Problem: Several chords are drawn in a circle of radius $1$, and each diameter of the circle intersects no more than four of them. Prove that the sum of their lengths does not exceed 13.
I couldn't make any progress. I just tried putting an upper bound on the number of diagonals (possibly 6?). This is supposed to be a pigeonhole principle problem but I have no clue where and how to use the principle in this problem Someone please help me. Hints would also be great.
For a chord of length $l$, one gets two arcs each of length greater than $l$. The other arc is the one diametrically opposite to the given arc.
Now if the sum of the lengths of the chords $> 13$ then the sum of the lengths of the arcs "covered" $> 26$ > four times the circumference.
So, at least one point must belong to the arcs or diametrically opposite arcs of at least five chords. But this is a contradiction as the diameter formed by this point and center will pass through at least five chords.
So, the sum of the lengths of the chords cannot exceed 13.