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 in an equilateral triangle formed by the center and the first point lying on either side of the radius formed

The last point must lie within this equilateral triangle

Therefore the total probablity is 1*1/3*1/6 = 1/18 or (C)

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

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