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If there are 5 points on the surface of a sphere then there is a closed half sphere containing at least 4 of them.

It's in a pidgeonhole list of problems, but I think I have to use rotations in more than 1 dimension.


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An answer is given by Calvin Lin here. This is a duplicate in a way, but the topic of that other question is vastly different (a call for trick questions), so I am a bit reluctant to call it a duplicate. – Jyrki Lahtonen Feb 13 '14 at 22:50
So if you ever see a headline like "80% of Olympic Sites In The Last 20 Years Have Been In The ___ Hemisphere," it means bupkis. – BobStein-VisiBone Feb 14 '14 at 16:36
up vote 54 down vote accepted

Pick two distinct points out of your 5 (if all 5 are identical then they clearly all lie in a single hemisphere). These two points define at least one great circle (if they're antipodal, they define infinitely many); pick a great circle they define. This circle then cuts the sphere into two hemispheres. Now pigeonhole the other three points between these two hemispheres.

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Wow, this is the most beautiful paragraph I've seen this week. Thanks. – Carry on Smiling Feb 13 '14 at 22:55
But the answer is sort of demoralizing, made me feel bad for not thinking of that. – Carry on Smiling Feb 14 '14 at 3:19
@user4140 Don't be too demoralized! A lot of problems like this come down to whether you can recognize the specific 'trick' - the thing to pigeonhole on; it's easy to see in retrospect but often impossible to spot from the other side. – Steven Stadnicki Feb 14 '14 at 5:03
I was thinking does this still holds if the sphere degenerates into circle, seemingly not. – zinking Feb 14 '14 at 6:14
@zinking: Take 5 points on $\mathbb S^1$ such that the angle of any two neighboring points is $2\pi/5$. Then in every closed hemisphere there lie at most 3 of the points. – wspin Feb 14 '14 at 14:54

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