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

Question

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.

Answer 1

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

Answer 2

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]]

Answer 3

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

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