# Problem 7. (4 points)

Point $I$ is the center of the inscribed circle of triangle $A B C$, point $O$ is the center of the excircle opposite to vertex $B$, touching side $A C$, segments $A C$ and $O I$ intersect at point $K$.

It turns out that $O I=50, I K=18, A K=24$. Find the length of the angle bisector of angle $B$ in triangle $A B C$.

#
# Answer: $576 / 7$

## Solution:

The center of the inscribed circle lies on the bisector of angle $\angle B A C$, and the center of the excircle lies on the bisector of the external angle $\angle A$. These bisectors are perpendicular, so triangle $A I O$ is a right triangle with a right angle at point $A$. Note that in the problem statement, $A K^{2}=I K \cdot O K$. This equality in a right triangle is satisfied if and only if $A K$ is the altitude. Therefore, $A K \perp I O$, which means $A C \perp I O$. Since line $I O$ is the bisector of angle $\angle B$, triangle $A B C$ is isosceles $(A B=B C)$.

From the right triangle $A K I$, we find $\operatorname{tg} \angle K A I=\frac{I K}{A K}$.

Then $\operatorname{tg} \angle B A C=\operatorname{tg} 2 \angle K A I=\frac{2 \operatorname{tg} \angle K A I}{1-\operatorname{tg}^{2} \angle K A I}=\frac{\frac{2 I K}{A K}}{1-\left(\frac{I K}{A K}\right)^{2}}=\frac{2 \cdot I K \cdot A K}{A K^{2}-I K^{2}}$

The length of the bisector we are looking for is $B K=A K \cdot \operatorname{tg} \angle B A C=\frac{2 \cdot I K \cdot A K^{2}}{A K^{2}-I K^{2}}$. Substituting the numbers from the problem statement into this formula, we get the answer.