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>
  
==Solution 1==
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==Solutions==
Drawing the diagram[SOMEONE DRAW IT PLEASE], 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, <math>m\angle ECB=\frac{60^\circ}{2}=30^\circ</math>. So, in <math>\triangle ECD</math>, <math>m\angle ECB=m\angle EDA=30^\circ</math>, and <math>m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ</math>. Our answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
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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>.  
  
==Solution 2==
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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>.
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>.  
 
  
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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===Solution 2===
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We know that <math>\triangle{EAB}</math> is equilateral, because all of its sides are congruent radii. Because point <math>A</math> is the center of a circle, <math>C</math> is at the border of a circle, and <math>E</math> and <math>B</math> are points on the edge of that circle, <math>m\angle{ECB}=\frac{1}{2}\cdot m\angle{EAB}=\frac{1}{2}\cdot60^{\circ}=30^{\circ}</math>. Since <math>\triangle{CED}</math> is isosceles, angle <math>\angle{CED}=180^{\circ}-2\cdot30^{\circ}=\boxed{\text{(C)}\; 120}</math> degrees -SweetMango77.
  
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==Video Solution==
  
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https://youtu.be/NumhAGApJ7c - Happytwin
  
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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}}

Revision as of 21:36, 16 January 2021

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 $\boxed{\textbf{(C) }\ 120}$.

Solution 2

We know that $\triangle{EAB}$ is equilateral, because all of its sides are congruent radii. Because point $A$ is the center of a circle, $C$ is at the border of a circle, and $E$ and $B$ are points on the edge of that circle, $m\angle{ECB}=\frac{1}{2}\cdot m\angle{EAB}=\frac{1}{2}\cdot60^{\circ}=30^{\circ}$. Since $\triangle{CED}$ is isosceles, angle $\angle{CED}=180^{\circ}-2\cdot30^{\circ}=\boxed{\text{(C)}\; 120}$ degrees -SweetMango77.

Video Solution

https://youtu.be/NumhAGApJ7c - Happytwin

See Also

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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