Source: IMO-2014-P6
Category: Existence and Construction

Problem:
A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call these its finite regions. Prove that for all sufficiently large $n$, in any set of $n$ lines in general position it is possible to colour at least $\sqrt{n}$ lines blue in such a way that none of its finite regions has a completely blue boundary.

Solution:
Suppose we have colored $k$ of the lines blue, and that it is not possible to color any additional lines without creating a finite region with a completely blue boundary. This means that any of the $n-k$ non-blue lines is a side of some finite region with an otherwise entirely blue perimeter.

For each such non-blue line $\ell$, select one such finite region, and take the next counterclockwise vertex along the region's perimeter; this vertex $v$ is the intersection of two blue lines. We say that $\ell$ is the *eyelid* of $v$.

Every intersection of two blue lines has at most two eyelids. Since there are exactly $\binom{k}{2}$ such intersections of blue lines, we obtain the inequality
\[
n-k \le 2 \binom{k}{2} = k^{2} - k.
\]
This simplifies to $n \le k^{2}$, which implies $k \ge \sqrt{n}$. Thus, it is always possible to color at least $\sqrt{n}$ lines blue such that none of the finite regions has a completely blue boundary.


Grading guidelines:
(Partial) 1. Introduced the maximality setup: start with a blue set of size k such that no further line can be colored blue without creating a finite region whose boundary is all blue, and correctly deduced that every non-blue line lies on the boundary of some finite region whose other sides are all blue. 2. Defined a way to associate (injectively up to bounded multiplicity) each non-blue line to an intersection point of two blue lines (e.g. by choosing a witness finite region and taking a specific adjacent vertex on its boundary), and justified that the chosen point is indeed an intersection of two blue lines. 3. Proved the local bound that a fixed intersection point of two blue lines can be associated to at most two non-blue lines (e.g. by analyzing the cyclic order of lines through that vertex and the two possible incident finite regions). (Almost) 1. Obtained the inequality n-k ≤ 2·C(k,2) (or an equivalent bound n ≤ k^2+O(k)) from the association argument, but did not fully justify why each non-blue line can be assigned to a finite region with otherwise all-blue boundary. 2. Carried out the counting argument correctly to conclude k ≥ √n, but left a small gap in the “at most two eyelids per blue-blue intersection” claim (e.g. did not exclude a third candidate region or did not use general position correctly). 3. Solution is almost complete, but made minor mistakes which are not negligible.