## Task 3 - 341213

The circles in the figure can be filled with real numbers such that the given numbers appear in the outer circles and that in each of the seven inner circles, the arithmetic mean of the six adjacent circles is placed.

Determine which number $x$ stands in the central circle.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2349.jpg?height=409&width=397&top_left_y=2257&top_left_x=1315)
For $x$ and the neighboring numbers $a, b, c, d, e, f$ (labeled in the same clockwise direction as the numbers 1, $2, \ldots, 12$, starting with the number between $x$ and 1), the following holds:

\[
\begin{aligned}
\frac{1}{6}(15+b+f+x) & =a \\
\frac{1}{6}(9+c+a+x) & =b \\
\frac{1}{6}(15+d+b+x) & =c \\
\frac{1}{6}(21+e+c+x) & =d \\
\frac{1}{6}(27+f+d+x) & =e \\
\frac{1}{6}(33+a+e+x) & =f \\
\frac{1}{6}(a+b+c+d+e+f) & =x
\end{aligned}
\]

Adding (1) to (6) results in

\[
20+\frac{1}{3}(a+b+c+d+e+f)+x=a+b+c+d+e+f
\]

From this and (7), it follows that

\[
20+2 x+x=6 x \quad \Rightarrow \quad x=\frac{20}{3}
\]

2. A verification is not required for a complete solution to the problem, as the existence of a labeling of the described type is evident from the problem statement.