I want to 3D print a wireframe of a icosahedron. For that I need the coordinates of the vertices so that one triangle lies flat on the $z=0$ plane.
All I've found are descriptions for an icosahedron standing on one vertice or and edge but not a face.
Does anyone have those coordinates?
Wikipedia gives us the coordinates $(0,\pm1,\pm\phi)$, $(\pm\phi,0,\pm1)$, $(\pm1,\pm\phi,0)$, where $\phi$ is the golden ratio.
To move a face to the bottom, just move the center of a face to the bottom. This can be done with a reflection. One of the faces is formed by the points $(0,1,-\phi),(0,-1,-\phi),(\phi,0,-1)$, so the direction of the center of the face is given by the vector $\vec a=(\phi,0,-1-2\phi)$, whose norm is $\sqrt{\phi^2+(1+2\phi)^2}=\sqrt{6+9\phi}$. To map the vector $\vec a$ to the vector $\vec b=(0,0,-\sqrt{6+9\phi})$, apply a reflexion through the hyperplane orthogonal to $\vec a-\vec b$. With the unit vector $\vec u=\dfrac{\vec a-\vec b}{||\vec a-\vec b||}$, the matrix of the transformation is $A=I-2uu^T$.
Then apply this matrix to all your points (seen as vectors starting at the origin).
I didn't try to simplify the result, so here are numerical coordinates ($x,y,z$ on each line, ordered by increasing $z$ then increasing $y$):
Afterthought: it could be simplified, as I recognize the values
$$\phi,\dfrac{1}{\sqrt3},\dfrac{2}{\sqrt3},\dfrac{\phi}{\sqrt3},\dfrac{2\phi}{\sqrt3},\dfrac{\phi-1}{\sqrt3},\dfrac{\phi+1}{\sqrt3}$$