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.

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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