Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 10A Problems/Problem 20"
(→‎Solution)
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
  
−
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.
+
Fix a point <math>A</math> from which we 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 18:55, 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 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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_20&oldid=36863"