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Inscribed sphere of rhombic triacontahedron

Projective Geometry, Solid Geometry
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Grigorios Kostakos
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Inscribed sphere of rhombic triacontahedron

#1

Post by Grigorios Kostakos » Mon Nov 09, 2015 1:47 am

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
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Grigorios Kostakos
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Re: Inscribed sphere of rhombic triacontahedron

#2

Post by Grigorios Kostakos » Tue Nov 10, 2015 11:27 am

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
inscribed_sphere_r30edron_i.png (73.58 KiB) Viewed 1939 times
[/centre]

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
inscribed_sphere_r30edron_ii.png (73.84 KiB) Viewed 1938 times
[/centre]

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
\[r=\frac{\Phi^2}{2}\,\delta=\frac{\Phi^2}{2}\,\frac{2}{\sqrt{1+\Phi^2}}=\frac{\Phi^2}{\sqrt{1 + \Phi^2}} =\frac{3 + \sqrt{5}}{\sqrt{10 + 2\sqrt{5}}}\,.\]
Grigorios Kostakos
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