# Using your Head is Permitted

## May 2012 solution

There are no such *P* and *Q*. The area of a polygon *P* with
vertices in Z^{2} is completely determined by the number of its
interior and boundary points as follows:
*A*_{P}=*I*_{P}+*B*_{P}/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 Z^{2} 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
*A*_{X}=*A*_{Y},
*I*_{X}=*I*_{Y} and
*B*_{X}=*B*_{Y}, 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.

Q.E.D.

For Z^{3} the same does not hold. The
Reeve Tetrahedron
provides a simple counterexample.

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