The set {1, . . . , 9} is split in any way into two subsets. Prove that in at least one subset there are three numbers of which one is the arithmetic mean of the other two.
I tried a lot on this using the pigeon hole principle but was unable to do this. I was able to solve it using Brute-force, but can we solve it by choosing a general n or a better method than considering almost all possible ways? Thanks
This is kind of brute force, but it's not too terrible:
Consider the set $S$ with the number $5$. It cannot contain both elements of either of the pairs $(3,7),(6,7),(7,9)$. Thus, $S$ cannot contain $7$, for if it did then $3,6,9$ would all be in the other set $T$. Similarly, $S$ also cannot contain $3$.
As $S$ cannot contain both $1$ and $9$, we may without loss of generality assume that $T$ contains $1$. Then we have that, as $T$ now contains each of $1,3,7$, it cannot contain $2$ or $4$, so $2,4,5\in S$. This means $6\in T$, which results in a contradiction as now $8$ cannot be in either set.
For general $n\geq 9$, we can note that any partition of $\{1,\cdots,n\}$ into two subsets will necessarily partition $\{1,\cdots,9\}$, so there must be three numbers of which one is the arithmetic mean of the other two. For $n\leq 8$, consider the partition $\{1,2,5,6\},\{3,4,7,8\}$ as one that does not contain any 3-term arithmetic progressions.