2022 AIME I Problems/Problem 5

Problem

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 $D$ .

Solution 1

Define $m$ as the number of minutes they swam for.

Let their meeting point be $A$ . In an alternative reality, there would be no current. Then, had they swum facing the same direction that they had in the real universe, they would've met at a point west of $A$ . Precisely, since the water moves at $14$ meters per minute, this alternative reality meeting point would have been $14m$ meters to the left of $A$ .

So, our alternative reality is just a geometry problem now: $[asy] unitsize(0.02cm); draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); pair B = (198,264); dot(B); draw((0,0)--B,dashed); draw((550,0)--B,dashed); label("$60m$", (0,0)--B, E); label("$80m$", (550,0)--B, W); label("$264$", (0,0)--(0,264), W); label("$\frac{D}{2} - 14m$", (0,264)--B, N); label("$\frac{D}{2} + 14m$", B--(550,264), N); label("$D$", (0,0)--(550,0), S); [/asy]$ Note that while this diagram was drawn knowing the correct dimensions, we do not actually know that the triangle with sides $60m$ , $80m$ and $D$ is a right triangle yet, so we cannot use that information.

By Pythagorean, we have $\begin{align*} 264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2} \\ 264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}. \end{align*}$

Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that $\begin{align*}264^{2} + 36^{2} m^{2} &= 60m^{2} \\ 264^{2} &= 96 \cdot 24 \cdot m^{2} \\ 11^{2} &= 4 \cdot m^{2} \\ m &= \frac{11}{2}. \end{align*}$

So $D = 100m = \boxed{550}$ .

~ ihatemath123

Solution 2

We have the following diagram:

WILL BE AVAILABLE WITHIN THE NEXT DAY.

Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2022 AIME I Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See Also