# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_d04b19ad82f79b6d81e2g-21.jpg?height=469&width=671&top_left_y=2384&top_left_x=795)

Since VK is the common chord of the two given circles, $O_{1} O_{2}$ is the perpendicular bisector of VK. Therefore, $\angle O_{2} O_{1} K = \frac{1}{2} \angle B O_{1} K = \angle B A K$, as an inscribed angle relative to the central angle. Similarly, $\angle O_{1} O_{2} K = \frac{1}{2} \angle B O_{2} K = \angle B C K \Rightarrow$ triangle $O_{1} O_{2} K$ is similar to triangle $A B C$ (by two angles).

Since AB is a chord of the first circle, its radius is not less than half of any chord. Therefore, the similarity ratio of triangles $O_{1} O_{2} K$ and $A B C$ is not less than 0.5 $\Rightarrow$ $O_{1} O_{2}: A C \geq 1: 2 \Rightarrow$ the smallest possible length of $O_{1} O_{2} = 0.5 A C$, i.e., $O_{1} O_{2}$ is the midline of triangle $A B C \Rightarrow A C = 4$.

Then, by the corollary of the Law of Sines, $R = \frac{A C}{2 \sin \angle A B C} = \frac{4}{2 \sin 75^{\circ}} = \frac{2}{\sin 75^{\circ}} = 2 \sqrt{6} - 2 \sqrt{2}$.

Answer: $2 \sqrt{6} - 2 \sqrt{2}$ cm.

Grading Criteria for 10th Grade Problems

| Problem | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Total |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Points | 9 | 10 | 12 | 12 | 12 | 15 | 15 | 15 | 100 |
8. In an acute-angled triangle $\mathrm{ABC}$, angle $\mathrm{B}=75^{\circ}$. A point K is chosen on side $\mathrm{AC}$. Circles are circumscribed around triangles $\mathrm{ABK}$ and $\mathrm{CBK}$ with centers $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$, respectively. Find the radius of the circumcircle of triangle $\mathrm{ABC}$ if the smallest possible length of segment $\mathrm{O}_{1} \mathrm{O}_{2}=2$ cm.