14th APMO 2002 Problem 3 ABC is an equilateral triangle. M is the midpoint of AC and N is the midpoint of AB. P lies on the segment MC, and Q lies on the segment NB. R is the orthocenter of ABP and S is the orthocenter of ACQ. The lines BP and CQ meet at T. Find all possible values for angle BCQ such that RST is equilateral.
15 o . Solution Suppose CP  NS. Hence (treating BC as horizontal), R is below S. But T must be to the right of the midpoint of BC. Hence T is to the right of the perpendicular bisector of RS, so RST cannot be equilateral. Contradiction. Similarly if CP > BQ. So CP = BQ. Let L be the midpoint of BC. Put ∠CBP = x and ∠RAM = y. So RM = AM tan y, TL = BL tan x = AM tan x. But ∠APB = 60 o + x, so y = 30 o - x. So if x ≠ 15 o , then TL ≠ RM. However, RST equilateral implies TL = RM, so x and hence also ∠BCQ = 15 o . 14th APMO 2002 © John Scholes jscholes@kalva.demon.co.uk 28 Nov 2003 Last corrected/updated 28 Nov 03