Rubaev I.S.

There is a square grid paper with a size of $102 \times 102$ cells and a connected figure of unknown shape, consisting of 101 cells. What is the maximum number of such figures that can be guaranteed to be cut out from this square? A figure composed of cells is called connected if any two of its cells can be connected by a chain of its cells, in which any two adjacent cells share a common side.
Lemma. Any connected figure composed of 101 cells can be enclosed in a rectangle with sides \(a\) and \(b\) such that \(a + b = 102\).

Take two cells of our figure that share a common side. They form a rectangle \(1 \times 2\), the sum of whose sides is 3. Due to the connectivity of the given figure, there will be a cell in it that is adjacent to this rectangle along a side. Attach this cell to the rectangle. The resulting configuration of three cells can be enclosed in a rectangle with the sum of sides equal to 4, if one of the sides of the rectangle \(1 \times 2\) is extended by 1. We will repeat this procedure until the configuration includes all the cells of the figure. The procedure will be performed no more than 99 times, so the sum of the sides of the rectangle in which the figure will eventually be enclosed will be no more than 102.

![](https://cdn.mathpix.com/cropped/2024_05_06_6a190adc6812227a77adg-08.jpg?height=140&width=140&top_left_y=1491&top_left_x=959)

From the lemma, it immediately follows that four figures equal to the given one can always be cut out: for this, it is sufficient to enclose it in a rectangle with the sum of sides 102, and then cut out four such rectangles from the given square as shown in the figure. Now consider a figure in the shape of a cross, each arm of which consists of 25 cells. It contains \(4 \times 25 + 1 = 101\) cells.

If such a cross is cut out from a square, its center must lie outside a border 25 cells wide, adjacent to the boundary of the square. This means that this center must lie within a square of side 52, obtained after removing the border. Divide this square into four equal squares with side 26.

It is easy to see that if several non-overlapping crosses are cut out from a sheet of paper, then in each of these four squares, the center of only one cross can be located (otherwise, two crosses will overlap).

Therefore, it is impossible to cut out more than four crosses from the sheet, which completes the proof.

## Answer

4.00