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](https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4306).
I can only prove the statement for $m=2$.



Bichromatic pencils
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](https://arxiv.org/abs/1411.1038).


