Given a set which represents an equilateral triangle in a plane, find the point inside the set which maximizes the product of the distances to the vertices.

Optimizing the product of the distances through straight differentiation seems like a useless work (getting an equation system composed of two cubic 2-dimensional polynomials in the process), since getting an answer from there doesn't seem like an option. So I'm trying to think this problem through analytic geometry.

I feel it's pretty obvious the answer should be the center of the triangle, but I'm not sure how to prove any other point should have a product of distances which is bigger. The only thing I know about any non-center point is that there exists another point which the product of the distances is the same as it, given the symmetry axes of the equilateral triangle. Besides that, I have no clue in how to procede.

  • $\begingroup$ Any point at infinity will have an infinite product of distances. Is it perhaps required that the point be inside the triangle? $\endgroup$
    – Marconius
    Sep 13 '15 at 1:50
  • $\begingroup$ I supposed it was understood from the fact that the set represents an equilateral triangle in a plane, but I'll edit the post to clarify this. $\endgroup$
    – Rono
    Sep 13 '15 at 4:21
  • 1
    $\begingroup$ The product didn't maximize in any interior point (see this answer), you should look for the maximum on the perimeter. $\endgroup$ Sep 13 '15 at 4:38
  • $\begingroup$ Ok I got it now, thank you. $\endgroup$
    – Rono
    Sep 13 '15 at 14:46

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