# Find the area of an equilateral triangle given the distances from an interior point to the vertices

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.

• 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

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

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

• 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
• en.wikipedia.org/wiki/Equilateral_triangle – Riccardo.Alestra Jun 14 '13 at 13:46
• 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

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]]

let $x,y,z$ be the distances between the interior point and vertices. $$x+y+z = a\sqrt{3}$$ $$a = \frac{x+y+z}{\sqrt{3}}$$

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