Let's construct a square, given its center and one point on each of two adjacent sides.
I. Solution. We construct a square whose center is the given center $O$, and one of its vertices is the given point $P$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8d7b0a62d8d7cd43a7a9g-1.jpg?height=410&width=423&top_left_y=234&top_left_x=840)

The vertex $R$ of the thus obtained square $P R S T$ is connected to the other given point $Q$, and perpendiculars are dropped from $P$ and $S$ to $Q R$, yielding the vertices $A$ and $B$ of the sought square $A B C D$. If $Q$ is connected to $T$, then the square $E F G H$ is obtained.

(Konstantin Neubauer, Budapest.)

II. Solution. Consider the problem as solved. Let $A B C D$ be the sought square, and $A_{1}$ and $B_{1}$ be the two given points.

![](https://cdn.mathpix.com/cropped/2024_05_02_8d7b0a62d8d7cd43a7a9g-1.jpg?height=429&width=463&top_left_y=978&top_left_x=820)

$A_{1} O$ and $B_{1} O$ intersect the opposite sides at $A_{2}$ and $B_{2}$. Draw perpendiculars to $A_{1} A_{2}$ and $B_{1} B_{2}$ at $O$. This gives us the points $A_{3}, A_{4}$ and $B_{3}, B_{4}$ on the sides of the square. Since $O A_{1}=O A_{2}=O A_{3}=O A_{4}$ and $O B_{1}=O B_{2}=O B_{3}=O B_{4}$, the construction is as follows: from $O$, draw circles with radii $O A_{1}$ and $O B_{1}$, and draw perpendiculars to $O A_{1}$ and $O A_{2}$ at $O$, which yield the points $A_{3}, A_{4}$ on one circle and $B_{3}, B_{4}$ on the other. $A_{1} B_{3}$ and $B_{1} A_{3}$ intersect at the vertex $B$ of the sought square.

(Rezső Földes, Budapest.)

Number of solutions: 36.