Difference between revisions of "2011 AMC 10A Problems/Problem 20"

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== Problem 20 ==
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Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?
  
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<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2} </math>
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== Solution ==
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Fix a point <math>A</math> from which you draw a clockwise chord. In order for the clockwise chord from another point <math>B</math> to intersect that of point <math>A</math>, <math>A</math> and <math>B</math> must be no more than <math>r</math> units apart. By drawing the circle, we quickly see that <math>B</math> can be on <math>\frac{120}{360}=\boxed{\frac{1}{3}  \ \textbf{(D)}}</math> of the perimeter of the circle.

Revision as of 19:55, 11 February 2011

Problem 20

Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?

$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}$

Solution

Fix a point $A$ from which you draw a clockwise chord. In order for the clockwise chord from another point $B$ to intersect that of point $A$, $A$ and $B$ must be no more than $r$ units apart. By drawing the circle, we quickly see that $B$ can be on $\frac{120}{360}=\boxed{\frac{1}{3}  \ \textbf{(D)}}$ of the perimeter of the circle.