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.

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

Thanks Prof. Andreescu !

Nalin Pithwa

 

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