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| − | == Problem ==
| + | #REDIRECT [[2010_AMC_12A_Problems/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>, 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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| − | <math>\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ</math>
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| − | == Solution ==
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| − | <center>[[File:AMC 2010 12A Problem 8.png]]</center>
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| − | <asy>
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| − | pair A,B,C,D,E,F,G,H;
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| − | G=(0,10);
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| − | A=(0,3.464);
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| − | B=(6,0);
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| − | C=(0,0);
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| − | draw(A--B--C--cycle);
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| − | F=(1,1.73);
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| − | E=(2,0);
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| − | draw(C--F--E);
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| − | D=(1.5,2.6);
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| − | draw(C--D);
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| − | label("$A$",A,W);
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| − | label("$B$",B,S);
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| − | label("$C$",C,S);
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| − | label("$F$",F,N);
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| − | label("$D$",D,NE);
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| − | label("$E$",E,S);
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| − | draw(A--E);
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| − | draw(anglemark(E,A,B));
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| − | draw(anglemark(D,C,A));
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| − | </asy>
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| − | | |
| − | Let <math>\angle BAE = \angle ACD = x</math>.
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| − | <cmath>\begin{align*}\angle 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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| − | Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math>
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| − | | |
| − | == See also ==
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| − | {{AMC10 box|year=2010|num-b=13|num-a=15|ab=A}}
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| − | {{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}
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| − | | |
| − | [[Category:Introductory Geometry Problems]]
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| − | {{MAA Notice}}
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