Difference between revisions of "2016 AMC 8 Problems/Problem 23"
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===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. | ||
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+ | ~Elijahman~ | ||
==Video Solution== | ==Video Solution== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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Revision as of 10:26, 18 June 2024
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.
Video Solution
https://www.youtube.com/watch?v=UZqVG5Q1liA&ab_channel=Elijahman
~Elijahman~
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/NumhAGApJ7c - Happytwin
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=968
~ pi_is_3.14
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.