## Task 2 - 090812

Given in the plane is a circle $k_{0}$ and 6 circles of radius $r$, each of which touches exactly two of them and also the circle $k_{0}$ from the outside in the manner shown in the figure.

Determine the radius $r_{0}$ of the circle $k_{0}$!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0629.jpg?height=343&width=300&top_left_y=1473&top_left_x=1438)
Assume that $M_{0}$ is the center and $r$ is the radius of a circle $k_{0}$, to which 6 circles of the desired type exist; whose centers are $M_{1}, M_{2}, \ldots, M_{6}$. Then 6 congruent isosceles triangles arise

$$
\triangle M_{1} M_{0} M_{2}, \quad \triangle M_{2} M_{0} M_{3}, \quad \ldots \quad \triangle M_{6} M_{0} M_{1}
$$

thus $\angle M_{1} M_{0} M_{2} \cong \angle M_{2} M_{0} M_{3} \cong \ldots \cong \angle M_{6} M_{0} M_{1}$, and since the sum of these angles is $360^{\circ}$, each of them has a size of $60^{\circ}$. The triangles (1) are therefore equilateral, and it follows that $r_{0}+r=2 r$, thus $r_{0}=r$.

Conversely, for $r_{0}=r$ the conditions of the problem are satisfied. For if $M_{0}, M_{1}, \ldots, M_{6}$ are chosen such that $\triangle M_{1} M_{0} M_{2}, \quad \triangle M_{2} M_{0} M_{3}, \quad \ldots \triangle M_{5} M_{0} M_{6}$ are equilateral triangles with side length $2 r$, then $\triangle M_{6} M_{0} M_{1}$ is isosceles, with $\angle M_{6} M_{0} M_{1}=360^{\circ}-5 \cdot 60^{\circ}$, and thus also equilateral.

Therefore, the circles around $M_{0}, M_{1}, \ldots, M_{6}$ with radius $r$ touch each other in the prescribed manner.