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

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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>.
 
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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== Solution 2(Trig and Angle Chasing) ==
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Let <math>AB=2a, AC=a</math>. Let <math>\angle BAE=\angle ACD=x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle FCE=60</math>, so <math>\angle ACB=60+x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle CFE=60</math>. Angles <math>AFD</math> and <math>CFE</math> are vertical, so <math>\angle AFD=60</math>. By triangle <math>ADF</math>, we have <math>\angle ADF=120-x</math>, and because of line <math>AB</math>, we have <math>\angle BDC=60+x</math>. Because Of line <math>BC</math>, we have <math>\angle AEB=120</math>, and by line <math>CD</math>, we have <math>\angle DFE=120</math>. By quadrilateral <math>BDFE</math>, we have <math>\angle ABC=60-x</math>.
  
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By the Law of Sines, we have <math>\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies </math>\sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)<math>. By the sine addition formula(which states </math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)<math> by the way), we have </math>2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)<math>. Because cosine is an even function, and sine is an odd function, we have </math>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)<math>. We know that </math>\sin(60)=\frac{\sqrt{3}}{2}<math>, and </math>\cos(60)=\frac{1}{2}<math>, hence </math>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies <math>\tan(x)=\frac{\sqrt(3)}{3}</math>. The only value of <math>x</math> that satisfies <math>60+x<180</math>(because <math>60+x</math> is an angle of the triangle)<math> is </math>x=30^{\circ}<math>. We seek to find </math>\angle ACB<math>, which as we found before is </math>60+x<math>, which is </math>90$. The answer is (C)
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-vsamc
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}
 
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}

Revision as of 16:39, 10 March 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 $\frac{AC}{AB} = \frac{1}{2}$, 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 (C)

-vsamc

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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