TITLE: 21 points on circumference of a circle must have at least 100 pairs separated by 120+ degrees.
QUESTION [5 upvotes]: Prove that at least 100 of the arcs determined by the pairs of these points subtend an angle not exceeding 120 degrees at the center. How do I prove this? Induction? Help please. Thanks.

REPLY [2 votes]: Form a graph with 21 vertices corresponding to the points and edges between pairs of points more than 120 degrees apart.  
This graph is triangle-free.  
The number of edges in a triangle-free graph on 21 vertices is (uniquely) maximized for the bipartite graph on 10 + 11 vertices which has 110 edges.  Therefore, there are at least ${21 \choose 2} - 110 = 210 - 110 = 100$ pairs of vertices not joined by an edge. Those pairs correspond to pairs of points at most 120 degrees apart.
This also explains what the maximal arrangements of points look like.  All of them can be formed by taking two disjoint closed 120 degree arcs on the circle, and placing $10$ and $11$ points respectively inside the two open intervals that are the complement of those arcs.