Pigeonhole principle: Five points on an orange Five points are drawn on the surface of an orange. Prove that it is possible to cut the orange in half in such a way that at least four of the points are on the same hemisphere. (Any points lying along the cut count as being on both hemispheres.)
Through any two points, there is a great circle that passes through those two points. Such a cut will split the other 3 pigeons &mdash; oh, I mean points &mdash; among 2 halves. 

[You can now handle additional points being on the great circle on your own, I believe.]
Given a sphere with radius $1$.

Use the points


  $$
A_+(1,0,0), \quad A_-(-1,0,0), \quad B_+(0,1,0), \quad B_-(0,-1,0), \quad C(0,0,1)
$$


And you cannot make such a cut.

As for any great circle, $(A_+,A_-)$ and $(B_+,B_-)$ are never on the same hemisphere.

That is if we exclude points on both hemispheres as being cut through a point.