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

m (Solution 1)
m (Solution 1)
Line 4: Line 4:
  
 
==Solution 1==
 
==Solution 1==
 +
[asy]
 +
pair A=(0,0);
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pair B=(1,0);
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draw(circle((0,0),1));
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draw(circle((1,0),1));
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dot("A", A, dir(-45));
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dot("B", B, dir(225));
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pair C=(-1,0);
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pair D=(2,0);
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dot("C", C, dir(-45));
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dot("D", D);
 +
pair E=(.5,.86602540378);
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dot("E", E, dir(90));
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draw(E--A);
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draw(E--B);
 +
draw(A--B);
 +
draw(C--A);
 +
draw(B--D);
 +
draw(C--E);
 +
draw(E--D);
 +
[/asy]
 
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>.
 
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>.
  

Revision as of 15:01, 20 February 2018

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 $\angle CED$?

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

Solution 1

[asy] pair A=(0,0); pair B=(1,0); draw(circle((0,0),1)); draw(circle((1,0),1)); dot("A", A, dir(-45)); dot("B", B, dir(225)); pair C=(-1,0); pair D=(2,0); dot("C", C, dir(-45)); dot("D", D); pair E=(.5,.86602540378); dot("E", E, dir(90)); draw(E--A); draw(E--B); draw(A--B); draw(C--A); draw(B--D); draw(C--E); draw(E--D); [/asy] Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that $\triangle EAB$ is equilateral as each side is the radius of one of the two circles. Therefore, $\overarc{EB}=m\angle EAB-60^\circ$. Therefore, since it is an inscribed angle, $m\angle ECB=\frac{60^\circ}{2}=30^\circ$. So, in $\triangle ECD$, $m\angle ECB=m\angle EDA=30^\circ$, and $m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ$. Our answer is $\boxed{\textbf{(C) }\ 120}$.

Solution 2

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


2016 AMC 8 (ProblemsAnswer KeyResources)
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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