HMMT-Nov 2014 team, Problem 10

Let $A B C D E F$ be a convex hexagon with the following properties. (a) $\overline{A C}$ and $\overline{A E}$ trisect $\angle B A F$. (b) $\overline{B E} \| \overline{C D}$ and $\overline{C F} \| \overline{D E}$. (c) $A B=2 A C=4 A E=8 A F$. Suppose that quadrilaterals $A C D E$ and $A D E F$ have area 2014 and 1400, respectively. Find the area of quadrilateral $A B C D$.

Solution: From conditions (a) and (c), we know that triangles $A F E, A E C$ and $A C B$ are similar to one another, each being twice as large as the preceding one in each dimension. Let $\overline{A E} \cap \overline{F C}=P$ and $\overline{A C} \cap \overline{E B}=Q$. Then, since the quadrilaterals $A F E C$ and $A E C B$ are similar to one another, we have $A P: P E=A Q: Q C$. Therefore, $\overline{P Q} \| \overline{E C}$. Let $\overline{P C} \cap \overline{Q E}=T$. We know by condition (b) that $\overline{B E} \| \overline{C D}$ and $\overline{C F} \| \overline{D E}$. Therefore, triangles $P Q T$ and $E C D$ have their three sides parallel to one another, and so must be similar. From this we deduce that the three lines joining the corresponding vertices of the two triangles must meet at a point, i.e., that $P E, T D, Q C$ are concurrent. Since $P E$ and $Q C$ intersect at $A$, the points $A, T, D$ are collinear. Now, because $T C D E$ is a parallelogram, $\overline{T D}$ bisects $\overline{E C}$. Therefore, since $A, T, D$ are collinear, $\overline{A D}$ also bisects $\overline{E C}$. So the triangles $A D E$ and $A C D$ have equal area. Now, since the area of quadrilateral $A C D E$ is 2014 , the area of triangle $A D E$ is $2014 / 2=1007$. And since the area of quadrilateral $A D E F$ is 1400 , the area of triangle $A F E$ is $1400-1007=393$. Therefore, the area of quadrilateral $A B C D$ is $16 \cdot 393+1007=7295$, as desired. $\fbox{7295}$.

Answer: 7295