There are several circles of total length $10$ inside a square of side $1$. Show that there exists a straight line which intersects at least four of these circles.
If the total length of the circles is $10$, then the sum of their diameters must be $10/\pi > 3$. Consider any such set of circles with center $p_i= (x_i, y_i)$ and radius $r_i$. If we project the circle to the $x$-axis, then we see that the circle projects down to the interval $(x_i-r_i, x_i+r_i)$. We note that the sum of these intervals must be $10/\pi$ since each interval has length the diameter of that circle. The intervals must be between $0$ and $1$ as well, since all the circles are contained in the unit square. Thus we have $n$ overlapping intervals in $[0, 1]$ that sum to a number greater than $3$. By the pigeon hole principle, there must exist a point that is covered by more than $3$ intervals. The vertical line going through such a point thus intersects each interval, and each circle, and so is a vertical line that goes through at least $4$ circles.
Special thanks to:
Olivier Massicot (Champaign, IL) and Ankit Agarwal (Bay Area) for submitting a solution to this challenge problem!