A Bulgarian Mathematical Olympiad Problem

Problem:

Two unit squares $K_{1}$ ,, $K_{2}$ with centers M, N are situated in the plane so that $MN=4$ . Two sides of $K_{1}$ are parallel to the line MN, and one of the diagonals of $K_{2}$ lies on MN. Find the locus of the midpoint of XY as X, Y vary over the interior of $K_{1}$ , $K_{2}$ respectively.

(1997 Bulgarian mathematical olympiad)

Solution:

Introduce complex numbers with $M=-2$ , $N=2$ . Then, the locus is the set of points of the form $-(w+xi)+(y+zi)$ , where $|w|$ , $|x|<\frac{1}{2}$ , and $|x+y|$ , $|x-y| < \frac{\sqrt{2}}{2}$ . The result is an octagon with vertices $\frac{1+\sqrt{2}}{2}+\frac{i}{2}$ , $\frac{1}{2}+\frac{(1+\sqrt{2})i}{2}$ , and so on.