Difference between revisions of "2016 AMC 8 Problems/Problem 23"
(→Solution 1) |
(→Solution 1) |
||
| Line 12: | Line 12: | ||
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 | 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 | ||
| + | |||
| + | ========================================================================================== | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 09:52, 3 November 2019
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
Drawing the diagram:
[\asy]
label("", D, SE);
label("", E, N);
[\asy]
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
==============================================================================
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
