JEE ADVANCED 2016 (9)

Geometry Level 5

Let A B C ABC be a triangle with B C = 20 BC = 20 and C A = 16 CA = 16 , and let I I be its incenter. If the altitude from A A to B C BC , the perpendicular bisector of A C AC , and the line through I perpendicular to A B AB intersect at a common point, then the length A B AB can be written as m + n m+\sqrt{n} for positive integers m , n m,n . What is 100 m + n ? 100m+n?

Try my set JEE ADVANCED 2016


This problem is from the Spring OMO 2016

460 420 480 None of these 440

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1 solution

Mark Hennings
Mar 20, 2018

With a suitable coordinate system, A , B , C A,B,C have coordinates ( 16 , 0 ) , ( 20 cos θ , 20 sin θ ) , ( 0 , 0 ) (16,0)\,,\,(20\cos\theta,20\sin\theta)\,,\,(0,0) , where θ = A C B \theta = \angle ACB . The point of intersection of the perpendicular bisector of A C AC and the altitude from A A is X ( 8 , 8 cot θ ) X\; (8,8\cot\theta) while the incentre of A B C ABC is the point I I with coordinates 1 36 + L [ ( 20 ( 16 , 0 ) + 16 ( 20 cos θ , 20 sin θ ) + L ( 0 , 0 ) ] = 320 36 + L ( 1 + cos θ , sin θ ) \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 = A B L = AB . Since X I XI must be perpendicular to A B AB we deduce, taking the scalar product of X I \overrightarrow{XI} and A B \overrightarrow{AB} , that L = 4 ( 40 cos θ 23 ) 2 5 cos θ L \; = \; \frac{4(40\cos\theta - 23)}{2 - 5\cos\theta} Using the Cosine Rule, we must have L 2 = 2 0 2 + 1 6 2 640 cos θ L^2 = 20^2 + 16^2 - 640\cos\theta . This leads to a cubic equation for cos θ \cos\theta , which gives cos θ = 1 , 1 80 ( 49 65 ) , 1 80 ( 49 + 65 ) \cos\theta \; = \; -1\;,\;\tfrac{1}{80}(49 - \sqrt{65})\;,\; \tfrac{1}{80}(49 + \sqrt{65}) and the corresponding values of L L are L = 36 , 2 ( 1 + 65 ) , 2 ( 1 65 ) L \; = \; -36\;,\; 2(1 + \sqrt{65}) \;,\; 2(1- \sqrt{65}) Since L L must be positive, only the middle of these three solutions is possible, and so A B = L = 2 + 260 AB = L = 2 + \sqrt{260} , making the answer 2 × 100 + 260 = 460 2\times100 + 260 = \boxed{460}

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