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

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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 (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 (Without 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, $\overline{AE} = \boxed{\textbf{(E)}~20}$ .

Solution 3 (Trigonometry)

By Law of Sines $\frac{BC}{\sin 75^\circ}=\frac{EC}{\sin15^\circ}\rightarrow\frac{10}{\frac{\sqrt{2}+\sqrt{6}}{4}}=\frac{EC}{\frac{\sqrt{6}-\sqrt{2}}{4}}\rightarrow\frac{10}{EC}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\rightarrow EC=\frac{10}{2+\sqrt{3}}=20-10\sqrt{3}.$ Thus, $DE=20-(20-10\sqrt{3})=10\sqrt{3}.$

We see that $\triangle{ADE}$ is a $30-60-90$ triangle, leaving $\overline{AE}=\boxed{\textbf{(E)}~20}.$

Solution 4 (Measuring)

If we draw rectangle $ABCD$ , and whip out a protractor, we can draw a perfect $\overline{BE}$ , almost perfectly $15^\circ$ degrees 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.

@@ Line 17: / Line 17: @@
 We see that <math>\triangle{ADE}</math> is a <math>30-60-90</math> triangle, leaving <math>\overline{AE}=\boxed{\textbf{(E)}~20}.</math>
+==Solution 4 (Measuring) ==
+If we draw rectangle <math>ABCD</math>, and whip out a protractor, we can draw a perfect <math>\overline{BE}</math>, almost perfectly <math>15^\circ</math> degrees off of <math>\overline{BC}</math>. Then we can draw <math>\overline{AE}</math>, and use a ruler to measure it.
+We can clearly see that the <math>\overline{AE}</math> is <math>\boxed{\textbf{(E)}~20}</math>.
+NOTE: this method is a last resort, and is pretty risky. Answer choice <math>\textbf{(D)}~11\sqrt{3}</math> is also very close to <math>\textbf{(E)}~20</math>, meaning that we wouldn't be 100% sure of our answer.
 ==See Also==
 {{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}}
 {{MAA Notice}}

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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