IMO 1975 · Problem 5

Determine, with proof, whether or not one can find 1975 points on the circumference of a circle with unit radius such that the distance between any two of them is a rational number.

Official text of the problem. Official problem paper (PDF)

Answer

Yes, such 1975 distinct points exist.

Solution

For each integer \(k=1,\ldots,1975\), put \[ a_k=\frac{k^2-1}{k^2+1}, \qquad b_k=\frac{2k}{k^2+1}, \] and define \[ P_k=(a_k^2-b_k^2,\;2a_kb_k). \] We shall prove that these points have the required properties.

First, \[ a_k^2+b_k^2 =\frac{(k^2-1)^2+4k^2}{(k^2+1)^2}=1. \] Consequently, \[ (a_k^2-b_k^2)^2+(2a_kb_k)^2 =(a_k^2+b_k^2)^2=1, \] so every \(P_k\) lies on the unit circle.

For any \(i,j\), expanding the squared distance and using \(a_i^2+b_i^2=a_j^2+b_j^2=1\), we obtain \[ \begin{aligned} |P_iP_j|^2 &=2-2\bigl((a_i^2-b_i^2)(a_j^2-b_j^2) +4a_ib_ia_jb_j\bigr)\\ &=4(a_ib_j-a_jb_i)^2. \end{aligned} \] Moreover, \[ a_ib_j-a_jb_i =\frac{2j(i^2-1)-2i(j^2-1)} {(i^2+1)(j^2+1)} =\frac{2(i-j)(ij+1)} {(i^2+1)(j^2+1)}. \] It follows that \[ |P_iP_j| =\frac{4|i-j|(ij+1)} {(i^2+1)(j^2+1)}. \] This is rational for every pair \(i,j\), and it is strictly positive whenever \(i\ne j\). Thus the points are distinct and all their pairwise distances are rational, as required.

The solution was checked against John Scholes' IMO solution archive (Kalva).