2009 AMC 10B Problem 23

The picture shows a fixed piece of track, so turn "in the picture" into "during a certain window of time" for each runner.

1. The picture shows one-fourth of the track, centered on the start line. So it reaches one-eighth of a lap on each side of the line.
2. A runner is in the picture when they are within one-eighth of a lap of the start line. That happens from one-eighth of a lap time before they cross the line until one-eighth of a lap time after. Which way they run does not matter.
3. Work in seconds. The photo is taken at a random moment between $600$ and $660$ seconds.
4. Rachel: she crosses the start every $90$ seconds, at $\ldots, 540, 630, 720, \ldots$ One-eighth of $90$ is $11.25$. Around $630$ she is in the picture from $630 - 11.25 = 618.75$ to $630 + 11.25 = 641.25$. Her windows around $540$ and $720$ lie outside the minute from $600$ to $660$.
5. Robert: he crosses the start every $80$ seconds, at $\ldots, 560, 640, 720, \ldots$ One-eighth of $80$ is $10$. Around $640$ he is in the picture from $630$ to $650$. His windows around $560$ and $720$ lie outside the minute.
6. Both are in the picture when both windows overlap: from $630$ to $641.25$. That is $11.25$ seconds.
7. The photo time is equally likely to be anywhere in the $60$-second minute, so
$$\text{probability} = \frac{11.25}{60} = \frac{45}{240} = \frac{3}{16}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{3}{16}}$.

When a moment is chosen at random from a stretch of time, a probability is just "good time divided by total time." The key step is changing a piece of track into a piece of time: one-eighth of a lap takes one-eighth of the lap time. Center each window on the moment the runner crosses the start line and the numbers stay small.