3 [ Common Tangent to Two Circles ]

Circles $S 1$ and $S 2$ with centers $O 1$ and $O 2$ respectively touch each other externally; a line is tangent to the circles at different points $A$ and $B$ respectively. It is known that the intersection point of the diagonals of quadrilateral $O 1 A B O 2$ lies on one of the circles. Find the ratio of the radii of the circles.
Let the point $M$ of intersection of the diagonals $AO_2$ and $BO_1$ of the quadrilateral $O_1ABO_2$ lie on the circle $S_1$. The radii $O_1A = r$ and $O_2B = R$ are perpendicular to the common tangent $AB$ of the circles, so $O_1A \parallel O_2B$. Therefore, $\angle MBO_2 = \angle MO_1A$, and since $\angle BMO_2 = \angle O_1MA$ and triangle $MO_1A$ is isosceles, triangle $MBO_2$ is also isosceles. Consequently, $MB = O_2B = R$, so

\[
O_1B = O_1M + MB = r + R = O_1O_2.
\]

The height $O_1F$ of the isosceles triangle $BO_1O_2$ is its median, hence

\[
O_2F = FB = O_1A = r, \quad R = O_2B = 2r, \quad \frac{r}{l} = \frac{1}{2}
\]

## Answer

$1: 2$.