We consider the following figure:

![](https://cdn.mathpix.com/cropped/2024_04_17_96cf05b4155531978f40g-3.jpg?height=399&width=1285&top_left_y=2319&top_left_x=385)

We are looking for labels for the nine fields in the figure with the numbers 1, 2, ..., 9. Each of these numbers should be used exactly once. Furthermore, the six sums of three or four numbers along the drawn straight connections should be equal.

Provide such a labeling.

Show that in all such labelings, the same number is in the top field.

How many such labelings are there in total? (Two labelings are different if they differ in at least one field.)

(Walther Janous)
We denote the numbers in the fields of the figure as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_96cf05b4155531978f40g-4.jpg?height=388&width=1282&top_left_y=654&top_left_x=387)

If we denote the common value of the six sums by \( s \), we obtain from the problem statement the equations

\[
\begin{aligned}
a + b + f & = s, \\
a + c + g & = s, \\
a + d + h & = s, \\
a + e + i & = s, \\
b + c + d + e & = s, \\
f + g + h + i & = s, \\
a + b + c + d + e + f + g + h + i & = 1 + 2 + \ldots + 9 = 45.
\end{aligned}
\]

If we add equations (1) to (4) and subtract equations (5) and (6), we get \( 4a = 2s \), so \( a = \frac{s}{2} \). Adding equations (5) and (6), we get from (7)

\[
\frac{s}{2} + s + s = a + b + c + d + e + f + g + h + i = 45,
\]

which means \( s = 18 \) and \( a = 9 \). The equations (1) to (6) then become

\[
\begin{gathered}
b + f = c + g = d + h = e + i = 9 \\
b + c + d + e = f + g + h + i = 18
\end{gathered}
\]

From (8), it follows that \(\{\{b, f\}, \{c, g\}, \{d, h\}, \{e, i\}\} = \{\{1, 8\}, \{2, 7\}, \{3, 6\}, \{4, 5\}\}\).

If we first look for solutions with \(\{b, f\} = \{1, 8\}, \{c, g\} = \{2, 7\}, \{d, h\} = \{3, 6\}, \{e, i\} = \{4, 5\}\), we see that with \( b = 1, c = 2 \) the equation (9) is not solvable, and the same applies for \( b = 8, c = 7 \). Therefore, either \( b = 1, c = 7, f = 8, g = 2 \) or \( b = 8, c = 2, f = 1, g = 7 \) must hold. In both cases, we see that there is only one possibility for \((d, e, h, i)\) due to equation (9). From these considerations, we obtain the solutions \((a, b, c, d, e, f, g, h, i) = (9, 1, 7, 6, 4, 8, 2, 3, 5)\) and \((9, 8, 2, 3, 5, 1, 7, 6, 4)\). This results, for example, in the following labeling.

![](https://cdn.mathpix.com/cropped/2024_04_17_96cf05b4155531978f40g-5.jpg?height=394&width=1282&top_left_y=106&top_left_x=387)

It has already been shown that \( a = 9 \) must hold. All other labelings are obtained from this by swapping \((b, c, d, e)\) with \((f, g, h, i)\) (2 possibilities) or independently by swapping \((b, f), (c, g), (d, h), (e, i)\) among themselves (24 possibilities). Therefore, there are a total of 48 possible labelings.

(Gerhard Kirchner)