Draw $n$ lines in a plane so that there are no parallel lines and there are no three lines passing through the same point.
Each intersection point is colored red, green or blue. Prove that it is possible to color all intersection points in a “proper” way, so that any two adjacent points (like $A_i, A_j$) have different colors.
This also means that if you "travel" along an arbitrary line, you will cross $n-1$ intersection points, constantly changing colors from one intersection point to another.
My first (and last) try was to use induction. Obviously for two or three lines, we have one or three intersection points and with three colors available we have the base of induction proved. However, the induction step is more difficult. I was able to prove the induction step if in every possible arrangement of lines it was possible to find a line that divides the plane into two halves with one half having no intersection points. However, I was able to construct counter-examples where such line does not exist.
The last time I tried to solve a similar red-green-blue problem I discovered Ramsey theory. I wonder what I will discover this time :)