17$\cdot$162 $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ are equilateral triangles, and their corresponding sides are parallel. They share a common center (as shown in the figure). The distance between $B C$ and $B^{\prime} C^{\prime}$ is $\frac{1}{6}$ of the height of $\triangle A B C$. The ratio of the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ to the area of $\triangle A B C$ is
(A) $\frac{1}{36}$.
(B) $\frac{1}{6}$.
(C) $\frac{1}{4}$.
(D) $\frac{\sqrt{3}}{4}$.
(E) $\frac{9+8 \sqrt{3}}{36}$.
(39th American High School Mathematics Examination, 1988)
[Solution] Let the heights of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ be $h$ and $h^{\prime}$, respectively, and let their centers be $O$.
Then $O H^{\prime}=\frac{1}{3} h^{\prime}, \quad O H=\frac{1}{3} h$.
Also, $H^{\prime} H=\frac{1}{6} h$, so $O H^{\prime}=\frac{1}{6} h$.
Thus, $h^{\prime}: h=O H^{\prime}: O H=\frac{1}{6} h: \frac{1}{3} h=1: 2$.
Therefore, $S_{\triangle A^{\prime} B^{\prime} C^{\prime}}: S_{\triangle A B C}=1^{2}: 2^{2}=1: 4$.
Hence, the answer is $(C)$.