2014 AMC 10A Problems/Problem 22

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)

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)

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

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

Page

Toolbox

Search