Disdyakis triacontahedron

In geometry, a disdyakis triacontahedron, hexakis icosahedron, decakis dodecahedron or kisrhombic triacontahedron is a Catalan solid with 120 faces and the dual to the Archimedean truncated icosidodecahedron. As such it is face-uniform but with irregular face polygons. It slightly resembles an inflated rhombic triacontahedron: if one replaces each face of the rhombic triacontahedron with a single vertex and four triangles in a regular fashion, one ends up with a disdyakis triacontahedron. That is, the disdyakis triacontahedron is the Kleetope of the rhombic triacontahedron. It is also the barycentric subdivision of the regular dodecahedron and icosahedron. It has the most faces among the Archimedean and Catalan solids, with the snub dodecahedron, with 92 faces, in second place.

If the bipyramids, the gyroelongated bipyramids, and the trapezohedra are excluded, the disdyakis triacontahedron has the most faces of any other strictly convex polyhedron where every face of the polyhedron has the same shape.

Projected into a sphere, the edges of a disdyakis triacontahedron define 15 great circles. Buckminster Fuller used these 15 great circles, along with 10 and 6 others in two other polyhedra to define his 31 great circles of the spherical icosahedron.

Geometry
Being a Catalan solid with triangular faces, the disdyakis triacontahedron's three face angles $$\alpha_4, \alpha_6, \alpha_{10}$$ and common dihedral angle $$\theta$$ must obey the following constraints analogous to other Catalan solids:


 * $$\sin(\theta/2) = \cos(\pi/4) / \cos(\alpha_4/2)$$
 * $$\sin(\theta/2) = \cos(\pi/6) / \cos(\alpha_6/2)$$
 * $$\sin(\theta/2) = \cos(\pi/10) / \cos(\alpha_{10}/2)$$
 * $$\alpha_4 + \alpha_6 + \alpha_{10} = \pi$$

The above four equations are solved simultaneously to get the following face angles and dihedral angle:


 * $$\alpha_4 = \arccos \left(\frac{7-4\phi}{30} \right) \approx 88.992^{\circ}$$
 * $$\alpha_6 = \arccos \left( \frac{17-4\phi}{20} \right) \approx 58.238^{\circ}$$
 * $$\alpha_{10} = \arccos \left( \frac{2+5\phi}{12} \right) \approx 32.770^{\circ}$$
 * $$\theta = \arccos \left( -\frac{155 + 48\phi}{241} \right) \approx 164.888^{\circ}$$

where $$\phi = \frac{\sqrt{5}+1}{2} \approx 1.618$$ is the golden ratio.

As with all Catalan solids, the dihedral angles at all edges are the same, even though the edges may be of different lengths.

Cartesian coordinates




The 62 vertices of a disdyakis triacontahedron are given by:


 * Twelve vertices $$\left(0, \frac{\pm 1}{\sqrt{\phi+2}}, \frac{\pm \phi}{\sqrt{\phi+2}} \right)$$ and their cyclic permutations,


 * Eight vertices $$\left(\pm R, \pm R, \pm R\right)$$,


 * Twelve vertices $$\left(0, \pm R\phi, \pm \frac{R}{\phi}\right)$$ and their cyclic permutations,


 * Six vertices $$\left(\pm S, 0, 0\right)$$ and their cyclic permutations.


 * Twenty-four vertices $$\left(\pm \frac{S\phi}{2}, \pm\frac{S}{2}, \pm\frac{S}{2\phi}\right)$$ and their cyclic permutations,

where
 * $$R = \frac{5}{3\phi\sqrt{\phi+2}} = \frac{\sqrt{25 - 10\sqrt{5}}}{3} \approx 0.5415328270548438$$,
 * $$S = \frac{(7\phi - 6) \sqrt{\phi+2}}{11} = \frac{(2\sqrt{5} - 3) \sqrt{25 + 10\sqrt{5}}}{11} \approx 0.9210096876986302$$, and
 * $$\phi = \frac{\sqrt{5} + 1}{2} \approx 1.618$$ is the golden ratio.

In the above coordinates, the first 12 vertices form a regular icosahedron, the next 20 vertices (those with R) form a regular dodecahedron, and the last 30 vertices (those with S) form an icosidodecahedron.

Normalizing all vertices to the unit sphere gives a spherical disdyakis triacontahedron, shown in the adjacent figure. This figure also depicts the 120 transformations associated with the full icosahedral group Ih.

Symmetry
The edges of the polyhedron projected onto a sphere form 15 great circles, and represent all 15 mirror planes of reflective Ih icosahedral symmetry. Combining pairs of light and dark triangles define the fundamental domains of the nonreflective (I) icosahedral symmetry. The edges of a compound of five octahedra also represent the 10 mirror planes of icosahedral symmetry.

Orthogonal projections
The disdyakis triacontahedron has three types of vertices which can be centered in orthogonally projection:

Uses
The disdyakis triacontahedron, as a regular dodecahedron with pentagons divided into 10 triangles each, is considered the "holy grail" for combination puzzles like the Rubik's cube. Such a puzzle currently has no satisfactory mechanism. It is the most significant unsolved problem in mechanical puzzles, often called the "big chop" problem.

This shape was used to make 120-sided dice using 3D printing.

Since 2016, the Dice Lab has used the disdyakis triacontahedron to mass-market an injection-moulded 120-sided die. It is claimed that 120 is the largest possible number of faces on a fair die, aside from infinite families (such as right regular prisms, bipyramids, and trapezohedra) that would be impractical in reality due to the tendency to roll for a long time.

A disdyakis tricontahedron projected onto a sphere is used as the logo for Brilliant, a website containing series of lessons on STEM-related topics.

Related polyhedra and tilings
It is topologically related to a polyhedra sequence defined by the face configuration V4.6.2n. This group is special for having all even number of edges per vertex and form bisecting planes through the polyhedra and infinite lines in the plane, and continuing into the hyperbolic plane for any n ≥ 7.

With an even number of faces at every vertex, these polyhedra and tilings can be shown by alternating two colors so all adjacent faces have different colors.

Each face on these domains also corresponds to the fundamental domain of a symmetry group with order 2,3,n mirrors at each triangle face vertex. This is *n32 in orbifold notation, and [n,3] in Coxeter notation.