Putnam 2002, Problem A-2

Given any five points on a sphere, show that some four of them must lie on a closed hemisphere.

Solution:
Draw a great circle through two of the points. There are two closed hemispheres with this great circle as boundary, and each of the other three points lies in one of them. By the pigeonhole principle, two of those three points lie in the same hemisphere, and that hemisphere thus contains four of the five given points.  

Note: by a similar argument, one can prove that among any \( n+3 \) points on an \( n \)-dimensional sphere, some \( n+2 \) of them lie on a closed hemisphere. (One cannot get by with only \( n+2 \) points: put them at the vertices of a regular simplex.) Namely, any \( n \) of the points lie on a great sphere, which forms the boundary of two hemispheres; of the remaining three points, some two lie in the same hemisphere.