26th Putnam 1965 Problem A1 The triangle ABC has an obtuse angle at B, and angle A is less than angle C. The external angle bisector at A meets the line BC at D, and the external angle bisector at B meets the line AC at E. Also, BA = AD = BE. Find angle A. Solution
: 12 degrees. Let angle BAC = k. Then since BA = BE, angle BEA = k. Take B' on BA the opposite side of B to A. Then angle B'BE = 2k. Angle B'BC is bisected by BE, so angle CBE = 2k. Hence angle ACB = 3k. So angle DBA = 4k. But AD = BA, so angle BDA = 4k. But AD is the exterior bisector, so angle BAD = 90 - k/2. The angles in BAD must sum to 180 deg, so k = 12 deg. 26th Putnam 1965 © John Scholes jscholes@kalva.demon.co.uk 25 Mar 2002