2014 AMC 10A Problems/Problem 22


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 $10$, $2\cdot 10 = 20$. $\boxed{\textbf{(E)}}$.


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


Video Solution by Richard Rusczyk


See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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