Here’s a challenging problem with a surprisingly easy answer: can you show that for any 5 points placed on a sphere, some hemisphere must contain 4 of the points?<\/p>\n\n\n\n
How about an easier question: can you show that if you place 5 points in a square of sidelength 1, some pair of them must be within distance 3\/4 of each other?<\/p>\n\n\n\n
If you play with this problem for a while, you’ll realize quickly that the extreme case occurs when 4 points are at the corners of the square with a 5th point at the center. In this case adjacent points are exactly distance Sqrt[2]\/2 ~ 0.707 apart. But what may be harder to show is that this is really the extreme case; why can’t any other configuration be more extreme?<\/p>\n\n\n\n
The pigeonhole principle<\/em> is one of the simplest but most useful ideas in mathematics, and can rescue us here. A basic version says that if (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.<\/em> Thus if 5 pigeons occupy 4 holes, then there must be some hole with at least 2 pigeons. It is easy to see why: otherwise, each hole as at most 1 pigeon and the total number of pigeons couldn’t be more than 4. (This proof shows that it does not even matter if the holes overlap so that a single pigeon occupies 2 holes.)<\/p>\n\n\n\n So, if I divide up the square into 4 smaller squares by cutting through center, then by the pigeonhole principle, for any configuration of 5 points, one of these smaller squares must contain two points. But the diameter of the smaller square is Sqrt[2]\/2, so these two points must be closer than 3\/4, as claimed. The pigeonhole principle made what seemed like a slippery argument airtight.<\/p>\n\n\n\n The Math Behind the Fact:<\/strong> If you have N pigeons in K holes, and (N\/K) is not an integer, then then some hole must have strictly more than (N\/K) pigeons. So 16 pigeons occupying 5 holes means some hole has at least 4 pigeons.<\/p>\n\n\n\n If you have\u00a0infinitely many\u00a0pigeons in finitely many holes, then some hole must have infinitely many pigeons!<\/p>\n\n\n\n If you have an\u00a0uncountable\u00a0number of pigeons in a countable number of holes, then some hole has an uncountable number of pigeons!<\/p>\n\n\n\n Did you think about the challenge problem? Here is the answer: for any configuration of 5 points on a sphere, any two of them determine a great circle on the sphere (a circle whose center is the center of the sphere), and that great circle divides the sphere into 2 hemispheres. Considering the remaining 3 points, the pigeonhole principle says that one of the hemispheres must contain at least 2 of those 3 points. Together with the 2 points on the great circle, that hemisphere contains at least 4 points.<\/p>\n\n\n\n How to Cite this Page:<\/strong> References:<\/strong> Fun Fact suggested by: <\/strong> Here’s a challenging problem with a surprisingly easy answer: can you show that for any 5 points placed on a sphere,...<\/p>\n","protected":false},"author":9,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[9,3,12,44],"class_list":["post-109","page","type-page","status-publish","hentry","tag-combinatorics","tag-easy","tag-other","tag-sphere"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/109","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/comments?post=109"}],"version-history":[{"count":6,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/109\/revisions"}],"predecessor-version":[{"id":1571,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/109\/revisions\/1571"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/media?parent=109"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/tags?post=109"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
The pigeonhole principle has many generalizations. For instance:<\/p>\n\n\n\n
Su, Francis E., et al. “Pigeonhole Principle.” Math Fun Facts<\/em>. <http:\/\/www.math.hmc.edu\/funfacts>.<\/p>\n\n\n\n
The sphere problem comes from a recent Putnam examination.<\/p>\n\n\n\n
Francis Su<\/p>\n","protected":false},"excerpt":{"rendered":"