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I have been fascinated by the Fermat point (the point in a triangle that minimizes the sum of the distances to the corners), which is known to be the point that makes the three paths separate by 120 degrees. Right now, there is one main question that I would like to have answered. Thank you for reading.

I am familiar with some classical proofs and constructions. What I want to discuss is the existence of a certain kind of generalized Fermat point. When a triangle's largest angle grows past 120 degrees, the Fermat point becomes still and fixes on the large angle. I want to generalize the Fermat point so that it moves outside the triangle, in a smooth way.

Say $P$ is a point and $\Delta ABC$ a triangle. Call the distances between $P$ and $A,B,C$ to be $r_A, r_B, r_C$. Usually, "generalized Fermat point" is taken to mean the point $P$ that minimizes the weighted sum $w_Ar_A+w_Br_B+w_Cr_C$.

Since I want the Fermat point to move smoothly outside the triangle when the big angle exceeds 120 degrees, I think I should partition the outside of the triangle into three regions based on which vertex is closest (this would interestingly bring in the perpendicular bisectors and the circumcircle!). Then, I would put four different conditional weights:

$$(-1,1,1),\ (1,-1,1),\ (1,1,-1),\ (1,1,1)$$

For example, if $P$ is inside the triangle, we use the weights $(1,1,1)$ and the function we want to minimize is $r_A+r_B+r_C$. If, on the other hand, $P$ is outside the triangle and closest to $C$, then the function we want to minimize would be $r_A+r_B-r_C$.

I would like to prove a few things:

a) that this is a true generalization of the Fermat point, i.e. that when all the angles in a triangle are <120, the above conditionally weighted minimizer coincides with the Fermat point.

b) that the lines from the vertices to the generalized Fermat point are always separated by 120 degrees, even if $\Delta ABC$ is super obtuse.

c) that this generalized point moves around smoothly as the angles of the triangle change, and has no hiccups like the classical Fermat point.

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  • $\begingroup$ Hmm, I think I partially get what you're saying. Six $60^\circ$ angles sounds like an awful lot of symmetry, though. Isn't it only met in the equilateral case? Perhaps the three points you're talking about correspond to the usual generalized Fermat points with weights (-1,1,1) (1,-1,1) (1,1,-1)? $\endgroup$ – Aaron Goldsmith Aug 30 '16 at 1:54
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What you are looking for is the first isogonic center (Wikipedia, MathWorld). This coincides with the Fermat point when the triangle has no angle greater than $120^\circ$, and its method of construction makes it clear that it varies smoothly as the vertices of the triangle are moved.

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  • $\begingroup$ I will try to add some explanation of how this matches your definition when time permits... $\endgroup$ – user856 Aug 30 '16 at 5:33
  • $\begingroup$ Ah, of course! Thank you. I am also thinking. $\endgroup$ – Aaron Goldsmith Aug 30 '16 at 14:45
  • $\begingroup$ I think the proof works the exact same way, which is what I hoped for! Thanks again. $\endgroup$ – Aaron Goldsmith Aug 30 '16 at 15:12

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