$\underline{\text { Vasiliev N.B. }}$

How many maximum parts can the coordinate plane xOy be divided into by the graphs of 100 quadratic trinomials of the form

$y=a_{n} X^{2}+b_{n} x+c_{n}(n=1,2, \ldots, 100) ?$
We will prove by induction that $n$ parabolas of the specified type can divide the plane into no more than $n^{2}+1$ parts.

Base case. One parabola divides the plane into $2=1^{2}+1$ parts.

Inductive step. The $n$-th parabola intersects each of the others at no more than two points. These intersection points, of which there are $m \leq 2 n-2$, divide it into $m+1$ segments. Each such segment divides one of the "old" parts of the plane into two, so the number of parts increases by no more than $2 n-1$. By the inductive hypothesis, the number of parts divided by $n$ parabolas does not exceed $2 n-1+(n-1)^{2}+1=n^{2}+1$.

Now let's construct a partition of the plane by 100 parabolas into $100^{2}+1$ parts. Consider the parabolas $y=a_{n} x^{2}-n$. Here, we take $a_{1}=1$,

and the other $a_{n}>0$ will be constructed inductively so that the absolute value of the abscissas of the points of intersection of the $n$-th parabola with the x-axis is less than the absolute values of the abscissas of the points of intersection of all previous parabolas with each other and with the x-axis. Then, obviously, each pair of parabolas intersects at two points below the x-axis, and no three intersect at the same point. From the previous reasoning, it follows that the number of parts of the partition will be maximal.

## Answer

Into 10001 parts.