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==Problem 23==
==Problem==
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>?
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>?


<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math>
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math>


==Solution 1==
==Solution==
Observe that <math>\triangle{EAB}</math> is equilateral. Therefore, <math>m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}</math>. Since <math>CD</math> is a straight line, we conclude that <math>m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}</math>. Since <math>BE=BD</math> (both are radii of the same circle), <math>\triangle{BED}</math> is isosceles, meaning that <math>m\angle{BED}=m\angle{BDE}=30^{\circ}</math>. Similarly, <math>m\angle{AEC}=m\angle{ACE}=30^{\circ}</math>.  
 
Observe that <math>\triangle{EAB}</math> is equilateral (all are radii of congruent circles). Therefore, <math>m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}</math>. Since <math>CD</math> is a straight line, we conclude that <math>m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}</math>. Since <math>BE=BD</math> (both are radii of the same circle), <math>\triangle{BED}</math> is isosceles, meaning that <math>m\angle{BED}=m\angle{BDE}=30^{\circ}</math>. Similarly, <math>m\angle{AEC}=m\angle{ACE}=30^{\circ}</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>.
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==
== Video Solution by Elijahman==
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>.
https://youtu.be/UZqVG5Q1liA?si=LDc8tMTnj1FMMlZc
 
~Elijahman
 
== Video Solution by Education, The Study of Everything ==
https://youtu.be/iGG_Hz-V6lU
 
== Video Solution by Omega Learn==
https://youtu.be/FDgcLW4frg8?t=968
 
~ pi_is_3.14


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 by WhyMath ==
https://youtu.be/nLlnMO6D5ek


==Video Solution==
~savannahsolver
https://youtu.be/WJ0Hodj0h2o - Happytwin


==See Also==
==See Also==
{{AMC8 box|year=2016|num-b=22|num-a=24}}
{{AMC8 box|year=2016|num-b=22|num-a=24}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Geometry Problems]]

Latest revision as of 19:55, 23 October 2025

Problem

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

Observe that $\triangle{EAB}$ is equilateral (all are radii of congruent circles). 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 $\boxed{\textbf{(C) }\ 120}$.

Video Solution by Elijahman

https://youtu.be/UZqVG5Q1liA?si=LDc8tMTnj1FMMlZc

~Elijahman

Video Solution by Education, The Study of Everything

https://youtu.be/iGG_Hz-V6lU

Video Solution by Omega Learn

https://youtu.be/FDgcLW4frg8?t=968

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/nLlnMO6D5ek

~savannahsolver

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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