6. Three circles with radii $1,2,3$ touch each other externally. Find the radius of the circle passing through the three points of tangency of these circles.
6. Let the centers of the given circles be denoted as $O_{1}, O_{2}, O_{3}$. Let $K, M, N$ be the points of tangency of the given circles. Then $O_{1} K=O_{1} N=1, O_{2} K=O_{2} M=2, O_{3} M=O_{3} N=3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8367be215fc9a6414e64g-1.jpg?height=585&width=508&top_left_y=2058&top_left_x=411)

We will show that the desired circle coincides with the incircle of the triangle $O_{1}, O_{2}, O_{3}$. Indeed, if we inscribe a circle in the triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$, then the segments connecting its vertices to the points of tangency will be the radii of the corresponding circles. Let $r$ be the radius of this circle. We will write the area of the triangle $O_{1}, O_{2}, O_{3}$ in terms of the perimeter and $r$, and using Heron's formula.

By equating these expressions for the area, we get: $\frac{1}{2} \operatorname{Pr}=\sqrt{p(p-a)(p-b)(p-c)}$, where $\mathrm{P}$ is the perimeter, and $p=\frac{\mathrm{P}}{2} ; a, b, c$ are the sides of the triangle $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}$. We obtain: $\frac{1}{2} r(2+4+6)=\sqrt{6(6-3)(6-4)(6-5)} \Rightarrow 6 r=6 ; r=1$. Answer: $\{1\}$.