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

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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
==Solution (Trigonometry)==
+
==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>\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>
+
Note that <math>\tan 15^\circ=2-\sqrt{3}=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (It is important to memorize the sin, cos, and tan values of <math>15^\circ</math> and <math>75^\circ</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)==
+
==Solution 2 (No Trigonometry)==
  
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>.
+
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{BC}{BF} = \frac{CE}{EF}</math>. Since <math>\triangle BFC</math> is a <math>30-60-90</math> right triangle, <math>CF = \frac{10\sqrt{3}}{3}</math> and <math>BF = \frac{20\sqrt{3}}{3}</math>. Additionally, <cmath>CE + EF = CF = \frac{10\sqrt{3}}{3}</cmath>Now, substituting in the obtained values, we get <math>\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{CE}{EF} \Rightarrow \frac{2\sqrt{3}}{3}CE = EF</math> and <math>CE + EF = \frac{10\sqrt{3}}{3}</math>. Substituting the first equation into the second yields <math>\frac{2\sqrt{3}}{3}CE + CE = \frac{10\sqrt{3}}{3} \Rightarrow CE = 20 - 10\sqrt{3}</math>, so <math>DE = 10\sqrt{3}</math>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>AE = \boxed{\textbf{(E)}~20}</math>.
 +
 
 +
~edited by ripkobe_745
 +
 
 +
==Solution 3 Quick Construction (No Trigonometry)==
 +
 
 +
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>
 +
 
 +
== Solution 4 (No Trigonometry) ==
 +
 
 +
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>
 +
 
 +
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>.
 +
 
 +
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>
 +
 
 +
~ 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==
 +
First, divide all side lengths by <math>10</math> to make things easier. We’ll multiply our answer by <math>10</math> at the end.  
 +
Call side length <math>BE</math> <math>x</math>. Using the Pythagorean Theorem, we can get side <math>EC</math> is <math>\sqrt{x^2-1}</math>.
 +
 
 +
The double angle identity for sine states that: <cmath>\sin{2a} = 2 \sin{a}\cdot \cos{a}</cmath> So, <cmath>\sin 30 = 2\sin 15\cdot \cos 15</cmath> We know <math>\sin 30 = \frac{1}{2}</math>. In triangle <math>BEC</math>, <math>\sin 15 = \frac{\sqrt{x^2-1}}{x}</math> and <math>\cos 15 = \frac{1}{x}</math>. Substituting these in, we get our equation:  <cmath>\frac{1}{2} = 2 \cdot \frac{\sqrt{x^2-1}}{x} \cdot \frac{1}{x}</cmath>  which simplifies to  <cmath>x^4-16x^2+16 = 0</cmath>
 +
 
 +
Now, using the quadratic formula to solve for <math>x^2</math>. <cmath>x^2 = 16 \pm \frac{\sqrt{16^2-4\cdot16}}{2} = 8 \pm 4\sqrt3</cmath>
 +
Because the length <math>BE</math> must be close to one, the value of <math>x^2</math> will be <math>8-4\sqrt3</math>.
 +
We can now find <math>EC</math> = <math>\sqrt{x^2-1} = \sqrt{7-4\sqrt3} = 2-\sqrt3</math> and use it to find <math>DE</math>. <math>DE = 2-EC = \sqrt3</math>.
 +
To find <math>AE</math>, we can use the Pythagorean Theorem with sides <math>AD</math> and <math>DE</math>, OR we can notice that, based on the two side lengths we know, <math>ADE</math> is a <math>30-60-90</math> triangle. So <math>AE = 2\cdot AD = 2</math>.
 +
 
 +
Finally, we must multiply our answer by <math>10</math>, <math>2\cdot 10 = 20</math>. <math>\boxed{\textbf{(E)}}</math>.
 +
 
 +
~AWCHEN01
 +
 
 +
==Solution 6 (Pure Euclidian Geometry)==
 +
 
 +
[[File:Square.PNG|500px]]
 +
 
 +
We are going to use pure Euclidian geometry to prove <math>AE=AB</math>.
 +
 
 +
Reflect rectangle <math>ABCD</math> along line <math>CD</math>. Let the square be <math>ABFG</math> as shown. Construct equilateral triangle <math>\triangle EFH</math>.
 +
 
 +
Because <math>HF=EF</math>, <math>GF=BF=20</math>, and <math>\angle GFH=\angle BFE=15^{\circ}</math>, <math>\triangle GFH\cong \triangle BFE</math> by <math>SAS</math>.
 +
 
 +
So, <math>GH=BE</math>, <math>GH=HE=HF</math>.
 +
 
 +
 
 +
Because <math>GH=HE=HF</math>, <math>\angle GHF= \angle BEF=75^{\circ} + 75^{\circ} = 150^{\circ}</math>, <math>\angle GHE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}</math>, <math>\angle GHE=\angle GHF</math>.
 +
 
 +
<math>\triangle GHE \cong \triangle GHF</math> by <math>SAS</math>.
 +
 
 +
So, <math>GF=GE</math>. By the reflection, <math>AE=GE=GF=AB</math>. <math>AE=AB=\boxed{\textbf{(E)}~20}</math>
 +
 
 +
This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 7 (Pure Euclidian Geometry)==
 +
 
 +
[[File:Rectangle.PNG|700px]]
 +
 
 +
We are going to use pure Euclidian geometry to prove <math>AE=AB</math>.
 +
 
 +
Construct equilateral triangle <math>\triangle BEF</math>, and let <math>GF</math> be the height of <math>\triangle ABF</math>.
 +
 
 +
<math>\angle GBF=90^{\circ}-15^{\circ}-60^{\circ}=15^{\circ}</math>, <math>\angle GBF=\angle CBE</math>, <math>\angle BGF=\angle BCE=90^{\circ}</math>, <math>BF=BE</math>.
 +
 
 +
<math>\triangle BGF \cong \triangle BCE</math> by <math>AAS</math>.
 +
 
 +
 
 +
<math>BG=BC=10, AG=20-10=10</math>, <math>AG=BG</math>, <math>GF=GF</math>, by <math>HL</math> <math>\triangle AGF \cong \triangle BGF</math>.
 +
 
 +
So, <math>AF=BF=EF</math>.
 +
 
 +
 
 +
<math>\angle AFB=75^{\circ}+75^{\circ}=150^{\circ}</math>, <math>\angle AFE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}</math>, <math>\angle AFB=\angle AFE</math>, <math>AF=AF</math>, <math>BF=EF</math>.
 +
 
 +
<math>\triangle AFB \cong \triangle AFE</math> by <math>SAS</math>.
 +
 
 +
So, <math>AE=AB=\boxed{\textbf{(E)}~20}</math>
 +
 
 +
Note: Similar to previous Solution
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 8 (Trigonometry)==
 +
All trigonometric functions in this solution are in degrees. We know <cmath>\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)</cmath> so <cmath>\sin\left(15\right)=\sin\left(45-30\right)=\sin\left(45\right)\cos\left(-30\right)+\sin\left(-30\right)\cos\left(45\right)</cmath>
 +
<cmath>=\frac{\sqrt{2}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{-\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}</cmath>
 +
<cmath>=\frac{\sqrt{2}-\sqrt{6}}{4}</cmath>
 +
Let <math>EC=x</math>, then <math>BE=\sqrt{x^{2}+100}</math>. By the definition of sine,
 +
<cmath>\frac{x}{\sqrt{x^{2}+100}}=\frac{\sqrt{2}-\sqrt{6}}{4}</cmath>
 +
Squaring both sides,
 +
<cmath>\frac{x^{2}}{x^{2}+100}=\frac{\left(\sqrt{2}-\sqrt{6}\right)^{2}}{16}=\frac{2-2\sqrt{12}+6}{16}=\frac{8-4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}</cmath>
 +
Cross-multiplying,
 +
<cmath>4x^{2}=\left(2-\sqrt{3}\right)\left(x^{2}+100\right)=2x^{2}+200-\sqrt{3}x^{2}-100\sqrt{3}</cmath>
 +
Simplifying,
 +
<cmath>\left(2+\sqrt{3}\right)x^{2}=200-100\sqrt{3}</cmath>
 +
<cmath>x^{2}=\frac{200-100\sqrt{3}}{2+\sqrt{3}}=\frac{100\left(2-\sqrt{3}\right)}{2+\sqrt{3}}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}</cmath>
 +
Let <math>\frac{2-\sqrt{3}}{2+\sqrt{3}}=p</math>. Notice that <math>\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^{2}-\sqrt{3}^{2}=1</math> so <math>2-\sqrt{3}=\frac{1}{2+\sqrt{3}}</math>. <math>p</math> is then <cmath>\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\frac{1}{2+\sqrt{3}}}{2+\sqrt{3}}=\frac{1}{\left(2+\sqrt{3}\right)^{2}}</cmath>
 +
Recall that
 +
<cmath>x^{2}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}</cmath> which we now know is <cmath>100\cdot\frac{1}{\left(2+\sqrt{3}\right)^{2}}=\frac{100}{\left(2+\sqrt{3}\right)^{2}}=\left(\frac{10}{2+\sqrt{3}}\right)^{2}</cmath>
 +
Therefore <cmath>x=\frac{10}{2+\sqrt{3}}</cmath>
 +
Rationalizing the denominator,
 +
<cmath>\frac{10}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{20-10\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}</cmath>
 +
Which by difference of squares reduces to
 +
<cmath>20-10\sqrt{3}</cmath>
 +
so <math>EC=20-10\sqrt{3}</math>. <math>ED</math> is then <math>20-\left(20-10\sqrt{3}\right)=10\sqrt{3}</math> and since we know <math>AD=10</math>, by the Pythagorean theorem, <math>AE = 20</math>. The answer is <math>\boxed{\textbf{(E)}~20}</math>
 +
 
 +
An alternate way to finish: since we know the lengths of <math>AD</math> and <math>DE</math>, we can figure out that <math>m\angle AED=30^{\circ}</math> and therefore <math>m\angle BEA=75^{\circ}</math>. Hence <math>\triangle ABE</math> is isosceles and <math>AE=AB=\boxed{\textbf{(E)}~20}</math>.
 +
 
 +
~JH. L
 +
 
 +
== Video Solution by Richard Rusczyk ==
 +
 
 +
https://www.youtube.com/watch?v=-GBvCLSfTuo
  
 
==See Also==
 
==See Also==

Latest revision as of 05:43, 20 June 2022

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 $\tan 15^\circ=2-\sqrt{3}=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$. (It is important to memorize the sin, cos, and tan values of $15^\circ$ and $75^\circ$.) 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, $\frac{BC}{BF} = \frac{CE}{EF}$. Since $\triangle BFC$ is a $30-60-90$ right triangle, $CF = \frac{10\sqrt{3}}{3}$ and $BF = \frac{20\sqrt{3}}{3}$. Additionally, \[CE + EF = CF = \frac{10\sqrt{3}}{3}\]Now, substituting in the obtained values, we get $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{CE}{EF} \Rightarrow \frac{2\sqrt{3}}{3}CE = EF$ and $CE + EF = \frac{10\sqrt{3}}{3}$. Substituting the first equation into the second yields $\frac{2\sqrt{3}}{3}CE + CE = \frac{10\sqrt{3}}{3} \Rightarrow CE = 20 - 10\sqrt{3}$, so $DE = 10\sqrt{3}$. Because $\triangle ADE$ is a $30-60-90$ triangle, $AE = \boxed{\textbf{(E)}~20}$.

~edited by ripkobe_745

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

First, divide all side lengths by $10$ to make things easier. We’ll multiply our answer by $10$ at the end. Call side length $BE$ $x$. Using the Pythagorean Theorem, we can get side $EC$ is $\sqrt{x^2-1}$.

The double angle identity for sine states that: \[\sin{2a} = 2 \sin{a}\cdot \cos{a}\] So, \[\sin 30 = 2\sin 15\cdot \cos 15\] We know $\sin 30 = \frac{1}{2}$. In triangle $BEC$, $\sin 15 = \frac{\sqrt{x^2-1}}{x}$ and $\cos 15 = \frac{1}{x}$. Substituting these in, we get our equation: \[\frac{1}{2} = 2 \cdot \frac{\sqrt{x^2-1}}{x} \cdot \frac{1}{x}\] which simplifies to \[x^4-16x^2+16 = 0\]

Now, using the quadratic formula to solve for $x^2$. \[x^2 = 16 \pm \frac{\sqrt{16^2-4\cdot16}}{2} = 8 \pm 4\sqrt3\] Because the length $BE$ must be close to one, the value of $x^2$ will be $8-4\sqrt3$. We can now find $EC$ = $\sqrt{x^2-1} = \sqrt{7-4\sqrt3} = 2-\sqrt3$ and use it to find $DE$. $DE = 2-EC = \sqrt3$. To find $AE$, we can use the Pythagorean Theorem with sides $AD$ and $DE$, OR we can notice that, based on the two side lengths we know, $ADE$ is a $30-60-90$ triangle. So $AE = 2\cdot AD = 2$.

Finally, we must multiply our answer by $10$, $2\cdot 10 = 20$. $\boxed{\textbf{(E)}}$.

~AWCHEN01

Solution 6 (Pure Euclidian Geometry)

Square.PNG

We are going to use pure Euclidian geometry to prove $AE=AB$.

Reflect rectangle $ABCD$ along line $CD$. Let the square be $ABFG$ as shown. Construct equilateral triangle $\triangle EFH$.

Because $HF=EF$, $GF=BF=20$, and $\angle GFH=\angle BFE=15^{\circ}$, $\triangle GFH\cong \triangle BFE$ by $SAS$.

So, $GH=BE$, $GH=HE=HF$.


Because $GH=HE=HF$, $\angle GHF= \angle BEF=75^{\circ} + 75^{\circ} = 150^{\circ}$, $\angle GHE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$, $\angle GHE=\angle GHF$.

$\triangle GHE \cong \triangle GHF$ by $SAS$.

So, $GF=GE$. By the reflection, $AE=GE=GF=AB$. $AE=AB=\boxed{\textbf{(E)}~20}$

This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42.

~isabelchen

Solution 7 (Pure Euclidian Geometry)

Rectangle.PNG

We are going to use pure Euclidian geometry to prove $AE=AB$.

Construct equilateral triangle $\triangle BEF$, and let $GF$ be the height of $\triangle ABF$.

$\angle GBF=90^{\circ}-15^{\circ}-60^{\circ}=15^{\circ}$, $\angle GBF=\angle CBE$, $\angle BGF=\angle BCE=90^{\circ}$, $BF=BE$.

$\triangle BGF \cong \triangle BCE$ by $AAS$.


$BG=BC=10, AG=20-10=10$, $AG=BG$, $GF=GF$, by $HL$ $\triangle AGF \cong \triangle BGF$.

So, $AF=BF=EF$.


$\angle AFB=75^{\circ}+75^{\circ}=150^{\circ}$, $\angle AFE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$, $\angle AFB=\angle AFE$, $AF=AF$, $BF=EF$.

$\triangle AFB \cong \triangle AFE$ by $SAS$.

So, $AE=AB=\boxed{\textbf{(E)}~20}$

Note: Similar to previous Solution

~isabelchen

Solution 8 (Trigonometry)

All trigonometric functions in this solution are in degrees. We know \[\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)\] so \[\sin\left(15\right)=\sin\left(45-30\right)=\sin\left(45\right)\cos\left(-30\right)+\sin\left(-30\right)\cos\left(45\right)\] \[=\frac{\sqrt{2}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{-\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}\] \[=\frac{\sqrt{2}-\sqrt{6}}{4}\] Let $EC=x$, then $BE=\sqrt{x^{2}+100}$. By the definition of sine, \[\frac{x}{\sqrt{x^{2}+100}}=\frac{\sqrt{2}-\sqrt{6}}{4}\] Squaring both sides, \[\frac{x^{2}}{x^{2}+100}=\frac{\left(\sqrt{2}-\sqrt{6}\right)^{2}}{16}=\frac{2-2\sqrt{12}+6}{16}=\frac{8-4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}\] Cross-multiplying, \[4x^{2}=\left(2-\sqrt{3}\right)\left(x^{2}+100\right)=2x^{2}+200-\sqrt{3}x^{2}-100\sqrt{3}\] Simplifying, \[\left(2+\sqrt{3}\right)x^{2}=200-100\sqrt{3}\] \[x^{2}=\frac{200-100\sqrt{3}}{2+\sqrt{3}}=\frac{100\left(2-\sqrt{3}\right)}{2+\sqrt{3}}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}\] Let $\frac{2-\sqrt{3}}{2+\sqrt{3}}=p$. Notice that $\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^{2}-\sqrt{3}^{2}=1$ so $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$. $p$ is then \[\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\frac{1}{2+\sqrt{3}}}{2+\sqrt{3}}=\frac{1}{\left(2+\sqrt{3}\right)^{2}}\] Recall that \[x^{2}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}\] which we now know is \[100\cdot\frac{1}{\left(2+\sqrt{3}\right)^{2}}=\frac{100}{\left(2+\sqrt{3}\right)^{2}}=\left(\frac{10}{2+\sqrt{3}}\right)^{2}\] Therefore \[x=\frac{10}{2+\sqrt{3}}\] Rationalizing the denominator, \[\frac{10}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{20-10\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\] Which by difference of squares reduces to \[20-10\sqrt{3}\] so $EC=20-10\sqrt{3}$. $ED$ is then $20-\left(20-10\sqrt{3}\right)=10\sqrt{3}$ and since we know $AD=10$, by the Pythagorean theorem, $AE = 20$. The answer is $\boxed{\textbf{(E)}~20}$

An alternate way to finish: since we know the lengths of $AD$ and $DE$, we can figure out that $m\angle AED=30^{\circ}$ and therefore $m\angle BEA=75^{\circ}$. Hence $\triangle ABE$ is isosceles and $AE=AB=\boxed{\textbf{(E)}~20}$.

~JH. L

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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