Let n be an integer greater than or equal to 3. Prove that there is a set of n points in the plane such that the distance between any two points is irrational and each set of 3 points determines a non-degenerate triangle with rational area.
Solution
Let xn be the point with coordinates (n, n2) for n = 1, 2, 3, ... . We show that the distance between any two points is irrational and that the triangle determined by any 3 points has non-zero rational area.
Take n > m. |xn - xm| is the hypoteneuse of a triangle with sides n - m and n2 - m2 = (n - m)(n + m). So |xn - xm| = (n - m)√(1 + (n+m)2). Now (n + m)2 < (n + m)2 + 1 < (n + m + 1)2 = (n + m)2 + 1 + 2(n + m), so (n + m)2 + 1 is not a perfect square. Hence its square root is irrational. [For this we may use the classical argument. Let N' be a non-square and suppose √N' is rational. Since N' is a non-square we must be able to find a prime p such that p2a+1 divides N' but p2a+2 does not divide N' for some a ≥ 0. Define N = N'/p2a. Then √N = (√N')/pa, which is also rational. So we have a prime p such that p divides N, but p2 does not divide N. Take √N = r/s with r and s relatively prime. So s2N = r2. Now p must divide r, hence p2 divides r2 and so p divides s2. Hence p divides s. So r and s have a common factor. Contradiction. Hence non-squares have irrational square roots.]
Now take a < b < c. Let B be the point (b, a2), C the point (c, a2), and D the point (c, b2). Area xaxbxc = area xaxcC - area xaxbB - area xbxcD - area xbDCB = (c - a)(c2 - a2)/2 - (b - a)(b2 - a2)/2 - (c - b)(c2 - b2)/2 - (c - b)(b2 - a2) which is rational.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
18 Nov 1998