2022 AIME I Problems/Problem 5
A straight river that is meters wide flows from west to east at a rate of meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of meters downstream from Sherry. Relative to the water, Melanie swims at meters per minute, and Sherry swims at 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 .
Solution 1 (Euclidean)
Define as the number of minutes they swim for.
Let their meeting point be . Melanie is swimming against the current, so she must aim upstream from point , to compensate for this; in particular, since she is swimming for minutes, the current will push her meters downstream in that time, so she must aim for a point that is meters upstream from point . Similarly, Sherry is swimming downstream for minutes, so she must also aim at point to compensate for the flow of the current.
If Melanie and Sherry were to both aim at point in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are and meters. We can draw out this new scenario, with the dimensions that we have: (While it is indeed true that the triangle above with side lengths , and is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)
By the Pythagorean Theorem, we have
Subtracting the first equation from the second gives us , so . Substituting this into our first equation, we have that
Solution 2 (Euclidean)
Median and altitude are drawn in triangle . are known. Let's denote .
In the coordinate system associated with water, the movement is described by the scheme in the form of a triangle, the side on which Melanie floats is , where t is the time of Melanie's movement, the side along which Sherry floats is .
The meeting point floated away at a distance of from the midpoint between the starting points of Melanie and Sherry.
In the notation of the Claim, Hence, firstname.lastname@example.org, vvsss
Solution 3 (Vectors)
We have the following diagram: Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.
Let and be some positive numbers. We have the following table: Recall that so We subtract from to get from which Substituting this into either equation, we have
It follows that Melanie and Sherry both swim for minutes. Therefore, the answer is ~MRENTHUSIASM
Solution 4 (Vectors)
We can break down movement into two components: the -component and the -component. Suppose that Melanie travels a distance of in the -direction and a distance of in the -direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of in the -direction but a distance of in the -direction in one minute when there is no current. The current only affects the -components because it goes in the -direction.
Now, note that because Sherry travels 60 meters in a minute. Thus, because Melanie travels 80 meters in a minute. Also, the distance they travel with the current must be the same in one minute because they reach the point equidistant from them at the same time. That means or . So now we can plug that into the two equations to get: We can solve the system of equations to get and . From this, we can figure out that it must've taken them minutes to get to the other side, because . This means that there are lengths of in each person's travel. Also, must be equal to because there are lengths of between them, on each person's side. Since , we have , so the answer is ~Curious_crow
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~AMC & AIME Training
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