Problem
1988 AHSME Problem 16
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("A'", A, N); label("B'", B, SE); label("C'", C, SW); label("A", D, E); label("B", E, E); label("C", F, W); [/asy]](/static/aops/99d677e49262bbd174ccaccf2dc6b15986c0de3d.png)
ABC and A'B'C' are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side BC and side B'C' is \frac{1}{6} the altitude of \triangle ABC. The ratio of the area of \triangle A'B'C' to the area of \triangle ABC is
\textbf{(A)}\ \frac{1}{36}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{\sqrt{3}}{4}\qquad \textbf{(E)}\ \frac{9+8\sqrt{3}}{36}
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