The problem I'm considering is just for an equilateral triangle and I want to show that its center is a local minimum for the sum of inverse distances to the vertices. I don't know how to show this, but is there a general point for the minimum of sum of inverse distances to the vertices of a triangle, just like the Fermat point is for the sum?
Thank you in advance!
Minimise sum of inverse distances to vertices in a triangle
euclidean-geometrytriangles
Best Answer
Your post is not clear with respect to what exactly are you seeking to minimise. I'm assuming it is the sum of the inverse distances to the vertices. Since you are dealing with equilateral triangles, you can assume without loss of generality that the vertices are $$ A = (-1,0), \quad B=(1,0), \quad C = (0, \sqrt{3}) $$ Therefore, for any point $(x,y)$ inside the triangle, the sum of inverse distances is given by $$ f(x,y)=\frac{1}{\sqrt{(x+1)^2+y^2}}+\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{x^2+(y-\sqrt{3})^2}} $$
You can determine the stationary points inside the triangle (there are four of them) and, in fact, the center is one of them. If you compute the Hessian matrix at this stationary point, you'll see that it is a local minimum.
The global minimum, however, is attained at the midpoints of each side of the triangle.