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

Research paper by **Alexander Magazinov**

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

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#### 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).