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

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==Problem==
 
==Problem==
  
In rectangle <math>ABCD</math>, <math>AB=20</math> and <math>BC=10</math>. Let <math>E</math> be a point on <math>\overline{CD}</math> such that <math>\angle CBE=15^\circ</math>. What is <math>AE</math>?
+
In rectangle <math>ABCD</math>, <math>\overline{AB}=20</math> and <math>\overline{BC}=10</math>. Let <math>E</math> be a point on <math>\overline{CD}</math> such that <math>\angle CBE=15^\circ</math>. What is <math>\overline{AE}</math>?
  
 
<math> \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 </math>
 
<math> \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 </math>
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
==Solution==
+
==Solution 1 (Trigonometry)==
Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\frac{1-\cos a}{\sin a}</math>).  Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math>
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Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}</math>).  Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math>
  
==Solution 2 (non-trig)==
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==Solution 2 (No Trigonometry)==
  
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, <cmath>\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}</cmath>. 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 in <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>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>\overline{AE} = \boxed{\textbf{(E)}~20}</math>.
+
Let <math>F</math> be a point on line <math>\overline{CD}</math> such that points <math>C</math> and <math>F</math> are distinct and 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, <cmath>\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}</cmath>Now, substituting 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>. Substituting the first equation into the second 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>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>\overline{AE} = \boxed{\textbf{(E)}~20}</math>.
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 +
~edited by dolphin7
 +
 
 +
==Solution 3 Quick Construction (No Trigonometry)==
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 +
Reflect <math>\triangle{ECB}</math> over line segment <math>\overline{CD}</math>. Let the point <math>F</math> be the point where the right angle is of our newly reflected triangle. By subtracting <math>90 - (15+15) = 60</math> to find <math>\angle ABF</math>, we see that <math>\triangle{ABC}</math> is a <math>30-60-90</math> right triangle. By using complementary angles once more, we can see that <math>\angle{EAD}</math> is a <math>60^\circ</math> angle, and we've found that <math>\triangle{EAD}</math> is a <math>30-60-90</math> right triangle. From here, we can use the <math>1-2-\sqrt{3}</math> properties of a <math>30-60-90</math> right triangle to see that <math>\overline{AE}=\boxed{\textbf{(E)}~20}.</math>
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 +
== Solution 4 (No Trigonometry) ==
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 +
Let <math>F</math> be a point on <math>BC</math> such that <math>\angle{FEC}=60^{\circ}</math>. Then <cmath>\angle{BEF}=\angle{BEC}-\angle{FEC}=15^{\circ}</cmath> Since <math>\angle{BEF}=\angle{EBF}</math>, <math>\bigtriangleup{BFE}</math> is isosceles.
 +
 
 +
Let <math>CF=x</math>. Since <math>\bigtriangleup{FEC}</math> is <math>60^{\circ}-90^{\circ}-30^{\circ}</math>, we have <math>EF=\frac{2}{\sqrt{3}}x</math>
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 +
Since <math>\bigtriangleup{BFE}</math> is isosceles, we have <math>BF=EF=\frac{2}{\sqrt{3}}x</math>. Since <math>BF+FC=BF</math>, we have <cmath>\frac{2}{\sqrt{3}}x+x=10 \Longrightarrow x=20\sqrt{3}-30</cmath> Thus <math>EC=\frac{1}{\sqrt{3}}BC=20-10\sqrt{3}</math> and <math>DE=DC-EC=20-EC=10\sqrt{3}</math>.
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Finally, by the Pythagorean Theorem, we have <cmath>AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}</cmath>
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 +
~ Solution by Nafer
 +
 
 +
~ Edited by TheBeast5520
 +
 
 +
Note from williamgolly:
 +
When you find DE, note how ADE is congruent to a 30-60-90 triangle and you can easily find AE from there
 +
 
 +
==Solution 5(15-75-90 Triangle)==
 +
We notice a 15-75-90 triangle on <math>\triangle{ADE}</math>.
 +
Implying the formula, we will continue as follows to get <math>\boxed{\mathrm{(E)}}</math>. ~mathboy282
 +
===Note===
 +
Some people may not know this relationship; and this method is only for those who know it. If you do not know this relationship I highly suggest NOT using this solution and/or searching it up.
 +
Here it one https://www.quora.com/What-are-the-side-relationships-of-a-15-75-90-triangle
 +
 
 +
== Video Solution by Richard Rusczyk ==
 +
 
 +
https://www.youtube.com/watch?v=-GBvCLSfTuo
  
 
==See Also==
 
==See Also==

Revision as of 01:05, 7 December 2020

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 1 (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 (No 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, $\overline{AE} = \boxed{\textbf{(E)}~20}$.

~edited by dolphin7

Solution 3 Quick Construction (No Trigonometry)

Reflect $\triangle{ECB}$ over line segment $\overline{CD}$. Let the point $F$ be the point where the right angle is of our newly reflected triangle. By subtracting $90 - (15+15) = 60$ to find $\angle ABF$, we see that $\triangle{ABC}$ is a $30-60-90$ right triangle. By using complementary angles once more, we can see that $\angle{EAD}$ is a $60^\circ$ angle, and we've found that $\triangle{EAD}$ is a $30-60-90$ right triangle. From here, we can use the $1-2-\sqrt{3}$ properties of a $30-60-90$ right triangle to see that $\overline{AE}=\boxed{\textbf{(E)}~20}.$

Solution 4 (No Trigonometry)

Let $F$ be a point on $BC$ such that $\angle{FEC}=60^{\circ}$. Then \[\angle{BEF}=\angle{BEC}-\angle{FEC}=15^{\circ}\] Since $\angle{BEF}=\angle{EBF}$, $\bigtriangleup{BFE}$ is isosceles.

Let $CF=x$. Since $\bigtriangleup{FEC}$ is $60^{\circ}-90^{\circ}-30^{\circ}$, we have $EF=\frac{2}{\sqrt{3}}x$

Since $\bigtriangleup{BFE}$ is isosceles, we have $BF=EF=\frac{2}{\sqrt{3}}x$. Since $BF+FC=BF$, we have \[\frac{2}{\sqrt{3}}x+x=10 	\Longrightarrow x=20\sqrt{3}-30\] Thus $EC=\frac{1}{\sqrt{3}}BC=20-10\sqrt{3}$ and $DE=DC-EC=20-EC=10\sqrt{3}$.

Finally, by the Pythagorean Theorem, we have \[AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}\]

~ Solution by Nafer

~ Edited by TheBeast5520

Note from williamgolly: When you find DE, note how ADE is congruent to a 30-60-90 triangle and you can easily find AE from there

Solution 5(15-75-90 Triangle)

We notice a 15-75-90 triangle on $\triangle{ADE}$. Implying the formula, we will continue as follows to get $\boxed{\mathrm{(E)}}$. ~mathboy282

Note

Some people may not know this relationship; and this method is only for those who know it. If you do not know this relationship I highly suggest NOT using this solution and/or searching it up. Here it one https://www.quora.com/What-are-the-side-relationships-of-a-15-75-90-triangle

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=-GBvCLSfTuo

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 10 Problems and Solutions

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