Difference between revisions of "2003 AMC 12B Problems/Problem 25"
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The next point must lie within <math>60</math> degrees of arc on either side, a total of <math>120</math> degrees possible, giving a total <math>\frac{1}{3}</math> chance. The last point must lie within <math>60</math> degrees of both. | The next point must lie within <math>60</math> degrees of arc on either side, a total of <math>120</math> degrees possible, giving a total <math>\frac{1}{3}</math> chance. The last point must lie within <math>60</math> degrees of both. | ||
| − | The minimum area of freedom we have to place the third point is a <math>60</math> degrees arc(if the first two are <math>60</math> degrees apart), with a | + | The minimum area of freedom we have to place the third point is a <math>60</math> degrees arc(if the first two are <math>60</math> degrees apart), with a <math>\frac{1}{6}</math> probability. |
The maximum amount of freedom we have to place the third point is a <math>120</math> degree arc(if the first two are the same point), with a <math>\frac{1}{3}</math> probability. | The maximum amount of freedom we have to place the third point is a <math>120</math> degree arc(if the first two are the same point), with a <math>\frac{1}{3}</math> probability. | ||
Revision as of 21:10, 26 November 2013
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
The first point anywhere on the circle, because it doesn't matter where it is chosen.
The next point must lie within 
degrees of arc on either side, a total of 
degrees possible, giving a total 
chance. The last point must lie within degrees of both.
The minimum area of freedom we have to place the third point is a degrees arc(if the first two are degrees apart), with a 
probability.
The maximum amount of freedom we have to place the third point is a degree arc(if the first two are the same point), with a probability.
As the second point moves farther away from the first point, up to a maximum of degrees, the probability changes linearly (every degree it moves, adds one degree to where the third could be).
Therefore, we can average probabilities at each end to find 
to find the average probability we can place the third point based on a varying second point.
Therefore the total probability is 
or 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
