7. In the corners $B$ and $C$ of triangle $ABC$, circles with centers $O_{1}$ and $O_{2}$ of equal radius are inscribed, and point $O$ is the center of the circle inscribed in triangle $ABC$. These circles touch side $BC$ at points $K_{1}, K_{2}$, and $K$ respectively, with $B K_{1}=4, C K_{2}=8$, and $B C=18$.

a) Find the length of the segment $C K$.

b) Let the circle with center $O_{1}$ touch side $AB$ at point $K_{3}$. Find the angle $ABC$, if it is known that point $O_{1}$ is the center of the circle circumscribed around triangle $O K_{1} K_{3}$.
Answer. a) $C K=12$, b) $\angle A B C=60^{\circ}$.

Solution. a) The lines $\mathrm{CO}_{2}$ and $\mathrm{BO}_{1}$ are the bisectors of angles $C$ and $B$ of the triangle, so they intersect at point $O$ - the center of the inscribed circle. Let the radii of the circles with centers $O_{1}$ and $O_{2}$ be $r$, and the radius of the inscribed circle be $R$. Triangles $O K B$ and $O_{1} K_{1} B$ are similar, with the similarity ratio equal to $\frac{R}{r}$, so $B K=\frac{4 R}{r}$. Similarly, $C K=\frac{8 R}{r}$, hence $\frac{12 R}{r}=18, C K=12$.

b) From the condition, it follows that $O_{1} O=O_{1} K_{1}=r$. Drop a perpendicular $O_{1} H$ from point $O_{1}$ to segment $O K$. Then $O H=R-r=\frac{1}{2} r, \angle O B C=\angle O O_{1} H=\arcsin \frac{O H}{O O_{1}}=\arcsin \frac{1}{2}=30^{\circ}$. Therefore, $\angle A B C=2 \cdot 30^{\circ}=60^{\circ}$.