On the table lies a rectangular sheet of paper. Sasha cuts it along a straight line into two pieces and puts the pieces on the table. Then he takes one of the pieces, cuts it again along a straight line into two pieces, and puts the pieces back on the table. Then he takes one piece from the table and cuts it again, and so on. What is the minimum number of cuts Sasha needs to make so that there are at least 252 undecagons on the table?
Notice that after one cut, the total number of vertices increases by two (if the cut passes through two vertices), by three (if the cut passes through a vertex and a point inside a side), or by four (if the cut passes through the interior points of two sides). Suppose $k$ cuts have been made, resulting in $k+1$ pieces, in which the total number of vertices is no more than $4k+4$.

Assume that among the resulting pieces, there are 252 undecagons (11-sided polygons). Each of the remaining pieces will have at least three vertices, and the total number of vertices will be no less than $11 \cdot 252 + 3(k+1-252) = 2772 + 3k - 753$

$= 3k + 2019 \leq 4k + 4$. Therefore, $k \geq 2015$.

We will show that 2015 cuts are sufficient. For example, first, cut the original rectangle into 252 rectangles (251 cuts). Now, to obtain 252 undecagons, it is enough to cut off a triangle from each rectangle seven times, increasing the number of vertices by one each time. This will require an additional $7 \cdot 252 = 1764$ cuts. In total, we will have $251 + 1764 = 2015$ cuts.

## Answer

2015 cuts.