Co-ordinates of the vertices an icosahedron relative to its centroid. 
This is a picture of an icosahedron. I need to know the coordinates of the vertices of the icosahedron relative to it's centroid in order to programme a projection of one on a three dimensional plane. (By the way, it has twelve vertices, so it is going to be a very long process)
Unfortunately, I have not been able to find one vertices. 

With $20$ faces, each face has an area of $\frac\pi5$ steradians. That means that the spherical excess in each face is $\frac\pi5$ radians. Thus, each angle in each spherical triangular face has an angle of $\frac\pi3+\frac\pi{15}=\frac{2\pi}5$.
We can use the Spherical Law of Cosines for Angles to get the angle subtended by each side. Since $\cos(\mathrm{A})=\cos(\mathrm{B})=\cos(\Gamma)=\frac{-1+\sqrt5}4$, we get
$$
\begin{align}
\cos(\gamma)
&=\frac{\cos(\Gamma)+\cos(\mathrm{A})\cos(\mathrm{B})}{\sin(\mathrm{A})\sin(\mathrm{B})}\\
&=\frac1{\sqrt5}
\end{align}
$$
Since $\cos\left(\frac{2\pi}5\right)=\frac{-1+\sqrt5}4$ and $\sin\left(\frac{2\pi}5\right)=\sqrt{\frac{5+\sqrt5}8}$ and $\cos\left(\frac{4\pi}5\right)=\frac{-1-\sqrt5}4$ and $\sin\left(\frac{4\pi}5\right)=\sqrt{\frac{5-\sqrt5}8}$, the set of vertices would be
$$
\begin{align}
a_1&=(1,0,0)\\
a_2&=\textstyle\left(\frac1{\sqrt5},\frac2{\sqrt5},0\right)\\
a_3&=\textstyle\left(\frac1{\sqrt5},\frac{5-\sqrt5}{10},\sqrt{\frac{5+\sqrt5}{10}}\right)\\
a_4&=\textstyle\left(\frac1{\sqrt5},\frac{-5-\sqrt5}{10},\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_5&=\textstyle\left(\frac1{\sqrt5},\frac{-5-\sqrt5}{10},-\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_6&=\textstyle\left(\frac1{\sqrt5},\frac{5-\sqrt5}{10},-\sqrt{\frac{5+\sqrt5}{10}}\right)\\
a_7&=\textstyle(-1,0,0)\\
a_8&=\textstyle\left(-\frac1{\sqrt5},-\frac2{\sqrt5},0\right)\\
a_9&=\textstyle\left(-\frac1{\sqrt5},\frac{-5+\sqrt5}{10},-\sqrt{\frac{5+\sqrt5}{10}}\right)\\
a_{10}&=\textstyle\left(-\frac1{\sqrt5},\frac{5+\sqrt5}{10},-\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_{11}&=\textstyle\left(-\frac1{\sqrt5},\frac{5+\sqrt5}{10},\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_{12}&=\textstyle\left(-\frac1{\sqrt5},\frac{-5+\sqrt5}{10},\sqrt{\frac{5+\sqrt5}{10}}\right)
\end{align}
$$
$a_1$ shares edges with $a_k$ for $2\le k\le6$.
For $2\le k\le6$, $a_k$ shares edges with $a_1,a_{2+\text{mod}(k-1,5)},a_{2+\text{mod}(k+2,5)},a_{8+\text{mod}(k,5)},a_{8+\text{mod}(k+1,5)}$.
$a_7$ shares edges with $a_k$ for $8\le k\le12$.
For $8\le k\le12$, $a_k$ shares edges with $a_7,a_{8+\text{mod}(k-2,5)},a_{8+\text{mod}(k+1,5)},a_{2+\text{mod}(k-1,5)},a_{2+\text{mod}(k,5)}$.  

Animation of the Icosahedron Generated above

