Prove that 3 sleep at a time 5 mathematicians (say A,B,C,D,E) went to see a movie. While watching, each of them slept exactly twice. For each pair of mathematicians, there is some time during the movie at which both those mathematicians are asleep.  I.e. there is at least one point during the movie at which both A and B (and perhaps others) are asleep; there is at least one point during the movie at which both A and C (and perhaps others) are asleep; etc.  Prove that there exists a time when 3 of them are asleep.
 A: Let $A_1,A_2,B_1,B_2,\ldots,E_1,E_2$, be the vertices of a graph $G$, representing the $10$ naps.  Connect an edge between any two naps if they overlapped.
I claim that a cycle in this graph represents a time when $3$ or more mathematicians slept simultaneously.  Suppose $v_1,\ldots,v_n$ is a cycle in $G$, $n\ge 3$.  If any nap represented by $v_i$ in this cycle began and ended during either nap $v_{i-1}$ or $v_{i+1}$, then it is clear that naps $v_{i-1}$, $v_i$, and $v_{i+1}$ all overlapped.  If no nap was contained in any other nap, then without loss of generality we may assume that $v_1$ is the nap that began first in the cycle $v_1,\dots,v_n$.  By our condition, $v_2$ must end after $v_1$ ends.  Now since $v_n$ must overlap with map $v_1$ by the rules of our graph, it follows that naps $v_1,v_2$, and $v_n$ all overlap at the end of nap $v_1$.
For contradiction, assume that there is no time when $3$ or more mathematicians are simultaneously asleep.  Then our graph can have no cycles, and is thus a tree.  Therefore, it has at most $10-1=9$ edges, which is not enough to create the $\binom{5}{2}=10$ pairs of nappers.
