Problem Statement
Let $f(n)$ be maximal such that there exists a set $A$ of $n$ points in $\mathbb{R}^2$ in which every $x\in A$ has at least $f(n)$ points in $A$ equidistant from $x$.
Is it true that $f(n)\leq n^{o(1)}$? Or even $f(n) < n^{O(1/\log\log n)}$?
Is it true that $f(n)\leq n^{o(1)}$? Or even $f(n) < n^{O(1/\log\log n)}$?
Categories:
Geometry Distances
Progress
This is a stronger form of the unit distance conjecture (see [90]).The set of lattice points imply $f(n) > n^{c/\log\log n}$ for some constant $c>0$. Erdős offered \$500 for a proof that $f(n) \leq n^{o(1)}$ but only \$100 for a counterexample. This latter prize is downgraded to \$50 in [ErFi97].
It is trivial that $f(n) \ll n^{1/2}$. A result of Pach and Sharir (Theorem 4 of [PaSh92]) implies $f(n) \ll n^{2/5}$. Hunter has observed that the circle-point incidence bound of Janzer, Janzer, Methuku, and Tardos [JJMT24] implies\[f(n) \ll n^{4/11}.\]Fishburn (personal communication to Erdős, later published in [ErFi97]) proved that $6$ is the smallest $n$ such that $f(n)=3$ and $8$ is the smallest $n$ such that $f(n)=4$, and suggested that the lattice points may not be best example.
See also [754].
Source: erdosproblems.com/92 | Last verified: January 13, 2026