Piercing Diametral Disks Induced by Edges of Maximum Spanning Tree

TL;DR

This paper proves that a single point can pierce all diametral disks induced by the edges of a maximum spanning tree of a point set, and provides a linear-time method to find this point.

Contribution

It establishes that the set of diametral disks from a maximum spanning tree is Helly and identifies the smallest enclosing circle's center as a universal piercing point.

Findings

One point pierces all disks in the set.

The set of disks is Helly.

The piercing point can be found in linear time.

Abstract

Let $P$ be a set of points in the plane and let $T$ be a maximum-weight spanning tree of $P$. For an edge $(p,q)$, let $D_{pq}$ be the diametral disk induced by $(p,q)$, i.e., the disk having the segment $\overline{pq}$ as its diameter. Let $\cal{D_T}$ be the set of the diametral disks induced by the edges of $T$. In this paper, we show that one point is sufficient to pierce all the disks in $\cal{D_T}$, thus, the set $\cal{D_T}$ is Helly. Actually, we show that the center of the smallest enclosing circle of $P$ is contained in all the disks of $\cal{D_T}$, and thus the piercing point can be computed in linear time.