Part (a) Color each ray in blue and extend each ray backwards in red. Therefore, each ray will now be a line consisting of a red ray, a blue ray and an initial point. Let us call these $2n$ double rays - light sabers. It suffices to find a line that intersect the light sabers at $n$ blue points and $n$ red points. Now, imagine a big circle $\mathcal{C}$ covering all the initial points of these $2n$ light sabers, pick a point $P\in\mathcal{C}$ and draw a tangent line $\ell_P$. Make sure to choose $P$ so that $\ell_P$ is not parallel to any of the light sabers. Now, move $P$ clockwise along the perimeter of $\mathcal{C}$ (hence $\ell_P$ is a moving tangent), and stop when $P$ reaches the diametrically opposite point of the starting position. Notice that since $\mathcal{C}$ covers all the initial points, if $\ell_P$ meets a particular light saber at blue point in initial position, then it will meet that light saber at red point in final position (and vice versa). Therefore, the difference between red and blue intersections has different signs for initial and final positions, and thus the difference is zero somewhere in-between (this uses the fact that no two light sabers are parallel). At that point, the tangent line will intersect exactly $n$ light sabers in red and exactly $n$ in blue. $\square$