The Napoleonic Triangle: A Classic Result

August 15, 2021 Shreyas Comments 0 Comment

Introduction

Given any triangle $\triangle ABC$ , suppose we take each side $AB, BC,$ and $AC$ and construct an equilateral triangle on it. You may notice that given a line segment, there are two ways to construct an equilateral triangle on it (one on each side); we will eventually consider both ways, but for now suppose each triangle is constructed on the outside of $\triangle ABC$ , i.e. they do not overlap with $\triangle ABC$ .

We now consider finding the centers of each of these equilateral triangles, as shown above $(D, E, F)$ . These points together form a triangle themselves. What stands out about this triangle? It clearly looks like all of its side lengths are the same, i.e. $\triangle DEF$ is equilateral. Is this a coincidence, or can we prove this in general?

Label the sides $BC, CA, AB$ as $a, b, c$ respectively. We will examine the side lengths of $DEF$ and prove that they are the same. But how do we find the length of, say, $DF$ ? Note that we can show

$\begin{equation*} DB = \frac{AB}{\sqrt{3}} = \frac{c}{\sqrt{3}} \end{equation*}$

by using the fact that $\angle DBA = 30^{\circ}$ (since $D$ is the center of $\triangle ABC'$ ), details are left to the reader. Similarly, $BF = \frac{a}{\sqrt{3}}$ . If we knew $\angle DBF$ , we could use the law of cosines to derive information about side $DF$ . Fortunately,

$\begin{equation*} $\angle DBF = \angle DBA + \angle ABC + \angle CBF = +60^{\circ}$ \end{equation*}$

Thus, by the law of cosines,

$\begin{equation*} $DF^2 = \left(\frac{a}{\sqrt{3}}\right)^2 + \left(\frac{c}{\sqrt{3}}\right)^2 - 2\left(\frac{a}{\sqrt{3}}\right)\left(\frac{c}{\sqrt{3}}\right)\cos(B + 60^{\circ})$ \end{equation*}$

$\begin{equation*} =\frac{a^2}{3} + \frac{c^2}{3} - \frac{2}{3}ac\left(\cos(B)\cos(60^{\circ}) - \sin(B)\sin(60^{\circ})\right) \end{equation*}$

using the cosine angle-sum formula. We could derive similar expressions for $DE^2$ and $EF^2$ , but note that by the symmetry of the problem, all we must do is replace every occurence of $(a, c)$ $(b, c)$ or $(a, b)$ and similarly with the angles. Thus, it is sufficient to prove that our expression for $DF^2$ is symmetric with respect to $(a, b, c)$ . We break this sum into two parts:

$\begin{equation*} $\left(\frac{a^2 + c^2 - 2ac\cos(B)\cos(60^{\circ})}{3}\right) + \left(\frac{2}{3}ac\sin(B)\sin(60^{\circ})\right)$ \end{equation*}$

Note that the second part of the sum is a constant multiple of $\frac{1}{2}ac\sin(B)$ , which is equal to the area of the triangle and thus must necessarily be symmetric in $(a, b, c)$ . We simplify the first part by substituting $\cos(60^{\circ}) = \frac{1}{2}$ and using the law of cosines on $\triangle ABC$ to substitute for $\cos(B)$ :

$\begin{equation*} \frac{a^2 + c^2 - 2ac\cos(B)\cos(60^{\circ})}{3} = \frac{a^2 + c^2 - 2ac\left(\frac{a^2 - b^2 + c^2}{2ac}\right)(\frac{1}{2})}{3} \end{equation*}$

$\begin{equation*} =\frac{a^2 + c^2 - \frac{a^2 - b^2 + c^2}{2}}{3} = \frac{a^2 + b^2 + c^2}{6} \end{equation*}$

Clearly, this is symmetric in $(a, b, c),$ so we are done!. We have proven that $\triangle DEF$ is always equilateral.

But what about the other part of the original question that we haven’t addressed? What if the equilateral triangles all point inward instead of outward? In this case, it is easy to see that we can do all the same things we did in the first case, but the angle $A + 60^{\circ}$ is replaced with $A - 60^{\circ}$ . This means that the appearance of $\cos(A)\cos(60^{\circ}) - \sin(A)\sin(60^{\circ})$ is replaced with $\cos(A)\cos(60^{\circ}) + \sin(A)\sin(60^{\circ})$ . However, the second term of this sum only winds up being a part of the second part of our partition, and crucially it is still a simple constant multiple of $\frac{1}{2}ac\sin(B)$ . The first part, meanwhile, is unchanged. So the exact same argument can be applied, and we see that the resultant triangle is still equilateral.