5. On a coordinate plane with the origin at point $O$, a parabola $y=x^{2}$ is drawn. Points $A, B$ are marked on the parabola such that $\angle A O B$ is a right angle. Find the smallest possible value of the area of triangle $A O B$.
5. Answer: 1. Solution. Let $A\left(a, a^{2}\right), B\left(b, b^{2}\right)$ be arbitrary points on the parabola. Then, by the Pythagorean theorem, $a^{2}+a^{4}+b^{2}+b^{4}=(a-b)^{2}+\left(a^{2}-b^{2}\right)^{2} \Leftrightarrow a b=-1$. From this, we find the expression for the area of the triangle: $2 S=\sqrt{a^{2}+a^{4}} \cdot \sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}=\sqrt{2+\left(a^{2}+\frac{1}{a^{2}}\right)} \geq 2$ (the last inequality follows from the inequality between the arithmetic mean and the geometric mean). It is clear that the value 1 is achieved for an isosceles triangle with vertices $O$, $A(1,1), B(-1,1)$.

Grading criteria. Correct solution - 7 points. The necessary inequality is proven, but it is not indicated that there exists a triangle with such an area - 5 points. The relationship between the coordinates of the points on the parabola is obtained, but there is no further progress - 1 point. The relationship between the coordinates is obtained and the formula for the area is written, but the estimate for the area is not made - 2 points. In all other cases - 0 points.