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

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(Solution 2(Trig and Angle Chasing))
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<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>
 
<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 ==
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== Solution 1 ==
  
 
<center>[[File:AMC 2010 12A Problem 8.png]]</center>
 
<center>[[File:AMC 2010 12A Problem 8.png]]</center>
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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>.
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== Solution 2(Trig and Angle Chasing) ==
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Let <cmath>AB=2a, AC=a</cmath>
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Let <cmath>\angle BAE=\angle ACD=x</cmath>
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Because <math>\triangle CFE</math> is equilateral, we get <math>\angle FCE=60</math>, so <math>\angle ACB=60+x</math>
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Because <math>\triangle CFE</math> is equilateral, we get <math>\angle CFE=60</math>.
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Angles <math>AFD</math> and <math>CFE</math> are vertical, so <math>\angle AFD=60</math>.
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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>.
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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>.
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By quadrilateral <math>BDFE</math>, we have <math>\angle ABC=60-x</math>.
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By the Law of Sines:
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<cmath>\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies 2\sin(60-x)=\sin(60+x)</cmath>
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By the sine addition formula(<math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)</math>):
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<cmath>2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)</cmath>
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 +
Because cosine is an even function, and sine is an odd function, we have <cmath>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)</cmath>
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We know that <math>\sin(60)=\frac{\sqrt{3}}{2}</math>, and <math>\cos(60)=\frac{1}{2}</math>, hence
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<cmath>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</cmath>
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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) 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</math>. The answer is <math>90, \text{or} \textbf{(C)}</math>
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-vsamc
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== Solution 3 (Similar Triangles) ==
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Notice that <math>\angle AEB=\angle AFC = 120^{\circ}</math> and <math>\angle ACF=\angle AEB</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2BC</math>, <math>AE=2CF=2FE</math>, as triangle CFE is equilateral. Therefore, <math>AF=FE=FC</math>, and since <math>\angle AFC=120^{\circ}</math>, <math>x=30</math>. Thus, the measure of <math>\angle ACE</math> equals to <math>\angle FCE+\angle ACF=90^{\circ}, \text{or} \textbf{(C)}</math>
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-HarryW
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==Video Solution by the Beauty of Math==
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https://youtu.be/kU70k1-ONgM?t=785
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2010|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}
 
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:46, 17 January 2022

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 1

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: \[\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies 2\sin(60-x)=\sin(60+x)\]

By the sine addition formula($\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$): \[2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)\]

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

Solution 3 (Similar Triangles)

Notice that $\angle AEB=\angle AFC = 120^{\circ}$ and $\angle ACF=\angle AEB$. Hence, triangle AEB is similar to triangle CFA. Since $AB=2BC$, $AE=2CF=2FE$, as triangle CFE is equilateral. Therefore, $AF=FE=FC$, and since $\angle AFC=120^{\circ}$, $x=30$. Thus, the measure of $\angle ACE$ equals to $\angle FCE+\angle ACF=90^{\circ}, \text{or} \textbf{(C)}$ -HarryW

Video Solution by the Beauty of Math

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

See also

2010 AMC 10A (ProblemsAnswer KeyResources)
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
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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