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Difference between revisions of "2003 AMC 12B Problems/Problem 25"
(→‎Solution: improved solution greatly, making it more fluent and easier to understand and read.)
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==Solution==
 
==Solution==
  
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First: One can choose the first point anywhere on the circle.
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The first point anywhere on the circle, because it doesn't matter where it is chosen.
 
   
 
   
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Secondly: The Next point must lie within <math>60</math> degrees of arc on either side, a total of <math>120</math> degrees possible, <math>\frac{1}{3}</math> chance.
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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.  
  
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The last point must lie within <math>60</math> degrees  of both.  This ranges from <math>60</math> degrees arc to sit on (if the first two are <math>60</math> degrees apart) and a <math>\frac{1}{6}</math> probability, to <math>120</math> degrees (if they are negligibly apart) and a <math>\frac{1}{3}</math> chance.   
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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 of <math>\frac{1}{6}</math> probability.
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As the second point moves from <math>60</math> degrees away to the first point, the probability changes linearly (every degree it moves, adds one degree to where the third could be), the probabilities at each end can be averaged to find <math>\frac{1}{4}</math>.
+
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.   
 +
 
 +
As the second point moves farther away from the first point, up to a maximum of <math>60</math> 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 <math>\frac{1}{4}</math> to find the average probability we can place the third point based on a varying second point.
  
 
Therefore the total probability is <math>1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}</math> or <math>\boxed{\text{(D)}}</math>
 
Therefore the total probability is <math>1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}</math> or <math>\boxed{\text{(D)}}</math>

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?

$\mathrm{(A)}\ \dfrac{1}{36} \qquad\mathrm{(B)}\ \dfrac{1}{24} \qquad\mathrm{(C)}\ \dfrac{1}{18} \qquad\mathrm{(D)}\ \dfrac{1}{12} \qquad\mathrm{(E)}\ \dfrac{1}{9}$

Solution

The first point anywhere on the circle, because it doesn't matter where it is chosen.

The next point must lie within $60$ degrees of arc on either side, a total of $120$ degrees possible, giving a total $\frac{1}{3}$ chance. The last point must lie within $60$ degrees of both.

The minimum area of freedom we have to place the third point is a $60$ degrees arc(if the first two are $60$ degrees apart), with a of $\frac{1}{6}$ probability. The maximum amount of freedom we have to place the third point is a $120$ degree arc(if the first two are the same point), with a $\frac{1}{3}$ probability.

As the second point moves farther away from the first point, up to a maximum of $60$ 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 $\frac{1}{4}$ to find the average probability we can place the third point based on a varying second point.

Therefore the total probability is $1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$ or $\boxed{\text{(D)}}$

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

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