8th Balkan 1991 Problem 1 The circumcircle of the acute-angled triangle ABC has center O. M lies on the minor arc AB. The line through M perpendicular to OA cuts AB at K and AC at L. The line through M perpendicular to OB cuts AB at N and BC at P. MN = KL. Find angle MLP in terms of angles A, B and C. Solution
: angle C. Solution by Bazavan Eduard ∠MKN = ∠AKL (opposite angles) = 90 o - ∠OAB = 1/2 ∠AOB = ∠C. A similar argument shows that ∠MNK = ∠C, so MNK is isosceles and MN = MK. Also triangles AKL and ACB are similar, so AK/KL = AC/CB (*). Extend ML to meet the circumcircle again at R. Then since OA is perpendicular to MR, we have AM = AR. Triangles AKR and MKB are similar, because ∠AKR = ∠MKB (opposite angles) and ∠ARK = ∠ARM (same angle) = ∠ABM (ARBM cyclic) = ∠MBK (same angle). So AK/AR = MK/MB. Hence MK/AK = MB/AR = MB/AM. Multiplying by (*) we get MK/KL = (AC/BC) (BM/AM) (**). But MK = MN = KL. Hence (AC/BC) (BM/AM) = 1. But a similar argument to that used to show (**) gives that MN/NP = (BC/AC) (AM/BM). Hence MN = NP. So N is the midpoint of MP and K is the midpoint of ML. Hence ∠MLP = ∠MKN = ∠C. 8th Balkan 1991 © John Scholes jscholes@kalva.demon.co.uk 4 Sep 2002