2003 AMC 12B Problems/Problem 25

Revision as of 17:04, 1 February 2012 by Andrewf (talk | contribs) (Solution)

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, 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 1/6 probability, to 120 degrees (if they are negligibly apart) and a 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 1/4.

Therefore the total probability is 1*1/3*1/4 = 1/12 or (D) (checked on amc.maa.org)

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
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