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Topic: Diophantine Triangle Centre (Read 871 times) 

JocK
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Diophantine Triangle Centre
« on: Nov 23^{rd}, 2005, 1:41pm » 
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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...! ) NB: all distances are to be taken as the normal (Euclidean) distances within the plane.

« Last Edit: Nov 26^{th}, 2005, 3:40am by JocK » 
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
x^{y}  y = x^{5}  y^{4}  y^{3} = 20; x>0, y>0.



JocK
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Re: Equilateral Triangle Centroid
« Reply #1 on: Nov 24^{th}, 2005, 2:29pm » 
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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 3(a^{4} + b^{4} + c^{4} + s^{4}) = (a^{2} + b^{2} + c^{2} + s^{2})^{2}

« Last Edit: Nov 24^{th}, 2005, 2:33pm by JocK » 
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
x^{y}  y = x^{5}  y^{4}  y^{3} = 20; x>0, y>0.



Barukh
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Re: Diophantine Triangle Centre
« Reply #2 on: Nov 26^{th}, 2005, 9:26am » 
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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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Diophantus
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Re: Diophantine Triangle Centre
« Reply #3 on: Nov 30^{th}, 2005, 10:59am » 
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Yeah, let's keep it up. Working on it. Just want to report that (570/989, 571/989, 572/989) does do the job unfortunately only for a triangle with edges 0.9999999999953.


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