Truncated pentachoron

The truncated pentachoron, or tip, also commonly called the truncated 5-cell, is a convex uniform polychoron that consists of 5 regular tetrahedra and 5 truncated tetrahedra. 1 tetrahedron and three truncated tetrahedra join at each vertex. As the name suggests, it can be obtained by truncating the pentachoron.

Vertex coordinates
The vertices of a truncated pentachoron of edge length 1 are given by:


 * $$\left(\frac{3\sqrt{10}}{20},\,-\frac{\sqrt6}{12},\,\frac{\sqrt3}{3},\,±1\right),$$
 * $$\left(\frac{3\sqrt{10}}{20},\,-\frac{\sqrt6}{12},\,-\frac{2\sqrt3}{3},\,0\right),$$
 * $$\left(\frac{3\sqrt{10}}{20},\,-\frac{\sqrt6}{4},\,0,\,±1\right),$$
 * $$\left(\frac{3\sqrt{10}}{20},\,\frac{\sqrt6}{4},\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(\frac{3\sqrt{10}}{20},\,-\frac{5\sqrt6}{12},\,\frac{\sqrt3}{6},\,±\frac12\right),$$
 * $$\left(\frac{3\sqrt{10}}{20},\,-\frac{5\sqrt6}{12},\,-\frac{\sqrt3}{3},\,0\right),$$
 * $$\left(-\frac{\sqrt{10}}{10},\,\frac{\sqrt6}{6},\,\frac{\sqrt3}{3},\,±1\right),$$
 * $$\left(-\frac{\sqrt{10}}{10},\,\frac{\sqrt6}{6},\,-\frac{2\sqrt3}{3},\,0\right),$$
 * $$\left(-\frac{\sqrt{10}}{10},\,-\frac{\sqrt6}{2},\,0,\,0\right),$$
 * $$\left(-\frac{7\sqrt{10}}{20},\,\frac{\sqrt6}{12},\,\frac{\sqrt3}{6},\,±\frac12\right),$$
 * $$\left(-\frac{7\sqrt{10}}{20},\,\frac{\sqrt6}{12},\,-\frac{\sqrt3}{3},\,0\right),$$
 * $$\left(-\frac{7\sqrt{10}}{20},\,-\frac{\sqrt6}{4},\,0,\,0\right).$$

Much simpler coordinates can be given in five dimensions, as all permutations of:


 * $$\left(\sqrt2,\,\frac{\sqrt2}{2},\,0,\,0,\,0\right).$$

Representations
A truncated pentachoron has the following Coxeter diagrams:


 * x3x3o3o (full symmetry)
 * xux3oox3ooo&#xt (A3 axial, tetrahedron-first)
 * xuxo oxux3ooox&#xt (A2×A1 axial, edge-first)

Semi-uniform variant
The truncated pentachoron has a semi-uniform variant of the form x3y3o3o that maintains its full symmetry. This variant uses 5 tetrahedra of size y and 5 semi-uniform truncated tetrahedra of form x3y3o as cells, with 2 edge lengths.

With edges of length a (surrounded by truncated tetrahedra only) and b (of tetrahedra), its circumradius is given by $$\sqrt{\frac{2a^2+3b^2+3ab}{5}}$$ and its hypervolume is given by $$(a^4+8a^3b+24a^2b^2+32ab^3+11b^4)\frac{\sqrt5}{96}$$.