4A. Around a given rectangle with sides $a$ and $b$, another rectangle is described whose area is $m^{2}$. For what values of the parameter $m$ does the problem have a solution?
Solution. Let $\varphi$ be the angle between the sides of the rectangles $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$. For the lengths of the sides of the circumscribed rectangle $A_{1} B_{1} C_{1} D_{1}$, we have:

$$
\begin{aligned}
& \overline{A_{1} D_{1}}=\overline{A A_{1}}+\overline{A D_{1}}=a \sin \varphi+b \cos \varphi \\
& \overline{A_{1} B_{1}}=\overline{B A_{1}}+\overline{B B_{1}}=a \cos \varphi+b \sin \varphi
\end{aligned}
$$

From the area condition, we have:

$$
m^{2}=(a \sin \varphi+b \cos \varphi)(a \cos \varphi+b \sin \varphi)
$$

and by simplifying the expression, we get

![](https://cdn.mathpix.com/cropped/2024_06_05_84a5520b68919fd51f4cg-08.jpg?height=486&width=488&top_left_y=865&top_left_x=996)

$$
\sin \varphi \cos \varphi=\frac{m^{2}-a b}{a^{2}+b^{2}}
$$

or equivalently

$$
\sin 2 \varphi=\frac{2\left(m^{2}-a b\right)}{a^{2}+b^{2}}
$$

Then the condition for the problem to have a solution will be $0 \leq \sin 2 \varphi \leq 1$, or solving the inequality, we get the desired condition $\sqrt{a b} \leq m \leq \frac{a+b}{\sqrt{2}}$.