Difference between revisions of "2016 AMC 8 Problems/Problem 23"
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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== | + | ==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>. | 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>. | 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. | 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}} | {{AMC8 box|year=2016|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:02, 16 January 2021
Contents
[hide]Problem
Two congruent circles centered at points and each pass through the other circle's center. The line containing both and is extended to intersect the circles at points and . The circles intersect at two points, one of which is . What is the degree measure of ?
Solutions
Solution 1
Observe that is equilateral. Therefore, . Since is a straight line, we conclude that . Since (both are radii of the same circle), is isosceles, meaning that . Similarly, .
Now, . Therefore, the answer is .
Solution 2
We know that is equilateral, because all of its sides are congruent radii. Because point is the center of a circle, is at the border of a circle, and and are points on the edge of that circle, . Since is isosceles, angle degrees -SweetMango77.
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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