Let be a triangle with and , and let be its incenter. If the altitude from to , the perpendicular bisector of , and the line through I perpendicular to intersect at a common point, then the length can be written as for positive integers . What is
Try my set JEE ADVANCED 2016
This problem is from the Spring OMO 2016
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
With a suitable coordinate system, A , B , C have coordinates ( 1 6 , 0 ) , ( 2 0 cos θ , 2 0 sin θ ) , ( 0 , 0 ) , where θ = ∠ A C B . The point of intersection of the perpendicular bisector of A C and the altitude from A is X ( 8 , 8 cot θ ) while the incentre of A B C is the point I with coordinates 3 6 + L 1 [ ( 2 0 ( 1 6 , 0 ) + 1 6 ( 2 0 cos θ , 2 0 sin θ ) + L ( 0 , 0 ) ] = 3 6 + L 3 2 0 ( 1 + cos θ , sin θ )
where L = A B . Since X I must be perpendicular to A B we deduce, taking the scalar product of X I and A B , that L = 2 − 5 cos θ 4 ( 4 0 cos θ − 2 3 ) Using the Cosine Rule, we must have L 2 = 2 0 2 + 1 6 2 − 6 4 0 cos θ . This leads to a cubic equation for cos θ , which gives cos θ = − 1 , 8 0 1 ( 4 9 − 6 5 ) , 8 0 1 ( 4 9 + 6 5 ) and the corresponding values of L are L = − 3 6 , 2 ( 1 + 6 5 ) , 2 ( 1 − 6 5 ) Since L must be positive, only the middle of these three solutions is possible, and so A B = L = 2 + 2 6 0 , making the answer 2 × 1 0 0 + 2 6 0 = 4 6 0