A square has two adjacent vertices lying on a unit circle. What are the possible minimum and maximum distances of the other two vertices from the center of the circle?

---

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
I. solution. The other two vertices of the square are at the same distance from the center of the circle due to symmetry. The key to the simple solution is the following rephrasing of the problem:

Given a unit radius circle $k$ and a point $A$ on it. The positively oriented square $A B C D$ has its vertex $B$ on the circle $k$. What are the possible minimum and maximum distances of the vertex $D$ from the center $O$ of the circle?

The vertex $D$ is obtained by a positive $90^{\circ}$ rotation around $A$ from $B$. Therefore, as the vertex $B$ traverses the points of the circle $k$ different from $A$, the geometric locus of the vertex $D$ consists of the points of the circle $k^{\prime}$ with center $O^{\prime}$, different from $A$, which is created from $k$ by the same positive $90^{\circ}$ rotation around $A$ (Figure 1). The extreme positions of the $D$ points lie on the line $O O^{\prime}$. Since the triangle $O A O^{\prime}$ is isosceles and right-angled, $O O^{\prime}=\sqrt{2}$. Therefore, the possible minimum distance of the point $D$ from $O$ is $\sqrt{2}-1$, and the possible maximum distance is $\sqrt{2}+1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0baa61f3aefe67d51f80g-1.jpg?height=384&width=618&top_left_y=548&top_left_x=735)

Figure 1

II. solution. Let us set up a Cartesian coordinate system such that the origin is the center $O$ of the given circle, and the coordinates of the two vertices of the square on the circle are $A(\cos t, \sin t)$ and $B(\cos t, -\sin t)$, where $0 < t \leq \pi / 2$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_0baa61f3aefe67d51f80g-1.jpg?height=442&width=596&top_left_y=1194&top_left_x=751)

Figure 2

Then the sides of the square are parallel to the coordinate axes, and their length is $2 \sin t$. Therefore, the coordinates of the other two vertices are either $C_{j}(\cos t + 2 \sin t, -\sin t)$ and $D_{j}(\cos t + 2 \sin t, \sin t)$, or $C_{b}(\cos t - 2 \sin t, -\sin t)$ and $D_{b}(\cos t - 2 \sin t, \sin t)$, depending on whether the vertices are on the same side as $O$ or on the opposite side of the line $AB$. Clearly, $O C_{j} = O D_{j} \geq O C_{b} = O D_{b}$, so we need to find the maximum distance among the segments of type $O C_{j}$ and the minimum distance among the segments of type $O D_{b}$. Since squaring preserves the order of positive numbers, it is sufficient to determine the extreme values of the squares of the distances. Thus, on the one hand, we need to find the maximum of the function

$$
O C_{j}^{2} = (0 - (\cos t + 2 \sin t))^2 + (0 + \sin t)^2 = c(t)
$$

and on the other hand, the minimum of the function

$$
O D_{b}^{2} = (0 - (\cos t - 2 \sin t))^2 + (0 - \sin t)^2 = d(t)
$$

both over the interval $0 < t \leq \pi / 2$.

Using known trigonometric identities, we get

$$
\begin{aligned}
c(t) & = (\cos t + 2 \sin t)^2 + \sin^2 t = 1 + 4 \sin^2 t + 4 \sin t \cos t = \\
& = 1 + 4 \sqrt{2} \sin t \left( \frac{\sqrt{2}}{2} \sin t + \frac{\sqrt{2}}{2} \cos t \right) = 1 + 4 \sqrt{2} \sin t \sin (t + \pi / 4) = \\
& = 1 + 2 \sqrt{2} (\cos (t - (t + \pi / 4)) - \cos (t + (t + \pi / 4))) = \\
& = 1 + 2 \sqrt{2} \left( \frac{\sqrt{2}}{2} - \cos (2 t + \pi / 4) \right) = 3 - 2 \sqrt{2} \cos (2 t + \pi / 4)
\end{aligned}
$$

and

$$
\begin{aligned}
d(t) & = (\cos t - 2 \sin t)^2 + \sin^2 t = 1 + 4 \sin^2 t - 4 \sin t \cos t = \\
& = 1 + 4 \sqrt{2} \sin t \left( \frac{\sqrt{2}}{2} \sin t - \frac{\sqrt{2}}{2} \cos t \right) = 1 + 4 \sqrt{2} \sin t \sin (t - \pi / 4) = \\
& = 1 - 2 \sqrt{2} (\cos (t + (t - \pi / 4)) - \cos (t - (t - \pi / 4))) = \\
& = 1 - 2 \sqrt{2} \left( \cos (2 t - \pi / 4) - \frac{\sqrt{2}}{2} \right) = 3 - 2 \sqrt{2} \cos (2 t - \pi / 4)
\end{aligned}
$$

Since for all $x$, $-1 \leq \cos x \leq 1$, it follows that

$$
c(t) \leq 3 + 2 \sqrt{2} \quad \text{and} \quad d(t) \geq 3 - 2 \sqrt{2}.
$$

If $0 < t \leq \pi / 2$, then equality holds exactly once in both inequalities, at $t = 3 \pi / 8$ and $t = \pi / 8$, respectively. Therefore, the maximum of $c(t)$ is $3 + 2 \sqrt{2} = (1 + \sqrt{2})^2$, and the minimum of $d(t)$ is $3 - 2 \sqrt{2} = (\sqrt{2} - 1)^2$.

Thus, the distance of the other two vertices of the square from $O$ is at least $\sqrt{2} - 1$ and at most $\sqrt{2} + 1$.

Remark. It is of course difficult to come up with the simple solution I. The "usual" solution to the problem is the second solution. In this solution, the extreme values of the functions $c(t)$ and $d(t)$ can also be determined by differentiation.