Small Displacements Revisited: The Fermat Point

Small Displacements Revisited: The Fermat Point

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1. Introduction

This post will continue my investigation into the “method” of small geometrical displacements that I discussed in my previous post. I will apply it to a famous problem regarding triangles: the Fermat point. The Fermat point of a triangle is defined as the point in the plane of the triangle that minimizes the sum of the distances to each of the vertices. As we learned in the last post, the method of small displacements can be very useful for statements regarding distances to fixed points/lines, and the problem of finding the Fermat point of a triangle certainly fits the bill. But how exactly can it be applied for the purposes of minimization? Before getting to that, let us get a general sense of how such a point should behave.

2. Inside or Outside?

One of the first questions we can ask about the Fermat point is whether it can lie outside the triangle in question. Some special points (like the orthocenter and circumcenter) can lie outside the triangle, but others (like the centroid and incenter) must always lie inside. Intuitively, we may expect that to minimize distances, we should choose a point inside the triangle.

Indeed, let us pick a point F outside triangle ABC. We expect it not to be the Fermat point, in other words, we expect there to be another point in the plane whose sum of distances to the vertices is less than the sum AF + BF + CF (the sum of the red lengths in the figure above). Let F' be the foot of the altitude of F onto the side AB:

We claim that the sum of the green lengths is less than the sum of the red lengths, so F cannot be the Fermat point. Indeed, F'A = \sqrt{FA^2 - FF'^2} < FA and likewise, F'B < FB. Also, because F is outside the triangle, it is on the opposite side of AB from C, it is clear that F' will be closer to C than F (this can be made rigorous by dropping a perpendicular from C to AB and using the Pythagorean Theorem). Convince yourselves that no matter which point F we choose outside the triangle, there is at least one line AB,\ BC,\ AC onto which we can drop an altitude and apply the same reasoning we did in the diagram. Hence, it is clear that the Fermat point cannot be outside a triangle.

3. Applying the Method

Now, we will apply what we learned in the last post to actually find the Fermat point. Pick an arbitrary point F inside (or on) the triangle ABC.

For simplicity, let us write \angle BFC = X,\ \angle AFC = Y,\ \angle AFB = Z. Furthermore, given any point P, define the functions \alpha(P) = AP, \beta(P) = BP, \gamma(P) = CP, and L(P) = AP + BP + CP = \alpha(P) + \beta(P) + \gamma(P). If F is the Fermat point of \triangle ABC, then displacing the point F by an infinitesimal amount dx in any direction of the plane should result in L(F) remaining unchanged, i.e. dL = d\alpha + d\beta + d\gamma = 0. This due to the fact that the function L is clearly continuous and the Fermat point F is a local minimum for L.

Suppose we displace F by dx in the direction of A. Then, we clearly have d\alpha = -dx, and we can find that d\beta = -dx\cos(Z) and d\gamma = -dx\cos(Y) using our results from the last post. Therefore, we have

    \begin{equation*} d\alpha + d\beta + d\gamma = -dx - dx\cos(Z) - dx\cos(Y) = 0 \rightarrow \cos(Y) + \cos(Z) + 1 = 0 \end{equation*}

Now, we can displace by dx in the direction of B instead of A and by symmetry we will get a similar equation, as well as when displacing towards C. The following equations are obtained:

    \begin{equation*} \cos(X) + \cos(Y) = -1 \end{equation*}

    \begin{equation*} \cos(X) + \cos(Z) = -1 \end{equation*}

    \begin{equation*} \cos(Y) + \cos(Z) = -1 \end{equation*}

These are simply linear equations in the cosines, and we can solve them using standard methods. Noticing the symmetry, we add the equations and divide by 2:

    \begin{equation*} \cos(X) + \cos(Y) + \cos(Z) = -\frac{3}{2} \rightarrow \end{equation*}

    \begin{equation*} \cos(X) + \left(\cos(Y) + \cos(Z)\right) = \cos(X) - 1 = -\frac{3}{2} \rightarrow \end{equation*}

    \begin{equation*} \cos(X) = -\frac{1}{2} \rightarrow X = 120^\circ \end{equation*}

Similarly, we have Y = Z = 120^\circ.

The Fermat point of ABC

We can now see that the Fermat point of such a triangle can be directly constructed with ruler and compass by constructing external equilateral triangles on each side of ABC, then constructing the circumcircles of those triangles and finding their common point of intersection. This has a straightforward proof using the fact that opposite angles in cyclic quadrilaterals add to 180^\circ, left to the reader.

So are we done? Have we completely characterized the Fermat point of any triangle? Can we finally wrap up today’s post?

Well, not quite. Consider a triangle like the one above, with one angle larger in magnitude than 120^\circ. In this case, there is no point F inside ABC such that \angle BFC = 120^\circ! Convince yourself that all such angles \angle BFC will be greater than 120^\circ. This means that the Fermat point F cannot possibly be inside the triangle, since the function L(F) defined earlier is continuous and differentiable everywhere inside \triangle ABC, and nowhere inside \triangle ABC does it attain a local minimum, i.e. no matter where you are inside the triangle, you can move to somewhere else with a smaller L.

So if the Fermat point cannot be inside the triangle, or outside the triangle as we showed previously, then it must be somewhere on the triangle!. We can do casework on which side of the triangle it is on: for example, if F is on AB, then we must minimize FA + FB + FC = AB + FC, equivalent to minimizing FC. If we were free to move across the entire line AB, then FC would be minimized at the projection of C onto AB. Since we are restricted to the segment, the best we can do is vertex A. Similarly, if F were on AC, then it would be optimal to place it at vertex A. If F is on BC, then the foot of the perpendicular of A onto BC is contained within the segment, so it is achievable.

Which is the Fermat point: A or A'?

We finally conclude today’s post with a proof that A, not A', is indeed the Fermat point. Observe that if \angle A \geq 120^\circ, then we must have \angle B + \angle C \leq 60^\circ since the angles in a triangle add to 180^\circ. Without loss of generality, let AA'=1. Then, L(A) = AB + BC = \csc(B) + \csc(C), which can be shown with elementary trigonometry. Also, L(A') = A'B + A'C + A'A = \cot(B) + \cot(C) + 1. Hence we would like to prove that

    \begin{equation*} \csc(B) + \csc(C) < \cot(B) + \cot(C) + 1 \Longleftrightarrow \left(\csc(B) - \cot(B)\right) + \left(\csc(C) - \cot(C)\right) < 1 \end{equation*}

However, observe that for any x, we have

    \begin{equation*} csc(x) - \cot(x) = \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \frac{1 - \cos(x)}{\sin(x)} = \tan\left(\frac{x}{2}\right) \end{equation*}

So we would like to prove that for any positive B,\ C with B + C \leq 60^\circ, we have

    \begin{equation*} \tan\left(\frac{B}{2}\right) + \tan\left(\frac{C}{2}\right) < 1 \end{equation*}

Clearly, the left side is maximized when we allow B + C = 60^\circ, which we will henceforth assume. Now, notice that it is also maximized when we let one angle (say B) equal the entire 60^\circ and C = 0^\circ. This is due to the fact that the function \tan\left(\frac{x}{2}\right) is convex over the desired interval (in other words, its graph curves upward). We can imagine setting B = C = 30^\circ, and then slowly moving B to 60^\circ and C to 0^\circ. Each variable would move the same amount in opposite directions, since their sum is constant. However, due to convexity, notice that \tan\left(\frac{B}{2}\right) will increase faster than \tan\left(\frac{C}{2}\right) will decrease, so the net sum will increase as B and C move farther apart. Finally, we plug in these maximal conditions and see that

    \begin{equation*} \tan\left(\frac{60^\circ}{2}\right) + \tan\left(\frac{0^\circ}{2}\right) = \frac{1}{\sqrt{3}} + 0 < 1 \end{equation*}

And we are finally done! We have now characterized the Fermat point for all triangles: if the triangle has no angle larger than 120^\circ, then the Fermat point is the one that splits the angles to each vertex into equal 120^\circ pieces, which we can construct fairly simply with ruler and compass as described above. If any angle is larger than 120^\circ, then the Fermat point is the vertex of that largest angle. I hope you all enjoyed today’s post, see you soon!

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