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

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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>.
 
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>.
  
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==Solution 2==
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==Solution 2 (Trig)==
 
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>.
 
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>.
  
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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>.
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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>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 22:52, 21 August 2020

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

See Also

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 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

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