# Difference between revisions of "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.

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

Note: Similar to previous Solution

## 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