HMMT November

Let $[X Y Z]$ denote the area of $\triangle X Y Z$. Solution 1: Let $A M=\ell$, let $D E=d$, and let the midpoint of $\overline{D E}$ be $F$. Since $\frac{A D}{A B}=\frac{A E}{A C}=\frac{2}{3}$ by the angle bisector theorem, $F$ lies on $\overline{A M}$ and $\triangle A D E$ is similar to $\triangle A B C$. Note that $\angle D M E$ is formed by angle bisectors of $\angle A M B$ and $\angle A M C$, which add up to $180^{\circ}$. Thus $\angle D M E$ is right, so both $\triangle D M F$ and $\triangle E M F$ are isosceles. This implies that $F M=\frac{d}{2}$. Applying the similarity between $\triangle A D E$ and $\triangle A B C$, we get $\frac{d}{\ell}=\frac{A D}{A B}=\frac{2}{3}$. Since $A F=2 F M,[A D E]=2[D M E]$. Finally, since $\frac{[A D E]}{[A B C]}=\frac{4}{9}, \frac{[D M E]}{[A B C]}=\frac{1}{2} \cdot \frac{4}{9}=\frac{2}{9}$. Solution 2: We compute the ratio $\frac{[D M E]}{[A B C]}$ by finding $1-\frac{[A D E]}{[A B C]}-\frac{[D B M]}{[A B C]}-\frac{[E M C]}{[A B C]}$. Since $\frac{A D}{D B}=$ $\frac{A E}{E C}=\frac{A M}{\frac{1}{2} B C}=2$ by the angle bisector theorem, we see that $\frac{[A D E]}{[A B C]}=\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)=\frac{4}{9}$. Also, since $B M=C M=\frac{1}{2} B C, \frac{[D B M]}{[A B C]}=\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)=\frac{1}{6}$, and $\frac{[E M C]}{[A B C]}=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{1}{6}$. Thus, $\frac{[D M E]}{[A B C]}=1-\frac{[A D E]}{[A B C]}-$ $\frac{[D B M]}{[A B C]}-\frac{[E M C]}{[A B C]}=1-\frac{4}{9}-\frac{1}{6}-\frac{1}{6}=\frac{2}{9}$.