Truncated icosahedron

Truncated icosahedron
(3D model) (OFF file)
Rank	3
Type	Uniform
Space	Spherical
Notation
Bowers style acronym	Ti
Coxeter diagram	o5x3x ()
Elements
Faces	12 pentagons, 20 hexagons
Edges	30+60
Vertices	60
Vertex figure	Isosceles 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 angles	6–5: $\arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263^\circ$
	6–6: $\arccos\left(-\frac{\sqrt5}{3}\right) ≈ 138.18968^\circ$
Central density	1
Number of pieces	32
Level of complexity	3
Related polytopes
Army	Ti
Regiment	Ti
Dual	Pentakis dodecahedron
Conjugate	Truncated great icosahedron
Abstract properties
Flag count	360
Euler characteristic	2
Topological properties
Surface	Sphere
Orientable	Yes
Genus	0
Properties
Symmetry	H3, order 120
Convex	Yes
Nature	Tame

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 $\frac{5a^3+30a^2b+60ab^2+30b^3}{4}+(5a^3+30a^2b+60ab^2+34b^3)\frac{\sqrt5}{12}$.

Related polyhedra

o5o3o truncations

Name	OBSA	Schläfli symbol	CD diagram	Picture
Dodecahedron	doe	{5,3}	x5o3o
Truncated dodecahedron	tid	t{5,3}	x5x3o
Icosidodecahedron	id	r{5,3}	o5x3o
Truncated icosahedron	ti	t{3,5}	o5x3x
Icosahedron	ike	{3,5}	o5o3x
Small rhombicosidodecahedron	srid	rr{5,3}	x5o3x
Great rhombicosidodecahedron	grid	tr{5,3}	x5x3x
Snub dodecahedron	snid	sr{5,3}	s5s3s

External links

• Bowers, Jonathan. "Polyhedron Category 2: Truncates" (#14).

• Klitzing, Richard. "ti".

• Quickfur. "The Truncated Icosahedron".

• Wikipedia Contributors. "Truncated icosahedron".
• McCooey, David. "Truncated Icosahedron"

• Hi.gher.Space Wiki Contributors. "Icosahedral truncate".