Berolov S.L.

Given a parallelogram $A B C D$ with angle $A$ equal to $60^{\circ}$. Point $O$ is the center of the circumscribed circle of triangle $A B D$. Line $A O$ intersects the bisector of the external angle $C$ at point $K$. Find the ratio $A O$ : OK.
The angle at vertex $A$ of triangle $ABC$ is acute, so vertex $A$ and the center $O$ of the circumscribed circle of this triangle lie on the same side of the line $BD$. $\angle BOD = 2 \angle BAD = 120^\circ = 180^\circ - \angle BCD$. Therefore, quadrilateral $OBDC$ is cyclic. Since the inscribed angles $BCO$ and $DCO$ subtend equal chords $OB$ and $OD$, $CO$ is the angle bisector of $\angle BCD$, and since $CK$ is the angle bisector of the adjacent angle, $\angle OCK = 90^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_11215e94233515718ac4g-21.jpg?height=318&width=629&top_left_y=567&top_left_x=726)

Let $O_1$ be the center of the circumscribed circle of triangle $CDB$. Note that triangles $CDB$ and $ABD$ are symmetric with respect to the center of the parallelogram, so segments $AO$ and $O_1C$ are equal and parallel.

The perpendicular bisector of segment $OC$ passes through point $O_1$ and is parallel to $CK$. Therefore, it passes through the midpoint $M$ of segment $OK$, and $O_1CKM$ is a parallelogram. Thus, $OM = MK = O_1C = AO$.

## Answer

$2: 1$.