Problem Statement
Let $x_1,\ldots,x_n\in \mathbb{R}^2$ and let $R(x_i)=\#\{ \lvert x_j-x_i\rvert : j\neq i\}$, where the points are ordered such that\[R(x_1)\leq \cdots \leq R(x_n).\]Let $\alpha_k$ be minimal such that, for all large enough $n$, there exists a set of $n$ points with $R(x_k)<\alpha_kn^{1/2}$. Is it true that $\alpha_k\to \infty$ as $k\to \infty$?
Categories:
Geometry Distances
Progress
It is trivial that $R(x_1)=1$ is possible, and that $R(x_2) \ll n^{1/2}$ is also possible, but we always have\[R(x_1)R(x_2)\gg n.\]Erdős originally conjectured that $R(x_3)/n^{1/2}\to \infty$ as $n\to \infty$, but Elekes proved that for every $k$ and $n$ sufficiently large there exists some set of $n$ points with $R(x_k)\ll_k n^{1/2}$.Source: erdosproblems.com/652 | Last verified: January 16, 2026