A pencil is a collection of some lines through a point, called the center of the pencil. If the points of the plane are colored, then call a pencil bichromatic if there is a color that is present on all the lines of the pencil such that this color is different from the color of the center of the pencil.

Given any non-monochromatic coloring of the plane with finitely many colors, and $m$ directions, $\alpha_1,\ldots, \alpha_m$, is it true that there is a point $p$ and an angle $\varphi$ such that the pencil determined by the lines of direction $\alpha_1+ \varphi,\ldots, \alpha_m+ \varphi$ through $p$ is bichromatic?

This is related to polymath16, see why here. I can only prove the statement for $m=2$.

up vote 4 down vote accepted

Observe that we are done if there is a monochromatic circle $C$, say blue: If a point inside the circle is not blue, then any pencil with that centre is bichromatic, so we may assume that the whole disk bounded by $C$ is blue. If a point outside is close enough and not blue, then we can find a bichromatic pencil with that centre. With this, we can extend blue to the whole plane or find a bichromatic pencil.

We now find a monochromatic circle:

Let the angles determined by cyclically consecutive lines be $\beta_1,\dots,\beta_{m-1}$, and $S=\{b_1,\dots,b_m\}$ a set of points on a circle $C$ such that for any $x\in C$ the angle of the line through $x$ and $b_1$ and through $x$ and $b_{i+1}$ is $\beta_i$ for any $1\leq i \leq m-1$. If every $b_i$ is blue, and $x\in C$ is not blue, then there is a bichromatic pencil centered at $x$. So it would be enough to find a monochromatic set similar to $S$. But this exists by Gallai's theorem.

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