2014 AMC 10B Problem 19

Let $O$ be the common center and fix the first point $P$ on the outer circle. The second point $X$ is uniform on the outer circle, so we need the fraction of that circle for which chord $PX$ crosses the inner circle.

Draw the two tangent lines from $P$ to the inner circle. If one touches at $T$, then $OT = 1$, $OP = 2$, and $OT \perp PT$, so triangle $OTP$ is a $30$-$60$-$90$ triangle and $\angle OPT = 30^\circ$. Extend the tangent to meet the outer circle again at $Q$; triangle $OPQ$ is isosceles with base angles $30^\circ$, so $\angle POQ = 120^\circ$.

The two tangents therefore hit the outer circle at points $Q_1, Q_2$ each $120^\circ$ from $P$, on opposite sides. The chord $PX$ swings inside the inner circle exactly when $X$ lies on the arc $Q_1 Q_2$ opposite $P$, which measures $360^\circ - 2 \cdot 120^\circ = 120^\circ$.

The probability is $\frac{120}{360} = \frac13$.

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

With two independent uniform points on a circle, fixing the first loses nothing by symmetry, and the question becomes a length-of-arc ratio. The boundary between "hits" and "misses" is tangency, so the tangents from the fixed point are the natural auxiliary lines; the $1{:}2$ ratio of radii makes the angles clean.

If the central angle between the two points is $\theta$, the chord's distance from $O$ is $2\cos\frac{\theta}{2}$. It is less than $1$ exactly when $\frac{\theta}{2} > 60^\circ$, i.e. $\theta > 120^\circ$. The minor central angle is uniform on $[0^\circ, 180^\circ]$, so the probability is $\frac{60}{180} = \frac13$.

• Measuring the favorable arc on the near side of the first point, or using the 60-degree half-angle instead of the full 120-degree arc, giving $\frac16$.
• Treating the chord's distance from the center as uniformly distributed, which gives $\frac12$.