8,9  [ Properties and characteristics of an isosceles triangle.]

Quadrilateral \(ABCD\) is inscribed in a circle, \(M\) is the intersection point of its diagonals, \(O_1\) and \(O_2\) are the centers of the inscribed circles of triangles \(ABM\) and \(CMD\) respectively, \(K\) is the midpoint of the arc \(AD\) not containing points \(B\) and \(C\), \(\angle O_1 K O_2 = 60^\circ\), \(K O_1 = 10\). Find \(O_1 O_2\).
Point $K$ is the midpoint of arc $AD$, so $BK$ is the bisector of the inscribed angle $ABD$, which means point $O_{1}$ lies on segment $BK$. Similarly, point $O_{2}$ lies on segment $CK$. The rays $MO_{1}$ and $MO_{2}$ are the bisectors of the vertical angles

$AMB$ and $CMD$, so point $M$ lies on segment $O_{1}O_{2}$ and $\angle BMO_{1} = \angle DMO_{2} = \angle CMO_{2}$.

The inscribed angles $ABD$ and $ACD$ are equal. Therefore, $\angle MBO_{1} = \angle MCO_{2}$.

By the exterior angle theorem of a triangle, $\angle KO_{1}O_{2} = \angle MBO_{1} + \angle BMO_{1} = \angle MCO_{2} + \angle CMO_{2} = \angle KO_{2}O_{1}$, so triangle $O_{1}K_{2}$ is isosceles, and since one of its angles is $60^{\circ}$, this triangle is equilateral. Therefore, $O_{1}O_{2} = KO_{1} = 10$.

## Answer

10.