# Truncated icosahedron

Truncated icosahedron Rank3
TypeUniform
SpaceSpherical
Notation
Bowers style acronymTi
Coxeter diagramo5x3x (     )
Elements
Faces12 pentagons, 20 hexagons
Edges30+60
Vertices60
Vertex figureIsosceles triangle, edge lengths (1+5)/2, 3, 3 Measures (edge length 1)
Circumradius$\sqrt{\frac{29+9\sqrt5}{8}} ≈ 2.47802$ Volume$\frac{125+43\sqrt5}{4} ≈ 55.28773$ Dihedral angles6–5: $\arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263^\circ$ 6–6: $\arccos\left(-\frac{\sqrt5}{3}\right) ≈ 138.18968^\circ$ Central density1
Number of pieces32
Level of complexity3
Related polytopes
ArmyTi
RegimentTi
DualPentakis dodecahedron
ConjugateTruncated great icosahedron
Abstract properties
Flag count360
Euler characteristic2
Topological properties
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryH3, order 120
ConvexYes
NatureTame

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 is the shape of a soccer ball.

## Vertex coordinates

A truncated icosahedron of edge length 1 has vertex coordinates given by all even permutations and all changes of sign of:

• $\left(0,\,±\frac12,\,±3\frac{1+\sqrt5}{4}\right),$ • $\left(±\frac12,\,±\frac{5+\sqrt5}{4},\,±\frac{1+\sqrt5}{2}\right),$ • $\left(±\frac{1+\sqrt5}{4},\,±1,\,±\frac{2+\sqrt5}{2}\right).$ ## 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 $\frac{5a^3+30a^2b+60ab^2+30b^3}{4}+(5a^3+30a^2b+60ab^2+34b^3)\frac{\sqrt5}{12}$ .