A1. The numbers 1, 2, 3, 4, 5, 6, 7, and 8 must be distributed over the circles in the figure such that the sum of the three numbers on each of the four sides of the square is the same.

Provide a solution where this sum is minimized.

![](https://cdn.mathpix.com/cropped/2024_04_17_0b81ff2ef64511bfe8eeg-1.jpg?height=243&width=243&top_left_y=1078&top_left_x=1569)
A1. $1+2+3+4+5+6+7+8=36$. If $S$ is the sum of the three numbers on each side, then $4S=36+$ the four numbers at the corners. To make $S$ as small as possible, the sum of the numbers at the four corners must be as small as possible. This cannot be achieved with $1,2,3,4$ because then $4S=36+10=46$ and 46 is not divisible by 4. The smallest value we can try for $S$ is thus 12, because $48=4 \times 12$.

This gives the possibilities: $48=36+1+2+3+6$ or $48=36+1+2+4+5$.

1 and 2 cannot lie on the same side because you would then need 9 to get 12. Therefore, 1 and 2 must lie on a diagonal, and so must 3 and 6. This gives the

$$
183
$$

solution $5 \quad 7$, apart from rotations and reflections.

$\begin{array}{lll}6 & 4 & 2\end{array}$

The second possibility 1,2,4,5 does not yield a solution.