How to calculate the dimensions of the required 20 regular hexagons and 12 regular pentagons for a sphere of given diameter (the soccer ball issue)

Question

I wish to calculate the dimensions of the required 20 hexagons and 12 pentagons to tesselate a sphere of given diameter (the soccer ball issue) I need the hexagons and pentagons to be plane figures. The sphere of a given dimension can be inscribed (all corners touching the surface, or super-scribed ( centres of all shapes touching the surface.) Thanks

Answer 1

The "soccer ball" polyhedron is formally called a truncated icosahedron. According to that Wikipedia article, if the sides are of length $a$, then the circumscribing sphere has a radius of $$ \frac{a}{2} \sqrt{ 1 + 9 \phi^2} \approx a \cdot 2.47802... $$ where $\phi = (1 + \sqrt{5})/2$ is the golden ratio.

There's a subtlety that arises if we try to inscribe a sphere inside this polyhedron. While the vertices are all the same distance from the origin, the centroids of the faces are not; see Eq. (4–5) of MathWorld's page on the truncated icosahedron. Specifically, the centers of the pentagons are a distance $$ r_5 = \frac{a}{2} \sqrt{ \frac{125+41\sqrt{5}}{10}} \approx a \cdot 2.32744... $$ from the origin, while the centers of the hexagons are a distance $$ r_6 = \frac{a}{2} \sqrt{ \frac{3(7 + 3\sqrt{5})}{2}} \approx a \cdot 2.26728... $$ from the origin. Since $r_6 < r_5$, the largest sphere that would fit inside a truncated icosahedron would only be tangent to the hexagonal faces, and would not touch the pentagonal faces.

Answer 2

The soccer ball polytope is a truncated icosahedron. If you look up that term, you’ll find everything you need. For example, the Wikipedia page gives the coordinates of the vertices.