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Topic: Question re Nick Hobson's #140 (Read 517 times) |
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ecoist
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Question re Nick Hobson's #140
« on: Jan 11th, 2007, 5:27pm » |
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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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markr
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Re: Question re Nick Hobson's #140
« Reply #1 on: Jan 12th, 2007, 12:16am » |
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Wouldn't 4 towns be sqrt(2) when arranged as the vertices of a square?
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Grimbal
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Re: Question re Nick Hobson's #140
« Reply #2 on: Jan 12th, 2007, 12:54am » |
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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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balakrishnan
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Re: Question re Nick Hobson's #140
« Reply #3 on: Jan 12th, 2007, 5:34am » |
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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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| « Last Edit: Jan 12th, 2007, 5:35am by balakrishnan » |
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ecoist
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Re: Question re Nick Hobson's #140
« Reply #4 on: Jan 12th, 2007, 11:26am » |
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Right you are, markr! Question (a) was dumb.
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