5. (3 points) Three circles with radii $\sqrt{13}, \sqrt{13}$, and $2 \sqrt{13-6 \sqrt{3}}$ are arranged such that the triangle formed by their centers is an equilateral triangle with side length $\sqrt{39}$. Find the radius of the circumscribed circle around the triangle, each vertex of which is the point of intersection of two of these circles, farthest from the center of the third circle.

Answer: $4 \sqrt{39}-6 \sqrt{13}$
Solution:

Let $A, B, C$ be the points of intersection of the first and second, second and third, and first and third circles, respectively, and $O_{1}, O_{2}, O_{3}$ be the centers of these circles.

The radii of the first and second circles are equal, so point $A$ will lie on the perpendicular bisector of $O_{1} O_{2}$. The triangle formed by the centers of the circles is equilateral, so point $O_{3}$ will also lie on this perpendicular bisector. Therefore, the entire picture is symmetric with respect to this perpendicular bisector, in particular, points $B$ and $C$ are symmetric to each other.

Triangle $O_{1} A O_{2}$ is isosceles with vertex $A$; moreover, its sides are $\sqrt{13}, \sqrt{13}$, and $\sqrt{39}$, from which its angles can be easily found: $30^{\circ}, 30^{\circ}$, and $120^{\circ}$.

Let $\angle O_{1} A C$ be $\alpha$, and $\angle C O_{1} O_{3}$ be $\beta$. In triangle $C O_{1} O_{3}$, we know all the sides, so we can find

$$
\cos \beta=\frac{13+39-4 \cdot 13-6 \sqrt{3}}{2 \cdot \sqrt{13} \cdot \sqrt{39}}=\frac{12}{13}
$$

Accordingly, $\sin \beta=\frac{5}{13}$. Further, $\angle A O_{1} C=\angle A O_{1} O_{2}+\angle O_{2} O_{1} O_{3}+\angle O_{3} O_{1} C=30^{\circ}+60^{\circ}+\beta=90^{\circ}+\beta$

Triangle $C O_{1} A$ is isosceles with angles $\alpha, 90^{\circ}+\beta$, and $\alpha$, so $\alpha=\frac{90^{\circ}-\beta}{2}$, from which

$$
\cos \alpha=\sqrt{\frac{1+\cos \left(90^{\circ}-\beta\right)}{2}}=\sqrt{\frac{1+\sin \beta}{2}}=\sqrt{\frac{9}{13}}=\frac{3}{\sqrt{13}}
$$

Accordingly, $\sin \alpha=\sqrt{\frac{4}{13}}=\frac{2}{\sqrt{13}}$. Moreover, $A C=2 O_{1} A \cos \alpha=2 \cdot \sqrt{13} \cdot \frac{3}{\sqrt{13}}=6$. $\angle B A C=120-2 \alpha=30+\beta$, so,

$$
\begin{aligned}
\sin \angle B A C & =\frac{1}{2} \cdot \frac{12}{13}+\frac{\sqrt{3}}{2} \cdot \frac{5}{13}=\frac{12+5 \sqrt{3}}{26} \\
\cos \angle B A C & =\frac{\sqrt{3}}{2} \cdot \frac{12}{13}-\frac{1}{2} \cdot \frac{5}{13}=\frac{12 \sqrt{3}-5}{26} \\
\sin \angle A B C=\cos \frac{\angle B A C}{2} & =\sqrt{\frac{1+\cos \angle B A C}{2}}=\sqrt{\frac{21+12 \sqrt{3}}{52}}=\frac{2 \sqrt{3}+3}{2 \sqrt{13}}
\end{aligned}
$$

By the Law of Sines,

$$
R=\frac{A C}{2 \sin \angle A B C}=\frac{6}{2 \cdot \frac{2 \sqrt{3}+3}{2 \sqrt{13}}}=\frac{6 \sqrt{13}}{2 \sqrt{3}+3}=\frac{6 \sqrt{13}(2 \sqrt{3}-3)}{3}=4 \sqrt{39}-6 \sqrt{13}
$$