# 1995 AHSME Problems/Problem 19

## Problem

Equilateral triangle $DEF$ is inscribed in equilateral triangle $ABC$ such that $\overline{DE} \perp \overline{BC}$. The ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is $\mathrm{(A) \ \frac {1}{6} } \qquad \mathrm{(B) \ \frac {1}{4} } \qquad \mathrm{(C) \ \frac {1}{3} } \qquad \mathrm{(D) \ \frac {2}{5} } \qquad \mathrm{(E) \ \frac {1}{2} }$

## Solution $[asy]draw((0,0)--(3,0)); draw((0,0)--(1.5, 2.59807621)); draw((3,0)--(1.5, 2.59807621)); draw((1,0)--(1, 1.73205081)); draw((1,0)--(2.5, 0.866025404)); draw((1, 1.73205081)--(2.5, 0.866025404));[/asy]$

Let's take one of the smaller right triangles. Without loss of generality, let the smaller leg be $1$. Since the triangle is a 30-60-90 right triangle, then the other leg is $\sqrt{3}$ and the hypotenuse is $2$. The side length of the bigger triangle is $1 + 2 = 3$ and the side length of the smaller triangle is $\sqrt{3}$. The ratio of the areas of two similar triangles is the square of the ratio of two corresponding side lengths, so the ratio of the area of triangle DEF to the area of triangle ABC is $\left(\frac{\sqrt{3}}{3}\right)^2 = \dfrac{1}{3} \rightarrow (C)$.

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