9,10,11 |
| :---: | :---: | :---: |
|  | Tangent Circles |  |
|  | Investigation of Quadratic Trinomials |  |
|  | Induction in Geometry |  |

Authors: Meddnikov L.Z., Evodokimov M.A.

Inside the parabola \( y = x^2 \), there are non-coincident circles \( \omega_1, \omega_2, \omega_3, \ldots \) such that for each \( n > 1 \), the circle \( \omega_n \) is tangent to the branches of the parabola and externally tangent to the circle \( \omega_{n-1} \) (see figure). Find the radius of the circle \( \sigma_{1998} \), given that the diameter of \( \omega_1 \) is 1 and it is tangent to the parabola at its vertex.

![](https://cdn.mathpix.com/cropped/2024_05_06_1cbe8f655c608884328cg-16.jpg?height=409&width=349&top_left_y=1373&top_left_x=863)
We will prove by induction that the radius $r_{n}$ of the circle $\omega_{n}$ is $n-1 / 2$, and its center is the point $(0, n^{2}-n+1 / 2)$. The base case is given in the problem statement.

Inductive step. The upper point of the circle $\omega_{n}$ has an ordinate of $n^{2}-n+1 / 2+n-1 / 2=n^{2}$. Therefore, the center of the circle $\omega_{n+1}$ with radius $n+1 / 2$, which is tangent to $\sigma_{n}$ at the upper point, has coordinates $\left(0, n^{2}+n+1 / 2\right)$. Thus, the equation of the circle $\sigma_{n+1}$ is

$x^{2}+\left(y-n^{2}-n-1 / 2\right)^{2}=(n+1 / 2)^{2}$ or $x^{2}+\left(y-n^{2}\right)^{2}-2\left(y-n^{2}\right)(n+1 / 2)=0$. Substituting $y$ for $x^{2}$, we get $y-n^{2}+\left(y-n^{2}\right)^{2}-2\left(y-n^{2}\right)(n+1 / 2)+n^{2}=0$, which simplifies to $\left(y-n^{2}-n\right)^{2}=0$. Therefore, this equation has a unique solution $y=n^{2}+n$. Hence, the circle $\omega_{n+1}$ intersects the parabola $y=x^{2}$ at exactly two points (with the specified ordinate). This means that it is tangent to the parabola.

## Answer

$r_{1998}=1997.5$