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Topic: Between six towns (Read 1541 times) |
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BMAD
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Between six towns
« on: May 23rd, 2014, 3:08pm » |
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he 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.
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BMAD
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Re: Between six towns
« Reply #1 on: May 31st, 2014, 9:09pm » |
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Bump
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dudiobugtron
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Re: Between six towns
« Reply #2 on: Jun 1st, 2014, 4:07am » |
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One way to work this out would be to find the configuration which has the greatest ratio m:M, and show that even then it's not as great as 1:sqrt(3).
I couldn't initially figure out what that best configuration would be, though, so I gave up! But since you bumped it, I'll think some more about it.
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pex
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Re: Between six towns
« Reply #3 on: Jun 1st, 2014, 4:45am » |
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Apparently sqrt(3) is not even a tight bound. According to this page (spoiler alert: shows optimal configuration), the smallest possible value for M/m is sqrt( (5 + sqrt(5)) / 2 ) or approximately 1.90; sqrt(3) is approximately 1.73. Moreover, that page claims that the result is "trivial", but I have to admit I'm not seeing that.
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| « Last Edit: Jun 1st, 2014, 4:47am by pex » |
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BMAD
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Re: Between six towns
« Reply #4 on: Jun 1st, 2014, 5:57am » |
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Wouldn't the sqrt (3) be tighter since it is smaller?
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pex
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Re: Between six towns
« Reply #5 on: Jun 1st, 2014, 6:31am » |
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on Jun 1st, 2014, 5:57am, BMAD wrote:| Wouldn't the sqrt (3) be tighter since it is smaller? |
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No. That page claims that any configuration must have M/m > 1.9. If that's true, it immediately implies that any configuration also has M/m > sqrt(3), but not the other way around.
(By your reasoning, the bound M/m > 1 would be even tighter - but that's much easier to prove!)
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BMAD
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Re: Between six towns
« Reply #6 on: Jun 1st, 2014, 6:33am » |
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Oops. I got the inequality backwards in my head.
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Grimbal
wu::riddles Moderator
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Re: Between six towns
« Reply #7 on: Jun 1st, 2014, 2:38pm » |
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I think the optimal configuration is a pentagon with one city in the center.
The max distance is M=2*sin(2*pi/5) assuming m=1.
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