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Title: Question re Nick Hobson's #140 Post by ecoist on Jan 11th, 2007, 5:27pm 140. The smallest distance between any two of six towns is m miles. The largest distance between any two of the towns is M miles. Show that M/m>sqrt(3). Assume the land is flat. (a) Same conclusion for 4 towns, right? (b) M/m is greater or equal twice the Golden Section for 5 or 6 towns? (Not sure if this is true or false) |
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Title: Re: Question re Nick Hobson's #140 Post by markr on Jan 12th, 2007, 12:16am Wouldn't 4 towns be sqrt(2) when arranged as the vertices of a square? |
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Title: Re: Question re Nick Hobson's #140 Post by Grimbal on Jan 12th, 2007, 12:54am I fail to see how to actually achieve a M/m ratio of sqrt(3). The best ratio I get is 2·sin(2·pi/5). I can not prove that is this the minimum though. |
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Title: Re: Question re Nick Hobson's #140 Post by balakrishnan on Jan 12th, 2007, 5:34am Grimbal is right, for 6 points, 2*sin(2*pi/5) For 7 points---> it is 2...6 points on a regular hexagon and 1 on its center For 8 points..It is a regular heptagon having 7 points and the 8th point is its center so it is cos(pi/14)/sin(pi/7)=2.247 I wrote a computer program and inferred the following: For 9 points,I get 2.6349 For 12 points I get sqrt(10) PS:I am not able to identify a pattern for the 9 point one(as given out by my code) For large N, the ratio will be ~ sqrt(N)*sqrt(2sqrt(3)/pi) Here are some figures: https://nrich.maths.org/discus/messages/67613/114012.html?1158282646 |
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Title: Re: Question re Nick Hobson's #140 Post by ecoist on Jan 12th, 2007, 11:26am Right you are, markr! Question (a) was dumb. |
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