[ Extreme properties of a triangle (miscellaneous).]

In a triangle, two sides $a$ and $b$ are known. What should the third side be so that the smallest angle of the triangle has the greatest possible magnitude?
Let's assume for definiteness that $a \geq b$. Then the smallest angle of the triangle is angle $B$ or angle $C$. Consider a semicircle $S$ with radius $a$. Let $C$ be the center of this semicircle, and $B$ be a point on the extension of the diameter such that $CB = a$. Draw a tangent $BA_1$ from point $B$ to the semicircle $S$. If $\angle A_1CB \geq \angle A_1BC$, then the smallest angle of the triangle will be the largest if vertex $A$ is in the position of $A_1$. Indeed, angle $B$ is always no greater than angle $A_1BC$. The inequality $\angle A_1CB \geq \angle A_1BC$ is equivalent to $b \leq c = \sqrt{a^2 - b^2}$, i.e., $a \geq \sqrt{2}b$. Thus, if $a \geq \sqrt{2}b$, the third side must be equal to $\sqrt{a^2 - b^2}$.

Now suppose that $a < \sqrt{2}b$. Then, in particular, $a < 2b$, so the perpendicular bisector of segment $BC$ intersects the semicircle at some point $A_2$. The smallest angle of the triangle will be the largest if vertex $A$ is in the position of $A_2$. Indeed, suppose first that point $A$ moves along the semicircle from the position of $A_2$ such that its projection on $BC$ moves towards point $B$. Then the smallest angle will be angle $C$, and it will decrease. Suppose now that point $A$ moves such that its projection moves away from point $B$. Then the smallest angle will be angle $B$, and it will also decrease. (This follows from the fact that in the considered situation $\angle CBA_2 \leq \angle CBA_1$). Thus, if $b \leq a < \sqrt{2}b$, the third side must be equal to $b$.

A square sheet of paper was cut along a straight line into two parts. One of the resulting parts was again cut into two parts, and so on many times. What is the minimum number of cuts required to ensure that among the resulting parts there are exactly 100 twenty-sided polygons?

## Solution

Answer: It is possible to obtain exactly 100 twenty-sided polygons by making 1699 cuts, and it is impossible to obtain 100 twenty-sided polygons with fewer cuts. With each cut, the total number of paper pieces increases by 1 (since one piece disappears and two new ones appear), so after $n$ cuts, there will be $(n+1)$ pieces of paper. Now let's count how many vertices can be in all the pieces together after $n$ cuts. With each cut, the total number of vertices increases by either 2 (if cutting through two vertices), 3 (if cutting through a vertex and a side), or 4 (if cutting through two sides). Since there were initially 4 vertices, after $n$ cuts, there will be no more than $4n + 4$ vertices in all the pieces together. Suppose that after $N$ cuts, 100 twenty-sided polygons were obtained. Since the total number of resulting pieces will be $N+1$, there will be $N+1-100$ additional pieces. Each of these pieces will have at least three vertices, so the total number of vertices in all the pieces will be at least $100 \cdot 20 + (N-99) \cdot 3$. As proven earlier, this number is no more than $4N + 4$. Therefore, $4N + 4 \geq 100 \cdot 20 + (N-99) \cdot 3 = 3N + 1703$, from which $N \geq 1699$. Thus, we have proven that it is impossible to obtain 100 twenty-sided polygons with fewer than 1699 cuts. This is the main and most difficult part of the proof.

Now let's show how it is possible to obtain 100 twenty-sided polygons with 1699 cuts. Here is one way: cut the square into 100 rectangles (99 cuts) and convert each rectangle into a twenty-sided polygon with 16 cuts by cutting off triangles from the corners (1600 cuts). In total, there will be 1699 cuts.

## Answer

It is possible to obtain exactly 100 twenty-sided polygons by making 1699 cuts, and it is impossible to obtain 100 twenty-sided polygons with fewer cuts.