A rectangularly paper is divided in polygons areas in the following way: at every step one of the existing surfaces is cut by a straight line, obtaining two new areas. Which is the minimum number of cuts needed such that between the obtained polygons there exists $251$ polygons with $11$ sides?
Indeed, we can obtain the polygons from $2007$ cuts: we first obtain $251$ quadrilaterals from the intial one with $250$ cuts. Observe that each $n$-gone can be decomposed in a $(n+1)$-gone and a triangle (cut the interior of two consecutive sides). So each quadrilateral can be processed up to a $11$-gon with $7$ cuts. We have a total of $250+251 \cdot 7 = 2007$ cuts.
For the second part of the proof (that $2007$ is minimum), denote $n$ the minimum number of cuts needed to obtain the polygons, $v$ the total number of vertices and $k$ the number of polygons which are not $11$-gons. So $n = 250+k$, and each of the $n$ cuts adds at most $4$ vertices, hence $4n+4 \geq v \geq 3k+251 \cdot 11$, which leads to $4n+4 \geq 3 \cdot (n-250)+251 \cdot 11$ $\Leftrightarrow$ $n \geq 2007$. This completes our proof.