Consider a sphere which has 5 points placed randomly on its surface. Demonstrate that at least 4 points lie in the same hemisphere (this is a past Putnam problem).
Select a great circle going through two of the points. As there are at most three points not on this circle, one…
Yep, this is the answer! I kind of thought you would get it ;) This is one of my favorite problems. Whenever I need to talk to non-math people about math, this is an exercise I turn to, because it’s beautifully simple. Once you’ve used something like this to convince them the pigeonhole principle is true, you can go on to tell them how to prove there is someone on earth who has the exact number of hairs on their head as them, etc…
What if the five points happen to be equally spaced on the same great circle, so that they’re the five vertices of an inscribed regular pentagon of maximal area? I claim you can’t find a hemisphere that contains four of these points. spinor’s answer doesn’t cover this case. Was this really a Putnam problem?
edit: OH WAIT if they mean closed hemisphere then this is trivial…nevermind. Why was I thinking in terms of open hemispheres?
