10th Irish 1997 Problem B2 The quadrilateral ABCD has an inscribed circle. ∠A = ∠B = 120 o , ∠C = 30 o and BC = 1. Find AD.
½(√3 - 1) Solution Let the sides AB, BC, CD, DA touch the circle at P, Q, R, S. Let the center of the circle be O. Evidently ∠D - 90o and ORDS is a square. Let the inradius be r. Considering the triangle OAS, we see that AS = r/ √3. Note that BQ = BP = AP = AS. Using the formula tan 2k = 2 tan k/(1 - tan 2 k) and the fact that tan 30 o = 1/√3, we find that tan 15 o = 2 - √3. So considering triangle COQ, we find that CQ = r(2 + √3). So 1 = BC = BQ + CQ = r/√3 + r(2+√3). Hence r = (2√3-3)/2. Hence AD = AS + DS = r/√3 + r = ½(√3 - 1). 10th Irish 1997 © John Scholes jscholes@kalva.demon.co.uk 17 Dec 2003 Last updated/corrected 17 Dec 03