1996 AHSME Problems/Problem 22

Problem

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 $AB$ intersects the chord $CD$?

$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{1}{3}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

Solution

Let $WXYZ$ be a convex cyclic quadrilateral inscribed in a circle. There are $\frac{\binom{4}{2}}{2} = 3$ ways to divide the points into two groups of two.

If you pick $WX$ and $YZ$, you have two sides of the quadrilateral, which do not intersect.

If you pick $XY$ and $ZW$, you have the other two sides of the quadrilateral, which do not intersect.

If you pick $WY$ and $XZ$, you have the diagonals of the quadrilateral, which do intersect.

Any four points on the original circle of $1996$ can be connected to form such a convex quadrilateral $WXYZ$, and only placing $A$ and $C$ as one of the diagonals of the figure will form intersecting chords. Thus, the answer is $\frac{1}{3}$, which is option $\boxed{\text{B}}$.

Notice that $1996$ is irrelevant to the solution of the problem; in fact, you may pick points from the entire circumference of the circle.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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