# 1988 AHSME Problems/Problem 16

## Problem $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("A'", A, N); label("B'", B, SE); label("C'", C, SW); label("A", D, E); label("B", E, E); label("C", F, W); [/asy]$ $ABC$ 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 $\textbf{(A)}\ \frac{1}{36}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{\sqrt{3}}{4}\qquad \textbf{(E)}\ \frac{9+8\sqrt{3}}{36}$

## Solution

Let $\triangle ABC$ have side length $s$ and $\triangle A'B'C'$ have side length $t$. Thus the altitude of $\triangle ABC$ is $\frac{s\sqrt{3}}{2}$. Now observe that this altitude is made up of three parts: the distance from $BC$ to $B'C'$, plus the altitude of $\triangle A'B'C'$, plus a top part which is equal to the length of the diagonal line from the bottom-left corner of $\triangle ABC$ to the bottom left corner of $\triangle A'B'C'$ (as an isosceles trapezium is formed with parallel sides $AB$ and $A'B'$, and legs $AA'$ and $BB'$). We drop a perpendicular from $B'$ to $BC$, which meets $BC$ at $D$. $\triangle BDB'$ has angles $30^{\circ}$, $60^{\circ}$, and $90^{\circ}$, and the vertical side is that distance from $BC$ to $B'C'$, which is given as $\frac{1}{6} \times \frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12}$, so that by simple trigonometry, the length of the diagonal line is $\frac{s\sqrt{3}}{12} \times \csc{30^{\circ}} = \frac{s\sqrt{3}}{6}.$ Thus using the "altitude in three parts" idea, we get $\frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12} + \frac{t\sqrt{3}}{2} + \frac{s\sqrt{3}}{6} \implies \frac{s\sqrt{3}}{4} = \frac{t\sqrt{3}}{2} \implies t = \frac{1}{2}s.$ Thus the sides of $\triangle A'B'C'$ are half as long as $\triangle ABC$, so the area ratio is $(\frac{1}{2}) ^ {2} = \frac{1}{4}$, which is $\boxed{\text{C}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 