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][1].
I can only prove the statement for $m=2$.


  [1]: https://dustingmixon.wordpress.com/2018/05/05/polymath16-fourth-thread-applying-the-probabilistic-method/#comment-4306