3. The graphs of the functions $y=x^{2}$ and $y=a x^{2}+b x+c$ intersect at points $A$ and $B$, which lie on opposite sides of the y-axis. Point $O$ is the origin. It turns out that $\angle A O B=90^{\circ}$. Find all possible values of $c$. (20 points)

International School Olympiad URFU "Emerald" 2022, 2nd stage
Solution: Let $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ be the coordinates of points $A$ and $B$ respectively, with $x_{1} \neq 0$ and $x_{2} \neq 0$. Then $x_{1}, x_{2}$ are the roots of the equation $(a-1)x^{2}+b x+c=0$. If $a=1$, the graphs touch at point $O$, and there cannot be two points of intersection. Otherwise, Vieta's formulas apply:

$$
x_{1} x_{2}=\frac{c}{a-1}
$$

By the Pythagorean theorem, $A B^{2}=A O^{2}+B O^{2}$, which in coordinate form becomes

$$
x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}
$$

which, after expanding and combining like terms, becomes

$$
x_{1} x_{2}=-y_{1} y_{2}
$$

Points $A$ and $B$ lie on the graph of the function $y=x^{2}$, so $y_{1}=x_{1}^{2}$ and $y_{2}=x_{2}^{2}$, and the last equation can be rewritten as

$$
x_{1} x_{2}=-x_{1}^{2} x_{2}^{2}
$$

from which $x_{1} x_{2}=-1$. Substituting into Vieta's formulas, we get $c=1-a$. Since $a$ can take any value except 1, $c$ can take any value except 0.

For any specified $c$, the roots of the equation $(a-1)x^{2}+b x+c=0$ exist, as the discriminant of the quadratic equation is $b^{2}-4(a-1)c=b^{2}+4c^{2}>0$. The equality $x_{1} x_{2}=-1$ ensures that these roots will have opposite signs.

Answer: $\boldsymbol{c} \neq \mathbf{0}$