11.5. a) A square of size 1 by 1 is divided into 25 not necessarily identical rectangles, each of which has the same perimeter \( \boldsymbol{p} \). Find the minimum and maximum possible value of \( \boldsymbol{p} \). b) Can a unit square be divided into 30 not necessarily identical rectangles with a perimeter of 2?
Answer. a) The minimum value of $p$ is $0.8=4/5$, the maximum value of $p$ is $2.08=2+2/25$. b) Yes, it is possible, the method is indicated in the solution.

Solution. a) One of the rectangles in the partition must have an area of at least $1/25$. Let the sides of this rectangle be $x$ and $y$. By the inequality of the arithmetic mean and the geometric mean, we have $\quad \frac{x+y}{2} \geq \sqrt{xy} \geq \sqrt{\frac{1}{25}}=\frac{1}{5}, \quad$ so the perimeter of this rectangle, which is $2(x+y)$, is not less than $\frac{4}{5}=0.8$. This value is achieved for the partition of the square into 25 identical squares with side length 0.2.

On the other hand, the partition of the square into 25 equal rectangles with sides 1 and $1/25$ gives an example of $p=2.08$, so the maximum $p$ is definitely greater than 2. In any partition of the unit square into 25 rectangles, there will be a rectangle with an area of no more than $1/25$, let its sides be $x \leq 1$ and $y \leq 1$, and $x+y=\frac{p}{2}>1$, and $x \geq \frac{p}{2}-y \geq \frac{p}{2}-1$. Therefore, there must be an $x$ in the interval $\left[\frac{p}{2}-1,1\right]$ such that $x\left(\frac{p}{2}-x\right) \leq \frac{1}{25}$. This function is quadratic with a negative leading coefficient, so its minimum on the interval is achieved at one of the endpoints of this interval. The corresponding values at the endpoints are $\frac{p}{2}-1$. Therefore,

$$
\frac{p}{2}-1 \leq \frac{1}{25} \quad \text{and} \quad p \leq 2.08
$$

b) We describe a method for partitioning the unit square into 30 rectangles with a perimeter of 2. Place four of them with sides $x<\frac{1}{2}$ and $1-x$ along the sides of the square so that a square with side $1-2x$ remains in the center. Divide this square into 26 equal rectangles with sides $1-2x$ and $\frac{1-2x}{26}$, their perimeters are $\frac{1-2x}{13}+2-4x$, which should equal 2, from which $x=\frac{1}{54}$. Note that in this partition, there are only two types of rectangles, and this method allows partitioning the square into any number of rectangles with a perimeter of 2, not less than 4.

Grading criteria. In part a), each estimate is worth 2 points. If one or both of the exact examples (one or both) of achieving the estimates are not provided, deduct 1 point. Part b) is worth 3 points.

If correct examples of partitions for the correct minimum and maximum values of "p" (both!) are provided: 1 point.

If there are attempts to justify the minimality and maximality of these values
using considerations such as: "the maximum area of a rectangle with a fixed perimeter is achieved for a square" and "the minimum area of a rectangle with a fixed perimeter is achieved for the most elongated rectangle, when one of its sides is $1$": another 1 point.

And, if the principle of Dirichlet was used to find rectangles with areas not less than $1/25$ and not more than $1/25$ in the partition, and these considerations were correctly applied to estimate the minimum and maximum values of "p": another 2 points.

If there are inaccuracies: deduct 1-2 points.