## Task 3 - 130523

The natural numbers from 1 to 12, each exactly in one of the fields $A, B, C, D, E, F, G, H, K, M, N, P$ of the figure, should be entered such that the sum of the numbers in the fields $A, B, C, D$ is 22, as well as the sum of the numbers in the fields $D, E, F, G$, the sum of the numbers in the fields $G, H, K, M$, and the sum of the numbers in the fields $M, N, P, A$.

a) Provide one such entry of numbers!

b) Investigate which numbers stand in the fields $A, D, G$ and $M$ in every such entry!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0063.jpg?height=454&width=463&top_left_y=1572&top_left_x=1296)
a) The figure shows an example of how the required entry can sound.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0063.jpg?height=431&width=448&top_left_y=2126&top_left_x=290)
b) Suppose there is an entry and let $a, b, c, d, e, f, g, h, k, m, n, p$ be the numbers in the fields $A, B, C, D, E, F, G, H, K, M, N, P$ in this order. Furthermore, let

$$
\begin{array}{lr}
s_{1}=e+b+c+d & s_{2}=d+e+f+g \\
s_{3}=g+h+k+m & s_{4}=m+n+p+a
\end{array}
$$

According to the problem, $s_{1}=s_{2}=s_{3}=s_{4}=22$. Therefore, $s_{1}+s_{2}+s_{3}+s_{4}=88$.

The sum of the natural numbers from 1 to 12 is 78. Thus, it is 10 less than the sum $s_{1}+s_{2}+s_{3}+s_{4}$.

However, the numbers in the corner fields $A, D, G, M$ are counted twice in the formation of the four sums. Therefore, the sum of these numbers must be 10.

If $a, g, d, m$ were not the numbers 1, 2, 3, 4, then at least one of them would be greater than 4, and the others would not be less than 1, 2, 3, so their sum would be greater than 10.

Therefore, in every correct entry of the specified type, the numbers 1, 2, 3, and 4 and no others must stand in the corner fields.