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2022 AIME I Problems/Problem 5

Revision as of 19:14, 17 February 2022 by MRENTHUSIASM (talk | contribs)

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.

Let $x$ and $y$ be some positive numbers. We have the following table: \[\begin{array}{c||c|c|c} & \textbf{Net Velocity Vector (m/min)} & \textbf{Natural Velocity Vector (m/min)} & \textbf{Natural Speed (m/min)} \\ \hline \hline &&& \\ [-2.25ex] \textbf{Melanie} & \langle -x,y\rangle & \langle -x-14,y\rangle & 80 \\ \hline &&& \\ [-2.25ex] \textbf{Sherry} & \langle x,y\rangle & \langle x-14,y\rangle & 60 \end{array}\] Recall that $|\text{velocity}|=\text{speed}.$ It follows that \begin{align*} (-x-14)^2 + y^2 &= 80^2, \\ (x-14)^2 + y^2 &= 60^2. \end{align*} ~MRENTHUSIASM

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions
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