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

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 $\boxed{\text{(D)}}$

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.

Page

Toolbox

Search

2003 AMC 12B Problems/Problem 25

Problem

Solution

See Also