Difference between revisions of "2010 AMC 10A Problems/Problem 14"

m (Problem)
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== Solution ==
 
== Solution ==
 +
<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>
 +
 +
 
Let <math>\angle BAE = \angle ACD = x</math>.
 
Let <math>\angle BAE = \angle ACD = x</math>.
  

Revision as of 13:19, 13 August 2017

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

[asy] pair A,B,C,D,E,F,G,H; G=(0,10); A=(0,3.464); B=(6,0); C=(0,0); draw(A--B--C--cycle); F=(1,1.73); E=(2,0); draw(C--F--E); D=(1.5,2.6); draw(C--D); label("$A$",A,W); label("$B$",B,S); label("$C$",C,S); label("$F$",F,N); label("$D$",D,NE); label("$E$",E,S); draw(A--E); draw(anglemark(E,A,B)); draw(anglemark(D,C,A)); [/asy]


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

\begin{align*}\angle 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 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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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