Difference between revisions of "2022 AIME I Problems/Problem 5"
Ihatemath123 (talk | contribs) |
Ihatemath123 (talk | contribs) |
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So, our alternative reality is just a geometry problem now: | So, our alternative reality is just a geometry problem now: | ||
| − | + | <asy> | |
unitsize(0.02cm); | unitsize(0.02cm); | ||
draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); | draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); | ||
| Line 16: | Line 16: | ||
draw((550,0)--B,dashed); | draw((550,0)--B,dashed); | ||
| − | label(" | + | label("$60m$", (0,0)--B, E); |
| − | label(" | + | label("$80m$", (550,0)--B, W); |
| − | label(" | + | label("$264$", (0,0)--(0,264), W); |
| − | label(" | + | label("$\frac{D}{2} - 14m$", (0,264)--B, N); |
| − | label(" | + | label("$\frac{D}{2} + 14m$", B--(550,264), N); |
| − | label(" | + | 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 <math>60m</math>, <math>80m</math> and <math>D</math> is a right triangle yet, so we cannot use that information. | 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. | ||
Revision as of 16:01, 17 February 2022
Problem
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
Define 
as the number of minutes they swam for.
Let their meeting point be 
. 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 . Precisely, since the water moves at meters per minute, this alternative reality meeting point would have been 
meters to the left of .
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]](http://latex.artofproblemsolving.com/1/4/f/14fbef7b98d21fa4a9b30317f64f1acb393c04ee.png)
Note that while this diagram was drawn knowing the correct dimensions, we do not actually know that the triangle with sides 
, 
and 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 
, so 
. 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 
.
~ ihatemath123
