# Catalan solid

The solids above (dark) shown together with their duals (light). The visible parts of the Catalan solids are regular pyramids.

In mathematics, a Catalan solid, or Archimedean dual, is a polyhedron that is dual to an Archimedean solid. There are 13 Catalan solids. They are named after 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, 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.

Eleven of the 13 Catalan solids have the Rupert property: a copy of the solid, of the same or larger shape, can be passed through a hole in the solid. [1]

## List of Catalan solids and their duals

Conway name Archimedean dual Face
polygon
Orthogonal
wireframes
Pictures Face angles (°) Dihedral angle (°) Midradius[2] Faces Edges Vert Sym.
triakis tetrahedron
"kT"
truncated tetrahedron Isosceles

V3.6.6
112.885
33.557
33.557
129.521 1.0607 12 18 8 Td
rhombic dodecahedron
"jC"
cuboctahedron Rhombus

V3.4.3.4
70.529
109.471
70.529
109.471
120 0.8660 12 24 14 Oh
triakis octahedron
"kO"
truncated cube Isosceles

V3.8.8
117.201
31.400
31.400
147.350 1.7071 24 36 14 Oh
tetrakis hexahedron
"kC"
truncated octahedron Isosceles

V4.6.6
83.621
48.190
48.190
143.130 1.5000 24 36 14 Oh
deltoidal icositetrahedron
"oC"
rhombicuboctahedron Kite

V3.4.4.4
81.579
81.579
81.579
115.263
138.118 1.3066 24 48 26 Oh
disdyakis dodecahedron
"mC"
truncated cuboctahedron Scalene

V4.6.8
87.202
55.025
37.773
155.082 2.2630 48 72 26 Oh
pentagonal icositetrahedron
"gC"
snub cube Pentagon

V3.3.3.3.4
114.812
114.812
114.812
114.812
80.752
136.309 1.2472 24 60 38 O
rhombic triacontahedron
"jD"
icosidodecahedron Rhombus

V3.5.3.5
63.435
116.565
63.435
116.565
144 1.5388 30 60 32 Ih
triakis icosahedron
"kI"
truncated dodecahedron Isosceles

V3.10.10
119.039
30.480
30.480
160.613 2.9271 60 90 32 Ih
pentakis dodecahedron
"kD"
truncated icosahedron Isosceles

V5.6.6
68.619
55.691
55.691
156.719 2.4271 60 90 32 Ih
deltoidal hexecontahedron
"oD"
rhombicosidodecahedron Kite

V3.4.5.4
86.974
67.783
86.974
118.269
154.121 2.1763 60 120 62 Ih
disdyakis triacontahedron
"mD"
truncated icosidodecahedron Scalene

V4.6.10
88.992
58.238
32.770
164.888 3.7694 120 180 62 Ih
pentagonal hexecontahedron
"gD"
snub dodecahedron Pentagon

V3.3.3.3.5
118.137
118.137
118.137
118.137
67.454
153.179 2.0971 60 150 92 I

## 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). The 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 ${\displaystyle \theta }$ , and denoting the face angle at the vertices where ${\displaystyle p}$ faces meet by ${\displaystyle \alpha _{p}}$, we have

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

This can be used to compute ${\displaystyle \theta }$ and ${\displaystyle \alpha _{p}}$, ${\displaystyle \alpha _{q}}$, ... , from ${\displaystyle p}$, ${\displaystyle q}$ ... 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 ${\displaystyle \alpha _{p}}$, ${\displaystyle \alpha _{q}}$ and ${\displaystyle \alpha _{r}}$ can be computed in the following way. Put ${\displaystyle a=4\cos ^{2}(\pi /p)}$, ${\displaystyle b=4\cos ^{2}(\pi /q)}$, ${\displaystyle c=4\cos ^{2}(\pi /r)}$ and put

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

Then

${\displaystyle \cos(\alpha _{p})={\frac {S}{2bc}}-1}$,
${\displaystyle \sin(\alpha _{p}/2)={\frac {-a+b+c}{2{\sqrt {bc}}}}}$.

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

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

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

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 ${\displaystyle \alpha _{p}}$can be computed by the following formula:

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

From this, ${\displaystyle \alpha _{q}}$, ${\displaystyle \alpha _{r}}$ and the dihedral angle can be easily computed. Alternatively, put ${\displaystyle a=4\cos ^{2}(\pi /p)}$, ${\displaystyle b=4\cos ^{2}(\pi /q)}$, ${\displaystyle c=4\cos ^{2}(\pi /p)+4\cos(\pi /q)\cos(\pi /r)}$. Then ${\displaystyle \alpha _{p}}$ and ${\displaystyle \alpha _{q}}$ can be found by applying the formulas for the triangular case. The angle ${\displaystyle \alpha _{r}}$ can be computed similarly of course. The faces are kites, or, if ${\displaystyle q=r}$, rhombi. Applying this, for example, to the deltoidal icositetrahedron (${\displaystyle p=4}$, ${\displaystyle q=3}$ and ${\displaystyle r=4}$), we get ${\displaystyle \cos(\alpha _{4})={\frac {1}{2}}-{\frac {1}{4}}{\sqrt {2}}}$.

### 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 ${\displaystyle \alpha _{p}}$can be computed by solving a degree three equation:

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

### Metric properties

For a Catalan solid ${\displaystyle {\bf {C}}}$ let ${\displaystyle {\bf {A}}}$ be the dual with respect to the midsphere of ${\displaystyle {\bf {C}}}$. Then ${\displaystyle {\bf {A}}}$ is an Archimedean solid with the same midsphere. Denote the length of the edges of ${\displaystyle {\bf {A}}}$ by ${\displaystyle l}$. Let ${\displaystyle r}$ be the inradius of the faces of ${\displaystyle {\bf {C}}}$, ${\displaystyle r_{m}}$ the midradius of ${\displaystyle {\bf {C}}}$ and ${\displaystyle {\bf {A}}}$, ${\displaystyle r_{i}}$ the inradius of ${\displaystyle {\bf {C}}}$, and ${\displaystyle r_{c}}$ the circumradius of ${\displaystyle {\bf {A}}}$. Then these quantities can be expressed in ${\displaystyle l}$ and the dihedral angle ${\displaystyle \theta }$ as follows:

${\displaystyle r^{2}={\frac {l^{2}}{8}}(1-\cos \theta )}$,
${\displaystyle r_{m}^{2}={\frac {l^{2}}{4}}{\frac {1-\cos \theta }{1+\cos \theta }}}$,
${\displaystyle r_{i}^{2}={\frac {l^{2}}{8}}{\frac {(1-\cos \theta )^{2}}{1+\cos \theta }}}$,
${\displaystyle r_{c}^{2}={\frac {l^{2}}{2}}{\frac {1}{1+\cos \theta }}}$.

These quantities are related by ${\displaystyle r_{m}^{2}=r_{i}^{2}+r^{2}}$, ${\displaystyle r_{c}^{2}=r_{m}^{2}+l^{2}/4}$ and ${\displaystyle r_{i}r_{c}=r_{m}^{2}}$.

As an example, let ${\displaystyle {\bf {A}}}$ be a cuboctahedron with edge length ${\displaystyle l=1}$. Then ${\displaystyle {\bf {C}}}$ is a rhombic dodecahedron. Applying the formula for quadrilateral faces with ${\displaystyle p=4}$ and ${\displaystyle q=r=3}$ gives ${\displaystyle \cos \theta =-1/2}$, hence ${\displaystyle r_{i}=3/4}$, ${\displaystyle r_{m}={\frac {1}{2}}{\sqrt {3}}}$, ${\displaystyle r_{c}=1}$, ${\displaystyle r={\frac {1}{4}}{\sqrt {3}}}$.

All vertices of ${\displaystyle {\bf {C}}}$ of type ${\displaystyle p}$ lie on a sphere with radius ${\displaystyle r_{c,p}}$ given by

${\displaystyle r_{c,p}^{2}=r_{i}^{2}+{\frac {2r^{2}}{1-\cos \alpha _{p}}}}$,

and similarly for ${\displaystyle q,r,\ldots }$.

Dually, there is a sphere which touches all faces of ${\displaystyle {\bf {A}}}$ which are regular ${\displaystyle p}$-gons (and similarly for ${\displaystyle q,r,\ldots }$) in their center. The radius ${\displaystyle r_{i,p}}$ of this sphere is given by

${\displaystyle r_{i,p}^{2}=r_{m}^{2}-{\frac {l^{2}}{4}}\cot ^{2}(\pi /p)}$.

These two radii are related by ${\displaystyle r_{i,p}r_{c,p}=r_{m}^{2}}$. Continuing the above example: ${\displaystyle \cos \alpha _{3}=-1/3}$ and ${\displaystyle \cos \alpha _{4}=1/3}$, which gives ${\displaystyle r_{c,3}={\frac {3}{8}}{\sqrt {6}}}$, ${\displaystyle r_{c,4}={\frac {3}{4}}{\sqrt {2}}}$, ${\displaystyle r_{i,3}={\frac {1}{3}}{\sqrt {6}}}$ and ${\displaystyle r_{i,4}={\frac {1}{2}}{\sqrt {2}}}$.

If ${\displaystyle P}$ is a vertex of ${\displaystyle {\bf {C}}}$ of type ${\displaystyle p}$, ${\displaystyle e}$ an edge of ${\displaystyle {\bf {C}}}$ starting at ${\displaystyle P}$, and ${\displaystyle P^{\prime }}$ the point where the edge ${\displaystyle e}$ touches the midsphere of ${\displaystyle {\bf {C}}}$, denote the distance ${\displaystyle PP^{\prime }}$ by ${\displaystyle l_{p}}$. Then the edges of ${\displaystyle {\bf {C}}}$ joining vertices of type ${\displaystyle p}$ and type ${\displaystyle q}$ have length ${\displaystyle l_{p,q}=l_{p}+l_{q}}$. These quantities can be computed by

${\displaystyle l_{p}={\frac {l}{2}}{\frac {\cos(\pi /p)}{\sin(\alpha _{p}/2)}}}$,

and similarly for ${\displaystyle q,r,\ldots }$. Continuing the above example: ${\displaystyle \sin(\alpha _{3}/2)={\frac {1}{3}}{\sqrt {6}}}$, ${\displaystyle \sin(\alpha _{4}/2)={\frac {1}{3}}{\sqrt {3}}}$, ${\displaystyle l_{3}={\frac {1}{8}}{\sqrt {6}}}$, ${\displaystyle l_{4}={\frac {1}{4}}{\sqrt {6}}}$, so the edges of the rhombic dodecahedron have length ${\displaystyle l_{3,4}={\frac {3}{8}}{\sqrt {6}}}$.

The dihedral angles ${\displaystyle \alpha _{p,q}}$between ${\displaystyle p}$-gonal and ${\displaystyle q}$-gonal faces of ${\displaystyle {\bf {A}}}$ satisfy

${\displaystyle \cos \alpha _{p,q}={\frac {l^{2}}{4}}{\frac {\cot(\pi /p)\cot(\pi /q)}{r_{m}^{2}}}-{\frac {r_{i,p}r_{i,q}}{r_{m}^{2}}}={\frac {l_{p}l_{q}-r_{m}^{2}}{r_{c,p}r_{c,q}}}}$.

Finishing the rhombic dodecahedron example, the dihedral angle ${\displaystyle \alpha _{3,4}}$ of the cuboctahedron is given by ${\displaystyle \cos \alpha _{3,4}=-{\frac {1}{3}}{\sqrt {3}}}$.

### Construction

The face of any Catalan polyhedron may be obtained from the vertex figure of the dual Archimedean solid using the Dorman Luke construction.[3]

### 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 ${\displaystyle \cos(\alpha _{3})={\frac {1}{4}}-{\frac {1}{4}}{\sqrt {5}}}$, or ${\displaystyle \alpha _{3}=108^{\circ }}$. This is not surprising: it is possible to cut off both apexes in such a way as to obtain a regular dodecahedron.