3rd Centromerican 2001 Problem A2 C and D are points on the circle diameter AB such that ∠AQB = 2 ∠COD. The tangents at C and D meet at P. The circle has radius 1. Find the distance of P from its center.
2/√3 Solution ∠AQB = ∠ACB + ∠CBQ = 90 o + ∠CBQ = 90 o + ½∠COD = 90 o + ¼∠AQB. Hence ∠AQB = 120 o , and ∠COD = 60 o . So OP = 1/cos 30 o = 2/√3. 3rd OMCC 2001 © John Scholes jscholes@kalva.demon.co.uk 30 January 2004 Last corrected/updated 30 Jan 04