2. Given three positive numbers: $a, b, c$. Petya wrote on the board the numbers $\frac{1}{a}+b c, \frac{1}{b}+a c, \frac{1}{c}+a b, a$ and Vasya wrote the numbers $2 a^{2}, 2 b^{2}, 2 c^{2}$. It turned out that both wrote the same three numbers (possibly in a different order). What is the product $a b c?$ (N. Agakhanov)
Answer. 1. First solution. Multiplying the expressions $\frac{1}{a}+b c, \frac{1}{b}+a c, \frac{1}{c}+a b$, we obtain the product $\frac{(a b c+1)^{3}}{a b c}$, while multiplying $2 a^{2}, 2 b^{2}, 2 c^{2}-8(a b c)^{2}$. According to the condition, the obtained products are equal. Equating them and multiplying both sides of the equation by $a b c$, we get $8(a b c)^{3}=(a b c+1)^{3}$, from which $2 a b c=a b c+1$ and $a b c=1$. Second solution. Let $a \geq b \geq c$ (other cases are similar). Then, as it is easy to see, $\frac{1}{a}+b c \leq \frac{1}{b}+a c \leq \frac{1}{c}+a b$. On the other hand, $2 a^{2} \geq 2 b^{2} \geq 2 c^{2}$. Hence, $\frac{1}{c}+a b=2 a^{2}, \frac{1}{b}+a c=2 b^{2}, \frac{1}{a}+b c=2 c^{2}$. Multiplying the first equation by $c$, the second by $b$, and the third by $a$, we get that $a b c+1=2 a^{2} c=2 b^{3}=2 c^{2} a$, from which $a=c=b$ and $a b c+1=a^{3}+1=2 a^{3}$, that is, $a=c=b=1$ and $a b c=1$.