2014 AMC 10A Problem 22

Everything hinges on the length $CE$ in right triangle $BCE$, where $BC = 10$ and $\angle CBE = 15^\circ$: $CE = 10\tan 15^\circ$.

To get $\tan 15^\circ$ without a table, start with a $30$-$60$-$90$ triangle $PQR$, right angle at $Q$, with $PQ = 1$, $QR = \sqrt3$, $PR = 2$, so $\angle PRQ = 30^\circ$. Extend $\overline{RQ}$ past $R$ to a point $S$ with $RS = RP = 2$. Triangle $PRS$ is isosceles with apex angle $180^\circ - 30^\circ = 150^\circ$, so $\angle PSR = 15^\circ$. In right triangle $PQS$,
$$\tan 15^\circ = \frac{PQ}{QS} = \frac{1}{2 + \sqrt3} = 2 - \sqrt3.$$

Hence $CE = 10(2 - \sqrt3) = 20 - 10\sqrt3$, and
$$DE = CD - CE = 20 - (20 - 10\sqrt3) = 10\sqrt3.$$

Finally, in right triangle $ADE$ with $AD = 10$,
$$AE = \sqrt{10^2 + (10\sqrt3)^2} = \sqrt{100 + 300} = 20.$$

(As a check, $DE : AD = \sqrt3$ makes $\angle DAE = 60^\circ$, so $\angle BAE = 30^\circ$ and triangle $ABE$ is isosceles with $AE = AB = 20$.)

The answer is $\boxed{\textbf{(E)}\ 20}$.

A $15^\circ$ angle is half of $30^\circ$, and halving an angle is exactly what the exterior-angle property of an isosceles triangle does: attaching a segment equal to the hypotenuse produces a triangle whose small angle is half the original. This gives exact values such as $\tan 15^\circ = 2 - \sqrt3$ from pure geometry. Once one leg of the rectangle is known, the rest is the Pythagorean theorem.

Angle bisector instead of trig: let $F$ be on $\overline{CD}$ with $\angle CBF = 30^\circ$, so $CF = \frac{10}{\sqrt3}$ and $BF = \frac{20}{\sqrt3}$. Then $\overline{BE}$ bisects $\angle CBF$, and the angle bisector theorem gives $\frac{CE}{EF} = \frac{BC}{BF} = \frac{\sqrt3}{2}$. So $CE = \frac{\sqrt3}{2 + \sqrt3}\cdot\frac{10}{\sqrt3} = \frac{10}{2 + \sqrt3} = 10(2 - \sqrt3)$, and the computation finishes as above.

• Not knowing tan 15 degrees and guessing it from the choices, or using CE = 10 tan 75 = 10(2 + root 3), which is longer than CD.
• Stopping at $DE = 10\sqrt3$ and selecting choice (B), which is a side of the right triangle rather than its hypotenuse $AE$.
• Computing $AE$ with the leg $CE$ instead of $DE$, giving $\sqrt{100 + (20 - 10\sqrt3)^2}$, which does not simplify to a choice.