A Pentagon, Hexagon And A Decagon Walk Into A Bar

The formula for the length of a chord of a triangle is $c=2r\sin\left(\frac{\theta}{2}\right)$ . For an n-sided shape, we have $\theta=\frac{2π}{n}$ . Therefore with $r=1$ , our formula for the side length of an inscribed n-gon is $s=2\sin\left(\frac{π}{n}\right)$

Side length of the pentagon: $2\sin\left(\frac{π}{5}\right)$

Side length of the hexagon: $2\sin\left(\frac{π}{6}\right)$

Side length of the decagon: $2\sin\left(\frac{π}{10}\right)$

A little known identity:

$\sin\left(\frac{π}{10}\right)^2+\sin\left(\frac{π}{6}\right)^2=\sin\left(\frac{π}{5}\right)^2$

This makes the triangle a pythagorean triple, simplifying the problem greatly.

$A=\frac{1}{2}bh$

$A=\frac{1}{2}\left(2\sin\left(\frac{π}{6}\right)\right)\left(2\sin\left(\frac{π}{10}\right)\right)$

$A=\frac{1}{2}\left(1\right)\left(\frac{\sqrt{5}-1}{2}\right)$

$A=\frac{\sqrt{5}-1}{4}$

Therefore our final answer must be $5+1+4=\boxed{10}$ .

This is a proof sketch

Using basic trig we find the side length of an n-gon to be $2\sin\left(\frac{π}{n}\right)$

So our 3 sides are of length: $2\sin\left(\frac{π}{5}\right)=\frac{\sqrt{10-\sqrt{20}}}{2}, \ \ 2\sin\left(\frac{π}{6}\right)=1, \ \ 2\sin\left(\frac{π}{10}\right)=\frac{\sqrt{5}-1}{2}$

We then use this version of herons formula for ease:

$\large A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$

$=\frac{1}{4}\sqrt{\left(\frac{5-\sqrt{5}}{2}+1+\frac{3-\sqrt{5}}{2}\right)^2-2\left(\frac{15-5\sqrt{5}}{2}+1+\frac{7-3\sqrt{5}}{2}\right)}$

$\large =\frac{1}{4}\sqrt{ \left(5-\sqrt{5}\right)^2- 24+8\sqrt{5}}=\frac{1}{4}\sqrt{6-2\sqrt{5}}=\frac{1}{4}\sqrt{5-2\sqrt{5}+1}$

$\large =\frac{1}{4}\sqrt{\left(\sqrt{5}-1\right)^2}=\boxed{\frac{\sqrt{5}-1}{4}}$

$\therefore a+b+c=5+1+4=\boxed{10}$

A little more intuitive approach to come up with the (true) Pythagorean identity:

First calculate all the sides, we have: side of (pentagon, hexagon, decagon) = $(a,b,c)=(\sqrt{\frac{5}{2} - \frac{\sqrt{5}}{2}},1,\sqrt{\frac{3}{2} - \frac{\sqrt{5}}{2}})$ . Note: The formula for the sides is in $cosine$ rather than $sine$ , so it's $side = r\sqrt{2-2\cos{\frac{360^\circ}{n}}}$

You will automatically think of Heron's formula for area given the 3 known sides. However, you will see that without knowing the alternate form of Heron's formula, it's gonna be a gore manipulation if not impossible. Anyhow, it's gonna be tedious. So let's take a closer look at the sides, you will see that $a^2 - 1 = b^2$ and $1$ can be replaced by $c^2$ . Thus, $a^2 = b^2 + c^2$ and $\angle(b,c) = 90^\circ$ and hence a right triangle.