Solve it in Friday 13th - part 1

Since $z^{13} = e^{\frac{2\pi}{13}i} = 1$ are the 13 $^{th}$ roots of unity. By Argand's diagram, we can see that:

$\cos{\frac{2\pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + \cos{\frac{10\pi}{13}} + \cos{\frac{12\pi}{13}} = -\frac{1}{2} \\ \cos{\frac{2\pi}{13}} + 2\cos^2{\frac{2\pi}{13}} - 1 + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + 2\cos^2{\frac{\color{#D61F06}{5}\pi}{13}} - 1 + 2\cos^2{\frac{6\pi}{13}} - 1 = -\frac{1}{2} \\ \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + 2\left( \cos^2{\frac{2\pi}{13}} + \cos^2{\frac{6\pi}{13}} + \cos^2{\frac{\color{#D61F06}{8}\pi}{13}} \right) - 3 = -\frac{1}{2} \\ \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + 2\left( \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} \right)^2 \\ \quad \quad - 4 (\color{#3D99F6}{\cos{\frac{2\pi}{13}}\cos{\frac{6\pi}{13}} + \cos{\frac{2\pi}{13}} \cos{\frac{8\pi}{13}} + \cos{\frac{6\pi}{13}} \cos{\frac{8\pi}{13}}}) = \frac{5}{2} \quad \color{#3D99F6}{[\text{Note 1}]} \\ \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + 2\left( \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} \right)^2 - 4 (\color{#3D99F6}{-\frac{1}{4}}) = \frac{5}{2} \\ 2\left( \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} \right)^2 + \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} - \frac{3}{2} = 0 \\ 4\left( \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} \right)^2 + 2(\cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}}) - 3 = 0$

$\begin{aligned} \Rightarrow \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} & = \frac {-2 \pm \sqrt{4+48}}{8} \\ & = \frac{\sqrt{13}-1}{4} \quad \color{#3D99F6}{[\text{Note 2}]} \end{aligned}$

$\Rightarrow a + b + c = 13+1+4 = \boxed{18}$

$\color{#3D99F6}{\text{Note 1: }}$ See solution of Part 2 .

$\color{#3D99F6}{\text{Note 2: }}$ $\cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} = \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} - \cos{\frac{5\pi}{13}}$ . We know that $\cos{\frac{2\pi}{13}} > \cos{\frac{5\pi}{13}} > \cos{\frac{6\pi}{13}} > 0 \quad \Rightarrow \cos{\frac{2\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} > 0$ .

Let $a=\cos{\dfrac{2\pi}{13}} + \cos{\dfrac{6\pi}{13}} + \cos{\dfrac{8\pi}{13}}$ and $b=\cos{\dfrac{4\pi}{13}} + \cos{\dfrac{10\pi}{13}} + \cos{\dfrac{12\pi}{13}}$ .

Also, let $w=e^{\frac{2\pi i}{13}}$ , so using $w^k+w^{13-k}=2\cos\left(\dfrac{2\pi k}{13}\right)$ , we get:

$a=\dfrac{w+w^{12}+w^3+w^{10}+w^4+w^9}{2} \\ b=\dfrac{w^2+w^{11}+w^5+w^8+w^6+w^7}{2}$ .

Then, try to find $a+b$ and $ab$ , the first was easy using $w+w^2+w^3+\cdots+w^{12}=-1$ , which follows inmediatly from factoring $w^{13}=1$ :

$a+b=-\dfrac{1}{2}$

But to find $ab$ let's find the whole product and use the fact that $w^{13}=1$ , we end with $ab=-\frac{3}{4}$ .

Finally, by the fact that $a>0$ and $b<0$ , use Vieta's formulas to find that $a=\dfrac{-1+\sqrt{13}}{4}$ and $b=\dfrac{-1-\sqrt{13}}{4}$ .

So, $\cos\left(\dfrac{2\pi}{13}\right)+\cos\left(\dfrac{6\pi}{13}\right)+\cos\left(\dfrac{8\pi}{13}\right)=\dfrac{-1+\sqrt{13}}{4}$

On comparing we get $\alpha=13,\beta=1,\gamma=4$ , thus the final answer is $13+1+4=\boxed{18}$ .