2016 AMC 8 Problems/Problem 23: Difference between revisions
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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>. | ||
==Video Solution== | |||
https://youtu.be/WJ0Hodj0h2o - Happytwin | |||
{{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 16:42, 12 May 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 
.
Video Solution
https://youtu.be/WJ0Hodj0h2o - Happytwin
| 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 | ||
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