User:Mennotk/sandbox

$$$$ Let $$\xi\approx 0.94315125924388$$ be the real zero of the polynomial $$x^3+2x^2-\phi^2$$, where $$\phi$$ is the golden ratio. Let the point $$p$$ be given by
 * $$p=

\begin{pmatrix} \phi^2-\phi^2\xi \\ -\phi^3+\phi\xi+2\phi\xi^2 \\ \xi \end{pmatrix} $$. Let the matrix $$M$$ be given by
 * $$M=

\begin{pmatrix} 1/(2\phi) & -\phi/2 & 1/2 \\ \phi/2  & 1/2     & 1/(2\phi) \\ -1/2    & 1/(2\phi) & \phi/2 \end{pmatrix} $$. $$M$$ is the rotation around the axis $$(0, 1, \phi)$$ over an angle of $$2\pi/5$$, counterclockwise. Let the linear transformations $$T_0, \ldots, T_{11}$$ be the transformations which send a point $$(x, y, z)$$ to to the even permutations of $$(\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$$ are the vertices of a snub dodecahedron. The coordinates of the vertices are integral linear combinations of $$1$$, $$\phi$$, $$\xi$$, $$\phi\xi$$, $$\xi^2$$ and $$\phi\xi^2$$. The edge length equals $$2\xi\sqrt{1-\xi}$$. Negating all coordinates gives the mirror image of this snub dodecahedron.