6. (4 points) Given three circles with radii 3, 5, and 7, 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: $\frac{7}{4}+\frac{7}{3 \sqrt{6}}+\frac{7}{\sqrt{30}}$.
# Solution:

Let's introduce the following notations: for the first circle, the center is point $O_{1}$, the radius is $a$; for the second circle, the center is point $O_{2}$, the radius is $b$; for the third circle, the center is point $O_{3}$, the radius is $c$. Since points $A, B, C$ are the points of tangency of two circles out of 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 8, 10, and 12. 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: $r=\frac{\sqrt{15 \cdot 3 \cdot 5 \cdot 7}}{15}=\sqrt{7}$.

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 $O O_{1}=\sqrt{B O^{2}+B O_{1}^{2}}=\sqrt{7+3^{2}}=$ $\sqrt{16}=4$.

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 $O_{1} M_{1}=\frac{B O_{1}^{2}}{O O_{1}}=\frac{9}{4}$, and thus $O M_{1}=O O_{1}-O_{1} M_{1}=$ $=4-\frac{9}{4}=\frac{7}{4}$

Similarly, we find $O O_{2}=\sqrt{5+5^{2}}=\sqrt{30}$, and $O M_{2}=\frac{7}{\sqrt{30}} ;$
$O O_{3}=\sqrt{5+7^{2}}=\sqrt{54}$, and $O M_{2}=\frac{7}{\sqrt{54}}=\frac{7}{3 \sqrt{6}}$.