My friend sent me the original question. Then he asked if 1-12 can be arranged in a way so that any three consecutive numbers have a sum that is not larger than 20. We guess the answer is no, since we tried many times and could not find a solution. However, we wonder why it is. Any help would be appreciated!
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The arrangement is indeed impossible.
As the answer to the question you linked mentions, the sums of each segment of three adjacent numbers all themselves have to add up to $3 (1 + 2 + \cdots + 12) = 234$. So the mean of the sums-of-three is 19.5. If no individual three-number segment has a sum above 20, then at least six of the segments must have sums of 20. Moreover, having two consecutive equal sums-of-three is impossible: if one part of the wheel has the numbers $a\,b\,c\,d$ in order where $a + b + c = b + c + d$, then $a = d$, but the wheel can't repeat numbers. So sums-of-three have to alternate between 20 and 19.
Now suppose the wheel has seven adjacent numbers $a\,b\,c\,d\,e\,f\,g$ where $a + b + c = c + d + e = e + f + g = 20$ and $b + c + d = d + e + f = 19$. Then $d = a - 1$ (by subtracting $b + c + d = 19$ from $a + b + c = 20$) and likewise $d = g - 1$ (by subtracting $d + e + f = 19$ from $e + f + g = 20$). So $a = g$; that is, we have a repeated number, a contradiction.
(As a final note, this method generalizes to show that any wheel with numbers $1$ through $2n$, $n \geq 4$, must have three adjacent numbers that add up to at least $3n + 3$.)
