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riddles >> medium >> Between six towns
(Message started by: BMAD on May 23rd, 2014, 3:08pm)

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Title: Between six towns
Post by BMAD on May 23rd, 2014, 3:08pm
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.

Title: Re: Between six towns
Post by BMAD on May 31st, 2014, 9:09pm
Bump

Title: Re: Between six towns
Post by dudiobugtron on Jun 1st, 2014, 4:07am
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.

Title: Re: Between six towns
Post by pex on Jun 1st, 2014, 4:45am
Apparently sqrt(3) is not even a tight bound. According to this page (http://www2.stetson.edu/~efriedma/maxmin/) (spoiler alert: shows optimal configuration), the smallest possible value for M/m is [hide]sqrt( (5 + sqrt(5)) / 2 )[/hide] 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.

Title: Re: Between six towns
Post by BMAD on Jun 1st, 2014, 5:57am
Wouldn't the sqrt (3) be tighter since it is smaller?

Title: Re: Between six towns
Post by pex on Jun 1st, 2014, 6:31am

on 06/01/14 at 05:57:19, BMAD wrote:
 Wouldn't the sqrt (3) be tighter since it is smaller?
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!)

Title: Re: Between six towns
Post by BMAD on Jun 1st, 2014, 6:33am
Oops. I got the inequality backwards in my head.

Title: Re: Between six towns
Post by Grimbal on Jun 1st, 2014, 2:38pm
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.