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\begin{document}
\title{Shortlist 2012 A2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 9}
\maketitle
\section*{Problem}
Decide whether there exists a partition of $S$ into three
non-empty subsets $A$, $B$, $C$ such that the sets
$A+B$, $B+C$, $C+A$ are disjoint, where (a) $S = \ZZ$, (b) $S = \QQ$.
\section*{Video}
\href{https://www.youtube.com/watch?v=pG4Ea3O88aA&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/pG4Ea3O88aA}}
\newpage
\section*{Solution}
For (a), just take $A = \{0 \bmod 3\}$, $B = \{1 \bmod 3\}$,
$C = \{2 \bmod 3\}$.
Actually, let's prove this construction for (a) is unique,
up to relabeling the sets.
We prove that:
\begin{claim*}
We have $A+B-C \subseteq C$.
\end{claim*}
\begin{proof}
The hypothesis states that
since if $a+b-c = a'$ for some $a,a' \in A$, $b \in B$, $c \in C$,
we get a contradiction, and similarly.
Similarly, we have $B+C-A \subseteq A$, $C+A-B \subseteq B$.
\end{proof}
Let $a$, $a'$ be any two elements of $A$.
Let $b_0$ be any fixed but arbitrary element of $B$.
Consider any $c \in C$.
Then \[ a + b_0 - c \in C \implies a' + b_0 - (a + b_0 - c)
= \boxed{c + (a'-a) \in C}. \]
This, together with symmetric variants,
implies that $A$, $B$, $C$ must be arithmetic progressions
with the same common difference
(by considering the smallest gap
between two consecutive elements in the same set).
Hence that common difference must be $3$ and the uniqueness is proved.
We show this implies the answer to (b) is negative.
Assume for contradiction such a partition exists.
Let $a_0$, $b_0$, $c_0$ be arbitrary elements of $A$, $B$, $C$.
Choose a large integer $n$ such that $na_0$, $nb_0$, $nc_0$
are all integers divisible by $3$.
Then consider $nA \cap \ZZ$, $nB \cap \ZZ$, $nC \cap \ZZ$.
It is a partition satisfying (a) and with each set nonempty,
but it is not the one we expected
(because all three sets have multiples of $3$).
This is a contradiction.
\end{document}