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

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 (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 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