TITLE: Name of this convex polyhedron?
QUESTION [8 upvotes]: 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!

REPLY [4 votes]: 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.