Difference between revisions of "2016 AMC 8 Problems/Problem 23"
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− | 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, | + | 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></math>m\angle{ACB}=\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{(A)}\; 120}$ degrees. |
{{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 17:28, 4 November 2020
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 ?
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 -SweetMango77
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, $$ (Error compiling LaTeX. ! Missing $ inserted.)m\angle{ACB}=\frac{1}{2}\cdot m\angle{EAB}=\frac{1}{2}\cdot60^{\circ}=30^{\circ}\triangle{CED}\angle{CED}=180^{\circ}-2\cdot30^{\circ}=\boxed{\text{(A)}\; 120}$ degrees.
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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