Two lines passing through point $M$, lying outside a circle with center $O$, touch the circle at points $A$ and $B$. The segment $OM$ is bisected by the circle. In what ratio is the segment $OM$ divided by the line $AB$?
In the right triangle $A M O$, the leg $O A$ is equal to half the hypotenuse $O M$. If $K$ is the midpoint of $O M$, then triangle $AOK$ is equilateral.

## Solution

The bisector of the isosceles triangle $A M B$, drawn from the vertex $M$, is the altitude. Therefore, $A B$ $\perp M O$. Let the circle intersect the segment $OM$ at point $K$. In the right triangle $A M O$, the leg $O A$ is equal to half the hypotenuse $M O$, so $\angle A M O=30^{\circ}$, and $\angle A O M=60^{\circ}$. Since the angle between the equal sides $O A$ and $OK$ of the isosceles triangle $A O K$ is $60^{\circ}$, the triangle is equilateral. Its height $A P$ is also the median, so $O P=K P=1 / 2 O K=1 / 4 O M$. Therefore, $O P: M P=1: 3$.

## Answer

$1: 3$ from point $O$.