Napoleon’s Theorem through the Discrete Fourier Transform

Napoleon’s Theorem states that if equilateral triangles are constructed externally on the three sides of any triangle, then the centroids of those three equilateral triangles form an equilateral triangle. Classical proofs often use angle chasing or trigonometric identities. There is, however, a more structural proof: the construction respects cyclic symmetry, and the discrete Fourier transform is exactly the language for separating the different cyclic symmetry components of a triangle.

Identify the Euclidean plane with the complex plane $\mathbb{C}$ . Let the vertices of an arbitrary triangle, labelled counterclockwise, be $(z_0,z_1,z_2\in\mathbb{C})$ , and collect them into the vector $Z=(z_0,z_1,z_2)^T\in\mathbb{C}^3$ . Cyclically relabelling the vertices is represented by the shift matrix $P(z_0,z_1,z_2)^T=(z_1,z_2,z_0)^T$ . Since applying this shift three times returns to the original labelling, $P^3=I$ . Thus $P$ represents the cyclic group $C_3$ .

Let $\omega=e^{2\pi i/3}$ , so that $\omega^3=1$ and $1+\omega+\omega^2=0$ . The three Fourier modes of $\mathbb{C}^3$ are the vectors $f_0=(1,1,1)^T$ , $f_+=(1,\omega,\omega^2)^T$ , and $f_-=(1,\omega^2,\omega)^T$ . They are eigenvectors of the cyclic shift: $Pf_0=f_0$ , $Pf_+=\omega f_+$ , and $Pf_-=\omega^2f_-$ . Every labelled triangle can therefore be written uniquely as

$\displaystyle Z=A f_0+B f_+ +C f_-$ .

The coefficient $A$ is the centroid of the original triangle. The mode $f_0$ therefore represents translation rather than shape. The vector $f_+$ represents a positively oriented equilateral triangle, while $f_-$ represents an equilateral triangle with the opposite orientation. Thus a triangle is equilateral exactly when, after translation is removed, it contains only one of these two nontrivial Fourier modes. A general triangle contains both.

Now construct outward equilateral triangles on the sides of the original triangle. Because the vertices are labelled counterclockwise, the exterior lies to the right of each directed edge. Rotating a complex vector through $60^\circ$ clockwise means multiplying it by $r=e^{-i\pi/3}=-\omega$ . The third vertex of the outward equilateral triangle erected on the side from $z_j$ to $z_{j+1}$ is therefore $t_j=z_j+r(z_{j+1}-z_j)$ . Its centroid, which we call $g_j$ , is the average of $z_j$ , $z_{j+1}$ , and $t_j$ . After simplifying, one gets

$\displaystyle g_j=\frac{2+\omega}{3}z_j+\frac{1-\omega}{3}z_{j+1}$ .

Thus the three centroids $G=(g_0,g_1,g_2)^T$ are obtained from $Z$ by a linear map $M$ , namely

$\displaystyle G=MZ,\quad M=\frac{2+\omega}{3}I+\frac{1-\omega}{3}P$ .

This is the crucial point. The Napoleon construction is a circulant linear transformation: it treats the three sides in the same cyclic manner, so it commutes with $P$ . Since the Fourier modes diagonalize $P$ , they also diagonalize every expression built from $P$ , including $M$ . We therefore only need to see what $M$ does to the three Fourier modes.

A short calculation gives

$\displaystyle Mf_0=f_0,\qquad Mf_+=e^{i\pi/3}f_+,\qquad Mf_-=0$ .

The interpretation is immediate. Translation survives unchanged, as it must: moving the original triangle moves every constructed centroid by the same amount. The positive equilateral mode survives, though it is rotated through $60^\circ$ . The negative equilateral mode is completely annihilated.

Applying $M$ to the Fourier decomposition of the original triangle gives

$\displaystyle MZ=A f_0+e^{i\pi/3}B f_+$ .

The $f_-$ contribution has disappeared. What remains is a translation component $A f_0$ together with a single positively oriented equilateral mode. But that is precisely the Fourier description of an equilateral triangle. Hence the three centroids $(g_0,g_1,g_2)$ form an equilateral triangle.

This is Napoleon’s Theorem.

The proof shows something stronger than a fortunate geometric coincidence. The original triangle has two nontrivial cyclic symmetry components, corresponding to the two possible orientations of an equilateral pattern. The Napoleon construction kills one of them exactly and preserves the other. Once only one orientation mode remains, an equilateral triangle is forced.

If one constructs equilateral triangles on the opposite sides of the original triangle, the roles of $f_+$ and $f_-$ are reversed. The resulting triangle is still equilateral, but its orientation is reversed. This accounts for the two classical versions of the Napoleon triangle.

The Napoleon–Barlotti Theorem

The same Fourier idea gives a generalization to $n$ -gons. Let $\mathcal P$ be an ordered $n$ -gon with vertices $z_0,\ldots,z_{n-1}$ , with indices understood modulo $n$ . On each directed side $z_jz_{j+1}$ , construct a regular $n$ -gon using the same orientation, and let $c_j$ be its center.

Theorem. The centers $c_0,\ldots,c_{n-1}$ , taken in side order, form a regular $n$ -gon if and only if $\mathcal P$ is an affine image of a regular $n$ -gon.

Let $\zeta=e^{2\pi i/n}$ , and let $S$ be the cyclic shift. Its Fourier modes are

$\displaystyle f_k=(1,\zeta^k,\zeta^{2k},\ldots,\zeta^{(n-1)k})^T,\qquad Sf_k=\zeta^k f_k.$

A real affine map of the plane has the form $w\mapsto a+bw+c\overline w$ . Applying it to the regular $n$ -gon $1,\zeta,\ldots,\zeta^{n-1}$ shows that $\mathcal P$ is affine-regular exactly when

$\displaystyle Z=A_0f_0+A_1f_1+A_{n-1}f_{n-1}.$

Choose the orientation for which $q=\zeta$ . Since the radius from $c_j$ to $z_j$ rotates by $2\pi/n$ to the radius from $c_j$ to $z_{j+1}$ , $\displaystyle z_{j+1}-c_j=q(z_j-c_j).$ Hence $\displaystyle c_j=\frac{z_{j+1}-qz_j}{1-q}.$

Thus, for $C=(c_0,\ldots,c_{n-1})^T$ , $\displaystyle C=MZ,\quad M=\frac{S-qI}{1-q}.$ The Fourier modes diagonalize $M$ : $\displaystyle Mf_k=\frac{\zeta^k-\zeta}{1-\zeta}f_k.$ Therefore $M$ kills exactly $f_1$ and is nonzero on every other Fourier mode. If $\mathcal P$ is affine-regular, then

$\displaystyle MZ=A_0f_0+\frac{\zeta^{-1}-\zeta}{1-\zeta}A_{n-1}f_{n-1}.$

This is a translation together with one regular Fourier mode, so the centers form a regular $n$ -gon.

Conversely, suppose that the centers form a regular $n$ -gon in side order. Such a polygon has only a translation mode and one of the two modes $f_1$ or $f_{n-1}$ . Since $M$ annihilates $f_1$ , the center polygon can contain only $f_0$ and $f_{n-1}$ . As $M$ is nonzero on $f_2,\ldots,f_{n-2}$ , the original polygon has no such modes. Hence