I was reading Sylvester-Gallai theorem and thought about the following question.
Let there be some finite number of points on a plane. Each point is either red or blue. Every straight line passing through two points of the same color must also pass through another point of the other color. Is it true that all points are on the same straight line?
My thinking was : if they are not on a single straight line then by Sylvester-Gallai, we know that there is a line passing through exactly two points. Then these points can't have the same color, so I'll call the red one $R_0$ and the blue one $B_0$. Now there has to be another point not on $R_0B_0$, so let this point be red (wlog) and call it $R_1$. Now $R_0R_1$ has to have a blue point $B_1\ne B_0$.
Then inductively $R_n$ is a blue point on $B_{n-2}B_{n-1}$ and $B_n$ is a blue point on $R_{n}R_{n-1}$. I think all these points are different but not sure how to prove it.
Best Answer
This is known as the Motzkin-Rabin theorem. Based on the existing proofs in the literature, I do not think this can be easily proved using the Sylvester-Gallai theorem as a lemma.
The earliest proofs of this theorem actually prove the dual version: given several lines in the plane colored red and blue, there exists a point where two or more lines meet, and all lines meeting there are the same color. This is equivalent since the entire problem can be formulated in the projective plane, and because of projective duality.
Motzkin's original proof (see page 130 in https://doi.org/10.1007/BF02112289) used the area of triangles in the projective plane as a monovariant. This can be formulated in terms of non-projective triangles alone, see below.
Chakerian published a proof (https://doi.org/10.2307/2317330) which relies on Euler's formula.
There also exists a proof, due to Pretorius and Swanepoel (https://doi.org/10.2307/4145133), which sticks to the original formulation of colored points, though it is quite subtle.
Here is a sketch of the version of Motzkin's proof for red and blue lines without projective geometry. Consider two intersecting blue lines, $B_1$ and $B_2$. We are done unless there exists a red line $R_1$ lying on their intersection, so assume this is so. Furthermore, assume there is some red line $R_2$ not passing through this intersection, to avoid trivial cases.
The entire configuration is either equivalent to situation 1 or 2 in the top row of the image below, defining a certain gray triangle $T$. If there is no blue line $B_3$ at the intersection of $R_1$ and $R_2$, we are done, so assume $B_3$ exists. In situation $1$, this creates a smaller version of situation 1, with the colors swapped, such that the corresponding triangle $T'$ has a smaller area. Iterating this, we cannot have smaller areas forever, so eventually the process stops.
In situation 2, there is a small hiccup where $T'$ can be larger than $T$. Iterating the process, $T'$ can only grow finitely many times, and once it shrinks, we go back to situation 1 where it will continue to shrink, so this process still must stop.