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

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==Solution 2 (non-trig)==
 
==Solution 2 (non-trig)==
  
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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>. 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}} \Rightarrow 2\sqrt{3}\overline{CE} = 3\overline{EF}</math> and <math>\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}</math>. Solving this system for <math>\overline{CE}</math> yields <math>\overline{CE} = 20 - 10\sqrt{3}</math>, so <math>\overline{DE} = 10\sqrt{3}</math> and <math>\overline{AE} = \boxed{\textbf{(E)}~20}</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>. 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}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}</math> and <math>\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}</math>. Plugging <math>\frac{2\sqrt{3}}{3}\overline{CE}</math> for <math>\overline{EF}</math> in the second equation yields <math>\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}</math>, so <math>\overline{DE} = 10\sqrt{3}</math> and <math>\overline{AE} = \boxed{\textbf{(E)}~20}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 13:56, 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 (non-trig)

Mark point $F$ on line $\overline{CD}$ such that $\angle EBF = 15^\circ$. By the angle bisector theorem, $\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}$. 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 $\frac{2\sqrt{3}}{3}\overline{CE}$ 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 $\overline{DE} = 10\sqrt{3}$ and $\overline{AE} = \boxed{\textbf{(E)}~20}$.

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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