Difference between revisions of "2016 AMC 8 Problems/Problem 23"
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==Problem== | ==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== |
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− | 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 (all are radii of congruent circles). 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>. | ||
− | == | + | == Video Solution by Elijahman== |
− | + | https://youtu.be/UZqVG5Q1liA?si=LDc8tMTnj1FMMlZc | |
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+ | ~Elijahman | ||
− | ==Video Solution== | + | == Video Solution by Education, The Study of Everything == |
https://youtu.be/iGG_Hz-V6lU | https://youtu.be/iGG_Hz-V6lU | ||
− | + | == Video Solution by Omega Learn== | |
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− | == Video Solution by | ||
https://youtu.be/FDgcLW4frg8?t=968 | https://youtu.be/FDgcLW4frg8?t=968 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
− | ==Video Solution== | + | == Video Solution by WhyMath == |
https://youtu.be/nLlnMO6D5ek | https://youtu.be/nLlnMO6D5ek | ||
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{{AMC8 box|year=2016|num-b=22|num-a=24}} | {{AMC8 box|year=2016|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 18:24, 25 June 2025
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
?
Solution
Observe that is equilateral (all are radii of congruent circles). 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 by Elijahman
https://youtu.be/UZqVG5Q1liA?si=LDc8tMTnj1FMMlZc
~Elijahman
Video Solution by Education, The Study of Everything
Video Solution by Omega Learn
https://youtu.be/FDgcLW4frg8?t=968
~ pi_is_3.14
Video Solution by WhyMath
~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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.