2009 AMC 10B Problem 23

The photo covers one-fourth of the track centered on the start, that is, one-eighth of a lap on each side of the start line. A runner is in the picture exactly when they are within one-eighth of a lap of the start line, which is within one-eighth of their lap time (in seconds) of a moment when they pass the line. Direction of travel does not matter.

Work in seconds; the picture is taken at a uniformly random time in $[600, 660]$.

Rachel passes the start every $90$ seconds, in particular at $630$ (seven laps). One-eighth of $90$ is $11.25$, so she is in the picture for $t \in [618.75, 641.25]$. Her other passings ($540$, $720$) have windows outside $[600, 660]$.

Robert passes the start every $80$ seconds, in particular at $640$ (eight laps). One-eighth of $80$ is $10$, so he is in the picture for $t \in [630, 650]$. His neighboring windows ($560 \pm 10$, $720 \pm 10$) are outside the minute.

Both are in the picture on the overlap $[630, 641.25]$, of length $11.25$ seconds. The probability is
$$\frac{11.25}{60} = \frac{45/4}{60} = \frac{45}{240} = \frac{3}{16}.$$

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

A uniformly random time turns the question into lengths of intervals: find each runner's "visible" window, intersect, and divide by the total. Converting a fraction of the track into a fraction of the lap time is the key translation, and centering the windows on the nearest passing of the start line keeps the arithmetic small.

• Using a quarter lap on each side of the start line instead of one-eighth, which doubles each window.
• Assuming the two events are independent and multiplying the individual probabilities $\frac{22.5}{60} \cdot \frac{20}{60} = \frac18$; the windows are fixed in time, so intersect them instead.