Partition of $S = \{1,2,\dots, 3n\}$ in to three subsets $A, B, C$ such that $|A| = |B| = |C| = n$ Let $n$ be a positive integer and consider the set $S = \{1,2,\dots ,3n\}$. Show that, for every partition of $S$ into 3 subsets $A, B, C$ such that:
i) $|A| = |B| = |C| = n$ (here $|X|$ denotes the number of elements of set $X$)
ii) $A \cap B = B \cap C = C \cap A = \varnothing$ and $A \cup B \cup C = S$
there exist 3 elements $a \in A, b \in B, c \in C$ such that one of $a, b, c$ equals sum of the two other elements.
 A: Suppose otherwise.
Write $A = \{a_1, \dotsc, a_n\}$, $B = \{b_1, \dotsc, b_n\}$, and $C = \{c_1, \dotsc, c_n\}$, each indexed increasingly, such that $a_1 < b_1 < c_1$.
Thus $a_1 = 1$. Set $k = b_1 - 1$, so that $\{1, \dotsc, k\} \subseteq A$.
Now, for any $b \in B$, if any of $b+1, \dotsc, b+k$ are in $C$, then we have a forbidden triple. Similarly for any $c \in C$, the succeeding $k$ elements cannot be in $B$. So between any elements of $B$ and of $C$ there must be at least $k$ elements of $A$.
Consider any $c \in C$ whose predecessor is not in $C$, i.e. so that $c - 1 \notin C$. Then before $c$ we have at least $k$ elements of $A$; in particular $c - k \in A$.
If $(c - k) + b_1 = c + 1$ is in $C$, then we have a forbidden triple.
Similarly, $(c - (k-1)) + b-1 = c + 2, \dotsc, (c-1)+b_1 = c+k$ cannot be in $C$.
So we can have at most one element of $C$ occur successively, whose $k$ predecessors and $k$ successors are all in $A$.
Thus, the most compactly that the elements of $C$ can occur is in a pattern like this:
$$
\underbrace{A\dotsm A}_k C \underbrace{A\dotsm A}_k C \dotsm\underbrace{A\dotsm A}_k C,
$$
where the last entry is $3n$ itself.
If any elements of $B$ were interspersed, even more elements of $A$ would have to appear.
For each element of $C$ we have at least $k$ elements of $A$, contradicting that $|C| = |A|$, unless $k = 1$.
Even then, the only way to have just $n$ elements of $A$ appear is with the pattern
$$
AC AC \dotsm AC
$$
where $c_n = 3n$, which contradicts that $1 \in A$.
