Truncated icosahedron

The truncated icosahedron, or ti, is one of the 13 Archimedean solids. It consists of 12 pentagons and 20 hexagons. Each vertex joins one pentagon and two hexagons. As the name suggests, it can be obtained by the truncation of the icosahedron. It can also be thought of as the shape of a soccer ball, but not all soccer balls look like this.

Vertex coordinates
A truncated icosahedron of edge length 1 has vertex coordinates given by all even permutations and all changes of sign of:
 * $$\left(0,\,±\frac12,\,±\frac{3+3\sqrt5}{4}\right),$$
 * $$\left(±\frac12,\,±\frac{5+\sqrt5}{4},\,±\frac{1+\sqrt5}{2}\right),$$
 * $$\left(±\frac{1+\sqrt5}{4},\,±1,\,±\frac{2+\sqrt5}{2}\right).$$

Representations
A truncated icosahedron has the following Coxeter diagrams:


 * o5x3x (full symmetry)
 * xuxuxfoo5oofxuxux&#xt (H2 axial, pentagon-first)
 * xuAxBfVoVofx3xfoVoVfBxAux&#xt (A2 axial, hexagon-first)

Semi-uniform variant
The truncated icosahedron has a semi-uniform variant of the form o5y3x that maintains its full symmetry. This variant has 12 pentagons of size y and 20 ditrigons as faces.

With edges of length a (between two ditrigons) and b (between a ditrigon and a pentagon), its circumradius is given by $$\sqrt{\frac{5a^2+12b^2+12ab+(a^2+4b^2+4ab)\sqrt5}{8}}$$ and its volume is given by $$\frac{5a^3+30a^2b+60ab^2+30b^3}{4}+\frac{\sqrt5}{12}(5a^3+30a^2b+60ab^2+34b^3)$$.