# On a conjecture by Eckhoff and Dolnikov concerning line transversals to Euclidean disks

Research paper by Alexander Magazinov

Indexed on: 30 Oct '17Published on: 30 Oct '17Published in: arXiv - Mathematics - Metric Geometry

#### Abstract

Let $K$ be a convex body in the Euclidean plane $\mathbb R^2$. We say that a point set $X \subseteq \mathbb R^2$ satsfies the property $T(K)$ if the family of translates $\{ K + x : x \in X \}$ has a line transversal. A weaker property, $T(K, s)$, of the set $X$ is that every subset $Y \subseteq X$ consisting of at most $s$ elements satisfies the property $T(K)$. The following question goes back to Gr\"unbaum: given $K$ and $s$, what is the minimal positive number $\lambda = \lambda(K, s)$ such that every finite point set in $\mathbb R^2$ with the property $T(K, s)$ also satisfies the property $T(\lambda K)$? The constant $\lambda_{disj}(K, s)$ is defined similarly, with the only additional assumption that the translates $x + K$ and $y + K$ are disjoint for every $x, y \in X$, $x \neq y$. One case of particular interest is $s = 3$ and $K = B$, where $B$ is a unit Euclidean ball. Namely, it was conjectured by Eckhoff and, independently, Dolnikov that $\lambda (B, 3) = \frac{1 + \sqrt{5}}{2}$. In this paper we propose a stronger conjecture, which, on the other hand, admits an algebraic formulation in a finite alphabet. We verify our conjecture numerically on a sufficiently dense grid in the space of parameters and thereby obtain an estimate $\lambda_{disj}(B, 3) \leq \lambda(B, 3) \leq 1.645$. This is an improvement on the previously known upper bounds $\lambda(B, 3) \leq \frac{1 + \sqrt{1 + 4\sqrt{2}}}{2} \approx 1.79$ (Jer\'onimo Castro and Rold\'an-Pensado, 2011) and $\lambda_{disj}(B, 3) \leq 1.65$ (Heppes, 2005).