, 2017

We will prove that given $n$ circles, there is some line intersecting more than $\frac{n}{46}$ of them. Let $S$ be the set of centers of the $n$ circles. We will first show that there is a line $\ell$ such that the projections of the points in $S$ lie in an interval of length at most $\sqrt{8068} < 90$ on $\ell$.

Let $A$ and $B$ be the pair of points in $S$ that are farthest apart and let the distance between $A$ and $B$ be $d$. Now consider any point $C \in S$ distinct from $A$ and $B$. The distance from $C$ to the line $AB$ must be at most $\frac{4034}{d}$ since triangle $ABC$ has area at most $2017$. Therefore if $\ell$ is a line perpendicular to $AB$, then the projections of $S$ onto $\ell$ lie in an interval of length $\frac{8068}{d}$ centered at the intersection of $\ell$ and $AB$. Furthermore, all of these projections must lie on an interval of length at most $d$ on $\ell$ since the largest distance between two of these projections is at most $d$. Since $\min(d, 8068/d) \le \sqrt{8068} < 90$, this proves the claim.

Now note that the projections of the $n$ circles onto the line $\ell$ are intervals of length $2$, all contained in an interval of length at most $\sqrt{8068} + 2 < 92$. Each point of this interval belongs to on average $\frac{2n}{\sqrt{8068}+2} > \frac{n}{46}$ of the subintervals of length $2$ corresponding to the projections of the $n$ circles onto $\ell$. Thus there is some point $x \in \ell$ belonging to the projections of more than $\frac{n}{46}$ circles. The line perpendicular to $\ell$ through $x$ has the desired property. Setting $n = 100$ yields that there is a line intersecting at least three of the circles. $\square$