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Difference between revisions of "1996 AHSME Problems/Problem 22"
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If you pick <math>WY</math> and <math>XZ</math>, you have the diagonals of the quadrilateral, which do intersect.
 
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{\text{B}}</math>.
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Any four points on the original circle of <math>1996</math> can be connected to form such a convex quadrilateral <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{\text{B}}</math>.
  
 
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.
 
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.
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[[Category:Introductory Probability Problems]]
 
[[Category:Introductory Probability Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 20:43, 11 April 2019

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 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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