Inverted snub dodecadodecahedron

In geometry, the inverted snub dodecadodecahedron (or vertisnub dodecadodecahedron) is a nonconvex uniform polyhedron, indexed as U60. 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
 * $$R = \frac12\sqrt{\frac{2\xi-1}{\xi-1}} \approx 0.8516302281174128$$

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

Medial inverted pentagonal hexecontahedron
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.