Difference between revisions of "2010 AMC 12A Problems/Problem 8"

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  \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath>
 
  \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath>
  
Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>.
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Since <math>\frac{AC}{AB} = \frac{1}{2}</math> and the angle between the hypotenuse and the shorter side is <math>60^\circ</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>.
  
 
== Solution 2(Trig and Angle Chasing) ==
 
== Solution 2(Trig and Angle Chasing) ==

Latest revision as of 18:07, 19 October 2020

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

AMC 2010 12A Problem 8.png


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 $\frac{AC}{AB} = \frac{1}{2}$ and the angle between the hypotenuse and the shorter side is $60^\circ$, triangle $ABC$ is a $30-60-90$ triangle, so $\angle BCA = \boxed{90^\circ\,\textbf{(C)}}$.

Solution 2(Trig and Angle Chasing)

Let $AB=2a, AC=a$. Let $\angle BAE=\angle ACD=x$. Because $\triangle CFE$ is equilateral, we get $\angle FCE=60$, so $\angle ACB=60+x$. Because $\triangle CFE$ is equilateral, we get $\angle CFE=60$. Angles $AFD$ and $CFE$ are vertical, so $\angle AFD=60$. By triangle $ADF$, we have $\angle ADF=120-x$, and because of line $AB$, we have $\angle BDC=60+x$. Because Of line $BC$, we have $\angle AEB=120$, and by line $CD$, we have $\angle DFE=120$. By quadrilateral $BDFE$, we have $\angle ABC=60-x$.

By the Law of Sines, we have $\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies \sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)$. By the sine addition formula(which states $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ by the way), we have $2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)$. Because cosine is an even function, and sine is an odd function, we have $2\sin(60)\cos(x)-2\cos(60)\sin(x)=\sin(60)\cos(x)+\cos(60)\sin(x) \implies \sin(60)\cos(x)=3\cos(60)\sin(x)$. We know that $\sin(60)=\frac{\sqrt{3}}{2}$, and $\cos(60)=\frac{1}{2}$, hence $\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}$. The only value of $x$ that satisfies $60+x<180$(because $60+x$ is an angle of the triangle) is $x=30^{\circ}$. We seek to find $\angle ACB$, which as we found before is $60+x$, which is $90$. The answer is $90, \text{or} \textbf{(C)}$

-vsamc

Video Solution

https://youtu.be/kU70k1-ONgM?t=785

~IceMatrix

See also

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

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