4. In triangle $A B C$ with area 51 , points $D$ and $E$ trisect $A B$ and points $F$ and $G$ trisect $B C$. Find the largest possible area of quadrilateral $D E F G$.
Solution: 17
Assume $E$ is between $D$ and $B$, and $F$ is between $G$ and $B$ (the alternative is to switch two points, say $D$ and $E$, which clearly gives a non-convex quadrilateral with smaller
area). If two triangles have their bases on the same line and the same opposite vertex, then it follows from the $\frac{1}{2} b h$ formula that their areas are in the same ratio as their bases. In particular (brackets denote areas),
$$
\frac{[D B G]}{[A B C]}=\frac{[D B G]}{[A B G]} \cdot \frac{[A B G]}{[A B C]}=\frac{D B}{A B} \cdot \frac{B G}{B C}=\frac{2}{3} \cdot \frac{2}{3}=\frac{4}{9},
$$
and similarly $[E B F] /[A B C]=1 / 9$. Subtracting gives $[D E F G] /[A B C]=1 / 3$, so the answer is $[A B C] / 3=17$.