April wrote: In a circle of diameter $ 1$, some chords are drawn. The sum of their lengths is greater than $ 19$. Prove that there is a diameter intersecting at least $ 7$ chords. It's quite funny this problem Mostly because $ 7$ is the wrong number, because the person who made the problem made an error in calculation. (All the team leaders did too, so it wasn't discovered, when we got it at the competition...!) Anyway divide the circle in $ n$ points $ P_1,P_2,..,P_n$ such that the lie on the circle an $ |P_1P_2| = |P_2P_3| = ... = |P_nP_1|$. And let $ n \to \infty$. Give every point $ P_i$ a value and let it be $ a_i$. Initially let it be $ a_1 = a_2 = ... = a_n = 0$. Then every time a chord is drawn between two points on a circle, then increment $ a_i$ by $ 1$ for every point $ P_i$ which's diameter intersect the chord. Then we easily obtain $ \frac{\sum_{i=1}^n a_i}{n} > 2\frac{19}{\pi} = \frac{38}{\pi} > 12$. Since $ \pi$ is the perimeter. So by dirichlets principle there exists $ a_i$ such that $ a_i > 12$, and thus $ a_i \ge 13$. So the diameter through $ P_i$ intersects at least $ 13$ chords!!