecahedron Compound.

According to Mathworld on compound polyhedron:

A polyhedron compound is an arrangement of a number of interpenetrating polyhedra, either all the same or of several distinct types, usually having visually attractive symmetric properties. Compounds of multiple Platonic and Archimedean solids can be especially attractive, as can compounds of these solids and their duals. [Note, the two intersecting figures are in fact duals.] [Italics, boldface, and brackets mine.]

See Mathworld on Cuboctahedron-Rhombic Dodecahedron Compound, and in particular, I believe the figure at very bottom, left: the Cuboctahedron-Rhombic Dodecahedron Compound- depicts the figure you've provided above. It's a compound of two (dual) Archimedean polyhedra: the cuboctahedron (pictured immediately below, to left) and the rhombic dodecahedron (image immediately below, to right).

$\qquad\qquad$ cuboctahedron $\qquad\qquad$ rhomic dodecahedron

EDIT: and actually, I think the figure below (right) is what you're after: it is described as the "Cuboctahedron-Rhombic Dodecahedron solid which is common to both polyhedra"; i.e. their intersection. Each figure (above) is the dual of the other. No specific name is given to the convex figure of intersection (below, to right), save for referring to it as a "Cuboctahedron-Rhombic Dodecahedron convex solid."

The page linked is worth a visit, since you can see the figures below, rotate them, and the page provides additional information on the polyhedra, the surface area, volume, etc. (edges, vertices,...). There are also links to other polyhedral compounds.

$\qquad\qquad$ cuboctahedron $\qquad\qquad\qquad$ rhombic-dodecahedron

  • $\begingroup$ Great! Thanks so much! Yes, this is exactly what I constructed. $\endgroup$ – Joseph O'Rourke Jun 19 '11 at 12:59
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    $\begingroup$ That's what I thought, when I came across this...I had trouble discerning, exactly, what the "common solid" (region of intersection of the two polyhedra) was, so I was pleased to find the figure to the right. Nice construction, by the way...it helps to justify (via its opaqueness) that the figure on the right is indeed the intersection! (I just wish I could have "scaled" the figure on the right to match the size of the figure on the left (i.e., shrink it a bit)...Glad to help! $\endgroup$ – amWhy Jun 19 '11 at 13:26
  • $\begingroup$ @Joseph: Also, did you visit the site I linked to (where you can find the figures, and "rotate" them? That's a nice feature; if you view the source code for that page, you can access the images, perhaps also the code for rotating...Can I ask what program you used to construct your image? $\endgroup$ – amWhy Jun 19 '11 at 13:29
  • $\begingroup$ Yes, a very useful page. Thanks again! $\endgroup$ – Joseph O'Rourke Jun 19 '11 at 15:07
  • $\begingroup$ Still not where I should be... I just saw that your answer turned into Community Wiki due to 12 edits (this happens automatically). If you want to undo that, flag it for moderator attention and file a request for undoing it by filling out the other field. Now I'm out of here for good (at least for a few hours). See you around! $\endgroup$ – t.b. Jun 21 '11 at 1:34

The intersection is a rectified cuboctahedron, which can be denoted $t_1\left\{3 \atop 4\right\}$. It can also be called a rhombicuboctahedron, but this name is usually reserved for the version which has been made "uniform" by turning the rectangular faces into squares. The two are isomorphic (i.e. combinatorially identical).

It could also be called a rectified rhombic dodecahedron; a cantellated cube $t_{0,2}\{4,3\}$; a cantellated octahedron; an expanded cube; an expanded octahedron; or a beveled cube.

Cantellation, or expansion, refers to the process of moving the facets of a polytope apart and filling in extra faces; in this case pulling apart the six faces of the cube (equivalently, shrinking them in-place) and adding rectangles where the edges used to be, and triangles where the vertices used to be. When you say "cantellated cube" it is often assumed that the process is done until the rectangles become square, and you have a rhombicuboctahedron; the rectified cuboctahedron is cantellated less than that. (So it is a cantellated cube, but not the cantellated cube.)

Rectification cuts all the vertices to the midpoints of edges, which is exactly what your polyhedron is. So "rectified cuboctahedron" definitely should mean this, non-uniform, figure. Unfortunately some people may assume that this has been made uniform, also, under a bias toward making everything uniform all the time whenever possible.

  • $\begingroup$ Perfect name for what I am constructing, since I needed the "uniform" version. $\endgroup$ – Engineer Aug 25 '18 at 6:39

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