7.4. Vera Aleksandrovna urgently needed to cut out three 20-sided polygons (not necessarily identical) from one rectangular sheet of paper. She can take this sheet and cut it along a straight line into two parts. Then take one of the resulting parts and cut it along a straight line. Then take one of the available pieces, cut it, and so on. What is the minimum number of cuts Vera Aleksandrovna will have to make to ensure that among the resulting pieces there are the three 20-sided polygons she needs?
Answer: 50 cuts.

Solution: With each cut, the total number of paper pieces increases by 1 (one piece turns into two new pieces), so after $n$ cuts, there will be $(n+1)$ pieces of paper. Let's calculate how many vertices all the pieces together can have after $n$ cuts. With each cut, the total number of vertices increases by 2 (if cutting through two vertices), 3 (if cutting through a vertex and a side), or 4 (if cutting through sides). Since there were initially 4 vertices, after $n$ cuts, the total number of vertices in all pieces together will not exceed $4 n+4$.

Suppose that after $n$ cuts, we have found three 20-sided polygons. Since the total number of pieces obtained will be $n+1$, there will be $n+1-3$ pieces in addition to these 20-sided polygons. Each of these pieces has at least three vertices, so the total number of vertices is at least $20 \cdot 3+3(n-2)=3 n+54$.

Thus, $4 n+4 \geqslant 3 n+54$, from which $n \geqslant 50$.

Now let's show how to get three 20-sided polygons with 50 cuts. Here is one way: cut the original sheet into three rectangles (2 cuts) and convert each rectangle into a 20-sided polygon with 16 cuts by cutting off triangles from the corners $(3 \times 16=48$ cuts $)$. In total, exactly 50 cuts.

Criteria: Only the answer - 0 points.

Example of 50 cuts - 2 points.

Proof of the minimality of 50 - 5 points. These points are combined from proving the following statements:

- After $n$ cuts, the number of vertices is no more than $4 n+4$ - 2 points;
- After $n$ cuts, the number of vertices is no less than $3 n+54$ - 2 points;
- From the two previous estimates, $n \geqslant 50$ - 1 point.