At time t = 0 a point starts to move clockwise around a regular n-gon from each vertex. Each of the n points moves at constant speed. At time T all the points reach vertex A simultaneously. Show that they will never all be simultaneously at any other vertex. Can they be together again at vertex A?
Solution
Solution by Demetres Christofides
Assume the sides of the n-gon are length 1. So in time T each point moves an integral distance. So in time nT each point returns to its original position. Suppose that T' is the first time (after the start) at which all points are at vertices. If all the points are at vertices at time T", then we may put T" = kT' + t, for some positive integer k and some 0 ≤ t < T'. But then each point travels an integral distance in time t < T', so we must have t = 0. Hence, in particular, T = mT' for some positive integer m. Clearly m ≤ n.
Label the points P0 = A, P1, P2, ... , Pn-1 moving anticlockwise around the n-gon. Suppose that at time T', point Pi has moved a distance ai mod n. Then we have m ai = i mod n. In particular, m a1 = 1 mod n, so m must be coprime to n. But m a0 = 0 mod n, so a0 = 0 mod n. Hence at any time T" at which all the points are at vertices, P0 is at vertex A, so they can never be together at some vertex distinct from A.
The motion is periodic with period nT, so they are regularly together at vertex A.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002