Given two points A and B, take C on the perpendicular bisector of AB. Define the sequence C1, C2, C3, ... as follows. C1 = C. If Cn is not on AB, then Cn+1 is the circumcenter of the triangle ABCn. If Cn lies on AB, then Cn+1 is not defined and the sequence terminates. Find all points C such that the sequence is periodic from some point on.
Solution
Answer: any C such that ∠ACB = 180o r/s, with r and s relatively prime integers and s not a power of 2.
Let ∠ACnB = xn, where the angle is measured clockwise, so that xn is positive on one side of AB and negative on the other side. Then xn uniquely identifies Cn on the perpendicular bisector.
We have xn+1 = 2xn. To make this work in all cases we have to take it mod 180o (so that if ACnB is obtuse, then Cn+1 lies on the other side of AB). If xn is eventually periodic then xm+1 = xn+1, for some n > m, so (2n - 2m)x1 = 0 mod 180. Hence x1 = 180 r/s for some relatively prime integers r, s. Also s cannot be a power of 2 for then we would have xk = 180r for some k, in which case the sequence would terminate rather than be periodic.
Conversely, suppose x1 = 180 r/s, with r and s relatively prime and s not a power of 2. Then xn+1 = 180 2nr/s cannot be 0 mod 180, so the sequence does not terminate. Put s = 2bc, with c odd. Let d = φ(c), where φ(m) is Euler's phi function, so that 2d = 1 mod c. Then xb+1 = 180 r/c mod 180 and 2b+d = 2b mod c, so xb+d+1 = 180 r/c mod 180. Hence the sequence is periodic.
© John Scholes
jscholes@kalva.demon.co.uk
24 Oct 2002