Difference between revisions of "2016 AMC 8 Problems/Problem 23"
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+ | ==Problem== | ||
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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== | + | ==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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/iGG_Hz-V6lU | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/FDgcLW4frg8?t=968 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/nLlnMO6D5ek | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==See Also== | ||
{{AMC8 box|year=2016|num-b=22|num-a=24}} | {{AMC8 box|year=2016|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:29, 24 July 2024
Contents
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
.
Video Solution
~Education, the Study of Everything
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.