Catalan solid

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 ${\displaystyle \cos(\alpha _{4})={\frac {1}{2}}-{\frac {1}{4}}{\sqrt {2}}}$.

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:

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

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 ${\displaystyle \cos \alpha _{3,4}=-{\frac {1}{3}}{\sqrt {3}}}$.

The face of any Catalan polyhedron may be obtained from the vertex figure of the dual Archimedean solid using the Dorman Luke construction.[2]

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.