2010 AMC 12A Problems/Problem 8

Revision as of 22:52, 1 February 2011 by Flyingpenguin (talk | contribs) (Solution)

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*}\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 $\frac{AC}{AB} = \frac{1}{2}$, triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90^\circ\ \textbf{(C)}}$

See also

2010 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AMC 12 Problems and Solutions