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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 pigeonhole list of problems. But, I think I have to use rotations in more than 1 dimension.

Regards

Asinomás
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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
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    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. – Bob Stein Feb 14 '14 at 16:36
  • @Bob Stein What is the maning of "bupkis" ? I never met this term... – Jean Marie Jun 22 '20 at 06:57
  • @JeanMarie "bupkis" means "nothing at all". According to yourdictionary it comes from Yiddish. I think it's especially funny when used in a string of synonyms. – Bob Stein Jun 22 '20 at 10:15
  • This was problem A2 from the 2002 Putnam Exam. See the solution here: https://www.youtube.com/watch?v=agJAWnfpNjY&ab_channel=MathsOlympiadTrainer – domotorp Nov 28 '21 at 17:33

1 Answers1

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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.

Steven Stadnicki
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    Wow, this is the most beautiful paragraph I've seen this week. Thanks. – Asinomás Feb 13 '14 at 22:55
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    But the answer is sort of demoralizing, made me feel bad for not thinking of that. – Asinomás Feb 14 '14 at 03:19
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    @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 05:03
  • I was thinking does this still holds if the sphere degenerates into circle, seemingly not. – zinking Feb 14 '14 at 06:14
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    @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. – Wolfgang Spindeler Feb 14 '14 at 14:54
  • @wspin exactly, but I didn't came up with the proof – zinking Feb 15 '14 at 08:52
  • Can your method be used to show that there is a (closed) hemisphere which contains exactly 4 points? I ask because there are 5*4/2=10 combinations of 2 points and we can just take a combination whose greater circle divides the sphere into 2 halves, one containing 2 points and the other containing 1 point. – Hemant Agarwal Mar 24 '21 at 07:39
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    Exactly 4 points? No, look at a regular pentagon inscribed in a great circle. The 2 hemispheres with that great circle as boundary contain all 5 points; any other hemisphere contains at most 3. – Mirlan Oct 01 '24 at 19:34