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.
Solution
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
$$\frac{\varDelta}{\delta}=\Phi=\frac{1+\sqrt{5}}{2} \tag{1}$$
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. )
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)
From the properties of dodecahedron, it is known that \[r=\frac{\Phi^2}{2}\,AB=\frac{\Phi^2}{2}\,\delta \tag{2}\] From Pythagoras's theorem in \(\triangle{AGH}\) we have that
$$\begin{aligned}
1 &=AG^2 \\
&= AH^2 + GH^2\\
&=\frac{AB^2}{4}+ \frac{GF^2}{4} \\
&= \frac{\delta ^2}{4}+\frac{\Delta ^2}{4}\\
&\overset{(1)}{=}\frac{\delta^2}{4}+ \frac{\Phi ^2 \delta^2}{4}= \frac{\delta ^2}{4}\left ( 1+\Phi ^2 \right )
\end{aligned}$$
Finally, from \((2)\) we have that
$$r=\frac{\Phi^2}{2}\,\delta=\frac{\Phi^2}{2}\,\frac{2}{\sqrt{1+\Phi^2}}=\frac{\Phi^2}{\sqrt{1 + \Phi^2}} =\frac{3 + \sqrt{5}}{\sqrt{10 + 2\sqrt{5}}}$$
and the exercise comes to and end.
The exercise can also be found in mathimatikoi.org
Solution
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
$$\frac{\varDelta}{\delta}=\Phi=\frac{1+\sqrt{5}}{2} \tag{1}$$
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. )
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)
$$\begin{aligned}
1 &=AG^2 \\
&= AH^2 + GH^2\\
&=\frac{AB^2}{4}+ \frac{GF^2}{4} \\
&= \frac{\delta ^2}{4}+\frac{\Delta ^2}{4}\\
&\overset{(1)}{=}\frac{\delta^2}{4}+ \frac{\Phi ^2 \delta^2}{4}= \frac{\delta ^2}{4}\left ( 1+\Phi ^2 \right )
\end{aligned}$$
Finally, from \((2)\) we have that
$$r=\frac{\Phi^2}{2}\,\delta=\frac{\Phi^2}{2}\,\frac{2}{\sqrt{1+\Phi^2}}=\frac{\Phi^2}{\sqrt{1 + \Phi^2}} =\frac{3 + \sqrt{5}}{\sqrt{10 + 2\sqrt{5}}}$$
and the exercise comes to and end.
The exercise can also be found in mathimatikoi.org
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