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.

…read the rest of this post!