How do I calculate the coordinates of the centers of the faces of a truncated icosahedron?

Question

I'd like to draw a truncated icosahedron in 3-space. Wikipedia gives formulas for the cartesian coordinates of all the vertices, but I'd like just the center coordinates to use as local coordinate systems, and I need to know which face is which. Either cartesian or polar is fine, since the transformation is trivial.

Edit: I just realized I will need the rotation of each face, too.

Answer 1

Since the faces are regular polygons their centers are the average of their vertices.

That answers the question in the title, given the data on the wikipedia page.

Then you want an answer to

Given face $n$, what is its position and rotation?

The faces don't come with any natural order, so asking about "face $n$" makes no sense. Perhaps what you really want is a list of the faces and their properties. One possible way to build that is to work with the dual, the pentakis dodecahedron. The vertices of that figure correspond to the faces of the truncated icosahedron.

A short search locates these sites. The first of these gives you the coordinates of the vertices.

http://dmccooey.com/polyhedra/PentakisDodecahedron.txt

http://mathworld.wolfram.com/PentakisDodecahedron.html (You might get more of what you need from mathematica code)

https://en.wikipedia.org/wiki/Pentakis_dodecahedron

https://rechneronline.de/pi/pentakis-dodecahedron.php