Snub icosidodecadodecahedron

Let ⁠${\displaystyle \rho =}$⁠ 1.324717957244746... be the real zero of the polynomial ⁠${\displaystyle x^{3}-x-1}$⁠. The constant ⁠${\displaystyle \rho }$⁠ is known as the plastic ratio. Denote by ⁠${\displaystyle \phi =}$⁠ 1.618033988749894... the golden ratio.

Let the point ⁠${\displaystyle p}$⁠ be given by $${\displaystyle p={\begin{pmatrix}\rho \\(\phi -\rho )/(\rho +1)\\(\rho -\phi )/\rho \end{pmatrix}}}$$ Let the matrix ⁠${\displaystyle M}$⁠ be given by $${\displaystyle M={\begin{pmatrix}1/2&-\phi /2&1/(2\phi )\\\phi /2&1/(2\phi )&-1/2\\1/(2\phi )&1/2&\phi /2\end{pmatrix}}}$$

${\displaystyle M}$ is the rotation around the axis ${\displaystyle (1,0,\phi )}$ by an angle of ${\displaystyle 2\pi /5}$, counterclockwise. Let the linear transformations ${\displaystyle T_{0},\ldots ,T_{11}}$ be the transformations which send a point ${\displaystyle (x,y,z)}$ to the even permutations of ${\displaystyle (\pm x,\pm y,\pm z)}$ with an even number of minus signs. The transformations ${\displaystyle T_{i}}$ constitute the group of rotational symmetries of a regular tetrahedron. The transformations ${\displaystyle T_{i}M^{j}}$ ${\displaystyle (i=0,\ldots ,11}$, ${\displaystyle j=0,\ldots ,4)}$ constitute the group of rotational symmetries of a regular icosahedron. Then the 60 points ${\displaystyle T_{i}M^{j}p}$ are the vertices of a snub icosidodecadodecahedron.

The edge length ${\displaystyle \lambda =2{\sqrt {(\rho ^{5}-\phi ^{2})\rho ^{-5}}}}$, the circumradius equals ${\displaystyle {\sqrt {(\rho ^{5}-\phi ^{2})(1+\rho ^{-5})}},}$ and the midradius ${\displaystyle {\sqrt {\rho ^{5}-\phi ^{2}}}.}$ To get a snub icosidodecadodecahedron with unit edge length, divide all above coordinates by ⁠${\displaystyle \lambda }$⁠.

The snub icosidodecadodecahedron was among the ten snub polyhedra discovered by Coxeter and Miller in the early 1930's, but published only in 1954. Its equation is in Table 4 of Uniform polyhedra.^[1]

The measures listed on David McCooey's Visual Polyhedra can be written in terms of ⁠${\displaystyle {\sqrt {5}}}$⁠ and plastic constant ⁠${\displaystyle \rho }$⁠.^[2]

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The snub icosidodecadodecahedron of unit edge length

element	size
circumscribed radius	${\displaystyle R={\tfrac {1}{2}}{\sqrt {\rho ^{5}+1}}}$
midscribed radius	${\displaystyle r={\tfrac {1}{2}}{\sqrt {\rho ^{5}}}}$
pentagon center radius	${\displaystyle r_{5}={\tfrac {1}{2}}{\sqrt {\rho ^{4}-{\tfrac {2}{\sqrt {5}}}}}}$
triangle center radius	${\displaystyle r_{3}={\tfrac {1}{2}}{\sqrt {\rho ^{5}-{\tfrac {1}{3}}}}}$
pentagram center radius	${\displaystyle r_{\tfrac {5}{3}}=-{\tfrac {1}{2}}{\sqrt {\rho ^{4}+{\tfrac {2}{\sqrt {5}}}}}}$
triangle-triangle angle	${\displaystyle \alpha _{33}=\arccos \left({\tfrac {1}{3}}(1-2\rho ^{2})\right)}$
pentagram-triangle angle	${\displaystyle \alpha _{{\tfrac {5}{3}}3}=\arccos \left({\sqrt {{\tfrac {1}{3}}(u-v)}}\right)}$
pentagon-triangle angle	${\displaystyle \alpha _{53}=\arccos \left(-{\sqrt {{\tfrac {1}{3}}(u+v)}}\right)}$

Close

With ${\displaystyle u=2\rho ^{2}-2\rho +1,}$ ${\displaystyle v=(\rho ^{2}-3\rho +1){\tfrac {2}{\sqrt {5}}}.}$

McCooey's metrics panel gives volume 4.620684..., but no formula. However, this solution (see Answer) is correct.

The skeleton of a snub icosidodecadodecahedron can be represented as a graph with 60 vertices and 180 edges, an Eulerian graph.^[3]

• Snubs Jonathan Bowers
• Uniform snubs Robert Ferréol
• Sided Richard Klitzing
• List of uniform polyhedra