6. (4 points) Given three circles with radii 3, 4, and 5, each pair of which touch each other at points $A, B$, and $C$. Find the sum of the distances from the center of the circumscribed circle of triangle $A B C$ to its sides.

Answer: $\frac{5}{\sqrt{14}}+\frac{5}{\sqrt{21}}+\frac{5}{\sqrt{30}}$.
Solution:

Let's introduce the following notations: for the first circle, the center is point $O_{1}$, and the radius is $a$; for the second circle, the center is point $O_{2}$, and the radius is $b$; for the third circle, the center is point $O_{3}$, and the radius is $c$. Since points $A, B, C$ are the points of tangency of two circles out of the three, these points lie on segments $O_{1} O_{2}, O_{1} O_{3}, O_{2} O_{3}$. Without loss of generality, we can assume that point $A$ lies on segment $O_{2} O_{3}$, point $B$ on $O_{1} O_{3}$, and point $C$ on $O_{1} O_{2}$. The center of the circumcircle of triangle $A B C$ is denoted as $O$.

The sides of triangle $O_{1} O_{2} O_{3}$ are 7, 8, and 9. Points $A, B$, and $C$ divide the sides of triangle $O_{1} O_{2} O_{3}$ in the same way as the points of tangency of the inscribed circle, so these are the points of tangency. Thus, the circumcircle of triangle $A B C$ is the incircle of triangle $O_{1} O_{2} O_{3}$. We calculate its radius by dividing the area of triangle $O_{1} O_{2} O_{3}$, computed using Heron's formula, by its semiperimeter: $r=\frac{\sqrt{12 \cdot 3 \cdot 4 \cdot 5}}{12}=\sqrt{5}$.

Triangles $O_{1} B C$ and $O B C$ are isosceles with base $B C$, and points $O$ and $O_{1}$ lie on the perpendicular bisector of $B C$. Then, by the Pythagorean theorem, we get $O O_{1}=\sqrt{B O^{2}+B O_{1}^{2}}=\sqrt{5+3^{2}}=\sqrt{14}$.

Let the midpoint of $B C$ be $M_{1}$. In the right triangle $O B O_{1}$, this point is the foot of the altitude dropped from the right angle, so $O_{1} M_{1}=\frac{B O_{1}^{2}}{O O_{1}}=\frac{9}{\sqrt{14}}$, and thus $O M_{1}=O O_{1}-O_{1} M_{1}=$ $=\sqrt{14}-\frac{9}{\sqrt{14}}=\frac{5}{\sqrt{14}}$.

Similarly, we find $O O_{2}=\sqrt{5+4^{2}}=\sqrt{21}$, and $O M_{2}=\frac{5}{\sqrt{21}}$; $O O_{3}=$ $\sqrt{5+5^{2}}=\sqrt{30}$, and $O M_{2}=\frac{5}{\sqrt{30}}$.