Truncated dodecahedron

{{Infobox polytope }} The truncated dodecahedron, or tid, is one of the 13 Archimedean solids. It consists of 20 triangles and 12 decagons. Each vertex joins one triangle and two decagons. As the name suggests, it can be obtained by truncation of the dodecahedron.
 * image = Truncated dodecahedron.png
 * 3d = Truncated dodecahedron.stl
 * off = Truncated dodecahedron.off
 * obsa = Tid
 * dim=3
 * type = Uniform
 * coxeter = x5x3o
 * symmetry = H3, order 120
 * army=Tid
 * reg=Tid
 * faces = 20 triangles, 12 decagons
 * edges = 30+60
 * vertices = 60
 * circum = $$\sqrt{\frac{37+15\sqrt5}{8}} ≈ 2.96945$$
 * volume = $$5\frac{99+47\sqrt5}{12} ≈ 85.03966$$
 * verf = Isosceles triangle, edge lengths 1, $\sqrt{(5+√5)/2}$, $\sqrt{(5+√5)/2}$
 * dih = 10–3: $$\arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263°$$
 * dih2 = 10–10: Math>\arccos\left(-\frac{\sqrt5]{5}) ≈ 116.56505°
 * pieces = 32
 * loc = 3
 * dual = Triakis icosahedron
 * conjugate = Quasitruncated great stellated dodecahedron
 * conv=Yes
 * orientable=Yes
 * nat=Tame

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

Representations
A truncated dodecahedron has the following Coxeter diagrams:


 * x5x3o (full symmetry)
 * xooxFVFx5xFVFxoox&#xt (H2 axial, decagon-first)
 * ooxFBVFxFVFx3xFVFxFVBFxoo&#xt (A2 axial, triangle-first)

Semi-uniform variant
The truncated dodecahedron has a semi-uniform variant of the form x5y3o that maintains its full symmetry. This variant has 20 triangles of size y and 12 dipentagons as faces.

With edges of length a (between two dipentagons) and b (between a dipentagon and a triangle), its circumradius is given by $$\sqrt{\frac{9a^2+12b^2+16ab+(3a^2+4b^2+8ab)\sqrt5}{8}}$$ and its volume is given by $$\frac{15a^3+60a^2b+60ab^2+30b^3}{4}+(21a^3+72a^2b+108ab^2+34b^3)\frac{\sqrt5}{12}$$.

Related polyhedra
The truncated dodecahedron can be augmented by attaching pentagonal cupolae onto its decagonal faces, with the squares of the pentagonal cupola adjacent to the triangles of the truncated dodecahedron. This leads to several Johnson solids:


 * Augmented truncated dodecahedron – One decagon is augmented.
 * Parabiaugmented truncated dodecahedron – Two opposite decagons are augmented.
 * Metabiaugmented truncated dodecahedron – Two non-adjacent, non-opposite decagons are augmented.
 * Triaugmented truncated dodecahedron – Three mutually non-adjacent decagons are augmented.

If the cupolae are gyrated so that the triangular faces of both solids are adjacent, these faces turn out coplanar, so they don't create any new Johnson solids.