Problem 1146

Solution:

The desired partition of the set $\{1,2, \ldots, 3 n\}$ corresponds to a partition of the vertices of a regular $3 n$-gon $P_{1} P_{2} \ldots P_{3 n}$ into triples $\left\{A_{i}, B_{i}, C_{i}\right\}$ such that the angles of each of the triangles $A_{i} B_{i} C_{i}$ are equal to $\frac{n-1}{3 n} \pi, \frac{n}{3 n} \pi$ and $\frac{n+1}{3 n} \pi$. By suitably labeling the vertices of the $3 n$-gon, we can arrange for the vertices $A_{1}, B_{1}, C_{1}$ to be exactly $P_{n}, P_{2 n-1}, P_{3 n}$. In other words, we lose no generality if we assume that among the triples $\{a, b, c\}$ into which the set $\{1,2, \ldots, 3 n\}$ is divided there is also the triple $\{n, 2 n-1,3 n\}$.

One of the remaining $n-1$ triples must contain two numbers from the interval $[2 n, 3 n-1]$, and these can only be $2 n$ and $3 n-1$. The only triple that contains these numbers and does not contain $n$ is $\{n-1,2 n, 3 n-1\}$.

All the other triples contain exactly one number from each of the intervals $[1, n-2],[n+1,2 n-2]$ and $[2 n+1,3 n-2]$. By applying the mapping $(a, b, c) \rightarrow(a, b-2, c-4)$ for $a<b<c$ we obtain the corresponding decomposition of the set $\{1,2, \ldots, 3(n-2)\}$ into triples. Since for $n=1$ such a decomposition is not possible, by simple induction we show that it is not possible for any odd $n$ either.

On the other hand, for even $n=2 m$ the triples $( 2 i-1,2 i+n, 2 i+2 n-1)$ and $(2 i, 2 i+n-1,2 i+2 n)$ for $i=1, \ldots, m$ satisfy the conditions.