Coming fall semester I’ll be teaching our geometry undergraduate course, and to prepare myself a little, I am going to try some of the ideas I’d like to explain. One emphasis is on simplicity. If we just concern ourselves with points in the (say Euclidean) plane, a first fundamental question is whether they can be in an interesting position, and if so, how we can tell?

There is (almost) nothing to say here for one or two points, but for three points it gets interesting. Three points can be collinear, a condition from projective geometry, about which I know what to say. Another natural choice is to place the three points at the vertices of an *equilateral triangle*.

That is already a fairly advanced concept. How to you explain what an equilateral triangle is to your friend’s children in elementary school? As a full orbit of a cyclic group of order 3? That is, in fact, the simplest definition I know, avoiding conceptually much more difficult measurements of lengths or angles. It also has the advantage that it generalizes to other geometries.

How do we tell? An elegant answer in the spirit of the orbit definition can be given be denoting the vertices A, B, C as complex numbers. They lie at the vertices of an equilateral triangle if and only if A+𝛇B+𝛇²C=0, where 𝛇 denotes a third root of unity: 𝛇 = (-1+i√3)/2. This is easy to see, because it holds for your favorite equilateral triangle, and all equilateral triangles are similar, i.e. differ by a complex linear transformation which changes the left hand side of the test equation by a nonzero factor. I know I will have to say a little more, but not much.

What is this good for? One amusing application is *Napoleon’s Theorem*:

*Let ABC be any triangle. Construct equilateral Napoleon triangles CBA’, ACB’ and BAC’ outside of ABC. Then the centroids A”, B”, C” of the **Napoleon** triangles form another equilateral triangle.*

The proof is very simple now: We just express the centroid A’’ of CBA’ as (C+B+A’)/3, and likewise the other two centroids as B’’=(A+C+B’)/3 and C’’=(B+A+C’)/3, and apply the test.

Could there possibly a better proof? Yes and no. Here is one that is more a revelation than a proof.

The first key observation is that there is a tiling of the plane, using the given triangle and the equilateral Napoleon triangles, as shown above. This tiling becomes periodic with a hexagonal period lattice.

This allows to overlay the Napoleon tiling with a tiling by equilateral triangles that have their vertices at the centroids of the Napoleon triangles.

The reader will find it amusing to fit this all together. How general is all this? For instance, can we do this in a plane over a field without a third root of unity?

It is not clear whether Napoleon has actually proven this theorem. He was interested in Mathematics, and discussed science with Lagrange and Laplace. The first known written appearance is an article by W.

Rutherford in 1825, which doesn’t mention Napoleon.