A regular hexagon $A B C D E F$ with side length of 2 units has circles drawn around its vertices $A, B$, and $C$, which mutually touch each other externally. What is the radius of the smallest circle?
We know that the centers of two tangent circles and the point of tangency lie on a straight line. If the two circles are externally tangent to each other, then the sum of their radii is equal to the distance between their centers.

![](https://cdn.mathpix.com/cropped/2024_05_02_07b915844683c1acbcceg-1.jpg?height=431&width=466&top_left_y=278&top_left_x=819)

Therefore, for the sides of triangle $ABC$, we have:

\[
\begin{aligned}
2 & = AB = r_A + r_B \\
2 \sqrt{3} & = AC = r_A + r_C \\
2 & = BC = r_B + r_C
\end{aligned}
\]

where $r_A, r_B, r_C$ are the radii of the circles centered at $A, B, C$, respectively. From the system of equations, it follows that $r_A = r_C = \sqrt{3}$ and $r_B = 2 - \sqrt{3}$. Therefore, the radius of the smallest circle is $2 - \sqrt{3}$.

Zsuzsanna Szabó (Mosonmagyaróvár, Kossuth L. Gymnasium, 2nd grade)