2011 AMC 10A Problems/Problem 20

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 1

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)

Solution 2

Do what Solution 1 did until the guessing part. We then realize that the chords and radii make an equilateral triangle of length $r$ . Therefore the arc degree is $60.$ The other arc degree is also $60.$ Therefore the sum is $120.$ Continue as follows.

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.

Page

Toolbox

Search

2011 AMC 10A Problems/Problem 20

Contents

Problem 20

Solution 1

Solution 2

See Also