2016 AMC 8 Problems/Problem 23: Difference between revisions

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

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

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Video Solution by WhyMath

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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

@@ Line 9: / Line 9: @@
 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 (SIMPLE)==
-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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