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Difference between revisions of "1996 AHSME Problems/Problem 22"
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<math> \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} </math>
 
<math> \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} </math>
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==Solution==
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Let <math>WXYZ</math> be a convex cyclic quadrilateral inscribed in a circle.  There are <math>\frac{\binom{4}{2}}{2} = 3</math> ways to divide the points into two groups of two.
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If you pick <math>WX</math> and <math>YZ</math>, you have two sides of the quadrilateral, which do not intersect.
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If you pick <math>XY</math> and <math>ZW</math>, you have the other two sides of the quadrilateral, which do not intersect.
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If you pick <math>WY</math> and <math>XZ</math>, you have the diagonals of the quadrilateral, which do intersect.
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Any four points on the original circle of <math>1996</math> can be connected to form such a convex quarilateral <math>WXYZ</math>, and only placing <math>A</math> and <math>C</math> as one of the diagonals of the figure will form intersecting chords.  Thus, the answer is <math>\frac{1}{3}</math>, which is option <math>\boxed{B}</math>.
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Notice that <math>1996</math> is irrelevant to the solution of the problem; in fact, you may pick points from the entire circumference of the circle.
  
 
==See also==
 
==See also==
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{{AHSME box|year=1996|num-b=21|num-a=23}}
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{{AHSME box|year=1996|num-b=21|num-a=2

Revision as of 14:09, 20 August 2011

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 quarilateral $WXYZ$, 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 $\boxed{B}$.

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

{{AHSME box|year=1996|num-b=21|num-a=2

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