Inscribed sphere of rhombic triacontahedron

[Grigorios Kostakos]

Consider a rhombic triacontahedron \(R\) with edge length \(1\) and the inscribed sphere \(S\) of \(R\) (tangent to each of the rhombic triacontahedron's faces). Prove that the radius \(r\) of \(S\) has length \[r=\frac{\Phi^2}{\sqrt{1 + \Phi^2}} =\frac{3 + \sqrt{5}}{\sqrt{10 + 2\sqrt{5}}}\,,\] where \(\Phi=\frac{1+\sqrt{5}}{2}\) is the golden ratio.

[Grigorios Kostakos]

A rhombic triacontahedron has \(30\) faces, all of which are golden rhombi. A golden rhombus is a rhombus such that the ratio of the long diagonal \(\varDelta\) to the short diagonal \(\delta\) is equal to the golden ratio \(\Phi\), ie
\[\frac{\varDelta}{\delta}=\Phi=\frac{1+\sqrt{5}}{2}\quad(1)\,.\]Also, the short diagonals \(\delta\)s are the edges of a dodecahedron \(T\), which is inscribed in the rhombic triacontahedron \(R\). ( In the first figure, the polygon \(ABCDEA\) is a face of the dodecahedron. )

inscribed_sphere_r30edron_i.png (73.58 KiB) Viewed 3962 times

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The sphere \(S\) which is inscribed in \(R\) (tangent to each of the rhombic triacontahedron's faces) touches the center of each golden rhombus, ie the point of intersection of its diagonals. Also, the same sphere \(S\) is circumscribed in dodecahedron \(T\) ( passes from dodecahedron's vertices ).
So, if \(O\) is the center of the sphere \(S\) and \(H\) is the center of the golden rhombus \(AFBG\), we have that \(r=OH\). (2nd figure)

inscribed_sphere_r30edron_ii.png (73.84 KiB) Viewed 3961 times

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From the properties of dodecahedron, it is known that \[r=\frac{\Phi^2}{2}\,AB=\frac{\Phi^2}{2}\,\delta\quad(2)\,.\] From the Pythagorean theorem in \(\triangle{AGH}\) we have that \begin{align*}
1=AG^2=&\;AH^2+GH^2=\frac{AB^2}{4}+\frac{GF^2}{4}=\frac{\delta^2}{4}+\frac{\varDelta^2}{4}\stackrel{(1)}{=\!=}\frac{\delta^2}{4}+\frac{\Phi^2\delta^2}{4}=\frac{\delta^2}{4}\,(1+\Phi^2)\quad\Rightarrow\\
& \delta=\frac{2}{\sqrt{1+\Phi^2}}\,.
\end{align*}
Finally, from \((2)\) we have that
\[r=\frac{\Phi^2}{2}\,\delta=\frac{\Phi^2}{2}\,\frac{2}{\sqrt{1+\Phi^2}}=\frac{\Phi^2}{\sqrt{1 + \Phi^2}} =\frac{3 + \sqrt{5}}{\sqrt{10 + 2\sqrt{5}}}\,.\]