2016 AMC 8 Problems/Problem 23: Difference between revisions

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

Solution 2 (SIMPLE)

Due to Thale's Thereom, $m\angle{CEB} and$ m\angle{AED} are both 90^{\circ} $. Because$ \triangle{AEB} $is equilateral (all all sides are radii of congruent circles),$ m\angle{AEB} is 60^{\circ} $. Thus,$ m\angle{CEA} and $m\angle{EBD} both equal 90^{\circ}$ - 60^{\circ} $= 30{\circ}$ . Therefore, m/angle{CED} = 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

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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 ==Solution 2 (SIMPLE)==
-Due to Thale's Thereom, m\angle{CEB} and m\angle{AED} are both 90^{\circ}<math>. Because </math>\triangle{AEB}<math> is equilateral (all all sides are radii of congruent circles), m\angle{AEB} is 60^{\circ}</math>. Thus, m\angle{CEA} and m\angle{EBD} both equal 90^{\circ}<math> - 60^{\circ}</math> = 30{\circ}<math>. Therefore, m/angle{CED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
+Due to Thale's Thereom, <math>m\angle{CEB} and </math>m\angle{AED} are both 90^{\circ}<math>. Because </math>\triangle{AEB}<math> is equilateral (all all sides are radii of congruent circles), </math>m\angle{AEB} is 60^{\circ}<math>. Thus, </math>m\angle{CEA} and <math>m\angle{EBD} both equal 90^{\circ}</math> - 60^{\circ}<math> = 30{\circ}</math>. Therefore, m/angle{CED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}<math>. Therefore, the answer is </math>\boxed{\textbf{(C) }\ 120}$.
 == Video Solution by Elijahman==

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