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2010 AMC 10A Problems/Problem 14: Difference between revisions

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== Solution ==
== 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 <math>\angle BAE = \angle ACD = x</math>.
Let <math>\angle BAE = \angle ACD = x</math>.


Revision as of 12:19, 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 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America.

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