Toroidal blend of 20 triangular hebesphenorotundae

The  is a Stewart toroid. It can be obtained by outer-blending twenty triangular hebesphenorotundae together at their square faces, leaving no squares behind. It has these squares, as well as pentagons and nonconvex pentambi, as pseudo-faces.

Vertex coordinates
The vertices of a, centered at the origin and with unit edge length, are given by all even permutations of:


 * $$\left(0,\,\pm\frac12,\,\pm3\frac{1+\sqrt5}{4}\right),$$
 * $$\left(\pm\frac12,\,\pm\frac{5+\sqrt5}{4},\,\pm\frac{1+\sqrt5}{2}\right),$$
 * $$\left(\pm\frac{1+\sqrt5}{4},\,\pm1,\,\pm\frac{2+\sqrt5}{2}\right),$$


 * $$\left(\pm\frac{\sqrt5-1}{4},\,\pm\frac{3+\sqrt5}{4},\,\pm\frac{5+3\sqrt5}{4}\right),$$
 * $$\left(0,\,\pm\frac{2+\sqrt5}{2},\,\pm3\frac{1+\sqrt5}{4}\right),$$
 * $$\left(\pm\frac12,\,\pm\frac{5+\sqrt5}{4},\,\pm\frac{3+\sqrt5}{2}\right),$$


 * $$\left(0,\,\pm\frac12,\,\pm\frac{7+3\sqrt5}{4}\right),$$
 * $$\left(\pm\frac{3+\sqrt5}{2},\,\pm\frac{2+\sqrt5}{2},\,\pm\frac{1+\sqrt5}{4}\right),$$
 * $$\left(\pm\frac{5+3\sqrt5}{4},\,\pm\frac{3+\sqrt5}{4},\,\pm\frac{3+\sqrt5}{4}\right),$$


 * $$\left(\pm\frac{1+\sqrt5}{4},\,0,\,\pm\frac{3+2\sqrt5}{2}\right),$$
 * $$\left(\pm\frac{1+\sqrt5}{2},\,\pm\frac12,\,\pm\frac{7+3\sqrt5}{4}\right),$$
 * $$\left(\pm\frac{2+\sqrt5}{2},\,\pm\frac{5+\sqrt5}{4},\,\pm\frac{3+\sqrt5}{2}\right).$$

Relations
Twice the 4–6 dihedral angle of a triangular hebesphenorotunda, plus the 6–6 dihedral angle of a truncated icosahedron, is 360°. This can be taken as an explanation of why the toroid forms: it is possible to blend the toroid with a truncated icosahedron placed in the center. Doing this will remove all tunnels (making the genus 0) and hexagonal faces.

The triangular hebesphenorotunda's relation to the small rhombicosidodecahedron can also provide some insight on the formation of the toroid. The two polyhedra have certain arrangements of faces in common.