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Given the distances from an interior point to the vertices of an equilateral triangle, find the area of that triangle.

I have already tried equating $\sqrt{3}\times a^2/4$ and sum of the area of three triangles interior to equilateral triangle formed by given lines. But that approach is making a hard equation to solve.

Any solution using Computer programming language may also help.

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Is it "distance", meaning one is considering the centre of the triangle, or distances, meaning the distance may vary between vertices? The latter problem would be significantly harder. –  Lord_Farin Jun 14 '13 at 13:31
    
Distance may vary...Let say an interior point is p of a triangle ABC then distances PA, PB, PC may be 5,3,4. –  r.bhardwaj Jun 14 '13 at 13:38
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4 Answers 4

up vote 2 down vote accepted

Almost the same question that was asked and answered about a month ago, that has specific lengths for the distances. I'm not sure how it changes if the distances are unspecified, but this other answer may provide useful information anyway: Equilateral triangle geometric problem

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The following relation holds: $$3(p^4+q^4+t^4+a^4)=(p^2+q^2+t^2+a^2)^2$$ where $p,q,t$ are the distances from the vertices and $a$ is the length of the side of the triangle

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As a hint this seems to be pretty....er...thrifty. Perhaps you could explain a little more this equality, and how could it be used to solve the problem. –  DonAntonio Jun 14 '13 at 13:43
    
    
knowing $p,q,t$ you get an equation in $a$ which can be solved using a little elementary algebra –  Riccardo.Alestra Jun 14 '13 at 13:47
    
Thanks @Riccardo, yet I can see in that link a much helpful (imo) relation two lines below the one you wrote down. Good +1 –  DonAntonio Jun 14 '13 at 13:48
    
Oh, not so little, @Riccardo...not even close: you get a rather nasty biquadratic , and the OP also had already a nasty equation. I don't think this one would be making his life much easier... –  DonAntonio Jun 14 '13 at 13:50
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According to the link given by @Greg (above), the area of the equilateral triangle ABC with an interior point P, will be

((sqrt(3)(PA^2+PB^2+PC^2)/4) + 3(area formed by triangle with 3 lines as PA, PB and PC))/2]]

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let x,y,z be the distances between the interior point and vertices.
x+y+z = a.sqrt(3)
a = x+y+z/sqrt(3)


area = sqrt(3)a2/4 = sqrt(3)(x+y+z)2/12

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