Disjoint Empty Convex Pentagons in Planar Point Sets

TL;DR

This paper proves new lower bounds on the number of disjoint empty convex pentagons in planar point sets, improving understanding of their combinatorial structure.

Contribution

It establishes that any sufficiently large planar point set contains multiple disjoint empty convex pentagons, with improved bounds on their minimum number.

Findings

Every set of 19 points contains two disjoint empty convex pentagons.

Sets of 2m+9 points can be partitioned into regions with a pentagon in one.

Lower bounds on disjoint empty convex pentagons are improved to approximately 5n/47 and 3n/28.

Abstract

Harborth [{\it Elemente der Mathematik}, Vol. 33 (5), 116--118, 1978] proved that every set of 10 points in the plane, no three on a line, contains an empty convex pentagon. From this it follows that the number of disjoint empty convex pentagons in any set of $n$ points in the plane is least $\lfloor\frac{n}{10}\rfloor$. In this paper we prove that every set of 19 points in the plane, no three on a line, contains two disjoint empty convex pentagons. We also show that any set of $2m+9$ points in the plane, where $m$ is a positive integer, can be subdivided into three disjoint convex regions, two of which contains $m$ points each, and another contains a set of 9 points containing an empty convex pentagon. Combining these two results, we obtain non-trivial lower bounds on the number of disjoint empty convex pentagons in planar points sets. We show that the number of disjoint empty convex pentagons in any set of $n$ points in the plane, no three on a line, is at least $\lfloor\frac{5n}{47}\rfloor$. This bound has been further improved to $\frac{3n-1}{28}$ for infinitely many $n$.