2011 AMC 10A Problem 20

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

Rotational symmetry lets one random point be fixed, reducing the problem to a single uniform variable. Whether two chords cross depends only on the cyclic order of their four endpoints, which turns a geometry question into a measurement of arc length. The answer $\frac13$ is $2 \cdot 60^\circ / 360^\circ$: the second point must fall within one chord-length of the first, in either direction.

• Counting only the case where the second point lies on the first chord's arc (60 degrees), forgetting the second chord's endpoint can land on the first arc, and answering 1/6.
• Treating the chords as subtending $90^\circ$ or $120^\circ$ by misplacing the equilateral triangle, or ignoring the "clockwise" rule and allowing both directions.