The Fermat Point of Triangles: A Correction

The Fermat Point of Triangles: A Correction

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1. Introduction

Previously, I had written a post detailing the Fermat point of a triangle: the unique point which minimizes the sum of distances to the triangle’s vertices. Recently, I realized a small oversight I had made in my proof of the case where all of the triangle’s angles are less than 120^\circ. In today’s post, we will be patching this oversight, so it is recommended to read through the original post before continuing (as a challenge, try to find the error yourself!).

2. The Correction

In my previous post, my approach to finding the Fermat point of triangle \triangle ABC was to define a function L(P) as the sum PA + PB + PC, and to minimize L using the “method of small displacements,” essentially an application of calculus. This works fine for almost any point P in the plane. However, I had overlooked the fact that the function L is not differentiable at the triangle’s vertices! Thus, the “Fermat Point” we found in the first post is actually a \textit{local} minimum of the function F, but we have not shown that it is the \textit{global} minimum.

More intuitively, the proof from last post relied on the assumption that the Fermat point P would be such that an infinitesimal displacement in any direction will not change the value of L. The reasoning behind this seems sound: if a displacement in any direction results in a decrease of L, then clearly P is not the Fermat point. If a displacement results in an \textit{increase} of L, then the diametrically opposite displacement should consequently result in a decrease, also ruling out P as the Fermat point. However, this latter result assumes that L is differentiable at P: if it weren’t, then there is no reason to believe that the opposite displacement should produce a decrease.

Thus, we have to specially consider the vertices A, B and C, and compare their distance sum to F, our hypothesized Fermat point from the previous post.

The Fermat point of ABC is F, A, B or C. Which one is it?

Let FA = x, FB = y, FC = z, AB = c, AC = b, BC = a. WLOG, let x\leq y \leq z, and scale the triangle so that x = 1. Then, by the law of cosines applied to \triangle BFC, \triangle CFA, \triangle AFB respectively, we have

    \begin{equation*} a^2 = y^2 + z^2 + yz \end{equation*}

    \begin{equation*}b^2 = x^2 + z^2 + xz = 1+z+z^2 \end{equation*}

    \begin{equation*}c^2 = x^2 + y^2 + xy=1+y+y^2 \end{equation*}

Since 1 = x\leq y \leq z, the above equations imply c\leq b \leq a. Thus, of the three vertices, the one with the minimum distance sum is A, which produces a sum of

    \begin{equation*} L(A) = 0 + b + c = \sqrt{1 + y + y^2} + \sqrt{1 + z + z^2} \end{equation*}

To ascertain the Fermat point, we must compare this to the distance sum of F, which is L(F) = 1 + y + z.

4. The Final Result

It turns out that F is still the true Fermat point. To see this, we define the function f(d) = \sqrt{1 + d + d^2}-d. Then,

    \begin{equation*} f(d) = \sqrt{1 + d + d^2} - d > \sqrt{\frac{1}{4} + d + d^2} - d = \sqrt{\left(\frac{1}{2} + d\right)^2} - d = \left(\frac{1}{2} + d\right) - d = \frac{1}{2} \end{equation*}

Applying this to the variables in our problem, we have

    \begin{equation*} \sqrt{1 + y + y^2} - y + \sqrt{1 + z + z^2} - z = f(y) + f(z) > \frac{1}{2} + \frac{1}{2} = 1 \rightarrow \end{equation*}

    \begin{equation*} 1+y+z < \sqrt{1+y+y^2}+\sqrt{1+z+z^2} \rightarrow L(F) < L(A) \end{equation*}

Now, we have fully ascertained the F is the Fermat point of \triangle ABC.

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