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Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42.
 
This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42.
  
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~isabelchen
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==Solution 7 (Pure Euclidian Geometry)==
 
==Solution 7 (Pure Euclidian Geometry)==
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So, <math>AE=AB=\boxed{\textbf{(E)}~20}</math>
 
So, <math>AE=AB=\boxed{\textbf{(E)}~20}</math>
  
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~isabelchen
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 08:23, 28 December 2021

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

~isabelchen

Video Solution by Richard Rusczyk

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

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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