14th Iberoamerican 1999 Problem B3 Given two points A and B, take C on the perpendicular bisector of AB. Define the sequence C 1 , C 2 , C 3 , ... as follows. C 1 = C. If C n is not on AB, then C n+1 is the circumcenter of the triangle ABC n . If C n lies on AB, then C n+1 is not defined and the sequence terminates. Find all points C such that the sequence is periodic from some point on. Solution
: any C such that ∠ACB = 180 o r/s, with r and s relatively prime integers and s not a power of 2. Let ∠AC n B = x n , where the angle is measured clockwise, so that x n is positive on one side of AB and negative on the other side. Then x n uniquely identifies C n on the perpendicular bisector. We have x n+1 = 2x n . To make this work in all cases we have to take it mod 180 o (so that if AC n B is obtuse, then C n+1 lies on the other side of AB). If x n is eventually periodic then x m+1 = x n+1 , for some n > m, so (2 n - 2 m )x 1 = 0 mod 180. Hence x 1 = 180 r/s for some relatively prime integers r, s. Also s cannot be a power of 2 for then we would have x k = 180r for some k, in which case the sequence would terminate rather than be periodic. Conversely, suppose x 1 = 180 r/s, with r and s relatively prime and s not a power of 2. Then x n+1 = 180 2 n r/s cannot be 0 mod 180, so the sequence does not terminate. Put s = 2 b c, with c odd. Let d = φ(c), where φ(m) is Euler's phi function, so that 2 d = 1 mod c. Then x b+1 = 180 r/c mod 180 and 2 b+d = 2 b mod c, so x b+d+1 = 180 r/c mod 180. Hence the sequence is periodic. 14th Ibero 1999 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 2002