2014 AMC 10B Problem 19

1. Fix the first point. The circle looks the same from every point on it, so we can put the first point anywhere; call it $P$, with the center $O$. Now ask: for which positions of the second point $X$ does chord $PX$ cross the inner circle? The answer is the fraction of the outer circle where $X$ works.
2. Find the boundary. If $X$ is close to $P$, the chord is short and stays near the edge, missing the inner circle. If $X$ is across from $P$, the chord passes through the center and hits it. The switch happens when the chord just touches the inner circle.
3. The touching chord. Let the chord from $P$ touch the inner circle at $T$. A radius meets a touching line at a right angle, so triangle $OTP$ has a right angle at $T$, with $OT = 1$ and $OP = 2$.
4. A right triangle whose hypotenuse is twice one leg is a $30$-$60$-$90$ triangle. The short leg $OT$ is across from the $30^\circ$ angle, so $\angle OPT = 30^\circ$ and $\angle POT = 60^\circ$.
5. The chord continues past $T$ to the outer circle at a point $Q$. The radius $OT$ cuts that chord into two equal halves, so $\angle TOQ = 60^\circ$ as well, and $\angle POQ = 120^\circ$.
6. There is one touching chord on each side of $P$. They end at points $120^\circ$ from $P$ in each direction. Those two sides use up $120^\circ + 120^\circ = 240^\circ$, leaving an arc of $360^\circ - 240^\circ = 120^\circ$ on the far side of the circle, across from $P$.
7. The chord crosses the inner circle exactly when $X$ is on that far arc. The probability is $\frac{120}{360} = \frac13$.

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

When two points are chosen at random on a circle, you can fix the first one without losing anything, and the question becomes "what fraction of the circle works for the second point?" The line between "hits" and "misses" is the chord that just touches the inner circle, so draw that chord first. The radii $1$ and $2$ were chosen to make a $30$-$60$-$90$ triangle, which is why the angles come out so clean.

Let the angle at the center between the two points be $\theta$, between $0^\circ$ and $180^\circ$. The distance from the center to the chord is $2\cos\frac{\theta}{2}$. The chord crosses the inner circle when that distance is less than $1$, which means $\cos\frac{\theta}{2} < \frac12$, so $\frac{\theta}{2} > 60^\circ$ and $\theta > 120^\circ$. The angle $\theta$ is equally likely to be anywhere from $0^\circ$ to $180^\circ$, so the chance is $\frac{180 - 120}{180} = \frac13$.