Difference between revisions of "2003 AMC 12B Problems/Problem 25"
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The last point must lie within this equilateral triangle | The last point must lie within this equilateral triangle | ||
| − | Therefore the total probablity is 1*1/3*1/6 = 1/18 | + | Therefore the total probablity is 1*1/3*1/6 = 1/18 or (C) |
==See Also== | ==See Also== | ||
{{AMC12 box|ab=B|year=2003|num-b=24|after=Last Problem}} | {{AMC12 box|ab=B|year=2003|num-b=24|after=Last Problem}} | ||
Revision as of 16:05, 24 August 2011
Problem
Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distance between the points are less than the radius of the circle?
Solution
First: One can choose the first point anywhere on the circle
Secondly: The Next point must lie in an equilateral triangle formed by the center and the first point lying on either side of the radius formed
The last point must lie within this equilateral triangle
Therefore the total probablity is 1*1/3*1/6 = 1/18 or (C)
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 24 |
Followed by Last Problem |
| 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 12 Problems and Solutions | |
