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Topic: Six points in the plane (Read 886 times) |
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NickH
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Six points in the plane
« on: Apr 20th, 2006, 6:03am » |
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Let the minimum distance between any two of six points in the plane be m, and the maximum distance between any two of the six be M. Find the minimum possible value of M/m.
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SMQ
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Re: Six points in the plane
« Reply #1 on: Apr 20th, 2006, 6:35am » |
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Well, as a quick upper limit, I see two obvious ways to achieve M/m = 2.
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SMQ
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Re: Six points in the plane
« Reply #3 on: Apr 20th, 2006, 7:39am » |
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Would that be 1/(2sin([pi]/12)) ~= 1.932?
Start with three points in an equilateral triangle. With each point as a center, construct the short arc which has the other two points as endpoints. Place the remaining three points at the midpoints of the arcs.
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NickH
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Re: Six points in the plane
« Reply #4 on: Apr 20th, 2006, 9:41am » |
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Or is it 2*sin(72°) ~= 1.902?
The vertices of a regular pentagon, together with its center.
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towr
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Re: Six points in the plane
« Reply #5 on: Apr 20th, 2006, 10:01am » |
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The latter, ~1.902
It seemed the most obvious to me, after a hexagon
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Wikipedia, Google, Mathworld, Integer sequence DB
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Oyibo
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Re: Six points in the plane
« Reply #6 on: Apr 21st, 2006, 10:32am » |
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How many points can be placed in a plane such that the ratio M/m does not exceed 3?
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Grimbal
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Re: Six points in the plane
« Reply #7 on: Apr 27th, 2006, 8:46am » |
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I can do 12. Fore more, I'll have to think.
(understand it on a regular triangular grid)
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