2014 AMC 10A Problems/Problem 22

Problem

In rectangle $ABCD$ , $AB=20$ and $BC=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $AE$ ?

$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$

Solution

Note that $\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$ . (If you do not know the tangent half-angle formula, it is $\frac{1-\cos a}{\sin a}$ ). Therefore, we have $DE=10\sqrt 3$ . Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}$

Solution 2 (non-trig)

Mark point $F$ on line $\overline{CD}$ such that $\angle EBF = 15^\circ$ . By the angle bisector theorem, $\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}$ . Let $\overline{CE}$ have length $x$ . Since $\triangle BFC$ is a $30-60-90$ right triangle, $\overline{CF} = \frac{10\sqrt{3}}{3}$ and $\overline{BF} = \frac{20\sqrt{3}}{3}$ . Additionally, $\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}$ . Now, plugging in the obtained values, we get $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow 2\sqrt{3}\overline{CE} = 3\overline{EF}$ and $\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}$ .

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2014 AMC 10A Problems/Problem 22

Contents

Problem

Solution

Solution 2 (non-trig)

See Also