2011 AMC 10A Problem 20

1. How much of the circle does a chord cover? A chord of length $r$, together with the two radii to its ends, makes a triangle with all three sides equal to $r$. That is an equilateral triangle, so the angle at the center is $60^\circ$. Each chord cuts off a $60^\circ$ arc.
2. Fix the first point. The circle looks the same no matter how you turn it, so we can put the first point at $0^\circ$ and measure clockwise in degrees. Its chord goes from $0^\circ$ to $60^\circ$. The second point is at some random position $\theta$ between $0^\circ$ and $360^\circ$, and its chord goes from $\theta$ to $\theta + 60^\circ$.
3. When do two chords cross? Two chords cross exactly when one of them has one end inside the other's arc and one end outside. So ask: does exactly one end of the second chord land on the arc from $0^\circ$ to $60^\circ$?
4. Check the positions of $\theta$.
- $\theta$ between $0^\circ$ and $60^\circ$: the start of the second chord is on the arc, and its end (between $60^\circ$ and $120^\circ$) is off it. They cross.
- $\theta$ between $300^\circ$ and $360^\circ$: the start is off the arc, but the end, $60^\circ$ later, has gone past $360^\circ$ and landed between $0^\circ$ and $60^\circ$. They cross.
- $\theta$ anywhere else (between $60^\circ$ and $300^\circ$): both ends are off the arc, so the chords don't meet.
5. Probability. The good positions make up $60^\circ + 60^\circ = 120^\circ$ out of $360^\circ$:
$$\frac{120}{360} = \frac{1}{3}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{3}}$.

Because a circle looks the same after any turn, you can pin one random point in place and let only the other move. Whether two chords cross depends only on the order of their four endpoints around the circle, so the question becomes "how much of the circle is a good landing spot?" Here the second point must land within one chord-width ($60^\circ$) of the first, on either side.

A chord of length $r$ together with the two radii to its endpoints forms an equilateral triangle, so the chord subtends a $60^\circ$ arc. Measure positions in degrees clockwise around the circle.

By symmetry, fix the first point $P$ at $0^\circ$; its chord runs from $0^\circ$ to $60^\circ$. Let the second point $Q$ be at angle $\theta$, uniform on $[0^\circ, 360^\circ)$; its chord runs from $\theta$ to $\theta + 60^\circ$.

Two chords intersect exactly when their endpoints alternate around the circle, i.e. exactly one endpoint of $QQ'$ lies on the arc from $0^\circ$ to $60^\circ$.
- If $0^\circ < \theta < 60^\circ$, then $Q$ is on the arc and $Q'$ (at $\theta + 60^\circ$, between $60^\circ$ and $120^\circ$) is not: the chords cross.
- If $300^\circ < \theta < 360^\circ$, then $Q$ is not on the arc but $Q' = \theta + 60^\circ$ is (it lies between $0^\circ$ and $60^\circ$): the chords cross.
- Otherwise neither endpoint is on the arc, and the chords are disjoint.

The favorable window is $120^\circ$ out of $360^\circ$, so the probability is $\dfrac{120}{360} = \dfrac{1}{3}$.

The answer is $\boxed{\textbf{(D)}\ \frac{1}{3}}$.