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

m (Solution 1)
m (Solution 1)
Line 6: Line 6:
 
Drawing the diagram:
 
Drawing the diagram:
  
<asy>
+
<asy
pair A, B, C, D, E;
+
 
A = (0,0);
+
label("<math>D</math>", D, SE);
B = (10,0);
+
label("<math>E</math>", E, N);
C = (-10,0);
 
D = (20,0);
 
E = (5, 8.75);
 
draw(Circle(A, 10));
 
draw(Circle(B, 10));
 
dot(A);
 
dot(B);
 
dot(C);
 
dot(D);
 
dot(E);
 
draw(C--D);
 
draw(A--E);
 
draw(B--E);
 
draw(C--E);
 
draw(D--E);
 
label("$A$", A, SW);
 
label("$B$", B, SE);
 
label("$C$", C, SW);
 
label("$D$", D, SE);
 
label("$E$", E, N);
 
 
</asy>
 
</asy>
  

Revision as of 12:08, 24 March 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

Drawing the diagram:

D</math>", D, SE);
label("<math>E</math>", E, N);
 (Error compiling LaTeX. 23e67869e1252b9ed56563a45ccdf2c266b7c77a.asy: 5.3: syntax error
error: could not load module '23e67869e1252b9ed56563a45ccdf2c266b7c77a.asy')

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

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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