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

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m (→‎Solution 4 (No Trigonometry))
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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>
 
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>
  
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== Solution 5 (No Trigonometry) ==
+
== 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>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.

Revision as of 22:10, 30 December 2019

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 $\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$. (If you do not know the tangent half-angle formula, it is $\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}$). 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, $\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}$. Since $\triangle BFC$ is a $30-60-90$ right triangle, $\overline{CF} = \frac{10\sqrt{3}}{3}$ and $\overline{BF} = \frac{20\sqrt{3}}{3}$. Additionally, \[\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}\]Now, substituting in the obtained values, we get $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}$ and $\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}$. Substituting the first equation into the second yields $\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}$, so $\overline{DE} = 10\sqrt{3}$. Because $\triangle ADE$ is a $30-60-90$ triangle, $\overline{AE} = \boxed{\textbf{(E)}~20}$. We see that $\triangle{ADE}$ is a $30-60-90$ triangle, leaving $\overline{AE}=\boxed{\textbf{(E)}~20}.$

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

See Also

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

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