# 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.

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.

Ref: Complex Numbers from A to …Z by Titu Andreescu and Dorin Andrica.

Thanks Prof. Andreescu !

Nalin Pithwa

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