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Difference between revisions of "2022 AIME I Problems/Problem 5"

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== Problem ==
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A straight river that is <math>264</math> meters wide flows from west to east at a rate of <math>14</math> meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of <math>D</math> meters downstream from Sherry. Relative to the water, Melanie swims at <math>80</math> meters per minute, and Sherry swims at <math>60</math> 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 <math>D</math>.
  
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== Solution ==
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Define <math>m</math> as the number of minutes they swam for.
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Let their meeting point be <math>A</math>. 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 <math>A</math>. Precisely, since the water moves at <math>14</math> meters per minute, this alternative reality meeting point would have been <math>14m</math> meters to the left of <math>A</math>.
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So, our alternative reality is just a geometry problem now:
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[asy]
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unitsize(0.02cm);
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draw((0,0)--(0,264)--(550,264)--(550,0)--cycle);
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pair B = (198,264);
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dot(B);
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draw((0,0)--B,dashed);
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draw((550,0)--B,dashed);
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label("<math>60m</math>", (0,0)--B, E);
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label("<math>80m</math>", (550,0)--B, W);
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label("<math>264</math>", (0,0)--(0,264), W);
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label("<math>\frac{D}{2} - 14m</math>", (0,264)--B, N);
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label("<math>\frac{D}{2} + 14m</math>", B--(550,264), N);
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label("<math>D</math>", (0,0)--(550,0), S);
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[/asy]
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(I skipped many steps here, but it should be somewhat self-explanatory where all these numbers came from)
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Note that while this diagram was drawn knowing the correct dimensions, we do not actually know that the triangle with sides <math>60m</math>, <math>80m</math> and <math>D</math> is a right triangle yet, so we cannot use that information.
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By Pythagorean, we have
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\begin{align*}
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264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2} \\
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264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}.
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\end{align*}Subtracting the first equation from the second gives us <math>28Dm = 2800m^{2}</math>, so <math>D = 100m</math>. Substituting this into our first equation, we have that
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\begin{align*}264^{2} + 36^{2} m^{2} &= 60m^{2} \\
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264^{2} &= 96 \cdot 24 \cdot m^{2} \\
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11^{2} &= 4 \cdot m^{2} \\
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m &= \frac{11}{2}.
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\end{align*}So <math>D = 100m = \boxed{550}</math>.
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 +
~ ihatemath123

Revision as of 17:01, 17 February 2022

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

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] (I skipped many steps here, but it should be somewhat self-explanatory where all these numbers came from) 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

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