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

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Let <math>CF=x</math>. Since <math>\bigtriangleup{FEC}</math> is <math>60^{\circ}-90^{\circ}-30^{\circ}</math>, we have <math>EF=\frac{2}{\sqrt{3}}x</math>
 
Let <math>CF=x</math>. Since <math>\bigtriangleup{FEC}</math> is <math>60^{\circ}-90^{\circ}-30^{\circ}</math>, we have <math>EF=\frac{2}{\sqrt{3}}x</math>
  
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Since <math>\bigtriangleup{BFE}</math> is isosceles, we have <math>BF=EF=\frac{2}{\sqrt{3}}x</math>. Since <math>BF+FC=BF</math>, we have <cmath>\frac{2}{\sqrt{3}}x+x=10 \Longrightarrow x=20\sqrt{3}-30</cmath> Thus <math>EC=\frac{1}{\sqrt{3}}CE=20-10\sqrt{3}</math> and <math>DE=DC-EC=20-EC=10\sqrt{3}</math>.
+
Since <math>\bigtriangleup{BFE}</math> is isosceles, we have <math>BF=EF=\frac{2}{\sqrt{3}}x</math>. Since <math>BF+FC=BF</math>, we have <cmath>\frac{2}{\sqrt{3}}x+x=10 \Longrightarrow x=20\sqrt{3}-30</cmath> Thus <math>EC=\frac{1}{\sqrt{3}}BC=20-10\sqrt{3}</math> and <math>DE=DC-EC=20-EC=10\sqrt{3}</math>.
  
 
Finally, by the Pythagorean Theorem, we have <cmath>AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}</cmath>
 
Finally, by the Pythagorean Theorem, we have <cmath>AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}</cmath>
  
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~ Nafer
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~ Solution by Nafer
 +
 
 +
~ Edited by TheBeast5520
  
 
==See Also==
 
==See Also==

Revision as of 19:23, 17 November 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 (Measuring)

If we draw rectangle $ABCD$ and whip out a protractor, we can draw a perfect $\overline{BE}$, almost perfectly $15^\circ$ off of $\overline{BC}$. Then we can draw $\overline{AE}$, and use a ruler to measure it. We can clearly see that the $\overline{AE}$ is $\boxed{\textbf{(E)}~20}$.

NOTE: this method is a last resort, and is pretty risky. Answer choice $\textbf{(D)}~11\sqrt{3}$ is also very close to $\textbf{(E)}~20$, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of $\triangle ADE$, we can clearly see that it is a $30-60-90$ triangle, which verifies our answer of $\boxed{20}$.

Solution 5 (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

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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