## Task 2.

Let $k$ and $\ell$ be natural numbers. Determine the smallest natural number $m$ for which it is possible to divide the square $A B C D$ into $m$ rectangles with sides parallel to the sides of the square such that every line parallel to $A B$ that intersects the interior of the square intersects at least $k$ rectangles, and every line parallel to $B C$ that intersects the interior of the square intersects at least $\ell$ rectangles.
## Solution.

If $k=1$, then the answer is $\ell$, and if $\ell=1$, then the answer is $k$.

For $k, \ell \geqslant 2$, we claim that the answer is $2(k+\ell-2)$.

The figure below shows an example that it is possible to satisfy the conditions of the problem by dividing into $2(k+\ell-2)$ rectangles.

![](https://cdn.mathpix.com/cropped/2024_05_30_fc99728c18f68446dfceg-21.jpg?height=800&width=880&top_left_y=862&top_left_x=588)

We will show that a smaller number of rectangles is not sufficient. The sides $\overline{A B}$ and $\overline{C D}$ must intersect (contain sides of) at least $k$ rectangles. Indeed, there exists a line parallel to $A B$ and so close to this line that it intersects at least $k$ rectangles whose one side lies on $\overline{A B}$. Similarly, the sides $\overline{B C}$ and $\overline{A D}$ must intersect (contain sides of) at least $\ell$ rectangles. Since $k, \ell \geqslant 2$, we have counted twice only those rectangles whose one vertex is also a vertex of the square $A B C D$, and these are exactly four different rectangles. Therefore, in a partition that satisfies the conditions of the problem, there must be at least $2 k+2 \ell-4$ rectangles.