2014 AMC 10B Problem 15

The answer is a ratio, so we may pick any convenient size. Let $AD = CB = 1$; then $DC = AB = 2$ and the rectangle's area is $2$.

1. The angles. Angle $ADC$ is $90^\circ$. Cutting it into three equal parts gives $30^\circ$ each. So $\angle ADE = 30^\circ$ and $\angle ADF = 60^\circ$ (measured from side $DA$).
2. Find $AE$. Triangle $DAE$ has a right angle at $A$ and a $30^\circ$ angle at $D$, so it is a $30$-$60$-$90$ triangle. The side next to the $30^\circ$ angle is $AD = 1$, and the side across from it is $AE$. In a $30$-$60$-$90$ triangle the side across from $30^\circ$ is the side next to it divided by $\sqrt3$:
$$AE = \frac{1}{\sqrt3}.$$
3. Find $AF$. Triangle $DAF$ has a $60^\circ$ angle at $D$. Now the side across from the $60^\circ$ angle is $AF$, which is $\sqrt3$ times the side $AD$ next to it:
$$AF = \sqrt3.$$
Both $\frac{1}{\sqrt3} \approx 0.58$ and $\sqrt3 \approx 1.73$ are less than $2$, so $E$ and $F$ really are on $AB$.
4. Base $EF$. $EF = AF - AE = \sqrt3 - \frac{1}{\sqrt3} = \frac{3}{\sqrt3} - \frac{1}{\sqrt3} = \frac{2}{\sqrt3}$.
5. Area of $DEF$. Its base $EF$ lies on $AB$, and the height from $D$ down to line $AB$ is $AD = 1$. So the area is $\frac12 \times \frac{2}{\sqrt3} \times 1 = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}$.
6. Ratio. $\dfrac{\sqrt3/3}{2} = \dfrac{\sqrt3}{6}$.

The answer is $\boxed{\textbf{(A)}\ \frac{\sqrt{3}}{6}}$.

Cutting a right angle into three equal parts creates $30^\circ$ and $60^\circ$ angles, so every triangle in the picture is a $30$-$60$-$90$ triangle with known side ratios. Because $E$ and $F$ are on the side $AB$, triangle $DEF$ has the same height as the rectangle, and only its base needs work. For a ratio question, choosing the short side to be $1$ keeps the numbers small.

Scale so that $CB = AD = 1$ and $DC = AB = 2$; the rectangle has area $2$.

The right angle $ADC$ is trisected, so $\angle ADE = 30^\circ$ and $\angle ADF = 60^\circ$. In right triangle $ADE$ with leg $AD = 1$,
$$AE = AD \tan 30^\circ = \frac{1}{\sqrt3}, \qquad AF = AD \tan 60^\circ = \sqrt3.$$
Both points land on $AB$ since $\sqrt3 < 2$.

Thus $EF = \sqrt3 - \frac{1}{\sqrt3} = \frac{3 - 1}{\sqrt3} = \frac{2}{\sqrt3}$. Triangle $DEF$ has base $EF$ on line $AB$ and height $AD = 1$, so its area is $\frac12 \cdot \frac{2}{\sqrt3} \cdot 1 = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}$.

The ratio is $\dfrac{\sqrt3/3}{2} = \dfrac{\sqrt3}{6}$.

The answer is $\boxed{\textbf{(A)}\ \frac{\sqrt{3}}{6}}$.