Quasitruncated rhombihedron

The quasitruncated rhombihedron, quasihyperhombicosicosahedron, quitar, or compound of five quasitruncated hexahedra is a uniform polyhedron compound. It consists of 40 triangles (which form coplanar pairs combining into 20 hexagrams) and 30 octagrams, with one triangle and two octagrams joining at each vertex. As the name suggests, it can be derived as the quasitruncation of the rhombihedron, the compound of five cubes.

Its quotient prismatic equivalent is the quasitruncated hexahedral pentachoroorthowedge, which is seven-dimensional.

Vertex coordinates
The vertices of a quasitruncated rhombihedron of edge length 1 can be given by all even permutations of:
 * $$\left(±\frac{\sqrt2-1}{2},\,±\frac{\sqrt2-1}{2},\,±\frac12\right),$$
 * $$\left(±\frac{\sqrt2+\sqrt{10}}{8},\,±\frac{-2+\sqrt2+2\sqrt5-\sqrt{10}}{8},\,±\frac{1+\sqrt5-\sqrt{10}}{4}\right),$$
 * $$\left(±\frac{\sqrt2}{4},\,±\frac{-2+3\sqrt2+2\sqrt5-\sqrt{10}}{8},\,±\frac{2-3\sqrt2+2\sqrt5-\sqrt{10}}{8}\right),$$
 * $$\left(±\frac{2-\sqrt2}{4},\,±\frac{4-3\sqrt2+\sqrt{10}}{8},\,±\frac{4-3\sqrt2-\sqrt{10}}{8}\right),$$
 * $$\left(±\frac{2-\sqrt2+2\sqrt5-\sqrt{10}}{8},\,±\frac{\sqrt2-\sqrt{10}}{8},\,±\frac{-1+\sqrt5-\sqrt{10}}{4}\right).$$