Waiting for 2016 - 3

Let $w=e^{2\pi i/13}$ be a primitive 13th root of unity. Then we know that $\dfrac{1}{2}(w^k+w^{13-k})=\cos\left(\dfrac{2\pi k}{13}\right)$ and $w+w^2+\cdots+w^{12}=-1$ .

So, $S=\dfrac{1}{2}(w+w^{12}+w^3+w^{10}+w^4+w^9)$ . Let's introduce a new variable $T$ such that $S+T=-\dfrac{1}{2}$ with the help of the identity above:

$T=\dfrac{1}{2}(w^2+w^{11}+w^5+w^8+w^6+w^7)$ . We are going to find now $ST$ , which is a very messy product to expand, here it ts (knowing that $w^{13}=1$ ):

$ST=\dfrac{3}{4}(w^{12}+w^{11}+w^{10}+w^9+w^8+w^7+w^6+w^5+w^4+w^3+w^2+w)\\ ST=-\dfrac{3}{4}$

Finally, use the identity $(S-T)^2=(S+T)^2-4ST$ and the fact that $S>0$ and $T<0$ :

$(S-T)^2=\left(-\dfrac{1}{2}\right)^2-4\left(-\dfrac{3}{4}\right)\\ (S-T)^2=\dfrac{13}{4}\\ S-T=\dfrac{\sqrt{13}}{2}$

Hence, $S=\dfrac{(S+T)+(S-T)}{2}=\dfrac{\sqrt{13}-1}{4}$ .

We get $a=13,b=1,c=4$ and $a+b+c=\boxed{18}$ .

Even I solved in the same way,except that after knowing the value of sum and product of S and T,we could get the quadratic equation whose roots are S and T and hence find both S and T.