1. The quadratic trinomials $P(x), Q(x)$, and $R(x)$ are such that each of the equations $P(x)+Q(x)=0, P(x)+R(x)=0, Q(x)+R(x)=0$, and $P(x)+Q(x)+R(x)=0$ has two roots. For each equation, the product of its roots was found. For the first three equations, these products are equal to $r, q$, and $p$ respectively. What is the product of the roots of the fourth equation?
Answer: $\frac{p+q+r}{3}$.

Solution: Let $P(x)=x^{2}+a_{1} x+a_{0}, Q(x)=x^{2}+b_{1} x+b_{0}, R(x)=x^{2}+c_{1} x+c_{0}$. Since $P(x)+Q(x)=2 x^{2}+\left(a_{1}+b_{1}\right) x+a_{0}+b_{0}$, it follows from Vieta's theorem that the product of the roots of this quadratic trinomial is $\frac{a_{0}+b_{0}}{2}$, so $r=\frac{a_{0}+b_{0}}{2}, a_{0}+b_{0}=2 r$. Similarly, we obtain that $a_{0}+c_{0}=2 q, b_{0}+c_{0}=2 p$. After adding these three equations, we get $2 a_{0}+2 b_{0}+2 c_{0}=2 p+2 q+2 r, a_{0}+b_{0}+c_{0}=p+q+r$. Since $P(x)+Q(x)+R(x)=3 x^{2}+\left(a_{1}+b_{1}+c_{1}\right) x+a_{0}+b_{0}+c_{0}$, the product of the roots of this trinomial is $\frac{a_{0}+b_{0}+c_{0}}{3}=\frac{p+q+r}{3}$.

Comment: Only the answer - 2 points.

Any of the equations $a_{0}+b_{0}=2 r, a_{0}+c_{0}=2 q, b_{0}+c_{0}=2 p$ obtained - 2 points.