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

Revision as of 03:02, 16 January 2021

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 $\angle CED$ ?

$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$

Solutions

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

We know that $\triangle{EAB}$ is equilateral, because all of its sides are congruent radii. Because point $A$ is the center of a circle, $C$ is at the border of a circle, and $E$ and $B$ are points on the edge of that circle, $m\angle{ECB}=\frac{1}{2}\cdot m\angle{EAB}=\frac{1}{2}\cdot60^{\circ}=30^{\circ}$ . Since $\triangle{CED}$ is isosceles, angle $\angle{CED}=180^{\circ}-2\cdot30^{\circ}=\boxed{\text{(C)}\; 120}$ degrees -SweetMango77.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
+==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>?
 <math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math>
-==Solution 1==
+==Solutions==
+===Solution 1===
 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>.
 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==
+===Solution 2===
 We know that <math>\triangle{EAB}</math> is equilateral, because all of its sides are congruent radii. Because point <math>A</math> is the center of a circle, <math>C</math> is at the border of a circle, and <math>E</math> and <math>B</math> are points on the edge of that circle, <math>m\angle{ECB}=\frac{1}{2}\cdot m\angle{EAB}=\frac{1}{2}\cdot60^{\circ}=30^{\circ}</math>. Since <math>\triangle{CED}</math> is isosceles, angle <math>\angle{CED}=180^{\circ}-2\cdot30^{\circ}=\boxed{\text{(C)}\; 120}</math> degrees -SweetMango77.
 {{AMC8 box|year=2016|num-b=22|num-a=24}}
 {{MAA Notice}}

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