# Problem 5. (3 points)

Circles $O_{1}, O_{2}$, and $O_{3}$ are located inside circle $O_{4}$ with radius 6, touching it internally, and touching each other externally. Moreover, circles $O_{1}$ and $O_{2}$ pass through the center of circle $O_{4}$. Find the radius of circle $O_{3}$.

#
# Answer: 2

## Solution:

We will solve the problem in general for all cases, specifically proving that if the radius of circle $O_{4}$ is $R$, then the radius of circle $O_{3}$ is $\frac{R}{3}$.

Since circles $O_{1}$ and $O_{2}$ pass through the center of circle $O_{4}$ and are internally tangent to it, for each of them, the center of circle $O_{4}$ is diametrically opposite to the point of tangency. This means that the radius of each of them is $\frac{R}{2}$. Additionally, they touch each other at the center of circle $O_{4}$, which means that the centers of all three circles lie on the same straight line.

Now consider the triangle formed by the centers of circles $O_{1}$ and $O_{2}$ (let's call them $A$ and $B$) and the point of tangency of circles $O_{2}$ and $O_{3}$ (let's call it $C$). By symmetry, it is clear that $A C = B C$. Moreover, point $O$ (the center of circle $O_{4}$) is the midpoint of side $A B$. Therefore, triangle $O A C$ is a right triangle, with $O A = \frac{R}{2}$ and $O C = R$.

Now consider point $X$ (the center of circle $O_{3}$). It lies on segment $O C$, so triangle $O A X$ is also a right triangle. Its legs are $O A = \frac{R}{2}$ and $O X = R - r$, and the hypotenuse is $A X = \frac{R}{2} + r$, where $r$ is the desired radius of circle $O_{3}$. By the Pythagorean theorem, $\frac{R^{2}}{4} + R^{2} - 2 R r + r^{2} = \frac{R^{2}}{4} + R r + r^{2} = R^{2}$, from which $3 R r = R^{2}$ and $r = \frac{R}{3}$.