 Solution

Construct $X\subseteq B$ greedily. Begin with one element of $B$. While some $b\in B$ satisfiesadjoin $b$ to $X$. The first translate contributes $|A|$ points to $A+X$, and each later translate contributes at least $|A|/2$ new points. Thereforeand hence

When the process stops, every $b\in B$ satisfies $|(A+b)\cap (A+X)|>|A|/2$. Each point in this intersection has the formand therefore gives a triple $(x,a,a^{\prime })\in X\times A\times A$ with $b=a-a^{\prime }+x$. Distinct intersection points give distinct $a^{\prime }$, so there are more than $|A|/2$ such triples.

For $b,b^{\prime }\in B$, the corresponding sets of possible $a^{\prime }\in A$ both have size greater than $|A|/2$ and hence intersect. Using a common $a^{\prime }$ givesso $b-b^{\prime }=a-c+x-y$. Thus