Source: IMO-2013-P2
Category: Extremal Problems

Problem:
A configuration of $4027$ points in the plane is called *Colombian* if it consists of $2013$ red points and $2014$ blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is *good* for a Colombian configuration if the following two conditions are satisfied:
\begin{enumerate}
\ii[(i)] No line passes through any point of the configuration.
\ii[(ii)] No region contains points of both colors.
\end{enumerate}
Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good arrangement of $k$ lines.

Solution:
The least value of $k$ is $2013$.

**Lower Bound:**
We first show that $k \ge 2013$ lines are necessary. Consider a configuration where $4026$ of the points form the vertices of a regular $4026$-gon, and the colors of these vertices alternate between red and blue. The $4027$th point is the remaining blue point, which can be placed anywhere such that no three points in the entire configuration are collinear. 

The perimeter of the regular $4026$-gon consists of $4026$ line segments, each connecting a red point and a blue point. In any good arrangement of lines, every such segment must be intersected by at least one line, because the endpoints of the segment are of different colors and must therefore belong to different regions. Since the $4026$-gon is strictly convex, any straight line can intersect its perimeter in at most two points. Therefore, a single line can intersect at most two of the $4026$ segments. To cut all $4026$ segments, we require at least $4026 / 2 = 2013$ lines. Thus, $k \ge 2013$.

**Upper Bound:**
We now prove that $k = 2013$ lines are always sufficient. Consider the convex hull of the $4027$ points. Since the points are not all collinear, the convex hull is a convex polygon with at least three vertices. We distinguish two cases based on the vertices of this convex hull:

*Case 1: The convex hull contains at least one red point.*
Let $R$ be a red point on the convex hull. Since $R$ is a vertex of the convex hull, we can draw a single line that separates $R$ from the other $4026$ points. 

There are $2012$ red points remaining. We group these $2012$ red points into $1006$ pairs arbitrarily. For each pair $\{A, B\}$, we draw two lines parallel to the segment $AB$ and sufficiently close to it such that the narrow strip between these two lines contains $A$ and $B$ but no other points of the configuration. (This is possible because no three points are collinear, so no other point lies on the line $AB$.)

We have drawn $1 + 2 \times 1006 = 2013$ lines. These lines partition the plane into several regions. We claim this is a good arrangement. Any point in the plane not on the drawn lines falls into one of the following categories:
- It lies in the half-plane containing $R$. Since this half-plane contains no other points of the configuration, any region here contains at most the red point $R$, and no blue points.
- It lies within one of the $1006$ narrow strips. By construction, the entire strip contains only the red points $A$ and $B$ and no other points of the configuration. Thus, any region within this strip contains no blue points.
- It lies outside the half-plane and outside all $1006$ strips. Since every red point is either $R$ or belongs to one of the pairs, all red points have been accounted for. Thus, any region here contains no red points (it may contain blue points or be empty).

In all cases, no region contains points of both colors. Thus, the arrangement is good.

*Case 2: The convex hull contains no red points.*
If there are no red points on the convex hull, then all vertices of the convex hull must be blue points. Since the convex hull has at least three vertices, there must exist two consecutive blue points on the perimeter of the convex hull, say $U$ and $V$. 

Since the segment $UV$ is an edge of the convex hull, we can draw a single line that separates $U$ and $V$ from the other $4025$ points. 

There are $2012$ blue points remaining. We group these $2012$ blue points into $1006$ pairs arbitrarily. For each pair, we draw two lines parallel to the segment connecting them, forming a narrow strip that contains exactly these two blue points and no other points of the configuration. 

Again, we have drawn $1 + 2 \times 1006 = 2013$ lines. By the exact same reasoning as in Case 1, any point not on the lines either lies in the half-plane containing $U$ and $V$ (which contains no red points), lies in one of the narrow strips (which contain only blue points), or lies outside all of these (which contains no blue points, and thus only red points or empty). Hence, no region contains points of both colors.

In all cases, a good arrangement of $2013$ lines can be found. Therefore, the least value of $k$ is $2013$.

Answer: 2013

Grading guidelines:
(Partial) 1. Established the lower bound idea by giving (or clearly describing) a configuration with alternating colors on the vertices of a convex 4026-gon and arguing that every red–blue edge must be crossed by the drawn lines, and that each line crosses the boundary of a strictly convex polygon at most twice, hence k≥2013. 2. Proved a correct geometric reduction for the upper bound: isolated one point (or two adjacent hull vertices) by a single line using a supporting line to the convex hull, ensuring that half-plane contains only those chosen point(s) of one color. 3. Gave the key construction step for the upper bound: for many same-colored points, paired them and for each pair drew two parallel lines close to the segment so that the strip contains exactly those points and no others (with a correct justification using ‘no three collinear’/small perturbation). (Almost) 1. Presented the full 2013-line construction (one separating line plus 1006 strips) but did not completely justify that the chosen strips can be made disjoint/placed so that each contains no other configuration points, leaving only a local existence/perturbation gap. 2. Handled one of the two convex-hull cases (a red hull vertex exists, or all hull vertices are blue and isolate two consecutive blue vertices) correctly and finished the region-color argument for that case, but omitted or inadequately treated the other case. 3. Proved both bounds k≥2013 and k≤2013 with correct overall structure, but made a small counting/edge-case slip (e.g. miscounting how many points remain to be paired, or not checking that no line passes through a configuration point), while the main idea is correct.