Problem 9.7. Given a quadratic trinomial $P(x)$, whose leading coefficient is 1. On the graph of $y=P(x)$, two points with abscissas 10 and 30 are marked. It turns out that the bisector of the first quadrant of the coordinate plane intersects the segment between them at its midpoint. Find $P(20)$.
Answer: -80.

Solution. The midpoint of this segment has coordinates $\left(\frac{10+30}{2}, \frac{P(10)+P(30)}{2}\right)$. Since it lies on the bisector of the first quadrant, i.e., on the line $y=x$, these coordinates are equal. From this, we get $P(10)+P(30)=40$.

Since $P(x)$ is a monic polynomial, it can be written as $P(x)=x^{2}+a x+b$. Then the condition $P(10)+P(30)=40$ can be rewritten as $100+10 a+b+900+30 a+b=40$, which implies $40 a+2 b=-960$ and $20 a+b=-480$.

Therefore, $P(20)=400+20 a+b=400-480=-80$.