Pigeonhole : Two circles of length $420$ There are two circles of length $420$. On one $420$ points are marked and on the other some arcs of circumference are painted red such that their total length adds up less than $1$. 
Show that there is a way to place one of the circles on top of the other so that no marked point is on the colored arc.
I still have no idea now, please give me some suggestions.
 A: Let $\theta_0, \theta_1, \theta_2, \dots, \theta_{419}$ be the angles at which the points are placed on the first circle. Placing one circle on top of the other requires choosing a point $q$ on the second circle where we place the point with angle $\theta_0$, determining the placeement of all the other points.
Let $R$ be the subset of the second circle that is colored red.
An obvious condition we have on $q$ is that $q \notin R$: if we chose $q \in R$, then the point with angle $\theta_0$ would overlap with $R$ when the circles are placed on top of each other.
Similarly, the point $q'$ which we get by rotating $q$ by $\theta_1 - \theta_0$ can't be in $R$: if that happened, then the point with angle $\theta_1$ would overlap with $R$. This is also a condition which limits where we can choose the point $q$.
Then there's $418$ other conditions along these lines. Can you describe the set of places where we shouldn't place $q$, or we'll violate one of the conditions? If you do, and show that it has total length less than $420$, then that proves that some way to overlap the circles exists.
Solution:

 Let $S$ be the union of all rotations of $R$ by $\theta_0 - \theta_k$ radians for any $k$. If we place the circles such that the point with angle $\theta_k$ is in $R$, then $q$ is in a $\theta_0 - \theta_k$ radian rotation of $R$, so it's in $S$. Conversely, if we place $q$ outside $S$, then there can be no overlaps. There are $420$ different rotations (including the identity rotation we get when $k=0$), so since $R$ has measure less than $1$, $S$ has measure less than $420$. Therefore we can place $q$ outside $S$, showing that there exists a placement with no overlaps.

Independent of any of the above, I also have a probabilistic argument:

 If we overlap the circles uniformly at random, then each point marked on the first circle has a probability of less than $\frac{1}{420}$ of ending up on a red arc. By linearity of expectation, the expected number of points that end up on a red arc is less than $420 \cdot \frac1{420} = 1$, so there must be an outcome in which the number of such points is less than $1$: that is, an outcome with no overlaps.

