31st IMO 1990 shortlist Problem 9 ABC is a triangle with incenter I. M is the midpoint of AC and N is the midpoint of AB. The lines NI and AC meet at X, and the lines MI and AB meet at Y. The triangles AXY and ABC have the same area. Find angle A. Solution Solution by Vivek Kumar Mehra
: A = 60 o . By similar triangles, h/r = 1 + z/k. But AI bisects angle NAX, so z/k = c/(2x). Hence h = r(1 + c/(2x) ). Similarly, if h' is the length of the perpendicular from M to AB, we have h' = r(1 + b/(2y) ). Hence (h/r - 1)(h'/r - 1) = 1/4 (*) We are given that xy = bc. Also r = (2 area ABC)/(a + b + c) = (bc sin A)/(a + b + c) But 2h = c sin A, and 2h' = b sin A. So substituting in (*) and simplifying we get a 2 = b 2 + c 2 - bc. Comparing with the cosine formula, we get A = 60 o . 31st IMO shortlist 1990 © John Scholes jscholes@kalva.demon.co.uk 17 Feb 2003 Last updated/corrected 17 Feb 2003