Using your Head is Permitted

May 2012 solution

There are no such P and Q. The area of a polygon P with vertices in Z2 is completely determined by the number of its interior and boundary points as follows: AP=IP+BP/2-1.

This result was first shown by Georg Pick in 1899, and is known as Pick's Theorem.

A simple argument for Pick's Theorem is the following. First, consider three simple polygons, X, Y and Z, such that Z is the disjoint union of X and Y. For the union to be disjoint, X and Y cannot share any area, and for Z to be simple, they must share exactly one path of edges. Considering that both X and Y share all Z2 points on this path, and that all except the first and last become interior points in Z, it is not difficult to ascertain that if Pick's Theorem holds for any two of the three polygons, it must also hold for the third. What's more, if X and Y have AX=AY, IX=IY and BX=BY, then if Pick's Theorem holds for Z, it must hold also for both X and Y.

Considering that any polygon can be triangulated, a corollary to the previous observation is that if Pick's Theorem holds for all triangles, it must also hold for all polygons.

Note now that by attaching axis-parallel right-angled triangles to any triangle, one can make a rectangle, and, furthermore, that any rectangle can be divided into two axis-parallel right-angled triangles (by splitting it across the diagonal). The conclusion is that if Pick's Theorem holds for all axis-parallel right-angled triangles, it also holds for any triangle, and therefore for every polygon.

Consider again the construction of the rectangle split into two axis-parallel right-angled triangles. Clearly, these triangles have the same area, number of boundary points and number of interior points, and their union clearly satisfies Pick's Theorem. This means that the theorem is also satisfied for the original triangles. Because this construction can be done with any axis-parallel right-angled triangle, this concludes the proof.


For Z3 the same does not hold. The Reeve Tetrahedron provides a simple counterexample.

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