Difference between revisions of "2011 AMC 10A Problems/Problem 20"
(→Problem 20) |
|||
| Line 3: | Line 3: | ||
<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> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == | ||
Revision as of 10:41, 13 August 2014
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?
Solution
Fix a point 
from which we draw a clockwise chord. In order for the clockwise chord from another point 
to intersect that of point , and must be no more than 
units apart. By drawing the circle, we quickly see that can be on 
of the perimeter of the circle.
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
