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Difference between revisions of "2003 AMC 12B Problems/Problem 25"

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First: One can choose the first point anywhere on the circle.
 
First: One can choose the first point anywhere on the circle.
 
   
 
   
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Secondly: The Next point must lie within 60 degrees of arc on either side, a total of 120 degrees possible, 1/3 chance.
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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 last point must lie within 60 degrees  of both.  This ranges from 60 degrees arc to sit on (if the first two are 60 degrees apart) and a 1/6 probability, to 120 degrees (if they are negligibly apart) and a 1/3 chance.   
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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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As the second point moves from 60 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 1/4.  
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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>.
  
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Therefore the total probability is 1*1/3*1/4 = 1/12 or (D)
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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>
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(checked on amc.maa.org)
 
  
 
==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 16:33, 1 February 2012

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

First: One can choose the first point anywhere on the circle.

Secondly: The Next point must lie within $60$ degrees of arc on either side, a total of $120$ degrees possible, $\frac{1}{3}$ chance.

The last point must lie within $60$ degrees of both. This ranges from $60$ degrees arc to sit on (if the first two are $60$ degrees apart) and a $\frac{1}{6}$ probability, to $120$ degrees (if they are negligibly apart) and a $\frac{1}{3}$ chance. As the second point moves from $60$ 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 $\frac{1}{4}$.

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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