The Generalized Fermat's Point

Abstract

In the 17th century, Fermat proposed and Torricelli solved the problem of determining a point P in a given acute triangle ABC such that the sum PA + PB + PC of the distances from P to the vertices of the triangle is the minimum. The first proof was published in 1659, and the point P is called Fermat 's point in the literature. There is a very natural generalization of Fermat's problem. Let 1, m, n be three positive real numbers. Find a point P in a given triangle ABC such that 1 PA + m a PB + n * PC is a minimum. This generalization was considered in two articles in this journal [2], [3]. The method in [2] is complicated, while the method in [3] does not use synthetic geometry. In this note, using the idea in the proof of the existence of Fermat's point [1], we give a proof for the generalized case. This proof is just a slight modification of the old one. Let us recall the proof of the existence of Fermat's point. On the sides of AB, BC, CA of triangle ABC, erect externally the equilateral triangles ABC', A'BC, and AB'C. Draw lines CC' and AA' to intersect at P. Then the point P is Fermat's point. The proof is based on the fact that a line segment is the shortest path between two points. Here is the outline of the proof in [1].