Consider 2015 lines in the plane, no two of which are parallel and no three of which are concurrent. Let $E$ be the set of their intersection points.

We want to assign a color to each point in $E$ such that any two points on the same line, whose segment connecting them contains no other point of $E$, are of different colors.

How many colors are needed at a minimum to achieve such a coloring?
The configuration contains at least one triangle. This can be proven, for example, by induction on the number of lines: three lines form a triangle, and if you add a line, it either leaves the triangle intact or splits it into a triangle and a quadrilateral. Consequently, it is impossible to perform a coloring with two colors.

We will now show that a coloring with three colors is possible: we order the points by increasing abscissa as before, and color them in this order. When it is time to color a point $P=\left(d_{i}\right) \cap\left(d_{j}\right)$, we have already colored at most one of its neighbors on $\left(d_{i}\right)$ (the one to the left) and one on $\left(d_{j}\right)$ (also the one to the left). Therefore, we have at most two forbidden colors and can always choose the third. The answer is therefore 3 colors.