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Difference between revisions of "2016 AMC 8 Problems/Problem 23"

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==Problem==
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Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?
 
Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?
  
 
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math>
 
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math>
  
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==Solution 1==
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==Solutions==
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Drawing the diagram:
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===Solution 1===
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Observe that <math>\triangle{EAB}</math> is equilateral. Therefore, <math>m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}</math>. Since <math>CD</math> is a straight line, we conclude that <math>m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}</math>. Since <math>BE=BD</math> (both are radii of the same circle), <math>\triangle{BED}</math> is isosceles, meaning that <math>m\angle{BED}=m\angle{BDE}=30^{\circ}</math>. Similarly, <math>m\angle{AEC}=m\angle{ACE}=30^{\circ}</math>.
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Now, <math>\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
  
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<asy>
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==Video Solution==
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pair A, B, C, D, E;
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https://youtu.be/iGG_Hz-V6lU
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A = (0,0);
 
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B = (10,0);
 
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C = (-10,0);
 
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D = (20,0);
 
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E = (5, 8.75);
 
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draw(Circle(A, 10));
 
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draw(Circle(B, 10));
 
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dot(A);
 
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dot(B);
 
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dot(C);
 
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dot(D);
 
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dot(E);
 
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draw(C--D);
 
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draw(A--E);
 
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draw(B--E);
 
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draw(C--E);
 
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draw(D--E);
 
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label("$A$", A, SW);
 
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label("$B$", B, SE);
 
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label("$C$", C, SW);
 
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label("$D$", D, SE);
 
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label("$E$", E, N);
 
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</asy>
 
  
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we see that <math>\triangle EAB</math> is equilateral as each side is the radius of one of the two circles. Therefore, <math>\overarc{EB}=m\angle EAB=60^\circ</math>. Therefore, since it is an inscribed angle
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~Education, the Study of Everything
  
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==Solution 2==
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== Video Solution by OmegaLearn ==
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As in Solution 1, observe that <math>\triangle{EAB}</math> is equilateral. Therefore, <math>m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}</math>. Since <math>CD</math> is a straight line, we conclude that <math>m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}</math>. Since <math>BE=BD</math> (both are radii of the same circle), <math>\triangle{BED}</math> is isosceles, meaning that <math>m\angle{BED}=m\angle{BDE}=30^{\circ}</math>. Similarly, <math>m\angle{AEC}=m\angle{ACE}=30^{\circ}</math>.
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https://youtu.be/FDgcLW4frg8?t=968
  
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Now, <math>\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
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~ pi_is_3.14
  
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==Video Solution==
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https://youtu.be/nLlnMO6D5ek
  
 +
~savannahsolver
  
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==See Also==
 
{{AMC8 box|year=2016|num-b=22|num-a=24}}
 
{{AMC8 box|year=2016|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:29, 24 July 2024

Problem

Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$. The circles intersect at two points, one of which is $E$. What is the degree measure of $\angle CED$?

$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$

Solutions

Solution 1

Observe that $\triangle{EAB}$ is equilateral. Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$. Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$. Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $m\angle{BED}=m\angle{BDE}=30^{\circ}$. Similarly, $m\angle{AEC}=m\angle{ACE}=30^{\circ}$.

Now, $\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}$. Therefore, the answer is $\boxed{\textbf{(C) }\ 120}$.

Video Solution

https://youtu.be/iGG_Hz-V6lU

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=968

~ pi_is_3.14

Video Solution

https://youtu.be/nLlnMO6D5ek

~savannahsolver

See Also

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AJHSME/AMC 8 Problems and Solutions

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