Difference between revisions of "2016 AMC 8 Problems/Problem 23"
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==Solution 1== | ==Solution 1== | ||
+ | [asy] | ||
+ | pair A=(0,0); | ||
+ | pair B=(1,0); | ||
+ | draw(circle((0,0),1)); | ||
+ | draw(circle((1,0),1)); | ||
+ | dot("A", A, dir(-45)); | ||
+ | dot("B", B, dir(225)); | ||
+ | pair C=(-1,0); | ||
+ | pair D=(2,0); | ||
+ | dot("C", C, dir(-45)); | ||
+ | dot("D", D); | ||
+ | pair E=(.5,.86602540378); | ||
+ | dot("E", E, dir(90)); | ||
+ | draw(E--A); | ||
+ | draw(E--B); | ||
+ | draw(A--B); | ||
+ | draw(C--A); | ||
+ | draw(B--D); | ||
+ | draw(C--E); | ||
+ | draw(E--D); | ||
+ | [/asy] | ||
Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that <math>\triangle EAB</math> is equilateral as each side is the radius of one of the two circles. Therefore, <math>\overarc{EB}=m\angle EAB-60^\circ</math>. Therefore, since it is an inscribed angle, <math>m\angle ECB=\frac{60^\circ}{2}=30^\circ</math>. So, in <math>\triangle ECD</math>, <math>m\angle ECB=m\angle EDA=30^\circ</math>, and <math>m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ</math>. Our answer is <math>\boxed{\textbf{(C) }\ 120}</math>. | Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that <math>\triangle EAB</math> is equilateral as each side is the radius of one of the two circles. Therefore, <math>\overarc{EB}=m\angle EAB-60^\circ</math>. Therefore, since it is an inscribed angle, <math>m\angle ECB=\frac{60^\circ}{2}=30^\circ</math>. So, in <math>\triangle ECD</math>, <math>m\angle ECB=m\angle EDA=30^\circ</math>, and <math>m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ</math>. Our answer is <math>\boxed{\textbf{(C) }\ 120}</math>. | ||
Revision as of 15:01, 20 February 2018
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
[asy] pair A=(0,0); pair B=(1,0); draw(circle((0,0),1)); draw(circle((1,0),1)); dot("A", A, dir(-45)); dot("B", B, dir(225)); pair C=(-1,0); pair D=(2,0); dot("C", C, dir(-45)); dot("D", D); pair E=(.5,.86602540378); dot("E", E, dir(90)); draw(E--A); draw(E--B); draw(A--B); draw(C--A); draw(B--D); draw(C--E); draw(E--D); [/asy] Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that is equilateral as each side is the radius of one of the two circles. Therefore, . Therefore, since it is an inscribed angle, . So, in , , and . Our answer is .
Solution 2
As in 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 .
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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