AMC10 2014 10B, Problem 15

In rectangle $ABCD$, $DC = 2 \cdot CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$? [asy] draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle); draw((0, 0)--(sqrt(3)/3, 1)); draw((0, 0)--(sqrt(3), 1)); label("A", (0, 1), N); label("B", (2, 1), N); label("C", (2, 0), S); label("D", (0, 0), S); label("E", (sqrt(3)/3, 1), N); label("F", (sqrt(3), 1), N); [/asy]

Solution: Let the length of $AD$ be $x$, so that the length of $AB$ is $2x$ and $\text{[}ABCD\text{]}=2x^2$. Because $ABCD$ is a rectangle, $\angle ADC=90^{\circ}$, and so $\angle ADE=\angle EDF=\angle FDC=30^{\circ}$. Thus $\triangle DAE$ is a $30-60-90$ right triangle; this implies that $\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}$, so $\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}$. Now drop the altitude from $E$ of $\triangle DEF$, forming two $30-60-90$ triangles. Because the length of $AD$ is $x$, from the properties of a $30-60-90$ triangle the length of $AE$ is $\frac{x\sqrt{3}}{3}$ and the length of $DE$ is thus $\frac{2x\sqrt{3}}{3}$. Thus the altitude of $\triangle DEF$ is $\frac{x\sqrt{3}}{3}$, and its base is $2x$, so its area is $\frac{1}{2}(2x)\left(\frac{x\sqrt{3}}{3}\right)=\frac{x^2\sqrt{3}}{3}$. To finish, $\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\fbox{\frac{\sqrt{3}}{6}}$.

Answer: \frac{\sqrt{3}}{6}