2016 AMC 8 Problems/Problem 23

Revision as of 17:44, 4 July 2019 by Qkddud (talk | contribs) (Solution 1)

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:

label("$D$", D, SE);
label("$E$", E, N);
 (Error compiling LaTeX. de252c60d1f2fe7247019fff9f040e897514e15b.asy: 5.6: no matching function 'label(string, <overloaded>, pair)'
de252c60d1f2fe7247019fff9f040e897514e15b.asy: 5.14: use of variable 'D' is ambiguous)

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
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