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

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== Problem 20 ==
 
== 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 <math>r</math> is drawn in a clockwise direction. What is the probability that the two chords intersect?
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Two points on the circumference of a circle of radius <math>r</math> are selected independently and at random. From each point a chord of length <math>r</math> is drawn in a clockwise direction. What is the probability that the two chords intersect?
  
 
<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>
 
<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>

Revision as of 15:39, 4 February 2018

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 we 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. (Imagine a regular hexagon inscribed in the circle)

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
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Problem 19 		Followed by
Problem 21
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All AMC 10 Problems and Solutions

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