Geometry question pertaining to $4$ points in the plane where $90$ degree projectors are on each point and we must illuminate the whole plane. Suppose we have $4$ points that can be positioned anywhere in $\mathbb{R}^2$. Now imagine each point has $90$ degree projectors coming out of them and you can rotate these projectors any way you would like. How can one prove that no matter where the four points are you can rotate the projectors so that the whole plane will be illuminated entirely? 
I would like to elaborate on illuminate. You see, if you can imagine to lines shooting out of the points where they form $90$ degree angles then the space contained inbetween the lines can be thought of as being illuminated. And so obviously if the points are positioned such that they look to be corners of a square, we can just rotate the projectors so that they look to be the sides of a square. And so then extending the projectors off to infinity we can illuminate the whole plane. This is just one case of course. I was not sure how to prove it no matter where the points are. My thought was wherever the points are try to draw an $xy$ axis such that each point is in a quadrant. Then one can just proceed to make the projectors so that they form a square like figure again. Its an interesting problem.
Heres a bonus question. Suppose there are $n$ projectors located at $n$ points where each projector illuminates $\displaystyle\frac{360}{n}$ degrees. Then one can rotate them until the whole plane is illuminated. 

Here is a way which works when $n$ is even. Choose a line where half of the points are on each side, and think of this line as horizontal. We first illuminate the upper half plane with the projectors in the lower half. Let $\theta=360/n$. Of the lower projectors, let $p_1$ be the one which is northmost when the plane is rotated clockwise by $\theta$. Aim $p_1$ so that the clockwise edge of its light is horizontal and to the right of $p_1$. In general, $p_k$ will be the projector not among $p_1,\dots,p_{k-1}$ which is northmost when the plane is rotated clockwise by $k\theta$, and the clockwise edge of $p_k$'s light should make an angle of $(k-1)\theta$ with the horizontal ray to the right of $p_k$. Because of the ``northmost when rotated by\dots" condition, the light cones of $p_k$ and $p_{k+1}$ will overlap in such a way that the entire upper half is covered.
Repeat this procedure with the projectors in the upper half plane, and the entire plane will be illuminated.
I believe a similar procedure can be applied when $n$ is odd, provided you can find a line which intersects one projector, divides the remaining projectors in half, and where this projector can be rotated so the line bisects its light and it illuminates no other projectors. 
