JEE ADVANCED 2016 (9)

With a suitable coordinate system, $A,B,C$ have coordinates $(16,0)\,,\,(20\cos\theta,20\sin\theta)\,,\,(0,0)$ , where $\theta = \angle ACB$ . The point of intersection of the perpendicular bisector of $AC$ and the altitude from $A$ is $X\; (8,8\cot\theta)$ while the incentre of $ABC$ is the point $I$ with coordinates $\tfrac{1}{36+L}\big[(20(16,0) + 16(20\cos\theta,20\sin\theta) + L(0,0)\big] \; =\; \tfrac{320}{36+L}(1+\cos\theta,\sin\theta)$

where $L = AB$ . Since $XI$ must be perpendicular to $AB$ we deduce, taking the scalar product of $\overrightarrow{XI}$ and $\overrightarrow{AB}$ , that $L \; = \; \frac{4(40\cos\theta - 23)}{2 - 5\cos\theta}$ Using the Cosine Rule, we must have $L^2 = 20^2 + 16^2 - 640\cos\theta$ . This leads to a cubic equation for $\cos\theta$ , which gives $\cos\theta \; = \; -1\;,\;\tfrac{1}{80}(49 - \sqrt{65})\;,\; \tfrac{1}{80}(49 + \sqrt{65})$ and the corresponding values of $L$ are $L \; = \; -36\;,\; 2(1 + \sqrt{65}) \;,\; 2(1- \sqrt{65})$ Since $L$ must be positive, only the middle of these three solutions is possible, and so $AB = L = 2 + \sqrt{260}$ , making the answer $2\times100 + 260 = \boxed{460}$