Kanel-Belov A.Y.

A unit square is divided into a finite number of smaller squares (the sizes of which may vary). Can the sum of the perimeters of the squares intersecting the main diagonal be greater than 1993? (If a square intersects the diagonal at a single point, this is also considered an intersection.)
Let's divide the unit square into four equal smaller squares. The squares that intersect the diagonal only at a vertex will be called squares of the first level. Each of the remaining squares will be divided into four smaller squares. The squares that intersect the diagonal at a vertex will be called squares of the second level. We will again divide each of the remaining squares into four smaller squares, and so on (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_1cbe8f655c608884328cg-31.jpg?height=380&width=466&top_left_y=1365&top_left_x=796)

We will perform this operation 500 times. We will get $2^{k}$ squares of the $k$-th level. The side length of each of them is $2^{-k}$. Therefore, their total perimeter is 4. Consequently, the total perimeter of all squares of all levels is $4 \cdot 500 > 1993$.

## Answer

Yes.