Editing Snub dodecahedron (section)

==Cartesian coordinates==
Let {{math|''ξ'' ≈ {{val|0.94315125924}}}} be the real zero of the [[cubic polynomial]] {{math|''x''<sup>3</sup> + 2''x''<sup>2</sup> − ''φ''<sup>2</sup>}}, where {{mvar|φ}} is the [[golden ratio]]. Let the point {{mvar|p}} be given by
<math display=block>p=
\begin{pmatrix}
        \varphi^2-\varphi^2\xi \\
        -\varphi^3+\varphi\xi+2\varphi\xi^2 \\
        \xi
\end{pmatrix}.
</math>
Let the [[rotation matrix|rotation matrices]] {{math|''M''<sub>1</sub>}} and {{math|''M''<sub>2</sub>}} be given by
<math display=block>
M_1=
\begin{pmatrix}
         \frac{1}{2\varphi}  & -\frac{\varphi}{2} & \frac{1}{2} \\
         \frac{\varphi}{2}   & \frac{1}{2}     & \frac{1}{2\varphi} \\
         -\frac{1}{2}     & \frac{1}{2\varphi} & \frac{\varphi}{2}
\end{pmatrix}, \quad M_2=
\begin{pmatrix}
         0  & 0 & 1 \\
         1  & 0 & 0 \\
         0  & 1 & 0
\end{pmatrix}.
</math>
{{math|''M''<sub>1</sub>}} represents the rotation around the axis {{math|(0, 1, ''φ'')}} through an angle of {{sfrac|2{{pi}}|5}} counterclockwise, while {{math|''M''<sub>2</sub>}} being a cyclic shift of {{math|(''x'', ''y'', ''z'')}} represents the rotation around the axis {{math|(1, 1, 1)}} through an angle of {{sfrac|2{{pi}}|3}}. Then the 60 vertices of the snub dodecahedron are the 60 images of point {{mvar|p}} under repeated multiplication by {{math|''M''<sub>1</sub>}} and/or {{math|''M''<sub>2</sub>}}, iterated to convergence. (The matrices {{math|''M''<sub>1</sub>}} and {{math|''M''<sub>2</sub>}} [[Generating set of a group|generate]] the 60 rotation matrices corresponding to [[icosahedral group|the 60 rotational symmetries]] of a [[regular icosahedron]].) The coordinates of the vertices are integral linear combinations of {{math|1, ''φ'', ''ξ'', ''φξ'', ''ξ''<sup>2</sup>}} and {{math|''φξ''<sup>2</sup>}}. The edge length equals
<math display=block>2\xi\sqrt{1-\xi}\approx 0.449\,750\,618\,41.</math>
Negating all coordinates gives the mirror image of this snub dodecahedron.

As a volume, the snub dodecahedron consists of 80 triangular and 12 pentagonal pyramids.
The volume {{math|''V''<sub>3</sub>}} of one triangular pyramid is given by:
<math display=block>
    V_3 = \frac{1}{3}\varphi\left(3\xi^2-\varphi^2\right) \approx 0.027\,274\,068\,85,
</math>
and the  volume {{math|''V''<sub>5</sub>}} of one pentagonal pyramid by:
<math display=block>
    V_5 = \frac{1}{3}(3\varphi+1)\left(\varphi+3-2\xi-3\xi^2\right)\xi^3 \approx 0.103\,349\,665\,04.
</math>
The total volume is
<math display=block>80V_3+12V_5 \approx 3.422\,121\,488\,76.</math>

The circumradius equals
<math display=block>\sqrt{4\xi^2-\varphi^2} \approx 0.969\,589\,192\,65.</math>
The [[Midsphere|midradius]] equals {{mvar|ξ}}. This gives an interesting geometrical interpretation of the number {{mvar|ξ}}. The 20 "icosahedral" triangles of the snub dodecahedron described above are coplanar with the faces of a regular icosahedron. The midradius of this "circumscribed" icosahedron equals 1. This means that {{mvar|ξ}} is the ratio between the midradii of a snub dodecahedron and the icosahedron in which it is inscribed.

The triangle–triangle dihedral angle is given by
<math display=block>
    \theta_{33} = 180^\circ - \arccos\left(\frac23\xi+\frac13\right) \approx 164.175\,366\,056\,03^\circ.
</math>

The triangle–pentagon dihedral angle is given by
<math display=block>\begin{align}
    \theta_{35} &= 180^\circ - \arccos\sqrt{\frac{-(4\varphi + 8)\xi^2 - (4\varphi + 8)\xi + 12\varphi + 19}{15}} \\[2pt]
    &\approx 152.929\,920\,275\,84^\circ.
\end{align}</math>