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2014 AMC 10A Problems/Problem 22

Revision as of 11:59, 28 December 2015 by Ghghghghghghghgh (talk | contribs) (→‎Solution 2 (non-trig))

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)

Let $F$ be a point on line $\overline{CD}$ such that points $C$ and $F$ are distinct and that $\angle EBF = 15^\circ$. By the angle bisector theorem, $\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}$. 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 $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}$ and $\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}$. Plugging in $\frac{2\sqrt{3}}{3}\overline{CE}$ for $\overline{EF}$ in the second equation yields $\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}$, so $\overline{DE} = 10\sqrt{3}$. Because $\triangle ADE$ is a $30-60-90$ triangle, $\overline{AE} = \boxed{\textbf{(E)}~20}$.

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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