  Let $F_{3,3}$ be the 3-graph on 6 vertices, labelled abcxyz, and 10 edges,
one of which is abc, and the other 9 of which are all triples that contain 1
vertex from abc and 2 vertices from xyz. We show that for all $n \ge 6$, the
maximum number of edges in an $F_{3,3}$-free 3-graph on $n$ vertices is
$\binom{n}{3} - \binom{\lfloor n/2 \rfloor}{3} - \binom{\lceil n/2 \rceil}{3}$.
This sharpens results of Zhou and of the second author and R\"odl.
