Find the smallest perimeter unit area rectangle $H$, for which there exists a rectangle $H_{1}$ whose perimeter is $50\%$ smaller than that of $H$, and whose area is $50\%$ larger than that of $H$.
Solution. Let the sides of rectangle $H$ be denoted by $a, b$, and those of rectangle $H_{1}$ by $c, d$. We know that $T_{H}=a b=1$, $K_{H}=2(a+b)$, $T_{H_{1}}=c d=1.5$ (50% larger than the area of $H$), and $K_{H_{1}}=2(c+d)=a+b$ (50% smaller than the perimeter of $H$). We seek the minimum of $K_{H}=2(a+b)=4(c+d)$ under these conditions. The values of $a, b, c, d$ are lengths of segments, so they are all positive. Thus, we can divide by $d$: $c=\frac{1.5}{d}$, and combining this with the previous relationships, we get $K_{H}=4(c+d)=\frac{6}{d}+4d$.

Using the arithmetic and geometric mean inequality, we can estimate $K_{H}$, as we know that

$$
\frac{6}{d}+4d \geq 2 \cdot \sqrt{\frac{6}{d} \cdot 4d}, \quad \text{that is} \quad \frac{6}{d}+4d \geq 4 \sqrt{6}
$$

and equality holds if and only if $\frac{6}{d}=4d$. Therefore, the perimeter of $H$ is at least $4 \sqrt{6}$. We will show that this estimate is sharp, i.e., there exists a rectangle $H$ with this perimeter that satisfies all the conditions.

We saw that equality can only hold if $\frac{6}{d}=4d$, which means $4d^2=6$. Since $d$ is positive, $d=\frac{\sqrt{6}}{2}$, and thus $c=\frac{1.5}{d}=\frac{\sqrt{6}}{2}$. Therefore, $H_{1}$ is a square with side length $\frac{\sqrt{6}}{2}$.

$$
\begin{aligned}
& \frac{H}{\sqrt{6}+\sqrt{5}} \sqrt{6}-\sqrt{5} \\
& t=1 \quad k=4 \sqrt{6}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_a36a4839ff7563b39309g-1.jpg?height=252&width=437&top_left_y=871&top_left_x=1191)

The sides of $H$, $a$ and $b$, are reciprocals of each other. If the perimeter of $H$ is $4 \sqrt{6}$, that is, $2\left(\frac{1}{b}+b\right)=4 \sqrt{6}$, then $2\left(1+b^{2}\right)=4 \sqrt{6} b$, and after rearranging, we get the quadratic equation $2 b^{2}-4 \sqrt{6} b+2=0$, whose roots are:

$$
b_{1,2}=\frac{4 \sqrt{6} \pm \sqrt{96-16}}{4}=\sqrt{6} \pm \sqrt{5}
$$

The product of the roots, by the known identity, is $6-5=1$, meaning the two roots define the same rectangle. Therefore, the sides of the minimal perimeter rectangle $H$ that satisfies the conditions are $\sqrt{6}+\sqrt{5}$ and $\sqrt{6}-\sqrt{5}$.

Remark. Instead of directly using the arithmetic and geometric mean inequality, some referred to a known result: if one rectangle's perimeter is half of the other, then their perimeters will be minimal at the same time. This reduces the problem to finding which rectangle has the minimal perimeter among those with a given area. This is the square (as also shown in the above solution, that $H_{1}$ is a square), which can be proven using the arithmetic and geometric mean inequality. Others arrived at the correct result by examining the discriminant of a quadratic equation.