In a convex 1950-gon, all diagonals are drawn. They divide it into polygons. Among them, we take the polygon with the largest number of sides. What is the maximum number of sides it can have?

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Answer: 1949. The same reasoning as in solving problem 1 for 7-8 grades shows that the resulting polygon has no more than 1950 sides, and if the number of its sides is 1950, then exactly two diagonals emanate from each vertex of the original polygon, bounding the resulting polygon. Let two diagonals $A_{1} A_{\mathrm{p}}$ and $A_{1} A_{\mathrm{q}}$ emanate from vertex $A_{1}$, bounding the resulting polygon. Then $A_{\mathrm{p}}$ and $A_{\mathrm{q}}$ are adjacent vertices, since otherwise there would be a diagonal inside the angle $A_{\mathrm{p}} A_{1} A_{\mathrm{q}}$ that would cut the resulting polygon. Indeed, a vertex lying between $A_{\mathrm{p}}$ and $A_{\mathrm{q}}$ would have to be connected to a vertex lying between $A_{1}$ and $A_{\mathrm{p}}$ or between $A_{1}$ and $A_{\mathrm{q}}$. By changing the direction of vertex numbering if necessary, we can assume that $q=p+1$ and $p \leq 1950 / 2=975$. If we exclude the diagonal $A_{1} A_{p}+1$, then any other diagonal bounding the resulting polygon connects one of the vertices numbered from 2 to $p$ with some vertex. Therefore, the resulting polygon can have no more than $1+974 \cdot 2=1949$ sides. To obtain an example of a 1950-gon, the cutting of which results in a 1949-gon, one can take a regular 1949-gon and cut off a small triangle, i.e., instead of vertex $A_{1}$, take two vertices $A_{1}$ ' and $A_{1950}$, located on sides $A_{1} A_{2}$ and $A_{1} A_{1949}$ near vertex $A_{1}$.

Inequality

$$
A a(B b+C c)+B b(C c+A a)+C c(A a+B b)>\frac{1}{2}\left(A B c^{2}+B C a^{2}+C A b^{2}\right)
$$

where $a>0, b>0, c>0$ are given numbers, holds for all $A>0, B>0, C>0$. Can a triangle be formed from segments $a, b, c$?

## Solution

Answer: yes, it can. Let $A=B=1, C=\varepsilon$. Then

$$
a(b+\varepsilon c)+b(\varepsilon c+a)+\varepsilon c(a+b)>\frac{1}{2}\left(c^{2}+\varepsilon a^{2}+\varepsilon b^{2}\right)
$$

Therefore, the inequality $2 a b \geq \frac{1}{2} c^{2}$ must hold, since otherwise inequality (1) would not hold for small $\varepsilon$. Thus, $c^{2} \leq 4 a b \leq (a+b)^{2}$. The numbers $a, b, c$ are positive, so $c \leq a+b$. The inequalities $a \leq b+c$ and $b \leq a+c$ are proved similarly.