2003 AMC 12B Problems/Problem 25

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 $\frac{1}{6}$ 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 $\boxed{\text{(D)}}$

2003 AMC 12B (Problems • Answer Key • Resources)
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2003 AMC 12B Problems/Problem 25

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