Can you find infinitely many points in the plane such that the distance between any two is rational and no three are collinear?
Solution
Answer: yes.
Let θ = cos-13/5. Take a circle center O radius 1 and a point X on the circle. Take Pn on the circle such that angle XOPn = 2nθ. We establish (A) that the Pn are all distinct and (B) that the distances PmPn are all rational.
We establish first that we can express 2 cos nx as a polynomial of degree n in (2 cos x) with integer coefficients and leading coefficient 1. For example, 2 cos 2x = (2 cos x)2 - 1, 2 cos 3x = (2 cos x)3 - 3 (2 cos x). We proceed by induction. Let the polynomial be pn(2 cos x). We have that p1(y) = y and p2(y) = y2 - 1. Suppose we have found pm for m ≤ n. Now cos(n+1)x = cos nx cos x - sin nx sin x, and cos(n-1)x = cos nx cos x + sin nx sin x, so cos(n+1)x = 2 cos x cos nx - cos(n-1)x. Hence pn+1(y) = y pn(y) - pn-1(y). Hence the result is also true for n+1.
It follows that (1) if cos x is rational, then so is cos nx, and (2) if cos x is rational, then x/π is irrational. To see (2), suppose that x/π = m/n, with m and n integers. Then nx is a multiple of π and hence cos nx = 0, so pn(2 cos x) = 0. Now we may write pn(y) = yn + an-1yn-1 + ... + a0. Now if also cos x = r/s, with r and s relatively prime integers, then we have, pn(2 cos x) = rn + an-1s rn-1 + ... + a0sn = 0. But now s divides all terms except the first. Contradiction.
Thus we cannot have cos mθ = cos nθ for any distinct integers m, n, for then θ/π would be rational as well as cos θ. So we have established (A).
We have also established that all cos nθ are rational. But since sin(n+1)x = sin nx cos x + cos nx sin x and sin θ = 4/5, it is a trivial induction that all sin nθ are also rational. Now PmPn = 2 |sin(m - n)θ|, so all the distances PmPn are rational, thus establishing (B).
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002