Find The Length

Geometry Level 5

Let △ A B C \triangle ABC be a triangle with side lengths B C = 9 , C A = 8 , A B = 6. BC = 9, CA= 8, AB= 6. Let O O and I I be the circumcenter and incenter of △ A B C \triangle ABC respectively. The incircle of △ A B C \triangle ABC touches B C BC at point D . D. Let X X be the second point of intersection of line A O AO and the circumcircle of △ A I D \triangle AID (apart from A A ). Given that A X = a b AX = \sqrt{\dfrac{a}{b}} for some coprime positive integers a , b , a,b, find a + b . a+b.

Details and assumptions

  • The image shown is not accurate.

  • This problem is adapted from a Russia 10th grade geometry problem.


The answer is 477.

Sreejato Bhattacharya by Sreejato Bhattacharya , 22, India

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1 solution

Since A I AI bisects ∠ B A C \angle BAC and ∠ A I D = 18 0 ∘ − ∠ X A D , \angle AID = 180^{\circ} - \angle XAD, A X AX and I D ID are antiparallel with respect to ∠ B A C , \angle BAC, which implies A X ∥ D I . AX \parallel DI. All cyclic trapezoids are isoceles, so we must have A X = I D . AX= ID. We just need to find the inradius of △ A B C , \triangle ABC, which is equal to 1 2 ( 9 + 8 − 6 ) ( 9 + 6 − 8 ) ( 8 + 6 − 9 ) 8 + 6 + 9 = 385 92 . \dfrac{1}{2} \sqrt{ \dfrac{(9+8-6)(9+6-8)(8+6-9)}{8+6+9}} = \sqrt{\dfrac{385}{92}}. Hence, a = 385 , b = 92 , a= 385, b= 92, and a + b = 385 + 92 = 477 . a+b= 385+92 = \boxed{477}.

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Can you please elaborate your solution

Ronak Agarwal - 6 years, 12 months ago

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Ok i will try

Vivek Singh - 6 years, 6 months ago

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