12. Given that the centroid of $\triangle A B C$ is $G, M$ is the midpoint of side $B C$, a line through $G$ parallel to $B C$ intersects side $A B$ at $X$ and side $A C$ at $Y$, and $X C$ intersects $G B$ at point $Q, Y B$ intersects $G C$ at point $P$. If the area of $\triangle A B C$ is 32 square units, find the area of $\triangle M P Q$.
12. As shown in the figure, extend $B G$ to intersect $A C$ at $N$, then $N$ is the midpoint of $A C$. From $X Y / / B C$, we know $\frac{A X}{X B}=\frac{A G}{G M}=2$, and $\frac{N C}{C A}=\frac{1}{2}$. Applying Menelaus' theorem to $\triangle A B N$ and the transversal $X Q C$, we have $\frac{A X}{X B} \cdot \frac{B Q}{Q N} \cdot \frac{N C}{C A}=2 \cdot \frac{B Q}{Q N} \cdot \frac{1}{2}=1$.
Thus, $B Q=Q N$. Therefore, $M Q / / A C$, and
$M Q=\frac{1}{2} C N=\frac{1}{4} A C$. Similarly, $M P / / A B$ and $M P=\frac{1}{4} A B$.
From this, we know that $\angle P M Q$ and $\angle B A C$ have their sides parallel and in opposite directions, thus $\angle P M Q=\angle B A C$, and $\frac{M P}{A B}=\frac{M Q}{A C}=\frac{1}{4}, \triangle M P Q \backsim \triangle A B C$, hence $S_{\triangle M P Q}=2$.