![](https://cdn.mathpix.com/cropped/2024_05_06_7900dea0235162852411g-20.jpg?height=286&width=2038&top_left_y=1495&top_left_x=-1)

Authors: Mednikov L.E., Evdokimov M.A.

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

![](https://cdn.mathpix.com/cropped/2024_05_06_7900dea0235162852411g-20.jpg?height=400&width=326&top_left_y=1998&top_left_x=866)
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 $\left(0, n^{2}-n+1 / 2\right)$. 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 $y=n^{2}+n$. Consequently, the circle $\omega_{n+1}$ has exactly two common points with the parabola $y=x^{2}$ (with the specified ordinate). This means that it is tangent to the parabola.

## Answer

$r_{1998}=1997.5$.