Soccer Ball problems

The area of a regular polygon of $n$ sides of unit side length is given by $A_n = \frac n4 \cot \frac \pi n$ (see Note 1). Then we have:

$\begin{aligned} F & = \frac {12A_5}{12A_5 + 20A_6} \\ & = \frac {15\cot \frac \pi 5}{15\cot \frac \pi 5 + 30 \color{#D61F06} \cot \frac \pi 6} & \small \color{#3D99F6} \text{Divide up and down by 15} \\ & = \frac {\cot \frac \pi 5}{\cot \frac \pi 5 + 2 \color{#D61F06} \sqrt 3} & \small \color{#3D99F6} \text{Multiply up and down by }\tan \frac \pi 5 \\ & = \frac 1{1 + 2 \sqrt 3 \tan \frac \pi 5} & \small \color{#3D99F6} \text{Note 2: }\cos \frac \pi 5 = \frac \varphi 2 \\ & = \frac 1{1 + 2 \sqrt 3 \cdot \frac {\sqrt{4-\varphi^2}}\varphi} & \small \color{#3D99F6} \implies \sin \frac \pi 5 = \frac {\sqrt{4-\varphi^2}}2 \\ & = \frac \varphi {\varphi + \sqrt{48-12\varphi^2}} \end{aligned}$

Therefore, $a-b = 48-12 = \boxed{36}$ .

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Notes:

1. A $n$ -side regular polygon of unit side length is made up of $n$ congruent isosceles triangles with the unit side as base. The angle between the equal sides is $\frac {2\pi}n$ . The area of a triangle is $A_\triangle = \frac 12 \times \frac {\frac 12}{\tan \frac \pi n} = \frac 14 \cot \frac \pi n$ . The area of the $n$ -side regular polygon $A_n = n A_\triangle = \frac n4 \cot \frac \pi n$ .

2. Considering the following:

$\begin{aligned} \cos \frac \pi 5 + \color{#3D99F6} \cos \frac 35 \pi & = \frac 12 & \small \color{#3D99F6} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ \cos \frac \pi 5 - \color{#3D99F6} \cos \frac 25 \pi & = \frac 12 \\ \cos \frac \pi 5 - \color{#3D99F6}2\cos^2 \frac \pi 5 + 1 & = \frac 12 \\ \implies 4\cos^2 \frac \pi 5 - 2 \cos \frac \pi 5 - 1 & = 0 \\ \implies \cos \frac \pi 5 & = \frac {1+\sqrt 5}2 = \frac \varphi 2 & \small \color{#3D99F6} \text{Note that }\cos \frac \pi 5 > 0 \end{aligned}$

The area of a pentagon with side $s$ is $A_p = \frac{1}{4}\sqrt{5(5 + 2\sqrt{5})}s^2$ and the area of a hexagon with side $s$ is $A_h = \frac{3\sqrt{3}}{2}s^2$ . The fraction of the area of the $12$ pentagons to the area of the $12$ pentagons and $20$ hexagons is then

$F = \frac{12A_p}{12A_p + 20A_h} = \frac{1}{1 + \frac{20A_h}{12A_p}} = \frac{\varphi}{\varphi + \frac{5 \varphi A_h}{3A_p}}$

The term $\frac{5 \varphi A_h}{3A_p}$ can be simplified to

$\frac{5 \varphi A_h}{3A_p} = \frac{5 \varphi \cdot \frac{3 \sqrt{3}}{2}s^2}{3 \cdot \frac{1}{4}\sqrt{5(5 + 2\sqrt{5})}s^2} = \frac{10 \varphi \sqrt{3}}{\sqrt{5(5 + 2\sqrt{5})}} = \sqrt{\frac{60 \varphi^2}{5 + 2\sqrt{5}}} = \sqrt{\frac{60 \varphi^2(5 - 2\sqrt{5})}{(5 + 2\sqrt{5})(5 - 2\sqrt{5})}} = \sqrt{12 \varphi^2(5 - 2\sqrt{5})}$ $= \sqrt{12 \frac{1}{2}(3 + \sqrt{5})(5 - 2\sqrt{5})} = \sqrt{6(5 - \sqrt{5})} = \sqrt{6(8 - (3 + \sqrt{5}))} = \sqrt{12(4 - \frac{1}{2}(3 + \sqrt{5}))} = \sqrt{48 - 12\varphi^2}$

Therefore,

$F = \frac{\varphi}{\varphi + \sqrt{48 - 12\varphi^2}}$

so $a = 48$ and $b = 12$ and $a - b = \boxed{36}$ .

The problem is how to show that cos 36 = phi/2 without resorting to calculators, square roots of 5 or assertion of "standard" results. After that the rest is straightforward. I shall assume only that phi is the positive root of x^2 -x - 1 =0. Now consider triangle ABC, with BC = 1 and angles ABC and ACB = 72, so angle BAC = 36. Then take D on AC so that DBC = 36, and therefore ABD = 36 and BDC = 72. Thus, from isosceles triangles, ABD and BDC, AD = BD = BC = 1. Let AB = x. Then by similar triangles ABC and BDC, AB/BD = BC/DC, i.e. x/1 = 1/(x-1), so x^2 -x -1 = 0, so x = phi. (This is all standard stuff, but I don't want to be accused of assuming too much!) Now let E be the mid-point of AB, so cos 36 = AE/AD = (phi/2)/1 = phi/2

The area of a hexagon with a edge of 1 is $\frac{3\sqrt{3}}{2}$ . The area of a pentagon with edge of 1 is $\frac{1}{4} \sqrt{5 \left(2 \sqrt{5}+5\right)}$ . $f = \frac{1}{1+2 \sqrt{15-6 \sqrt{5}}}$ . After multiplying by $1$ in the form $\frac{\phi}{\phi}$ and moving the $\phi$ inside the square root, $f = \frac{\phi }{\phi+\sqrt{48-12 \phi ^2}}$ results. The answer is 36.