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
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1$\begingroup$ 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. $\endgroup$– Jyrki LahtonenFeb 13, 2014 at 22:50
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5$\begingroup$ 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. $\endgroup$– Bob SteinFeb 14, 2014 at 16:36
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$\begingroup$ @Bob Stein What is the maning of "bupkis" ? I never met this term... $\endgroup$– Jean MarieJun 22, 2020 at 6:57
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$\begingroup$ @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. $\endgroup$– Bob SteinJun 22, 2020 at 10:15
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$\begingroup$ This was problem A2 from the 2002 Putnam Exam. See the solution here: youtube.com/… $\endgroup$– domotorpNov 28, 2021 at 17:33
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1 Answer
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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.
answered Feb 13, 2014 at 22:48
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7$\begingroup$ Wow, this is the most beautiful paragraph I've seen this week. Thanks. $\endgroup$– AsinomásFeb 13, 2014 at 22:55
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2$\begingroup$ But the answer is sort of demoralizing, made me feel bad for not thinking of that. $\endgroup$– AsinomásFeb 14, 2014 at 3:19
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4$\begingroup$ @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. $\endgroup$ Feb 14, 2014 at 5:03
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$\begingroup$ I was thinking does this still holds if the sphere degenerates into circle, seemingly not. $\endgroup$– zinkingFeb 14, 2014 at 6:14
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1$\begingroup$ @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. $\endgroup$ Feb 14, 2014 at 14:54