10th Irish 1997 Problem A2 ABC is an equilateral triangle. M is a point inside the triangle. D, E, F are the feet of the perpendiculars from M to BC, CA, AB. Find the locus of M such that ∠FDE = 90 o .
MDBF is cyclic, so ∠MDF = ∠MBF = 60 o - ∠MBC. Similarly, ∠MDE = 60 o - ∠MCB. Hence ∠FDE = 120 o - (∠MCB + ∠MBC). So (∠MCB + ∠MBC) = 30 o . Hence ∠BMC = 150 o . So the locus is just the arc of the circle through B and C such that ∠BMC = 150 o . 10th Irish 1997 © John Scholes jscholes@kalva.demon.co.uk 21 Dec 2003 Last updated/corrected 21 Dec 03