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.)

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.
