Beautiful pigeonhole Olympiad problem You have a set of disjoint arcs on a circle of circumference $1$ with sum of lengths greater than $1-\frac{1}{n}$. 
Show that there is a regular $n$-gon such that each of its vertices lies on one of the arcs.
 A: Average the number of covered vertices over the possible positions of the $n$-gon. The result is greater than $n(1-\frac1n)=n-1$. Since the number of covered vertices is an integer, it follows that there is at least one position with $n$ vertices covered.
A: Assume a regular n sided polygon at an arbitrary position and rotate it about centre of circle through angle (2*π)/n then atleast at some point of time during rotation all vertices should be on the given set of arcs if not atleast 1/n length of circumference is not covered by given set of arcs
A: Split the circle into $n$ equally-sized arcs of size $1/n$. And then line them up one on top of each other over the interval $[0, 1/n)$.
If there's a point in $[0, 1/n)$ such that all $n$ corresponding points in the circle are covered by our disjoint arc set, then those $n$ vertices form a regular $n$-gon on the original circle (as they are spaced $1/n$ apart).
But by pigeonhole there must be such a covered point.

Note: if we want to be extra rigorous here we can discretize: split the circle into $M$ parts with $n | M$; each $1/n$ arc is split into $M/n$ parts.
For large enough $M$, we will find that at least one full $1/M$-length part is covered in all $n$.
Proof: here each interval of length $\ell_i$ covers at least $\ell_i M - 2$ intervals. Now let $\varepsilon = \sum_i \ell_i - (1-1/n)$. Then the number $N$ of covered intervals is $\ge (\sum_i \ell_i) M - 2k = (1-1/n + \varepsilon)M - 2k$. Since $\varepsilon$ and $k$ are just constants, we can pick $M$ big enough so that the total number of covered intervals is $> (1-1/n)M$, and we can now apply usual discrete pigeonhole.
