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2023 MOAA Team P15

Source:

October 14, 2023
MOAA 2023

Problem Statement

Triangle ABCABC has circumcircle ω\omega. Let DD be the foot of the altitude from AA to BCBC and let ADAD intersect ω\omega at E≠AE \neq A. Let MM be the midpoint of ADAD. If ∠BMC=90∘\angle{BMC} = 90^\circ, AB=9AB = 9 and AE=10AE = 10, the area of △ABC\triangle{ABC} can be expressed in the form abc\frac{a\sqrt{b}}{c} where a,b,ca,b,c are positive integers and bb is square-free. Find a+b+ca+b+c.
Proposed by Andy Xu
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