China Southeastern Mathematical Olympiad

Denote the sum of labels of $A$ and $B$ by $x$ and $y$, respectively. Then $y=2x$. Thus, we have $$3x=x+y=1+2+\cdots +12=78,x=26.$$ Obviously, $1,2\in A$ and $11,12\in B$. Denote $A=\{1,2,a,b,c,d\}$, where $a<b<c<d$. Then $a+b+c+d=23$, and $a\ge 3$, $8\le d\le 10$ (if $d\le 7$, then $a+b+c+d\le 4+5+6+7=22$, which is a contradiction.)

(1) If $d=8$, then $A=\{1,2,a,b,c,8\}$, $c\le 7$, $a+b+c=15$. Thus, $(a,b,c)=(3,5,7)$ or $(4,5,6)$, that is, $A=\{1,2,3,5,7,8\}$ or $A=\{1,2,4,5,6,8\}$.

If $A=\{1,2,3,5,7,8\}$, then $B=\{4,6,9,10,11,12\}$. Since $B$ contains $11,4,6$ and $12$, there is only one tower that, in $A$, $8$ and $3$, $3$ and $1$, $1$ and $5$, $5$ and $7$ are adjacent.



If $A=\{1,2,4,5,6,8\}$, then $B=\{3,7,9,10,11,12\}$. Similarly, we see that, in $A$, $1$ and $2$, $5$ and $6$, $4$ and $8$ are adjacent, respectively. There are two arrangements, that is, there are two towers.



(2) If $d=9$, then $A=\{1,2,a,b,c,9\}$, $c\le 8$, $a+b+c=14$, where $(a,b,c)=(3,5,6)$ or $(3,4,7)$, that is, $A=\{1,2,3,5,6,9\}$ or $A=\{1,2,3,4,7,9\}$.

If $A=\{1,2,3,5,6,9\}$, then $B=\{4,7,8,10,11,12\}$. To obtain $4$, $10$ and $12$ in $B$, $1$, $3$, and $9$ in $A$ must be adjacent pairwise, it is impossible!



If $A=\{1,2,3,4,7,9\}$, then $B=\{5,6,8,10,11,12\}$. To obtain $6$, $8$ and $12$ in $B$, $2$ and $4$, $1$ and $7$, $9$ and $3$ must be adjacent in $A$, respectively. There are two arrangements, that is, there are two towers.

(3) If $d=10$, then $A=\{1,2,a,b,c,10\}$, where $c\le 9$, $a+b+c=13$. Thus, $(a,b,c)=(3,4,6)$, that is, $A=\{1,2,3,4,6,10\}$ and $B=\{5,7,8,9,11,12\}$. To obtain $8$, $9$, $11$ and $12$ in $B$, $6$ and $2$, $6$ and $3$, $10$ and $1$, $10$ and $2$ must be adjacent, respectively. There is only one tower.



Summing up, there are six different towers all together. ☐