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PostPosted: Fri, 9 May 2008 17:15:15 UTC 
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here is a classic problem. it's a good one if you haven't seen the solution.

Given four points at corner of a unit suqare, what is the shortest sum of the lengths of the set of line segments that connects them?

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Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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PostPosted: Thu, 12 Jun 2008 22:23:36 UTC 
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the case n=3 is the fermat point of the triangle.

the case n=4, I've seen a proof. I thought it was complete until I realized that the proof only attempts to do local minimization hinging on the fact that
Spoiler:
as with the fermat point, any steiner point (intersection point of line segs) must have degree 3 all of which have angle of 2pi/3
. it does not however argue why a local min is a global maximum as well, since there can be multiple ways of doing so

_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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PostPosted: Wed, 20 May 2009 20:36:18 UTC 
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Joined: Tue, 12 May 2009 17:57:45 UTC
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Location: Birmingham HS
Given a square with side n, the length of the "network" will n(Root 3 + 1).

The diagram looks a bit like a double "wishbone".
There are 2 nodes and 5 mini-segments, all of which connect at an
angle of 120 degrees. Start with a square of side 8 and use 30-60-90
triangles to get the sides of the "diagonal" segments. The middle one can
be gotten by the subtracting 2 short legs from the side of the square.
The result is 8Root 3 + 8 which can be factored and matches the above
expression. :wink:

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I always enjoy a good riddle or puzzle or problem -- one that is accessible with a grasp of high school/college mathematics.


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