11.3. On the $O x$ axis, points $0,1,2, \ldots, 100$ were marked and graphs of 200 different quadratic functions were drawn, each of which passes through two of the marked points and touches the line $y=-1$. For each pair of graphs, Oleg wrote on the board a number equal to the number of common points of these graphs. After that, he added all 19900 numbers written on the board. Could he have obtained the number 39699?
Answer: Could not.

Solution. Each of our 200 polynomials corresponds to two integer points $a$ and $b$ on the $O x$ axis. Without loss of generality, we will assume that $a < b$. The width of the polynomial is $w=b-a$, and the axis is $c=\frac{a+b}{2}$. If the width is $w>0$ and the axis is $c$, then it is written as $f(x)=\frac{4}{w^{2}}(x-c)^{2}-1$.

We will show that the graphs of two different polynomials of this form have exactly two common points when they have different widths and axes. If they have the same width or axis, then they have exactly one common point.

Indeed, $\frac{4}{w_{1}^{2}}\left(x-c_{1}\right)^{2}-1=\frac{4}{w_{2}^{2}}\left(x-c_{2}\right)^{2}-1$ is equivalent to $\left(x-c_{1}\right) w_{2}= \pm\left(x-c_{2}\right) w_{1}$. If $w_{1} \neq w_{2}$, then each of the two linear equations has roots, and they coincide only if $c_{1}=c_{2}$. If $w_{1}=w_{2}$, then $c_{1} \neq c_{2}$ (the polynomials are different) and one of the two linear equations has no roots, while the other has.

Note that the width of a polynomial can take a value from 1 to 100, and there will be no more than one polynomial with a width of 100. Let $x_{i}$ be the number of polynomials with width $i$. We estimate the number of pairs of polynomials with the same width:

$$
\begin{aligned}
\sum_{i=1}^{100} \frac{x_{i}\left(x_{i}-1\right)}{2} & =\sum_{i=1}^{100} \frac{\left(x_{i}^{2}-4 x_{i}+4\right)+3 x_{i}-4}{2}= \\
& =\sum_{i=1}^{100} \frac{\left(x_{i}-2\right)^{2}}{2}+\frac{3 \cdot 200-4 \cdot 100}{2} \geqslant 1+100
\end{aligned}
$$

In the last inequality, we used the following reasoning: since the sum of one hundred numbers $x_{i}$ is 200 and $x_{100} \neq 2$, there will be at least one more $x_{i} \neq 2$, therefore, $\sum_{i=1}^{100}\left(x_{i}-\right.$ $-2)^{2} \geqslant 2$.

The axis of a polynomial can be any integer or half-integer from $\frac{1}{2}$ to $99 \frac{1}{2}$, such numbers are 199, therefore, there will be at least one pair of polynomials with a common axis. This will be a previously unaccounted pair, since polynomials with the same width and axis coincide.

To find the number of intersection points of the graphs, we need to subtract the number of pairs with the same width or axis from twice the number of pairs of polynomials. Thus, the number of intersection points is no more than $2 \cdot \frac{200 \cdot 199}{2}-101-1=39698$, which is less than 39699.

Comment. The correct answer without explanation - 0 points.

The conditions under which a pair of graphs has exactly one common point are formulated - 1 point.

Proved the conditions under which a pair of graphs has exactly one common point - another 1 point.

Proved that the number on the board is no more than 39699 - 5 points (these points are not cumulative with the previous ones).