There is a circle and a point which is not on the circle. Construct a square, one of whose vertices is the given point, and one of whose sides is a chord of the circle.
I. Solution: Let the given point be denoted by $A$ and the given circle with center $O$ and radius $r$ be denoted by $k$. Let $A B C D$ be the desired square and let the side $B C$ be a chord of $k$. Imagine the problem is solved (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_c83877562015a30ac8a7g-1.jpg?height=542&width=529&top_left_y=266&top_left_x=782)

Figure 1

The diagonal $A C$ is the result of rotating the side $A B$ by $45^{\circ}$ around $A$ and then stretching it by a ratio of $\sqrt{2}: 1$. This transformation, however, maps the circle $k$ to another circle $k^{\prime}$ whose center $O^{\prime}$ is the result of applying the same transformation to the distance $A O$, and whose radius is $r \sqrt{2}$. Therefore, the vertex $C$ lies on both the circle $k$ and the circle $k^{\prime}$.

Accordingly, the construction process is as follows: The distance $A O$ is rotated by $45^{\circ}$ in both directions and stretched by a ratio of $\sqrt{2}: 1$, yielding the points $O^{\prime}$ and $O^{\prime \prime}$. (The simplest way is to measure the distance $O A=O O^{\prime}=O O^{\prime \prime}$ on the perpendicular to $O A$ at $O$ in both directions.) Circles with centers $O^{\prime}$ and $O^{\prime \prime}$ and radius $r \sqrt{2}$ (Figure 2) intersect the circle $k$ at points $C_{1}^{\prime}, C_{2}^{\prime}$ and $C_{1}^{\prime \prime}, C_{2}^{\prime \prime}$. (The construction of points $B$ and $D$ is trivial.) The latter two solutions are the reflections of the former two across the axis $O A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c83877562015a30ac8a7g-1.jpg?height=675&width=569&top_left_y=1289&top_left_x=775)

Figure 2

The condition for a solution to exist is that the circles $k$ and $k^{\prime}$ (and similarly $k$ and $k^{\prime \prime}$) have a common point, which means that $O A \leq r \sqrt{2} + r$ and, if $A$ is inside the circle $k$, $O A \geq r \sqrt{2} - r$. Therefore, a solution exists if

$$
r(\sqrt{2}-1) \leq O A \leq r(\sqrt{2}+1)
$$

If $O A$ does not satisfy either of the above inequalities, there is no solution. If $O A$ equals one of the bounds, there are 2 solutions; otherwise, there are 4 solutions, except when $O A = r$ (i.e., when $A$ is on the circle $k$), in which case $C_{1}^{\prime}$ and $C_{1}^{\prime \prime}$ coincide at the point opposite to $A$ (the two solutions degenerate into a square inscribed in $k$) and $C_{2}^{\prime}$ and $C_{2}^{\prime \prime}$ coincide at $A$ (the two solutions degenerate into a point).

Gyula Beke (Hatvan, Bajza József g. I. o. t.)

II. Solution: Imagine the problem is solved, but for simplicity, we omit the irrelevant point $D$ (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_c83877562015a30ac8a7g-2.jpg?height=406&width=575&top_left_y=125&top_left_x=756)

Figure 3

Let the second intersection points of the side $A B$ and the diagonal $A C$ with the circle $k$ be $C^{*}$ and $B^{*}$, respectively. The right angle at $B$ implies, by Thales' theorem, that $C C^{*}$ is a diameter of $k$, and thus, by Thales' theorem again, $C^{*} B^{*}$ is perpendicular to $A B^{*}$. Since $\angle B^{*} A C^{*} = 45^{\circ}$, the triangle $A B^{*} C^{*}$ is an isosceles right triangle (i.e., $B^{*} C^{*}$ is another solution to the problem), and thus $\angle B^{*} C^{*} B = 135^{\circ}$, which implies that $\angle B O B^{*} = 90^{\circ}$.

By constructing the square $B O B^{*} K$ and a circle with center $K$ and radius $r$, this circle not only passes through points $B$ and $B^{*}$ but also through point $A$, since the arc $B B^{*}$ subtends an angle of $45^{\circ}$ at $A$.

We use this construction for the case where the point $A$ lies inside the given circle. Based on $O K = r \sqrt{2}$ and $A K = r$, we construct the point $K$, for which we get two solutions: $K^{\prime}$ and $K^{\prime \prime}$ (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_c83877562015a30ac8a7g-2.jpg?height=639&width=637&top_left_y=1041&top_left_x=733)

Figure 4

Circles with centers $K^{\prime}$ and $K^{\prime \prime}$ and radius $r$ intersect the circle $k$ at points $B_{1}^{\prime}, B_{2}^{\prime}$ and $B_{1}^{\prime \prime}, B_{2}^{\prime \prime}$.