Monsky's Theorem

Summary

\(\newcommand{\RR}{\Bbb R} \newcommand{\QQ}{\Bbb Q} \newcommand{\ZZ}{\Bbb Z}\) For which \(n\) can you cut a square into \(n\) triangles of equal area? This question appears quite simple; it could have been posed to the Ancient Greeks. But like many good puzzles, it is a remarkably stubborn one. It was first solved in 1970, by Paul Monsky. Despite the completely geometric nature of the question, his proof relies primarily on number theory and combinatorics! There is a small amount of algebraic machinery involved, but his proof is quite accessible, and we will describe it below. If you have a napkin on hand, it should be straightforward to come up with a solution for \(n = 2\) and \(4\). A little more thought should yield solutions for any even \(n\). One such scheme is depicted below: But when \(n\) is odd, you will have considerably more trouble. Monsky’s theorem states that such a task is, in fact, impossible. Monsky's Theorem The unit square cannot be dissected into an odd number of triangles of equal area. The result clearly extends to squares of any size, and in fact, arbitrary parallelograms. There are two key ingredients here: Sperner’s Lemma 2-adic valuations Proof sketch: Color the vertices of the dissection using three colors Find a triangle with exactly one vertex of each color Show that such a triangle cannot have area \(1/n\) If the last step seems ridiculous to you, don’t worry. It’s completely non-obvious that the coloring of a triangle’s vertices could at all be related to its area. But once you see the trick, it will (hopefully) seem less mysterious. Just hang in there. Sperner’s Lemma Consider a polygon \(P\) in the plane, and some dissection of it into triangles \(T_i\). As promised in the previous section, color the vertices with three colors; we’ll use red, green, and blue. We will call a segment purple if it has one red and one blue endpoint. A triangle with exactly one corner of each color will be called trichromatic. (Great terminology, eh?) ...