# Ask Uncle Colin: how big do the patches on a football need to be?

Dear Uncle Colin,

I’m trying to sew a traditional football in the form of a truncated icosahedron. If I want a radius of 15cm, how big do the polygons need to be?

-- Plugging In Euler Characteristic’s Excessive

Hello, PIECE, and thank you for your message!

Getting an exact answer to that is a little tricky, but we can come up with a pretty good approximation: if we assume the ball’s surface area is the same as that of a sphere, we can work it out.

### The truncated icosahedron

Now, a truncated icosahedron is made of 12 regular pentagons and 20 regular hexagons, all of the same side length, which I’ll call $E$.

Finding the area of a regular polygon boils down to trigonometry: you can split any regular polygon with $n$ sides into $2n$ right-angled triangles with an apex angle of $\frac{\pi}{n}$ opposite a base of $\frac 12 E$. The area of each triangle is $\frac1 8 E^2 \cot\left(\frac{\pi}{n}\right)$, so the polygon area is $\frac{n}{4}E^2 \cot\left(\frac{\pi}{n}\right)$.

In particular, a regular pentagon has an area of $\frac{5}{4}E^2 \cot\left(\frac{\pi}{5}\right)$, and the hexagon is $\frac{6}{4}E^2 \cot\left(\frac{\pi}{6}\right)$, which is $\frac{3}{2}\sqrt{3}E^2$.

Multiplying the pentagon area by 12 and the hexagon area by 20 gives an unholy mess that gives a total surface area of about $72.607E^2$. Don’t tell the Mathematical Ninja.

### Compared to a sphere

The surface area of a sphere of radius 15cm is $4\pi r^2 = 900π \mathrm{cm}^2$.

So, we have $900 \pi \approx 72.607E^2$, and we can rearrange to find $E^2 \approx 38.94 \mathrm{cm}^2$ and $E\approx 6.24$cm((The accurate answer is around 6.05cm, but it’s even more of a pain to work out.)).

Hope that helps!

-- Uncle Colin