jred wrote: There are $12$ acrobats who are assigned a distinct number ($1, 2, \cdots , 12$) respectively. Half of them stand around forming a circle (called circle A); the rest form another circle (called circle B) by standing on the shoulders of every two adjacent acrobats in circle A respectively. Then circle A and circle B make up a formation. We call a formation a “tower” if the number of any acrobat in circle B is equal to the sum of the numbers of the two acrobats whom he stands on. How many heterogeneous towers are there? (Note: two towers are homogeneous if either they are symmetrical or one may become the other one by rotation. We present an example of $8$ acrobats (see attachment). Numbers inside the circle represent the circle A; numbers outside the circle represent the circle B. All these three formations are “towers”, however they are homogeneous towers.) It is clear that the acrobats on circle $B$ must have numbers totaling exactly double the acrobats on circle $A$. So since the total of all the numbers is $\frac{12*13}{2}=78$, the acrobats on circle $A$ must total $\frac{78}{3}=26$. A little trial and error will show there are only $7$ combinations for possible sets of numbers for circle $A$: Case (a): $1,2,3,4,5,11$ Case (b): $1,2,3,4,6,10$ Case (c): $1,2,3,4,7,9$ Case (d): $1,2,3,5,6,9$ Case (e): $1,2,3,5,7,8$ Case (f): $1,2,4,5,6,8$ Case (g): $1,3,4,5,6,7$ Case (a) is quickly ruled out, however, since it would require that acrobat $11$ be adjacent (within circle $A$) at least one acrobat with number at least $2$--resulting in a circle $B$ acrobat with number at least $13$--an impossibility. For case (b), to avoid an impossible acrobat $13$ or higher, we would require that acrobat $10$ be adjacent (within circle $A$) to both acrobats $1$ and $2$--creating the following partial sequence: Case (b): $1-10-2-??-??-??-1$ However acrobats $4$ and $6$ may not be adjacent (with circle $A$) since otherwise we'd have acrobat $10$ in circle $B$--leaving the following two cases--up to homogeneity: Case (b1): $1-10-2-6-3-4-1$ Case (b2): $1-10-2-4-3-6-1$ However, case (b2) results in two different acrobats $7$ in circle $B$ so it doesn't work. But case (b1) does satisfy the problem conditions. For case (c), the only way to produce an acrobat $8$ from the acrobats in circle $A$ is for acrobats $1$ and $7$ to be adjacent (in circle $A$). Moreover, to produce an acrobat $10$ in circle $B$ in would be required that in circle $A$, either acrobats $3$ and $7$ or acrobats $1$ and $9$ be adjacent. This gives the following two cases--up to homogeneity: Case (c1): $1-7-3-??-??-??-1$ Case (c2): $7-1-9-??-??-??-7$ For case (c1), acrobat $9$ must be adjacent (in circle $A$) to two acrobats both numbered $3$ or below--so the middle of the unknown acrobats must be acrobat $2$, resulting in the following two subcases: Case (c1a): $1-7-3-9-2-4-1$ Case (c1b): $1-7-3-4-2-9-1$ Case (c1a) satisfies the problem conditions; case (c1b) does not. Finally, for case (c2), acrobats $2$ and $3$ must be adjacent (in circle $A$) (to be able to produce an acrobat $5$) and one must be adjacent to acrobat $9$ (in circle $A$), leaving the following two subcases: Case (c2a): $7-1-9-2-3-4-7$ Case (c2b): $7-1-9-3-2-4-7$ Case (c2b) satisfies the problem conditions; case (c2a) does not. Case (d) doesn't lead to any solutions. To produce acrobat $4$ in circle $B$, acrobats $1$ and $3$ must be adjacent in circle $A$. This in turn requires that acrobats $9$ and $2$ be adjacent, and next to, acrobats $1$ and $3$ in circle $A$. This in turn requires that acrobats $5$ and $6$ be adjacent in circle $A$--leading to two acrobats $11$ in circle $B$. For case (e), to produce acrobat $4$ in circle $B$, acrobats $1$ and $3$ must be adjacent in circle $A$. Moreover, to produce acrobat $6$ in circle $B$, acrobats $1$ and $5$ must be adjacent in circle $A$. And acrobats $7$ and $8$ must not be adjacent in circle $A$. This leads to the following subcases: Case (e1): $3-1-5-7-2-8-3$ Case (e2): $3-1-5-8-2-7-3$ Case (e1) satisfies the problem conditions; case (e2) does not. By similar arguments, for case (f), we can show that there are two subcases satisfying the problem conditions: Case (f1): $1-2-5-6-4-8$ Case (f2): $1-2-8-4-5-6$ Case (g) does not lead to any solutions, as it would require that acrobat $1$ be adjacent in circle $A$ to at least one acrobat in the range $3$ through $6$, resulting in an acrobat in circle $B$ which is actually already in circle $A$. Up to homogeneity, therefore, only the following sequences of acrobats in circle $A$ satisfy the problem conditions: $1-2-5-6-4-8-1$ $1-2-8-4-5-6-1$ $1-3-8-2-7-5-1$ $1-4-2-9-3-7-1$ $1-4-3-6-2-10-1$ $1-7-4-2-3-9-1$ Once circle $A$ is known, of course, circle $B$ is uniquely determined. There are $6$ heterogeneous towers.