# Triangle centers as power sum minimizing points: Which is the locus of all these points?

Some of the classical triangle centers can be expressed as solutions to minimization problems: Given a triangle $A_1, A_2, A_3$ define $d_i, i=1,2,3$ to be the distance of a given point $P$ to $A_i$, and $f_q$ as the sum of the $q$-th power of these distances:$f_q = \sum_{i=1}^3 d_i^q$. I'm looking for the point $P$ which minimizes $f_q$. For $q=1$ this is the Steiner point, for $q=2$ the centroid, for $q \to \infty$ the circumcenter, for $q \to 0$ the point where the product of distances is minimized. An obvious question is to find the curve of all these points for reasonably general $q$ (e.g. $q \in \mathbb{R}_{>0}$). However, in the ressources for triangle centers, as e.g. http://faculty.evansville.edu/ck6/encyclopedia/ETC.html this problem seems not to be considered.

EDIT:

1) I would like to restrict the problem to $q \ge 1$ since then $f_q$ is convex and a unique minimum is guaranteed.

2) I would like to add a generalization of the question: Consider all continuous functions $f(d_1,d_2,d_3)$ that a) are invariant under permutations of $d_1, d_2, d_3$ and b) have a unique minimum. What can be said about the locus of all these minima?

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Don't you lose convexity when $q$ is small? It seems like you would get local minima near each corner in the zero limit. –  S. Carnahan May 30 '10 at 1:41
When $q \rightarrow \infty$, $P$ does not (generally) approach the circumcenter. Consider an obtuse isosceles triangle. If we let the obtuse angle approach $\pi$, the circumcircle radius grows to infinity, so the center grows arbitrarily far from the vertices of the triangle. Such a point clearly cannot minimize $f_q$. In this case, P approaches the midpoint of the side opposite the obtuse angle. –  Matthew Conroy Jun 29 '10 at 17:28

Your curve might end up looking quite strange. Consider what happens if $A_1 \rightarrow \infty$ along a ray connecting it to the origin. the Steiner point stays fixed, and the centroid goes to infinity. While I'm not sure, it seems like for any $q > 1$, the point will go to infinity.