The regular polygon shows that $\lfloor n/2\rfloor$ is the best possible here.

This would be implied if there was a vertex such that no three vertices of the polygon are equally distant to it, which was originally also conjectured by Erdős [Er46b], but this is false (see [97]).

Let $f(n)$ be the maximal number of such distances that are guaranteed. Moser [Mo52] proved that\[f(n) \geq \left\lceil\frac{n}{3}\right\rceil.\]This was improved by Erdős and Fishburn [ErFi94] to\[f(n) \geq \left\lfloor \frac{n}{3}+1\right\rfloor,\]then\[f(n) \geq \left\lceil \frac{13n-6}{36}\right\rceil\]by Dumitrescu [Du06b], and most recently\[f(n) \geq \left(\frac{13}{36}+\frac{1}{22701}\right)n-O(1)\]by Nivasch, Pach, Pinchasi, and Zerbib [NPPZ13].

In [Er46b] Erdős makes the even stronger conjecture that on every convex curve there exists a point $p$ such that every circle with centre $p$ intersects the curve in at most $2$ points. Bárány and Roldán-Pensado [BaRo13] noted that the boundary of any acute triangle is a counterexample.

Bárány and Roldán-Pensado prove that, for any planar convex body, there is a point $p$ on the boundary such that every circle with centre $p$ intersects the boundary in at most $O(1)$ (where the implied constant depends on the convex body). They conjecture that there this can be bounded by an absolute constant - that is, Erdős's conjecture is true if we replace $2$ by some larger constant $C$.

See also [93].

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This page was last edited 19 October 2025. View history