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2003 AMC 12B/Problem 8

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Problem

Equilateral $\triangle DEF$ is inscribed in equilateral $\triangle ABC$ with $\overline{DE}$ perpendicular to $\overline{BC}$ . What is the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ ?

$(\mathrm{A})\ \frac16 \qquad (\mathrm{B})\ \frac14 \qquad (\mathrm{C})\ \frac13 \qquad (\mathrm{D})\ \frac25 \qquad (\mathrm{...$

Solution

With some quick angle chasing, we can find that it is also true that $\overline{DE}$ is perpendicular to $\overline{AC}$ and $\overline{FD}$ is perpendicular to $\overline{AB}$ . Then we have $\triangle DEF$ and three congruent (by AAS congruency) triangles making up $\triangle ABC$ . So, letting $DE = 1$ , which is permissible since we only want the ratio of the areas, and all equilateral triangles are similar, we have that $BD = \frac{1}{\sin{60^\circ}}= \frac2{\sqrt3}$ and $CD = \frac{1}{\tan{60^\circ}} = \frac{1}{\sqrt3}$ . So we want:

$\left(\frac{ED}{BC}\right)^2 = \left(\frac{ED}{BD+DC}\right)^2= \left(\frac{1}{\frac2{\sqrt3}+\frac1{\sqrt3}}\right)^2 = \fra...$

It follows that the answer is $\mathrm{C}$ .

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Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

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