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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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  • $\begingroup$ 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. $\endgroup$
    – Lord_Farin
    Jun 14, 2013 at 13:31
  • $\begingroup$ Distance may vary...Let say an interior point is p of a triangle ABC then distances PA, PB, PC may be 5,3,4. $\endgroup$
    – r.bhardwaj
    Jun 14, 2013 at 13:38
  • $\begingroup$ 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: math.stackexchange.com/questions/329761/… $\endgroup$
    – user77970
    Jun 14, 2013 at 13:54

3 Answers 3

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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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  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Jun 14, 2013 at 13:43
  • $\begingroup$ en.wikipedia.org/wiki/Equilateral_triangle $\endgroup$
    – Riccardo.Alestra
    Jun 14, 2013 at 13:46
  • $\begingroup$ knowing $p,q,t$ you get an equation in $a$ which can be solved using a little elementary algebra $\endgroup$
    – Riccardo.Alestra
    Jun 14, 2013 at 13:47
  • $\begingroup$ Thanks @Riccardo, yet I can see in that link a much helpful (imo) relation two lines below the one you wrote down. Good +1 $\endgroup$
    – DonAntonio
    Jun 14, 2013 at 13:48
  • $\begingroup$ 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... $\endgroup$
    – DonAntonio
    Jun 14, 2013 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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  • $\begingroup$ Whatever link @Greg provided it has now disappeared. $\endgroup$
    – sammy gerbil
    Apr 23, 2020 at 18:38
  • $\begingroup$ As my try, the sqrt(3)*(PA^2+PB^2+PC^2)/4 should be sqrt(3)*(PA^2+PB^2+PC^2)/8 $\endgroup$
    – mayi
    Jan 25, 2022 at 5:10
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let $x,y,z$ be the distances between the interior point and vertices. $$x+y+z = a\sqrt{3}$$ $$a = \frac{x+y+z}{\sqrt{3}}$$

$$area = \sqrt{3}\frac{a^2}{4} = \sqrt{3}\frac{(x+y+z)^2}{12}$$

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  • $\begingroup$ First statement here is incorrect. User might be thinking of Viviani's Theorem which states that the sum of distances of an interior point from sides of an equilateral triangle equals the altitude of the triangle. $\endgroup$
    – sammy gerbil
    Apr 23, 2020 at 17:26

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