SFFT with a Triangle?

The key to solving this problem is to find $cos(C)$ , as once this is done, we will have enough information to determine all three side-lengths of this problem and swiftly use Heron's Formula to finish it off. Start by neatly drawing a triangle and label the points A, B, and C, as per the problem. Since we know that that $cos(C) = AB/AC$ , using the Law of Cosines is clearly the best approach. Let $AB = c$ and $AC = b$ . We then have $c^2 = a^2 + b^2 - 2abcos(C)$ . Furthermore, we know that $a = (b+c)/2$ from the problem. Substituting everything in, we are now ready to do some tedious (but extremely useful) algebra. So $c^2 = ((b+c)/2)^2 + b^2 - 2(b)((b+c)/2)cos(c)$ . Carefully rearranging everything into one side and substituting $cos(C) = AB/AC = c/b$ gives us a multivariable quadratic polynomial equation, namely $7c^2 - 5b^2 + 2bc = 0$ . Now we can clearly see by inspection that $c=-b$ is a root of this quadratic, so utilizing Simon's Favorite Factoring Trick (SFFT), we can factor this multivariable polynomial as $7(b+c)(c-(5b/7)) = 0$ . Hence we have $c=-b$ and $c=5b/7$ . However, if we plug in the former solution, we get $cos(C) = -1$ which implies the triangle is degenerate (and in fact a line). Hence we eliminate this extraneous solution and conclude that $c=5b/7$ , thus $cos(C) = 5/7$ . Now, we know that $c = 5$ and $b = 7$ . Since $a$ is defined to be the average of $b$ and $c$ , we have $a = (5+7)/2 = 6$ . Finally, the semi-perimeter $s$ of this triangle is $(6+7+5)/2 = 9$ , so plugging into Heron's Formula, denoted by $\sqrt{(s)(s-a)(s-b)(s-c)}$ , gives us $\sqrt{(9)(9-6)(9-7)(9-5)}$ which simplifies to $\sqrt{(9)(3)(2)(4)}$ which simplifies to $\sqrt{216}$ which finally simplifies to $6\sqrt{6}$ . Thus, $X = 6$ and $Y = 6$ , and $6 + 6 = 12$ , as desired.