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The Fermat–Torricelli point of a triangle is a point which minimizes the total distance from the point to the vertices.

The geometric method of finding the Fermat–Torricelli point for triangles is well known.

We may apply Lagrange Multipliers to find such a point for polygons.

Is there a geometric construction of Fermat–Torricelli point for polygons ??

Mohammad Riazi-Kermani
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  • Wikipedia mentions that Lagrange analysis is "a long tedious business" even for the case of triangles. To save the reader some of this trouble, could you provide results of Lagrange analysis for, say, the quadrilateral and pentagon cases? – Blue Mar 16 '18 at 15:01
  • If you have a polygon $A_1A_2...A_n$, then the Fermat point will be given by a point $T$ such that $\angle A_1TA_2 = \angle A_2TA_3=...=\angle A_nTA_1$, provided that such a point exists. This is a relatively simple result; however, I have never heard of a geometric construction on this subject. Not to mention, the existence of such a point is not guaranteed for a general $n$, like $n=4.$ – dezdichado Mar 16 '18 at 18:17
  • @dezdichado: If such a point exists then it will be the Fermat point, but the Fermat point always exists even if it does not satisfy your property. For example, the Fermat point of a rectangle is its centroid. –  Mar 16 '18 at 19:40
  • @Rahul that's literally what I said. – dezdichado Mar 16 '18 at 19:54
  • @dezdichado: Sorry, it wasn't obvious from your comment that if such a point does not exist then the Fermat point still does. I interpreted it in the same vein as "The sum of the series $1+a+a^2+\cdots$ is $1/(1-a)$, provided $|a|<1$". –  Mar 17 '18 at 16:03

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Not a purely geometric construction, but an efficient algorithm (due to Weiszfeld) is well-known in the literature. These notes by Nam are an excellent survey on the topic, in my opinion.

Jack D'Aurizio
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