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

(Redirected page to 2010 AMC 12A Problems/Problem 8)
(Tag: New redirect)
 
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== Problem ==
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#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>?
 
 
 
<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>
 
 
 
== Solution ==
 
 
 
<center>[[File:AMC 2010 12A Problem 8.png]]</center>
 
 
 
<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 <math>\angle BAE = \angle ACD = x</math>.
 
 
 
<cmath>\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*}</cmath>
 
 
 
Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math>
 
 
 
== See also ==
 
{{AMC10 box|year=2010|num-b=13|num-a=15|ab=A}}
 
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}
 
 
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 

Latest revision as of 12:24, 26 May 2020