# Catalan solid

In mathematics, a Catalan solid, or Archimedean dual, is a dual polyhedron to an Archimedean solid. There are 13 Catalan solids. They are named for the Belgian mathematician, Eugène Catalan, who first described them in 1865.

The Catalan solids are all convex. They are face-transitive but not vertex-transitive. This is because the dual Archimedean solids are vertex-transitive and not face-transitive. Note that unlike Platonic solids and Archimedean solids, the faces of Catalan solids are not regular polygons. However, the vertex figures of Catalan solids are regular, and they have constant dihedral angles. Being face-transitive, Catalan solids are isohedra.

Additionally, two of the Catalan solids are edge-transitive: the rhombic dodecahedron and the rhombic triacontahedron. These are the duals of the two quasi-regular Archimedean solids.

Just as prisms and antiprisms are generally not considered Archimedean solids, so bipyramids and trapezohedra are generally not considered Catalan solids, despite being face-transitive.

Two of the Catalan solids are chiral: the pentagonal icositetrahedron and the pentagonal hexecontahedron, dual to the chiral snub cube and snub dodecahedron. These each come in two enantiomorphs. Not counting the enantiomorphs, bipyramids, and trapezohedra, there are a total of 13 Catalan solids.

## Symmetry

The Catalan solids, along with their dual Archimedean solids, can be grouped in those with tetrahedral, octahedral and icosahedral symmetry. For both octahedral and icosahedral symmetry there are six forms. The only Catalan solid with genuine tetrahedral symmetry is the triakis tetrahedron (dual of the truncated tetrahedron). Rhombic dodecahedron and tetrakis hexahedron have octahedral symmetry, but they can be colored to have only tetrahedral symmetry. Rectification and snub also exist with tetrahedral symmetry, but they are Platonic instead of Archimedean, so their duals are Platonic instead of Catalan. (They are shown with brown background in the table below.)

## Geometry

All dihedral angles of a Catalan solid are equal. Denoting their value by $\theta$ , and denoting the face angle at the vertices where $p$ faces meet by $\alpha _{p}$ , we have

$\sin(\theta /2)=\cos(\pi /p)/\cos(\alpha _{p}/2)$ .

This can be used to compute $\theta$ and $\alpha _{p}$ , $\alpha _{q}$ , ... , from $p$ , $q$ ... only.

### Triangular faces

Of the 13 Catalan solids, 7 have triangular faces. These are of the form Vp.q.r, where p, q and r take their values among 3, 4, 5, 6, 8 and 10. The angles $\alpha _{p}$ , $\alpha _{q}$ and $\alpha _{r}$ can be computed in the following way. Put $a=4\cos ^{2}(\pi /p)$ , $b=4\cos ^{2}(\pi /q)$ , $c=4\cos ^{2}(\pi /r)$ and put

$S=-a^{2}-b^{2}-c^{2}+2ab+2bc+2ca$ .

Then

$\cos(\alpha _{p})={\frac {S}{2bc}}-1$ .

For $\alpha _{q}$ and $\alpha _{r}$ the expressions are similar of course. The dihedral angle $\theta$ can be computed from

$\cos(\theta )=1-2abc/S$ .

Applying this, for example, to the disdyakis triacontahedron ($p=4$ , $q=6$ and $r=10$ , hence $a=2$ , $b=3$ and $c=\phi +2$ , where $\phi$ is the golden ratio) gives $\cos(\alpha _{4})={\frac {2-\phi }{6(2+\phi )}}={\frac {7-4\phi }{30}}$ and $\cos(\theta )={\frac {-10-7\phi }{14+5\phi }}={\frac {-48\phi -155}{241}}$ .

Of the 13 Catalan solids, 4 have quadrilateral faces. These are of the form Vp.q.p.r, where p, q and r take their values among 3, 4, and 5. The angle $\alpha _{p}$ can be computed by the following formula:

$\cos(\alpha _{p})={\frac {2\cos ^{2}(\pi /p)-\cos ^{2}(\pi /q)-\cos ^{2}(\pi /r)}{2\cos ^{2}(\pi /p)+2\cos(\pi /q)\cos(\pi /r)}}$ .

From this, $\alpha _{q}$ , $\alpha _{r}$ and the dihedral angle can be easily computed. The faces are kites, or, if $q=r$ , rhombi. Applying this, for example, to the deltoidal icositetrahedron ($p=4$ , $q=3$ and $r=4$ ), we get $\cos(\alpha _{4})={\frac {1}{2}}-{\frac {1}{4}}{\sqrt {2}}$ .

### Pentagonal faces

Of the 13 Catalan solids, 2 have pentagonal faces. These are of the form Vp.p.p.p.q, where p=3, and q=4 or 5. The angle $\alpha _{p}$ can be computed by solving a degree three equation:

$8\cos ^{2}(\pi /p)\cos ^{3}(\alpha _{p})-8\cos ^{2}(\pi /p)\cos ^{2}(\alpha _{p})+\cos ^{2}(\pi /q)=0$ .

### Application to other solids

All of the formulae of this section apply to the Platonic solids, and bipyramids and trapezohedra with equal dihedral angles as well, because they can be derived from the constant dihedral angle property only. For the pentagonal trapezohedron, for example, with faces V3.3.5.3, we get $\cos(\alpha _{3})={\frac {1}{4}}-{\frac {1}{4}}{\sqrt {5}}$ , or $\alpha _{3}=108^{\circ }$ . This is not surprising: it is possible to cut off both apexes in such a way as to obtain a regular dodecahedron.