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 $m=2$.