17・148 In $\triangle ABC$, $AM$ and $CN$ are the medians to sides $BC$ and $AB$, respectively, and they intersect at point $O$. $P$ is the midpoint of $AC$, and $MP$ intersects $CV$ at $Q$. If the area of $\triangle OMQ$ is $n$, then the area of $\triangle ABC$ is
(A) $16 n$.
(B) $18 n$.
(C) $21 n$.
(D) $24 n$.
(E) $27 n$.
(17th American High School Mathematics Examination, 1966)
[Solution] Let $O$ be the centroid. Therefore, we have
$$
\begin{aligned}
O Q & =C O-C Q=\frac{2}{3} C N-\frac{1}{2} C N \\
& =\frac{1}{6} \mathrm{CN} .
\end{aligned}
$$

Let $h$ be the height from $\triangle O M Q$ to side $O Q$, then the height from $\triangle C N B$ to side $C N$ is $2h$.

Thus, $S_{\triangle O M Q}=\frac{1}{2} \cdot O Q \cdot h=\frac{1}{12} \cdot C N \cdot h=n$, and $S_{\triangle A B C}=2 S_{\triangle C N B}=2 \cdot \frac{1}{2} \cdot C N \cdot 2 h=2 C N \cdot h=24 n$.
Therefore, the answer is $(D)$.