2014 AMC 10B Problem 15

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

A trisected right angle produces $30^\circ$ and $60^\circ$ angles, so every triangle in sight is a $30$-$60$-$90$ triangle whose sides you can read off. Because $E$ and $F$ both lie on line $AB$, triangle $DEF$ shares the rectangle's height, and only the base $EF$ needs computing. Normalizing $CB = 1$ costs nothing since the answer is a ratio.

• Assuming the trisectors cut AB into three equal pieces, which gives 1/3 (choice D); equal angles do not give equal segments.
• Computing $AE$ and $AF$ with the tangents swapped ($AE = \sqrt3$), which puts $F$ off the rectangle and produces a wrong base.