9. Let $A B C D E F$ be a regular hexagon of area 1. Let $M$ be the midpoint of $D E$. Let $X$ be the intersection of $A C$ and $B M$, let $Y$ be the intersection of $B F$ and $A M$, and let $Z$ be the intersection of $A C$ and $B F$. If $[P]$ denotes the area of polygon $P$ for any polygon $P$ in the plane, evaluate $[B X C]+[A Y F]+[A B Z]-[M X Z Y]$.
Answer: 0
Let $O$ be the center of the hexagon. The desired area is $[A B C D E F]-[A C D M]-[B F E M]$. Note that $[A D M]=[A D E] / 2=[O D E]=[A B C]$, where the last equation holds because $\sin 60^{\circ}=\sin 120^{\circ}$. Thus, $[A C D M]=[A C D]+[A D M]=[A C D]+[A B C]=[A B C D]$, but the area of $A B C D$ is half the area of the hexagon. Similarly, the area of $[B F E M]$ is half the area of the hexagon, so the answer is zero.
10. For all real numbers $x$, let
$$
f(x)=\frac{1}{\sqrt[2011]{1-x^{2011}}} .
$$

Evaluate $(f(f(\ldots(f(2011)) \ldots)))^{2011}$, where $f$ is applied 2010 times.
Answer: $2011^{2011}$ Direct calculation shows that $f(f(x))=\frac{\sqrt[2011]{1-x^{2011}}}{-x}$ and $f(f(f(x)))=x$. Hence $(f(f(\ldots(f(x)) \ldots)))=x$, where $f$ is applied 2010 times. So $(f(f(\ldots(f(2011)) \ldots)))^{2011}=$ $2011^{2011}$.