Difference between revisions of "2014 AMC 10B Problems/Problem 19"
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| − | Pick any arbitrary point on the circle and it is clear that 120 degrees of the circle | + | Pick any arbitrary point on the circle, and it is clear that the second point must lie on the arc of 120 degrees, on the opposite end of the circle from the point. Therefore the probability is <math>120/360=1/3</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2014|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:02, 20 February 2014
Problem
Two concentric circles have radii 
and 
. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
Solution
Pick any arbitrary point on the circle, and it is clear that the second point must lie on the arc of 120 degrees, on the opposite end of the circle from the point. Therefore the probability is 
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 AMC 10 Problems and Solutions | ||
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