Table of Contents

Overview

Properties

\begin{eqnarray*} r &=& a * \sin{\frac{2\pi}{5}} \end{eqnarray*}

\begin{eqnarray*} r_{i} &=& \frac{\sqrt{3}}{12}(3 + \sqrt{5})a \end{eqnarray*}

\begin{eqnarray*} r_{m} &=& a * \cos{\frac{\pi}{5}} \end{eqnarray*}

\begin{eqnarray*} \delta &=& \arccos-\frac{\sqrt{5}}{3} \\ &\approx& 138.189685^{\circ} \end{eqnarray*}

Inner Angle between Faces

We want to determine $\alpha$ which is the inner angle between two segments that connect the centre of the icosahedron to the centre of two adjacent triangles.

In order to do that, we observe the kite formed by the points $OABC$:

In the yellow triangle, we observe that the segment $BC$ is the line from the side of triangle to the centroid and is equal to $\frac{h}{2}$ where, for an equilateral triangle, $h=\frac{a\sqrt{3}}{2}$. We thus know that:

\begin{eqnarray*} BC &=& a\frac{\sqrt{3}}{6} \end{eqnarray*}

We also observe that $OC$ is the line from the centre of the icosahedron to the centroid of the triangle, which is equal to the radius of the inscribed sphere $r_{i}$:

\begin{eqnarray*} OC &=& \frac{\sqrt{3}}{2}(3+\sqrt{5})a \end{eqnarray*}

Now, for the triangle $\triangle BMC$ we apply $\sin{\frac{\delta}{2}}$:

\begin{eqnarray*} \sin{\frac{\delta}{2}} &=& \frac{h}{a\frac{\sqrt{3}}{6}} \end{eqnarray*}

and extract $h$:

\begin{eqnarray*} h &=& a\frac{\sqrt{3}}{6}\sin{\frac{\delta}{2}} \end{eqnarray*}

For triangle $\triangle OMC$, we apply $\sin{\frac{\alpha}{2}}$ and substitute $h$:

\begin{eqnarray*} \sin{\frac{\alpha}{2}} &=& \frac{h}{\frac{\sqrt{3}}{2}(3+\sqrt{5})a} \\ &=& \arcsin{ \frac{a\frac{\sqrt{3}}{6}\sin{\frac{\delta}{2}}}{\frac{\sqrt{3}}{2}(3+\sqrt{5})a} } \\ &=& \arcsin{ \frac{\sqrt{3}\sin{\frac{\delta}{2}}}{\frac{\sqrt{3}}{2}(3+\sqrt{5}) } \\ &=& \arcsin{ \frac{ 2\sin{\frac{\delta}{2}} }{\sqrt{5}+3} } \end{eqnarray*}

To find out the numeric value, we substitute $\frac{\delta}{2}$ and obtain $\sin{\frac{\alpha}{2}}$:

\begin{eqnarray*} \frac{\alpha}{2} &\approx& 0.36486\text{radians} \\ &\approx&20.904940^{\circ} \end{eqnarray*}

Thus, $\alpha$ is:

\begin{eqnarray*} \alpha &\approx& 41.809880^{\circ} \end{eqnarray*}