4. Convex hexagon $A B C D E F$ is drawn in the plane such that $A C D F$ and $A B D E$ are parallelograms with area 168. $A C$ and $B D$ intersect at $G$. Given that the area of $A G B$ is 10 more than the area of $C G B$, find the smallest possible area of hexagon $A B C D E F$.
Answer: 196
Since $A C D F$ and $A B D E$ have area 168, triangles $A B D$ and $A C D$ (which are each half a parallelogram) both have area 84 . Thus, $B$ and $C$ are the same height away from $A D$, and since $A B C D E F$ is convex, $B$ and $C$ are on the same side of $A D$. Thus, $B C$ is parallel to $A D$, and $A B C D$ is a trapezoid. In particular, we have that the area of $A B G$ equals the area of $C D G$. Letting this quantity be $x$, we have that the area of $B C G$ is $x-10$, and the area of $A D G$ is $84-x$. Then notice that $\frac{[A B G]}{[C B G]}=\frac{A G}{G C}=\frac{[A D G]}{[C D G]}$. This means that $\frac{x}{x-10}=\frac{84-x}{x}$. Simplifying, we have $x^{2}-47 x+420=0$; this has solutions $x=12$ and $x=35$. The area of $A B C D E F$ is twice the area of trapezoid $A B C D$, or $2[x+(x-10)+(84-x)+x]=4 x+148$; choosing $x=12$, we get that the smallest possible area is $48+148=196$