Triakis tetrahedron, pentagonal icositetrahedron and disdyakis triacontahedron.

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 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, 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.

List of Catalan solids and their duals

Conway name	Archimedean dual	Face polygon	Orthogonal wireframes	Pictures	Face angles (°)	Dihedral angle (°)	Midradius[1]	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.)

Tetrahedral symmetry

Archimedean (Platonic)
Catalan (Platonic)

Octahedral symmetry

Archimedean
Catalan

Icosahedral symmetry

Archimedean
Catalan

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 ${\displaystyle \cos(\theta )={\frac {-10-7\phi }{14+5\phi ))={\frac {-48\phi -155}{241))}$.

Quadrilateral faces

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.^[2]

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.