### 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.

**Solution**

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