Great retrosnub icosidodecahedron explained

Cartesian coordinates

Let

\xi ≈ -1 . 8934600671194555 be the smallest (most negative) zero of the polynomial x ^{3+2x} ^{2-\phi} ^{-2} , where \phi is the golden ratio. Let the point p be given byp = \begin{pmatrix} \xi \\ \phi ^{-2} - \phi ^{-2} \xi \\ - \phi ^{-3} + \phi ^{-1} \xi +2 \phi ^{-1} \xi ^{2 \end{pmatrix}} .Let the matrix M be given byM = \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 great snub icosahedron. The edge length equals -2 \xi\sqrt{ 1- \xi} , the circumradius equals - \xi\sqrt{ 2- \xi} , and the midradius equals - \xi .

For a great snub icosidodecahedron whose edge length is 1,the circumradius is

R = (12 \sqrt{ (2- \xi)/(1- \xi))/() } \approx 0.5800015046400155 Its midradius isr = (1)/(\sqrt{ 2) (1)/(1- \xi) } \approx 0.2939417380786233

The four positive real roots of the sextic in,$$4096R^\; -\; 27648R^\; +\; 47104R^8\; -\; 35776R^6\; +\; 13872R^4\; -\; 2696R^2\; +\; 209\; =\; 0$$are the circumradii of the snub dodecahedron (U₂₉), great snub icosidodecahedron (U₅₇), great inverted snub icosidodecahedron (U₆₉), and great retrosnub icosidodecahedron (U₇₄).

See also

• List of uniform polyhedra
• Great snub icosidodecahedron
• Great inverted snub icosidodecahedron

References

1. Web site: 74: great retrosnub icosidodecahedron. Maeder. Roman. MathConsult.