Solved by Altman [Al63]. The stronger variant that says there is one point which determines at least $\lfloor \frac{n+1}{2}\rfloor$ distinct distances 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 (see [Er97e]) that this stronger variant remains true if we only assume that no three points are on a line, and proved this with the weaker bound of $n/3$.

See also [660].