Truncated tetrahedron atop truncated octahedron

Truncated tetrahedron atop truncated octahedron, or tutatoe, is a CRF segmentochoron (designated K-4.76 on Richard Klitzing's list). As the name suggests, it consists of a truncated tetrahedron and a truncated octahedron as bases, connected by 6 triangular prisms, 4 triangular cupolas, and 4 hexagonal prisms.

It can be constructed as a cap of the prismatorhombated pentachoron, with a truncated tetrahedron at the top.

Vertex coordinates
The vertices of a truncated tetrahedron atop truncated octahedron segmentochoron of edge length 1 are given by: Alternative coordinates can be obtained from those of the prismatorhombated pentachoron:
 * $$\left(\frac{3\sqrt2}{4},\,\frac{\sqrt2}{4},\,\frac{\sqrt2}{4},\,\frac{\sqrt{10}}{4}\right)$$ and all permutations and even sign changes of first three coordinates
 * $$\left(±\sqrt2,\,±\frac{\sqrt2}{2},\,0,\,0\right)$$ and all permutations of first 3 coordinates
 * $$\left(\frac{\sqrt{10}}{10},\,-\frac{\sqrt6}{6},\,\frac{\sqrt3}{6},\,±\frac32\right),$$
 * $$\left(\frac{\sqrt{10}}{10},\,\frac{\sqrt6}{6},\,-\frac{\sqrt3}{6},\,±\frac32\right),$$
 * $$\left(\frac{\sqrt{10}}{10},\,-\frac{\sqrt6}{6},\,\frac{2\sqrt3}{3},\,±1\right),$$
 * $$\left(\frac{\sqrt{10}}{10},\,\frac{\sqrt6}{6},\,-\frac{2\sqrt3}{3},\,±1\right),$$
 * $$\left(\frac{\sqrt{10}}{10},\,-\frac{\sqrt6}{6},\,-\frac{5\sqrt3}{6},\,±\frac12\right),$$
 * $$\left(\frac{\sqrt{10}}{10},\,\frac{\sqrt6}{6},\,\frac{5\sqrt3}{6},\,±\frac12\right),$$
 * $$\left(\frac{\sqrt{10}}{10},\,±\frac{\sqrt6}{2},\,0,\,±1\right),$$
 * $$\left(\frac{\sqrt{10}}{10},\,±\frac{\sqrt6}{2},\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(\frac{7\sqrt{10}}{20},\,-\frac{\sqrt6}{12},\,\frac{\sqrt3}{3},\,±1\right),$$
 * $$\left(\frac{7\sqrt{10}}{20},\,-\frac{\sqrt6}{12},\,-\frac{2\sqrt3}{3},\,0\right),$$
 * $$\left(\frac{7\sqrt{10}}{20},\,\frac{\sqrt6}{4},\,0,\,±1\right),$$
 * $$\left(\frac{7\sqrt{10}}{20},\,\frac{\sqrt6}{4},\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(\frac{7\sqrt{10}}{20},\,-\frac{5\sqrt6}{12},\,\frac{\sqrt3}{6},\,±\frac12\right),$$
 * $$\left(\frac{7\sqrt{10}}{20},\,-\frac{5\sqrt6}{12},\,-\frac{\sqrt3}{3},\,0\right),$$