60th Putnam 1999 Problem B1 The triangle ABC has AC = 1, angle ACB = 90 o , and angle BAC = φ. D is the point between A and B such that AD = 1. E is the point between B and C such that angle EDC = φ. The perpendicular to BC at E meets AB at F. Find lim φ→0 EF.
Considering ACD, angle ACD = angle ADC = 90 o - φ/2. Hence angle DCE = φ/2. So, using the sine rule on triangle CDE: CE/DE = sin φ/sin(φ/2) which tends to 2. Angle EBD = 90 o - φ, angle EDB = 90 o - φ/2. So DE/EB = cos φ/cos(φ/2) which tends to 1. Hence EB/CE tends to 1/2 or EB/BC tends to 1/3. But EF/AC = EB/BC, so EF tends to AC/3 = 1/3. 60th Putnam 1999 © John Scholes jscholes@kalva.demon.co.uk 14 Dec 1999