67th Romanian Mathematical Olympiad

If $X=\{a,b,c\}$, with $a<b<c$, then $a+b<a+c<b+c$ are distinct elements of $Y$. Therefore $Y=\{a+b,b+c,c+a\}$. Since $a<b<a+b<a+c<b+c$, the common element of $X$ and $Y$ can be only $c=a+b$.

If $c=3$, then $a=1,b=2$, therefore $X=\{1,2,3\},Y=\{3,4,5\}$ which fulfill the given conditions ($5\in Y$ and $X\cap Y=\{5\}$). If $b=3$, then $a\in \{1,2\}$. If $a=1$, then $c=a+b=4$, hence $X=\{1,3,4\},Y=\{4,5,7\}$. If $a=2$, then $c=5$, hence $X=\{2,3,5\},Y=\{5,7,8\}$. Both pairs of sets fulfill the given conditions. If $a=3$, then $b>3$ and $c>3$, therefore all the elements of $Y$ are larger than 5, so there are no solutions in this case.