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Difference between revisions of "2010 AMC 12A Problems/Problem 8"

(Created page with '== Problem 8 == Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, r…')
 
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== Problem 8 ==
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== Problem ==
 
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?
 
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>?
  
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Let <math>\angle BAE = \angle ACD = x</math>.
 
Let <math>\angle BAE = \angle ACD = x</math>.
  
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<math>\angle BCD = \angle AEC = 60^\circ</math>
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<cmath>\begin{align*}&ngle BCD = \angle AEC = 60^\circ\\
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&\angle EAC + \angle FCA + \angle ECF + \angle AEC = \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\
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&\angle EAC = 60^\circ - x\\
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&\angle BAC = \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath>
  
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<math>\angle EAC + \angle FCA + \angle ECF + \angle AEC = \angle EAC + x + 60^\circ + 60^\circ = 180^\circ</math>
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Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math>
  
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<math>\angle EAC = 60^\circ - x</math>
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== See also ==
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{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}
  
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<math>\angle BAC = \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ</math>
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[[Category:Introductory Geometry Problems]]
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Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math>
 

Revision as of 22:00, 25 February 2010

Problem

Triangle $ABC$ has $AB=2 \cdot AC$. Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$, respectively, such that $\angle BAE = \angle ACD$. Let $F$ be the intersection of segments $AE$ and $CD$, and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$?

$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ$

Solution

Let $\angle BAE = \angle ACD = x$.

\begin{align*}&ngle BCD = \angle AEC = 60^\circ\\ &\angle EAC + \angle FCA + \angle ECF + \angle AEC = \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ &\angle EAC = 60^\circ - x\\ &\angle BAC = \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}

Since $\frac{AC}{AB} = \frac{1}{2}$, $\angle BCA = \boxed{90^\circ\ \textbf{(C)}}$

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AMC 12 Problems and Solutions
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