2016 AMC 8 Problems/Problem 23

Revision as of 09:45, 24 October 2018 by Thalassic (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:

[asy] pair A, B, C, D, E; A = (0,0); B = (10,0); 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] 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
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All AJHSME/AMC 8 Problems and Solutions

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