35th Swedish 1995 Problem 5 A, B, C, D are points on a circle radius r in that order. AB = BC = CD = s < r and AD = s+r. Find the angles of the quadrilateral.
∠A = ∠D = 72 o , ∠B = ∠C = 108 o Solution Let the center of the circle be O. Let ∠AOB = 2θ. Then s = 2r sin θ, r+s = 2r sin 3θ. We have sin 3θ = 3 sin θ - 4 sin 3 θ, so putting k = sin θ, we have 8k 3 - 4k + 1 = 0. Factorising, (2k-1)(4k 2 +2k-1) = 0. k = 1/2 gives s = r, but we are given s  0, hence k = (√5 - 1)/4. If you are lucky, you will recognize that as sin 18 o ! 35th Swedish 1995 © John Scholes jscholes@kalva.demon.co.uk 16 January 2004 Last corrected/updated 16 Jan 04