Solved by Altman [Al63]. The stronger variant that says there is one point which determines at least $\lfloor \frac{n}{2}\rfloor$ distinct distances (see [982]) is still open. Fishburn in fact conjectures that if $R(x)$ counts the number of distinct distances from $x$ then\[\sum_{x\in A}R(x) \geq \binom{n}{2}.\]Szemerédi conjectured a stronger form in which the convexity is replaced by the assumption that there are no three points on a line - see [1082].

See also [660].

View the LaTeX source

This page was last edited 19 October 2025.