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

(Very simple yet risky method)
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==Solution 4 (Measuring) ==
 
==Solution 4 (Measuring) ==
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If we draw rectangle <math>ABCD</math>, and whip out a protractor, we can draw a perfect <math>\overline{BE}</math>, almost perfectly <math>15^\circ</math> degrees off of <math>\overline{BC}</math>. Then we can draw <math>\overline{AE}</math>, and use a ruler to measure it.  
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If we draw rectangle <math>ABCD</math> and whip out a protractor, we can draw a perfect <math>\overline{BE}</math>, almost perfectly <math>15^\circ</math> degrees off of <math>\overline{BC}</math>. Then we can draw <math>\overline{AE}</math>, and use a ruler to measure it.  
 
We can clearly see that the <math>\overline{AE}</math> is <math>\boxed{\textbf{(E)}~20}</math>.
 
We can clearly see that the <math>\overline{AE}</math> is <math>\boxed{\textbf{(E)}~20}</math>.
  
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NOTE: this method is a last resort, and is pretty risky. Answer choice <math>\textbf{(D)}~11\sqrt{3}</math> is also very close to <math>\textbf{(E)}~20</math>, meaning that we wouldn't be 100% sure of our answer.
+
NOTE: this method is a last resort, and is pretty risky. Answer choice <math>\textbf{(D)}~11\sqrt{3}</math> is also very close to <math>\textbf{(E)}~20</math>, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of <math>\triangle ADE, we can clearly see that it is a </math>30-60-90<math> triangle, which verifies our answer of </math>20$.
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==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}}
 
{{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:11, 13 January 2018

Problem

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

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

Solution (Trigonometry)

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 $\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}$). 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 (Without Trigonometry)

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, substituting 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 $\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}$. Substituting the first equation into the second 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, $\overline{AE} = \boxed{\textbf{(E)}~20}$.

Solution 3 (Trigonometry)

By Law of Sines\[\frac{BC}{\sin 75^\circ}=\frac{EC}{\sin15^\circ}\rightarrow\frac{10}{\frac{\sqrt{2}+\sqrt{6}}{4}}=\frac{EC}{\frac{\sqrt{6}-\sqrt{2}}{4}}\rightarrow\frac{10}{EC}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\rightarrow EC=\frac{10}{2+\sqrt{3}}=20-10\sqrt{3}.\]Thus, $DE=20-(20-10\sqrt{3})=10\sqrt{3}.$

We see that $\triangle{ADE}$ is a $30-60-90$ triangle, leaving $\overline{AE}=\boxed{\textbf{(E)}~20}.$

Solution 4 (Measuring)

If we draw rectangle $ABCD$ and whip out a protractor, we can draw a perfect $\overline{BE}$, almost perfectly $15^\circ$ degrees off of $\overline{BC}$. Then we can draw $\overline{AE}$, and use a ruler to measure it. We can clearly see that the $\overline{AE}$ is $\boxed{\textbf{(E)}~20}$.

NOTE: this method is a last resort, and is pretty risky. Answer choice $\textbf{(D)}~11\sqrt{3}$ is also very close to $\textbf{(E)}~20$, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of $\triangle ADE, we can clearly see that it is a$30-60-90$triangle, which verifies our answer of$20$.

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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