There are many 5-holes

TL;DR

This paper proves that any set of n points in general position on the plane contains at least on the order of n^{20/11} empty convex pentagons, improving previous bounds and moving closer to the conjectured quadratic lower bound.

Contribution

The authors establish a new lower bound of Ω(n^{20/11}) for the number of empty convex pentagons in point sets, without relying on computer-assisted proofs.

Findings

Improved lower bound of Ω(n^{20/11}) for 5-holes.

The result narrows the gap towards the conjectured Ω(n^2) bound.

Proof does not require computer assistance.

Abstract

Given a set P of points on the plane, a polygon with vertices in P is said to be empty if it contains no element of P in its interior. We show that every set of n points in general position on the plane determines at least $\Omega(n^{20/11})$ empty convex pentagons (also known as 5-holes). This result improves upon the previous bound of $\Omega(n\cdot(\log n)^{4/5})$ obtained by Aicholzer et al. [JCT A, 2020], and significantly narrows the gap with respect to the conjectured $\Omega(n^2)$ lower bound (which, if true, would be tight). Unlike some of the other works in this line of research, our proof does not require computer assistance.