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Difference between revisions of "2014 AMC 10A Problems/Problem 22"

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

Revision as of 13:11, 26 December 2015

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

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 $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}}$ and $\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}$.

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 10 Problems and Solutions

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