Question

A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D.$

Answer

Solution:

Put Sherry at the origin and Melanie at $(D, 0)$ on the south bank. A point on the north bank equidistant from both is $\left(\frac{D}{2}, 264\right).$ If both arrive at time $t,$ then each swimmer's velocity relative to the water is her ground velocity minus the current $(14, 0),$ so $\left(\frac{D}{2t} - 14\right)^2 + \left(\frac{264}{t}\right)^2 = 60^2, \qquad \left(-\frac{D}{2t} - 14\right)^2 + \left(\frac{264}{t}\right)^2 = 80^2.$

Subtracting, with $u = \frac{D}{2t}:$ $(u + 14)^2 - (u - 14)^2 = 56u = 6400 - 3600 = 2800,$ so $u = 50.$ Substituting back, $(50 - 14)^2 + \left(\frac{264}{t}\right)^2 = 3600$ gives $\frac{264}{t} = 48,$ so $t = \frac{11}{2}.$

Therefore $D = 2ut = 100t = 550.$

2022 AIME I Problem 5

Solution:

Problem 5 in Other Years