# Truncated dodecahedron

Truncated dodecahedron
Rank3
TypeUniform
Notation
Bowers style acronymTid
Coxeter diagramx5x3o ()
Conway notationtD
Stewart notationT5
Elements
Faces20 triangles, 12 decagons
Edges30+60
Vertices60
Vertex figureIsosceles triangle, edge lengths 1, (5+5)/2, (5+5)/2
Measures (edge length 1)
Circumradius${\displaystyle {\sqrt {\frac {37+15{\sqrt {5}}}{8}}}\approx 2.96945}$
Volume${\displaystyle 5{\frac {99+47{\sqrt {5}}}{12}}\approx 85.03966}$
Dihedral angles10–3: ${\displaystyle \arccos \left(-{\sqrt {\frac {5+2{\sqrt {5}}}{15}}}\right)\approx 142.62263^{\circ }}$
10–10: ${\displaystyle \arccos \left(-{\frac {\sqrt {5}}{5}}\right)\approx 116.56505^{\circ }}$
Central density1
Number of external pieces32
Level of complexity3
Related polytopes
ArmyTid
RegimentTid
DualTriakis icosahedron
ConjugateQuasitruncated great stellated dodecahedron
Abstract & topological properties
Flag count360
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryH3, order 120
ConvexYes
NatureTame

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.

## Vertex coordinates

A truncated dodecahedron of edge length 1 has vertex coordinates given by all even permutations of:

• ${\displaystyle \left(0,\,\pm {\frac {1}{2}},\,\pm {\frac {5+3{\sqrt {5}}}{4}}\right)}$,
• ${\displaystyle \left(\pm {\frac {1}{2}},\,\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {3+{\sqrt {5}}}{2}}\right)}$,
• ${\displaystyle \left(\pm {\frac {3+{\sqrt {5}}}{4}},\,\pm {\frac {1+{\sqrt {5}}}{2}},\,\pm {\frac {2+{\sqrt {5}}}{2}}\right)}$.

## 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 ${\displaystyle {\sqrt {\frac {9a^{2}+12b^{2}+16ab+(3a^{2}+4b^{2}+8ab){\sqrt {5}}}{8}}}}$ and its volume is given by ${\displaystyle {\frac {15a^{3}+60a^{2}b+60ab^{2}+30b^{3}}{4}}+(21a^{3}+72a^{2}b+108ab^{2}+34b^{3}){\frac {\sqrt {5}}{12}}}$.

It has coordinates given by all even permutations of:

• ${\displaystyle \left(0,\,\pm {\frac {a}{2}},\,\pm {\frac {a\varphi +2b}{2}}\varphi \right)}$,
• ${\displaystyle \left(\pm {\frac {b}{2}},\,\pm {\frac {a+b\varphi }{2}},\,\pm {\frac {a+b}{2}}\varphi ^{2}\right)}$,
• ${\displaystyle \left(\pm {\frac {a\varphi +b}{2}},\,\pm {\frac {a+b}{2}}\varphi ,\,\pm {\frac {a+b\varphi }{2}}\varphi \right)}$.

where ${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}}$.

## 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:

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.