Difference between revisions of "2022 AIME I Problems/Problem 5"
MRENTHUSIASM (talk | contribs) m (→Solution 4 (Vectors)) |
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label("$B$", B, N); | label("$B$", B, N); | ||
− | label(" | + | label("Downstream", (350,325), E); |
− | label(" | + | label("Upstream", (200,325), W); |
draw((225,325)--(325,325), Arrows); | draw((225,325)--(325,325), Arrows); | ||
</asy> | </asy> | ||
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Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white)); | Label L3 = Label("Current $(14)$", position=EndPoint, filltype=Fill(3,0,white)); | ||
Label L4 = Label("$a$", align=(-1,0), position=Relative(0.4)); | Label L4 = Label("$a$", align=(-1,0), position=Relative(0.4)); | ||
− | Label L5 = Label("$b | + | Label L5 = Label("$b+14$", align=(0,1), position=Relative(0.4)); |
Label L6 = Label("$a$", align=(1,0), position=Relative(0.4)); | Label L6 = Label("$a$", align=(1,0), position=Relative(0.4)); | ||
− | Label L7 = Label("$c | + | Label L7 = Label("$c-14$", align=(0,1), position=Relative(0.4)); |
draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); | draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); | ||
draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); | draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); | ||
Line 158: | Line 158: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
a^2 + c^2 &= 80^2, \\ | a^2 + c^2 &= 80^2, \\ | ||
− | a^2 + ( | + | a^2 + (c-28)^2 &= 60^2. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | We can solve the system of equations to get <math>a = 48</math> and <math> | + | We can solve the system of equations to get <math>a = 48</math> and <math>c = 64</math>. From this, we can figure out that it must've taken them <math>5.5</math> minutes to get to the other side, because <math>264/48 = 5.5</math>. This means that there are <math>5.5</math> lengths of <math>48</math> in each person's travel. Also, <math>D</math> must be equal to <math>11(b+14) = 11(c-14) </math> because there are <math>(5.5)2 = 11</math> lengths of <math>b-14</math> between them, <math>5.5</math> on each person's side. Since <math>c = 64</math>, we have <math>c-14 = 50</math>, so the answer is <cmath>D=11\cdot50=\boxed{550}.</cmath> |
~Curious_crow | ~Curious_crow | ||
− | == Video Solution == | + | == Video Solution 1 == |
https://youtu.be/MJ_M-xvwHLk?t=1487 | https://youtu.be/MJ_M-xvwHLk?t=1487 | ||
Line 174: | Line 174: | ||
~Chickenugget | ~Chickenugget | ||
+ | |||
+ | == Video Solution 3 == | ||
+ | |||
+ | https://youtu.be/XAe8AkmHexw | ||
+ | |||
+ | ~AMC & AIME Training | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=4|num-a=6}} | {{AIME box|year=2022|n=I|num-b=4|num-a=6}} |
Latest revision as of 14:35, 23 February 2023
Contents
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 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
So .
~ihatemath123
Solution 2 (Euclidean)
Claim
Median and altitude
are drawn in triangle
.
are known. Let's denote
.
Prove that
Proof
Solution
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,
vladimir.shelomovskii@gmail.com, 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
Video Solution 1
https://youtu.be/MJ_M-xvwHLk?t=1487
~ThePuzzlr
Video Solution 2
https://www.youtube.com/watch?v=s6nXXnBLTdA
~Chickenugget
Video Solution 3
~AMC & AIME Training
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |