Truncated icosicosahedron

The truncated icosicosahedron, te, or compound of ten truncated tetrahedra is a uniform polyhedron compound. It consists of 40 triangles (which form coplanar pairs combining into 20 hexagrams) and 40 hexagons, with one triangle and two hexagons joining at each vertex. As the name suggests, it can be derived as the truncation of the icosicosahedron, the compound of ten tetrahedra. It can alternatively be constructed as the compound of the two chiral forms of the truncated chiricosahedron.

Its quotient prismatic equivalent is the truncated tetrahedral decayottoorthowedge, which is twelve-dimensional.

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

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