Overview

Properties

  • The radius of the sphere circumscribing the icosahedron, spanning from the centre of the icosahedron to each vertex, is:

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

  • The radius of the inscribed sphere, spamming from the centre of the icosahedron to each face, is:

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

  • The midradius that spans from the centre of the icosahedron to the middle of each edge, is:

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

  • The dihedral angle (the angle between the sides of any two triangles on the icosahedron) of the icosahedron is:

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

Inner Angle between Faces

fuss_mathematics_geometry_icosahedron_regular_kite.svg

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 $OABC$:

fuss_mathematics_geometry_icosahedron_regular_kite_section.svg

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*}


fuss/mathematics/geometry/icosohedron/regular.txt · Last modified: 2017/02/22 18:30 (external edit)

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