1996 AMC 12 Problem 22

Question

Four distinct points,

A,

B,

C,

and

D,

are to be selected from

1996

points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord

\overline{AB}

intersects the chord

\overline{CD}?

\frac14

\frac13

\frac12

\frac23

\frac34

Answer: B

Concepts:geometric probability symmetry

Difficulty rating: 1520

Small hint:

Fix any four selected points and examine the three ways to pair them into two chords

Big hint:

Exactly one pairing joins alternating points around the circle

Solution:

For any fixed four points, there are three ways to partition their labels into two unordered chord pairs. Exactly one pairing joins alternating points around the circle and therefore crosses. The named pair

\overline{AB},\overline{CD}

is equally likely to be any of these three pairings, so the probability is

\frac{1}{3}.

The correct answer is B.