# Inverted snub dodecadodecahedron

Inverted snub dodecadodecahedron :
| Type | Uniform star polyhedron |
| Elements | F = 84, E = 150 V = 60 (χ = −6) |
| Faces by sides | 60{3}+12{5}+12{5/2} |
| Coxeter diagram | |
| Wythoff symbol | | 5/3 2 5 |
| Symmetry group | I, [5,3]$^+$, 532 |
| Index references | U$_{60}$, C$_{76}$, W$_{114}$ |
| Dual polyhedron | Medial inverted pentagonal hexecontahedron |
| Vertex figure | 3.3.5.3.5/3 |
| Bowers acronym | Isdid |
In geometry, the inverted snub dodecadodecahedron (or vertisnub dodecadodecahedron) is a nonconvex uniform polyhedron, indexed as U$_{60}$. It is given a Schläfli symbol sr{5/3,5}.

## Cartesian coordinates

Let $\xi \approx 2.109759446579943$ be the largest real zero of the polynomial $P=2x^{4}-5x^{3}+3x+1$. Denote by $\phi$ the golden ratio. Let the point $p$ be given by
    $p={\begin{pmatrix}\phi ^{-2}\xi ^{2}-\phi ^{-2}\xi +\phi ^{-1}\\-\phi ^{2}\xi ^{2}+\phi ^{2}\xi +\phi \\\xi ^{2}+\xi \end{pmatrix}}$.
Let the matrix $M$ be given by
    $M={\begin{pmatrix}1/2&-\phi /2&1/(2\phi )\\\phi /2&1/(2\phi )&-1/2\\1/(2\phi )&1/2&\phi /2\end{pmatrix}}$.
$M$ is the rotation around the axis $(1,0,\phi )$ by an angle of $2\pi /5$, counterclockwise. Let the linear transformations $T_{0},\ldots ,T_{11}$ be the transformations which send a point $(x,y,z)$ to the even permutations of $(\pm x,\pm y,\pm z)$ with an even number of minus signs. The transformations $T_{i}$ constitute the group of rotational symmetries of a regular tetrahedron. The transformations $T_{i}M^{j}$ $(i=0,\ldots ,11$, $j=0,\ldots ,4)$ constitute the group of rotational symmetries of a regular icosahedron. Then the 60 points $T_{i}M^{j}p$ are the vertices of a snub dodecadodecahedron. The edge length equals $2(\xi +1){\sqrt {\xi ^{2}-\xi }}$, the circumradius equals $(\xi +1){\sqrt {2\xi ^{2}-\xi }}$, and the midradius equals $\xi ^{2}+\xi$.
For a great snub icosidodecahedron whose edge length is 1, the circumradius is
    $R={\frac {1}{2}}{\sqrt {\frac {2\xi -1}{\xi -1}}}\approx 0.8516302281174128$
Its midradius is
    $r={\frac {1}{2}}{\sqrt {\frac {\xi }{\xi -1}}}\approx 0.6894012223976083$
The other real root of P plays a similar role in the description of the Snub dodecadodecahedron

## Related polyhedra
### Medial inverted pentagonal hexecontahedron

Medial inverted pentagonal hexecontahedron :
| Type | Star polyhedron |
| Face | |
| Elements | F = 60, E = 150 V = 84 (χ = −6) |
| Symmetry group | I, [5,3]$^+$, 532 |
| Index references | DU$_{60}$ |
| dual polyhedron | Inverted snub dodecadodecahedron |
The medial inverted pentagonal hexecontahedron (or midly petaloid ditriacontahedron) is a nonconvex isohedral polyhedron. It is the dual of the uniform inverted snub dodecadodecahedron. Its faces are irregular nonconvex pentagons, with one very acute angle.

## Proportions

Denote the golden ratio by $\phi$, and let $\xi \approx -0.236\,993\,843\,45$ be the largest (least negative) real zero of the polynomial $P=8x^{4}-12x^{3}+5x+1$. Then each face has three equal angles of $\arccos(\xi )\approx 103.709\,182\,219\,53^{\circ }$, one of $\arccos(\phi ^{2}\xi +\phi )\approx 3.990\,130\,423\,41^{\circ }$ and one of $360^{\circ }-\arccos(\phi ^{-2}\xi -\phi ^{-1})\approx 224.882\,322\,917\,99^{\circ }$. Each face has one medium length edge, two short and two long ones. If the medium length is $2$, then the short edges have length $1-{\sqrt {\frac {1-\xi }{\phi ^{3}-\xi }}}\approx 0.474\,126\,460\,54,$ and the long edges have length $1+{\sqrt {\frac {1-\xi }{\phi ^{-3}-\xi }}}\approx 37.551\,879\,448\,54.$ The dihedral angle equals $\arccos(\xi /(\xi +1))\approx 108.095\,719\,352\,34^{\circ }$. The other real zero of the polynomial $P$ plays a similar role for the medial pentagonal hexecontahedron.