IMO 1979 SL 17

Origin: NET

Problem

Inside an equilateral triangle ABC one constructs points P, Q and R such that \angleQAB = \anglePBA = 15◦, \angleRBC = \angleQCB = 20◦, \anglePCA = \angleRAC = 25◦. Determine the angles of triangle PQR.

Solution

Let K, L, and M be intersections of CQ and BR, AR and CP, and AQ and BP, respectively. Let \angleX denote the angle of the hexagon KQMPLR at the vertex X, where X is one of the six points. By an elementary calculation of angles we get \angleK = 140◦, \angleL = 130◦, \angleM = 150◦, \angleP = 100◦, \angleQ = 95◦, \angleR = 105◦. Since \angleKBC = \angleKCB, it follows that K is on the symmetry line of ABC through A. Analogous statements hold for L and M. Let KR and KQ be points symmetric to K with respect to AR and AQ, respectively. Since \angleAKQQ = \angleAKQKR = 70◦and \angleAKRR = \angleAKRKQ = 70◦, it follows that KR, R, Q, and KQ are collinear. Hence \angleQRK

2\angleR −180◦and \angleRQK = 2\angleQ − 180◦. We analogously get \anglePRL = 2\angleR−180◦, \angleRPL

2\angleP − 180◦, \angleQPM = 2\angleP −180◦and \anglePQM = 2\angleQ −180◦. From these formulas we easily get \angleRPQ = 60◦, \angleRQP = 75◦, and \angleQRP = 45◦. A B C K L M P Q R KQ KR 15o 15o 20o 20o 25o 25o