Tridecagonal Triangle

The angles are $A = \tfrac{6}{13}\pi$ , $B = \tfrac{5}{13}\pi$ and $C = \tfrac{2}{13}\pi$ . If we write $z = e^{\frac{2\pi i}{13}}$ , then $\cos A - \cos B + \cos C \; =\; \tfrac12\big(z^3 + z^{-3} + z^9 + z^{-9} + z + z^{-1}\big) \; = \; \tfrac12(z + z^{-1} + z^3 + z^{-3} + z^4 + z^{-4}\big)$ Note that $z$ is a root of $X^{13}-1=0$ which is not equal to $1$ , and hence $0 \; = \; z^{-6}\sum_{j=0}^{12}z^j \; = \; z^6 +z^{-6} + z^5 + z^{-5} + z^4 + z^{-4} + z^3 + z^{-3} + z^2 + z^{-2} + z + z^{-1} + 1$ If we define $\alpha \; = \; z + z^{-1} + z^3 + z^{-3} + z^4 + z^{-4} \hspace{2cm} \beta \; = \; z^2 + z^{-2} + z^5 + z^{-5} + z^6 + z^{-6}$ then it is easy to show that $\alpha + \beta \; = \; -1 \hspace{2cm} \alpha\beta \; = \; -3$ and hence $\alpha,\beta$ are the roots of the quadratic $X^2 + X - 3 = 0$ , and moreover $\alpha > 0 > \beta$ . Thus we deduce that $\alpha = \tfrac12(\sqrt{13}-1)$ , and hence $\cos A - \cos B + \cos C \; = \; \tfrac12\alpha \; = \; \tfrac14(\sqrt{13}-1)$ making the answer $13 + 1 + 4 = \boxed{18}$ .