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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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    $\begingroup$ Don't you lose convexity when $q$ is small? It seems like you would get local minima near each corner in the zero limit. $\endgroup$
    – S. Carnahan
    May 30, 2010 at 1:41
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    $\begingroup$ 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. $\endgroup$ Jun 29, 2010 at 17:28

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I posted an animation showing approximations of this curve for a sequence of triangles, and a range of q values, here.

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  • $\begingroup$ Looks like it's a private video. $\endgroup$
    – j.c.
    Jun 28, 2010 at 13:10
  • $\begingroup$ I believe it is now viewable by everyone. $\endgroup$ Jun 28, 2010 at 16:59
  • $\begingroup$ This is very nice. How did you produce it? Would it be possible to share some code. $\endgroup$ Jul 11, 2017 at 19:28
  • $\begingroup$ Thanks. I'll figure a good way share the code and get back to you. $\endgroup$ Jul 11, 2017 at 20:17
  • $\begingroup$ I've added an applet to Open Processing. Let me know if you have more questions. Cheers! openprocessing.org/sketch/440663 $\endgroup$ Jul 20, 2017 at 0:26
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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.

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