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

## 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

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 $\boxed{\textbf{(C) }\ 120}$.

## Solution 2 (Trig)

Let $r$ be the radius of both circles (we are given that they are congruent). Let's drop the altitude from $E$ onto segment $AB$ and call the intersection point $F$. Notice that $F$ is the midpoint of $A$ and $B$, which means that $AF = BF = \frac{r}{2}$. Also notice that $\triangle{EAB}$ is equilateral, which means we can use the Pythagorean Theorem to get $EF = \frac{r\sqrt{3}}{2}$.

Now let's apply trigonometry. Let $\theta = \angle{CEF}$. We can see that $\tan\theta = \frac{CF}{EF} = \frac{\frac{3r}{2}}{\frac{r\sqrt{3}}{2}} = \sqrt{3}$. This means $m\angle{CEF} = \frac{\pi}{3}$. However, this is not the answer. The question is asking for $m\angle{CED}$. Notice that $\angle{CEF}\cong\angle{DEF}$, which means $m\angle{CED} = 2m\angle{CEF}$. Thus, $\angle{CED} = 2\cdot\frac{\pi}{3} = \frac{2\pi}{3} = 120^{\circ}$. Therefore, the answer is $\boxed{\textbf{(C) }\ 120}$.

## Video Solution

https://youtu.be/WJ0Hodj0h2o - Happytwin