[Geometry]
Problem: $\quad$ Let $T=T N Y W R$. In parallelogram $A R M L$, points $P$ and $Q$ trisect $\overline{A R}$ and points $W, X, Y, Z$ divide $\overline{M L}$ into fifths (where $W$ is closest to $M$, and points $X$ and $Y$ are both between $W$ and $Z$ ). If $[A R M L]=T$, compute $[P Q W Z]$.

Solution: Let $h$ be the distance between $\overline{A R}$ and $\overline{M L}$, and for simplicity, let $A R=M L=15 n$. Then $[A R M L]=15 n h$, and $[P Q W Z]=(1 / 2)(P Q+W Z) h$. Note that $P Q=15 n / 3=5 n$ and $W Z=15 n-3 n-3 n=9 n$. Thus $[P Q W Z]=7 n h=(7 / 15) \cdot[A R M L]=7 T / 15$. With $T=15$, the answer is 7 .

Final answer: 7