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

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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

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, $\angle{CED} = 2\cdot\frac{\pi}{3} = \frac{2\pi}{3} = 120^{\circ}$ .

Video Solution

https://youtu.be/WJ0Hodj0h2o - Happytwin

@@ Line 1: / Line 1: @@
+==Problem 23==
 Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?
@@ Line 7: / Line 8: @@
 Now, <math>\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
+==Solution 2==
+Let <math>r</math> be the radius of both circles (we are given that they are congruent). Let's drop the altitude from <math>E</math> onto segment <math>AB</math> and call the intersection point <math>F</math>. Notice that <math>F</math> is the midpoint of <math>A</math> and <math>B</math>, which means that <math>AF = BF = \frac{r}{2}</math>. Also notice that <math>\triangle{EAB}</math> is equilateral, which means we can use the Pythagorean Theorem to get <math>EF = \frac{r\sqrt{3}}{2}</math>.
+Now let's apply trigonometry. Let <math>\theta = \angle{CEF}</math>. We can see that <math>\tan\theta = \frac{CF}{EF} = \frac{\frac{3r}{2}}{\frac{r\sqrt{3}}{2}} = \sqrt{3}</math>. This means <math>m\angle{CEF} = \frac{\pi}{3}</math>. However, this is not the answer. The question is asking for <math>m\angle{CED}</math>. Notice that <math>\angle{CEF}\cong\angle{DEF}</math>, which means <math>m\angle{CED} = 2m\angle{CEF}</math>. Thus, <math>\angle{CED} = 2\cdot\frac{\pi}{3} = \frac{2\pi}{3} = 120^{\circ}</math>.
 ==Video Solution==
 https://youtu.be/WJ0Hodj0h2o - Happytwin
+==See Also==
 {{AMC8 box|year=2016|num-b=22|num-a=24}}
 {{MAA Notice}}

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