55rd Ukrainian National Mathematical Olympiad - Third Round

Analysis. Let $ABCD$ be our trapezoid with right angles $A$ and $B$. Let $EFGH$ be the required square centered at $O$, and suppose $E\in AB$, $F\in BC$. In the quadrilateral $EBFO$ two opposite angles are right, hence, it's cyclic. This implies that $\angle EFO=\angle EBO=45^{\circ }$ as they intercept the same arc $EO$ (Fig. 4). Likewise, $\angle EAO=45^{\circ }$. Thus $O$ is the intersection point for bisectors of angles $A$ and $B$ in the trapezoid. Also, $E$ and $G$ are symmetric with respect to $O$.

Construction. Construct $O$ as the intersection point of two bisectors drawn from $A$ and $B$. Then we can construct a straight line $l$ symmetric to $AB$ with respect to $O$. The line $l$ intersects the segment $CD$ at a unique point $G$, which is a vertex of our square. Having the center and one vertex of the square, we can reconstruct it in a unique way.



Fig. 4