5. Three circles with radii 1, 2, 3 touch each other externally at three points. Find the radius of the circle passing through these three points.
Answer: 1.

Solution: Let $\mathrm{O}_{1}, \mathrm{O}_{2}$ and $\mathrm{O}_{3}$ be the centers of the given circles, K, M, N the points of tangency, such that $\mathrm{O}_{1} \mathrm{~K}=\mathrm{O}_{1} \mathrm{~N}=1, \mathrm{O}_{2} \mathrm{~K}=\mathrm{O}_{2} \mathrm{M}=2$ and $\mathrm{O}_{3} \mathrm{~N}=$ $\mathrm{O}_{3} \mathrm{M}=3$. It is known that when two circles touch, the point of tangency lies on the segment connecting their centers. Therefore, K, M, N are points on the sides of the triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$. Note that the desired circle coincides with the circle inscribed in the triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$. Indeed, since the equalities $\mathrm{O}_{1} \mathrm{~K}=\mathrm{O}_{1} \mathrm{~N}, \mathrm{O}_{2} \mathrm{~K}=\mathrm{O}_{2} \mathrm{M}$ and $\mathrm{O}_{3} \mathrm{~N}=\mathrm{O}_{3} \mathrm{M}$ hold, the points $\mathrm{K}, \mathrm{M}, \mathrm{N}$ are the points of tangency of the inscribed circle with the sides of the triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$.

Let $r$ be the desired radius, write the area of the triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ in terms of the perimeter and $r$, and using Heron's formula. We get the equation:

$$
\frac{1}{2}(3+4+5) r=\sqrt{6(6-3)(6-4)(6-5)} \Rightarrow 6 r=6 \Rightarrow r=1
$$

## ALL-RUSSIAN SCHOOL OLYMPIAD 2022/23

## MATHEMATICS  11th GRADE

Grading criteria: Correct and justified solution - 7 points, incorrect solution or only answer - $\mathbf{0}$ points.