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. 
 A: It is a well-known property of the regular icosahedron that three mutually perpendicular rectangles of aspect ratio $\varphi = (1+\sqrt{5})/2$, arranged symmetrically about fourfold axes of symmetry and sharing a common center, will describe $12$ vertices that coincide with those of the regular icosahedron.  Specifically, the rectangles formed by cyclic permutations of the vertices $$(\pm \varphi, \pm 1, 0)$$ where the signs may be taken in any combination, will give a regular icosahedron of circumradius $R = \sqrt{\varphi^2 + 1^2} = \sqrt{\varphi+2}$.  Explicitly, one may write out all $12$ vertices as
$$(\varphi,1,0) \\
(\varphi,-1,0) \\
(-\varphi,-1,0) \\
(-\varphi,1,0) \\
(1,0,\varphi) \\
(-1,0,\varphi) \\
(-1,0,-\varphi) \\
(1,0,-\varphi) \\
(0,\varphi,1) \\
(0,\varphi,-1) \\
(0,-\varphi,-1) \\
(0,-\varphi,1)$$
Here is what it looks like, with a little animation.

A: 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

A: PolyhedronData["Icosahedron", "VertexCoordinates"]

$$
\left(
\begin{array}{ccc}
 0 & 0 & -\frac{5}{\sqrt{50-10 \sqrt{5}}} \\
 0 & 0 & \frac{5}{\sqrt{50-10 \sqrt{5}}} \\
 -\sqrt{\frac{2}{5-\sqrt{5}}} & 0 & -\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
 \sqrt{\frac{2}{5-\sqrt{5}}} & 0 & \frac{1}{\sqrt{10-2 \sqrt{5}}} \\
 \frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2} &
   -\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
 \frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2} &
   -\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
 -\frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2} &
   \frac{1}{\sqrt{10-2 \sqrt{5}}} \\
 -\frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2} &
   \frac{1}{\sqrt{10-2 \sqrt{5}}} \\
 -\frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2}
   \sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & -\frac{1}{\sqrt{10-2 \sqrt{5}}}
   \\
 -\frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2}
   \sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & -\frac{1}{\sqrt{10-2 \sqrt{5}}}
   \\
 \frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2}
   \sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & \frac{1}{\sqrt{10-2 \sqrt{5}}}
   \\
 \frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2}
   \sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & \frac{1}{\sqrt{10-2 \sqrt{5}}}
\end{array}
\right)
$$
A: Let the centroid be $[0,0,0]$ and one vertex $[0,0,r]$.  Another vertex is $[0,0,-r]$.  There are five vertices equally spaced around a circle on some plane $z=$constant, say at
$[a \cos(2\pi j/5), a \sin(2\pi j/5), b]$, $j=0\ldots4$, where $a^2 + b^2 = r^2$, and the other five are the reflections of these through the origin, so
$[-a \cos(2\pi j/5, -a \sin(2\pi j/5), -b]$.  Determine $a$ and $b$ so that the
distances between nearest neighbours are equal.
A: Using spherical coordinates $ (\theta,\phi)$ , Vertex coordinates can be easily expressed.
The two Poles are at:
$$ ( any  , \pm 90^0 )$$
If one typical elevation angle $ E=\tan^{-1} \frac12 = 26.565051177078^0, $ then the ten Vertices are oscillating around axis of poles at constant azimuth interval $36^0$ :
$$ (0^0, E) ,  (36^0,-E), (72^0, E), (108^0, -E) , (144^0, E),(180^0,-E),(216^0, E) , (252^0,-E),(288^0, E),(324^0, -E). $$
