# Maximize the distance to vertices in an equilateral triangle

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.

• Any point at infinity will have an infinite product of distances. Is it perhaps required that the point be inside the triangle? Sep 13 '15 at 1:50
• 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.
– Rono
Sep 13 '15 at 4:21
• The product didn't maximize in any interior point (see this answer), you should look for the maximum on the perimeter. Sep 13 '15 at 4:38
• Ok I got it now, thank you.
– Rono
Sep 13 '15 at 14:46