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Title: Diophantine Triangle Centre Post by JocK on Nov 23rd, 2005, 1:41pm Given an equilateral triangle with edges of unit length. Can you select a point at rational distance to each of the three vertices that is as close as possible to the centre point of the triangle? How close can you get? (Get your computers ready...! 8) ) NB: all distances are to be taken as the normal (Euclidean) distances within the plane. |
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Title: Re: Equilateral Triangle Centroid Post by JocK on Nov 24th, 2005, 2:29pm For those of you who are contemplating how to approach this, the following might be useful: Given a equilateral triangle with edges s, the distances (a, b, c) from a point in the plane to the vertices of the triangle satisy [hide] 3(a4 + b4 + c4 + s4) = (a2 + b2 + c2 + s2)2 [/hide] |
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Title: Re: Diophantine Triangle Centre Post by Barukh on Nov 26th, 2005, 9:26am Just to keep this thread high (there are many new problems these days :) The good news is that theoretically, the point can be chosen arbitrarily close to the center (there exists a theorem stating that the set of all such points is dense in the plane). Practically, I am aware of a family of solutions given parametrically which approximates the rather disappointing distance [sqrt]57/24 = 0.3145... |
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Title: Re: Diophantine Triangle Centre Post by Diophantus on Nov 30th, 2005, 10:59am Yeah, let's keep it up. Working on it. Just want to report that (570/989, 571/989, 572/989) does do the job [hide] unfortunately only for a triangle with edges 0.9999999999953[/hide]. :( |
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