Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3 (Similar Triangles))
(→‎Solution 4)
 
(10 intermediate revisions by 8 users not shown)
Line 4: Line 4:
 
<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 ==
+
== Solution 1 ==
  
 
<center>[[File:AMC 2010 12A Problem 8.png]]</center>
 
<center>[[File:AMC 2010 12A Problem 8.png]]</center>
Line 19: Line 19:
  
 
== Solution 2(Trig and Angle Chasing) ==
 
== Solution 2(Trig and Angle Chasing) ==
−
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>.
+
Let <cmath>AB=2a, AC=a</cmath>
 +
Let <cmath>\angle BAE=\angle ACD=x</cmath>
 +
Because <math>\triangle CFE</math> is equilateral, we get <math>\angle FCE=60</math>, so <math>\angle ACB=60+x</math>
  
−
By the Law of Sines, we have <math>\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)</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 \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) 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>
+
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>.
 +
 
 +
By the Law of Sines:
 +
<cmath>\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies 2\sin(60-x)=\sin(60+x)</cmath>
 +
 
 +
By the sine addition formula(<math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)</math>):
 +
<cmath>2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)</cmath>
 +
 
 +
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>
 +
 
 +
We know that <math>\sin(60)=\frac{\sqrt{3}}{2}</math>, and <math>\cos(60)=\frac{1}{2}</math>, hence  
 +
<cmath>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies \tan(x)=\frac{\sqrt{3}}{3}</cmath>
 +
 
 +
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>
  
 
-vsamc
 
-vsamc
  
 
== Solution 3 (Similar Triangles) ==
 
== Solution 3 (Similar Triangles) ==
−
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>, since 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>  
+
Notice that <math>\angle AEB=\angle AFC = 120^{\circ}</math> and <math>\angle ACF=\angle BAE</math>. Hence, triangle AEB is similar to triangle CFA. Since <math>AB=2AC</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>  
 
-HarryW
 
-HarryW
 +
 +
 +
== Solution 4 ==
 +
Notice that <math>\triangle ADF \sim \triangle CDA</math> (by AA Similarity.) Since the corresponding angles of a pair of similar triangles are congruent, we have <math>\angle DAC = \angle DFA = 60^\circ.</math> Since <math>AB = 2 \cdot AC</math> and <math>\angle DFA = 60^\circ,</math> we have that <math>\triangle ABC</math> is congruent by SAS to a <math>30-60-90</math> right triangle, which gives the answer <math>\boxed{\textbf{(C)}}</math>.
 +
~clever14710owl
  
 
==Video Solution by the Beauty of Math==
 
==Video Solution by the Beauty of Math==
 
https://youtu.be/kU70k1-ONgM?t=785
 
https://youtu.be/kU70k1-ONgM?t=785
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/O_o_-yjGrOU?t=58
 +
 +
~ pi_is_3.14
  
 
== 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}}

Latest revision as of 19:33, 19 August 2024

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 $\angle ACF=\angle BAE$. Hence, triangle AEB is similar to triangle CFA. Since $AB=2AC$, $AE=2CF=2FE$, 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


Solution 4

Notice that $\triangle ADF \sim \triangle CDA$ (by AA Similarity.) Since the corresponding angles of a pair of similar triangles are congruent, we have $\angle DAC = \angle DFA = 60^\circ.$ Since $AB = 2 \cdot AC$ and $\angle DFA = 60^\circ,$ we have that $\triangle ABC$ is congruent by SAS to a $30-60-90$ right triangle, which gives the answer $\boxed{\textbf{(C)}}$. ~clever14710owl

Video Solution by the Beauty of Math

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

Video Solution by OmegaLearn

https://youtu.be/O_o_-yjGrOU?t=58

~ pi_is_3.14

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
2010 AMC 12A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_8&oldid=225342"