Problem 1000

I. Solution. a) The center of the square bisects the diagonals, so the vertices of the quadrilateral $O_{1} L O_{2} K=S$ bisect the sides of the quadrilateral $A M P C=T$ (Figure 1). Thus, $O_{1} L$ is the midline of the triangle $A P M$ parallel to $A P$, and $O_{1} L = A P / 2$. Similarly, from the triangle $A P C$, $K O_{2} \| A P$ and $K O_{2} = A P / 2$, and thus $O_{1} L \# K O_{2}$. Similarly, $O_{1} K \# L O_{2}$, which means $S$ is a parallelogram.

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Figure 1

$S$ is a square if and only if two adjacent sides are perpendicular and equal in length. It suffices to show that the diagonals $A P$ and $M C$ of $T$ are perpendicular and equal. By adding the vertex $B$ to the endpoints, the triangles $A B P$ and $M B C$ are congruent, as they can be rotated $90^{\circ}$ around $B$ to map one onto the other. Indeed, the circumferences of the squares $A B M N$ and $P B C Q$ are the same, so $B A$ can be rotated $90^{\circ}$ around $B$ to map to $B M$, just as $B P$ maps to $B C$. This rotation thus maps the triangle $A B P$ onto the triangle $M B C$. This is valid even if $\angle A B C = 90^{\circ}$, in which case the triangles $A B P$ and $M B C$ degenerate into right-angled isosceles triangles. Therefore, the segments $A P$ and $M C$ are perpendicular and equal, as a suitable $90^{\circ}$ rotation maps one onto the other. From this, as we have seen, the statement of part a) of the problem follows.

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Figure 2

The proof is valid whether the angle at $B$ in the triangle $A B C$