# Problem 4. (3 points)

Circles $O_{1}$ of radius $b$ and $O_{2}$ of radius $c$ touch at point $O$ - the center of circle $O_{3}$ of radius $a$. Point $A$ is one of the intersection points of circles $O_{1}$ and $O_{3}$. Circle $O_{4}$ touches circle $O_{1}$ at point $A$ and circle $O_{2}$ at point $B$. Point $C$ is a point on the line $O B$ such that triangles $O A B$ and $O C A$ are similar. Find $A C$.

All the tangencies mentioned in the problem occur externally.
Answer: $\frac{a^{2}}{2}\left(\frac{1}{b}+\frac{1}{c}\right)$.

Solution: Apply inversion with respect to circle $O_{3}$. Circle $O_{3}$ will map to itself, circles $O_{1}$ and $O_{2}$ will map to parallel lines $\ell_{1}$ and $\ell_{2}$, the first of which passes through point $A$, which maps to itself under the inversion.

Points $B$ and $C$ will map to each other under this inversion, since $a^{2}=O A^{2}=O B \cdot O C$ (this equality follows from the similarity of triangles $O A B$ and $O C A$; these triangles cannot be similar in any other way, as they share angle $O$, and points $B$ and $C$ are assumed to be distinct).

Circle $O_{4}$ will map to a circle that is tangent to lines $\ell_{1}$ and $\ell_{2}$ at points $A$ and $C$ respectively. Since the lines are parallel, this means that the length of segment $A C$ is equal to the distance between these lines.

Drop a perpendicular from point $O$ to line $\ell_{1}$. This perpendicular will intersect circle $O_{1}$ at point $X$, which is the inverse of the foot of the perpendicular $H$ and is diametrically opposite to $O$. This means that $O H=\frac{a^{2}}{O X}=\frac{a^{2}}{2 b}$.

Similarly, the distance from $O$ to $\ell_{2}$ is $O H=\frac{a^{2}}{O X}=\frac{a^{2}}{2 c}$. The desired distance between the two lines is the sum of these values, since point $O$ lies between lines $\ell_{1}$ and $\ell_{2}$, otherwise circles $O_{1}$ and $O_{2}$ would touch internally.