Difference between revisions of "1988 AHSME Problems/Problem 16"

Revision as of 02:00, 23 October 2014

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

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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Difference between revisions of "1988 AHSME Problems/Problem 16"

Revision as of 02:00, 23 October 2014

Problem

Solution

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