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 the length relationships in a 30-60-90 triangle,, the length of the diagonal line is $\frac{s\sqrt{3}}{12} \times 2 = \frac{s\sqrt{3}}{6}.$ 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 $\boxed{\text{C}}$ .

~Johnxyz1(minorEdits)

1988 AHSME (Problems • Answer Key • Resources)
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1988 AHSME Problems/Problem 16

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