Here is one strategy that you could use to find a solution:
Let the four numbers in the centre four cells be \( a \), \( b \), \( c \) and \( d \). Each of these numbers contributes to two line sums. Therefore the sum of the four line sums is:
$$ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + a + b + c + d $$As each line sum is the same this total must be divisible by 4.
Simplifying; \( 36 + a + b + c + d \) is a multiple of 4.
As 36 is a multiple of 4 then \( a + b + c + d \) must also be a multiple of 4.
Choose four numbers accordingly. The smallest four would be \( 1 + 2 + 3 + 6 \). Arrange these in the centre cells of the Spinsum windmill so that the two smallest are not adjacent.
Now calculate the combined total of the four line sums:
$$ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 1 + 2 + 3 + 6 = 48 $$So each individual line sum will be a quarter of this: \(48 \div 4 = 12 \)
The remaining four numbers can now be placed to achive line sums of 12. So here is one possible solution:
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