Take a square and divide it down a diagonal, dividing the square into two triangles. Drawing the opposite diagonal now divides it into four triangles. In these two examples, we divided a square into an even number of triangles, all with equal area. Can we divide a square into an *odd* number of nonoverlapping triangles, all with equal area? In this question, we do not require that all the triangles be congruent, as in the above examples.

It turns out, you can't. Paul Monksy proved this in a 1970 American Mathematical Monthly paper [1], though John Thomas proved this earlier when the vertices of the triangles are restricted to having rational coordinates.

The proof progresses in several steps. I won't go through every detail, but try and convey the flavour of the proof. The reader is invited to read the proof in its entirety, which is something I just did and I recommend it.

Let us consider a general region in space defined by gluing together nonoverlapping triangles $\{T_1,\dots, T_m\}$. Let's use the word vertex to refer to any vertex of a triangle in this arrangement. A face will refer to a face either of a triangle or of the boundary (which will be some kind of polygon). Let's say a face is of type $\Acl\Bcl$ if one end of the face has vertex $\Acl$ and the other end is a vertex of type $\Bcl$. The key to the proof is the following:

**Lemma**. Divide the vertices of the triangles into three disjoint classes: $\Acl, \Bcl,$ and $\Ccl$. If not face contains vertices of all three types and the boundary of the region contains an odd number of faces of type $\Acl\Bcl$, then some $T_i$ has vertices of all three types.

So, the strategy to prove that a square cannot be divided into an odd number of triangles all of equal area is to find some way to partition the set of vertices of the triangles dividing a square such that the hypotheses of the lemma hold. Then, try and prove that the number of dividing triangles is even given that at least one triangle has vertices of all three types. It's actually quite a conundrum, and the reason why Monsky's proof is so cool is that it actually manages to do this in a very elegant way.

Monsky's strategy is to first just assume that the square is $S = [0,1]\times [0,1]$ without loss of generality. Then, define a partition of $\R^2$ as follows:

- $(x,y)\in \Acl$ iff $\lVert x\rVert \lt 1$ and $\lVert y\rVert \lt 1$
- $(x,y)\in \Bcl$ iff $\lVert x\rVert \geq 1$ and $\lVert x\rVert\geq \lVert y\rVert$
- $(x,y)\in \Ccl$ iff $\lVert y\rVert\geq 1$ and $\lVert y\rVert\geq \lVert x\rVert$

In thes equations, $\lVert\cdot\rVert$ is the extension of the $2$-adic absolute value on the rational numbers to the real numbers. Now assume the square $S$ is divide into $m$ triangles, each with area $1/m$. I won't go into the proof that the vertices of the triangles satisfy the conditions of the lemma (for that, read the paper, but it's not long), but they do. Therefore, there exists a triangle $T_i$ that has vertices of all three types. Here is a short exercise that follows directly from the definition of the sets $\Acl, \Bcl,$ and $\Ccl$:

- Prove that if you take any point $X$ of a given type, translating it by a point of type $\Acl$ does not change the type of $X$.

With this, let us translate our triangle $T_i$ so that the vertex of type $\Acl$ is now the origin, and let $(x,y)$ and $(x',y')$ be the points of type $\Bcl$ and $\Ccl$ respectively. Then the area of the triangle is, up to sign, $\frac{1}{2}(xy' – x'y) = \pm 1/m$. Here is another exercise, again which follows directly from the definition of the point classes $\Acl,\Bcl$ and $\Ccl$:

- Prove that if $(x,y)$ and $(x',y')$ are of type $\Bcl$ and $\Ccl$ respectively, then $\lVert xy'\rVert \gt \lVert x'y\rVert$.

By this exercise, the 2-adic norm of the area of the triangle is

$$\lVert \frac 1 2\rVert\cdot \lVert xy'\rVert$$

since a nonarchimedean norm always satisfies $\lVert x + y\rVert \leq \max\{ \lVert x\rVert,\lVert y\rVert \}$ with equality whenever $\lVert x\rVert\not=\lVert y\rVert$. If $(x,y)\in \Bcl$ and $(x',y')\in \Ccl$ then we also know that $\lVert x\rVert\geq 1$ and $\lVert y\rVert\geq 1$ so that the 2-adic norm of the area of $T_i$ is greater than or equal to $\lVert\frac 1 2\rVert$, which is strictly greater than one. Therefore,

$$\lVert 1/m\rVert \gt 1.$$

This means that $m$ must be even.

[1] Monsky, Paul. On dividing a square into triangles. Amer. Math. Monthly 77 1970 161-164.