Subset $A \subset [0,8]^2$ with the largest measure satisfying $(A+(3,4) )\cap A=\emptyset$

Question

Find the subset $A \subset [0,8]^2$ with the largest (Lebesgue) measure such that $(A+(3,4) )\cap A=\emptyset$.

I tried first rotating the problem so that I'll only have to ensure that $A \cap (A+(5,0))=\emptyset$. But now I'm stuck again.

Answer 1

Rotating to make the vector horizontal is reasonable, because the situation then fits Fubini's theorem better. But I find it more convenient to keep the square as is, with the understanding that Fubuni's theorem can be applied to rotated axes, one of which is parallel to $(3,4)$. Every line parallel to $(3,4)$ crosses the square along a segment with length between $0$ and $10$. Indeed, the segments AI and CJ have length 10 and bound a parallelogram between them.

• If a segment has length less than $5$, we can include all of it into the set $A$.
• If it has length $10$, we can get a set of length at most $5$, because if the segment is cut in half and two halves are moved together, the set can't overlap itself.
• Therefore, segments of length between $5$ and $10$ can contribute at most $5$.

One example of a set that achieves the maximal measure was given by Mher in a comment: $A = [0,8]^2\setminus [3,8]\times[4,8]$. Another example is the convex hexagon BHFDEG.