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

(Solution 2)
m (Solution 1)
 
Line 4: Line 4:
  
 
==Solution 1==
 
==Solution 1==
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>.  
+
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>.
 
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>.

Latest revision as of 13:53, 10 November 2019

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

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

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

Invalid username
Login to AoPS