2014 AMC 10A Problem 22

$15^\circ$ is not a friendly angle, but it is half of $30^\circ$, and $30^\circ$ is very friendly. So draw a $30^\circ$ line and cut it in half.

Picture the rectangle with $A$ top-left, $B$ top-right, $C$ bottom-right and $D$ bottom-left, so $BC = 10$ is the right side and $CD = 20$ is the bottom.

1. A friendly helper point. Let $F$ be the point on $CD$ with $\angle CBF = 30^\circ$. Triangle $BCF$ is a $30$-$60$-$90$ triangle with the long leg $BC = 10$. So the short leg is $CF = \frac{10}{\sqrt3}$ and the hypotenuse is $BF = \frac{20}{\sqrt3}$ (twice the short leg).
2. $BE$ cuts the $30^\circ$ angle in half. Since $\angle CBE = 15^\circ$, point $E$ lies between $C$ and $F$, and $BE$ splits angle $CBF$ into two equal parts.
3. The angle bisector rule. When a line cuts an angle of a triangle in half, it divides the opposite side in the same ratio as the two sides next to that angle. Here
$$\frac{CE}{EF} = \frac{BC}{BF} = \frac{10}{20/\sqrt3} = \frac{\sqrt3}{2}.$$
4. Find $CE$. $CE$ and $EF$ are in the ratio $\sqrt3 : 2$, and together they make $CF$. So $CE$ is the fraction $\frac{\sqrt3}{\sqrt3 + 2}$ of $CF$:
$$CE = \frac{\sqrt3}{2 + \sqrt3} \cdot \frac{10}{\sqrt3} = \frac{10}{2 + \sqrt3}.$$
Multiply top and bottom by $2 - \sqrt3$; the bottom becomes $4 - 3 = 1$. So $CE = 10(2 - \sqrt3) = 20 - 10\sqrt3$.
5. Find $DE$. $DE = CD - CE = 20 - (20 - 10\sqrt3) = 10\sqrt3$.
6. Find $AE$. Triangle $ADE$ has a right angle at $D$, with $AD = 10$ and $DE = 10\sqrt3$:
$$AE = \sqrt{10^2 + (10\sqrt3)^2} = \sqrt{100 + 300} = \sqrt{400} = 20.$$

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

A $15^\circ$ angle is half of a $30^\circ$ angle, and the angle bisector rule turns "half the angle" into a ratio of lengths. After that, everything is the $30$-$60$-$90$ side ratios and the Pythagorean theorem. Notice also that $DE : AD = \sqrt3 : 1$, so triangle $ADE$ is itself a $30$-$60$-$90$ triangle with hypotenuse twice $AD$, which is a quick check on $20$.

Build $15^\circ$ directly from a $30$-$60$-$90$ triangle. Take such a triangle with legs $1$ and $\sqrt3$ and hypotenuse $2$. Extend the long leg past the $30^\circ$ corner by a length $2$ (equal to the hypotenuse). The new triangle formed there has two equal sides, so its two base angles are equal, and together they make the outside angle of $30^\circ$; each is $15^\circ$. The big right triangle now has legs $1$ and $2 + \sqrt3$, so the ratio "opposite over adjacent" for $15^\circ$ is $\frac{1}{2 + \sqrt3} = 2 - \sqrt3$. In triangle $BCE$ this ratio is $\frac{CE}{BC}$, so $CE = 10(2 - \sqrt3)$, and the rest is as above.