# 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 $\triangle ABC$? $(\mathrm{A})\ \frac16 \qquad (\mathrm{B})\ \frac14 \qquad (\mathrm{C})\ \frac13 \qquad (\mathrm{D})\ \frac25 \qquad (\mathrm{E})\ \frac12$

## 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: $$\left(\frac{ED}{BC}\right)^2 = \left(\frac{ED}{BD+DC}\right)^2= \left(\frac{1}{\frac2{\sqrt3}+\frac1{\sqrt3}}\right)^2 = \frac13$$

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

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