Problem:
Al walks down to the bottom of an escalator that is moving up and he counts steps. His friend, Bob, walks up to the top of the escalator and counts steps. If Al's speed of walking (in steps per unit time) is three times Bob's speed, how many steps are visible on the escalator at any given time? (Assume that this number is constant.)
Solution:
Let denote Al's speed (in steps per unit time) and his time. Similarly, let and denote Bob's speed and time. Moreover, let be the speed of the escalator, and let be the number of steps visible at any given time. Then, from the information given,
From it follows that
We also know that , from which we have , and hence
Therefore, from and , upon setting their right sides equal, we find that .
Assume that Al and Bob start at the same time from their respective ends of the escalator. Then the number of steps initially separating them is the same as the number of visible steps on the escalator.
Hence, to solve the problem, we must find the number of steps each of them takes until they meet, and then add these two numbers.
Since Al can take steps while Bob takes steps, it follows (from ) that Al walks down the escalator in of the time it takes Bob to walk up. Therefore, they meet of the way from the bottom of the escalator. To that point, Al takes steps, while Bob takes steps. As indicated above, the sum of these, , is the number of visible steps of the escalator.
The problems on this page are the property of the MAA's American Mathematics Competitions