Difference between revisions of "2018 AIME II Problems/Problem 1"
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Since Paul and Ina reach <math>B</math> at the same time, we know that <math>\frac{2(1800-x)}{5v} = \frac{x}{2v}</math> (notice that <math>v</math> cancels out on both sides). Solving the equation gives <math>x = \boxed{800}</math>. | Since Paul and Ina reach <math>B</math> at the same time, we know that <math>\frac{2(1800-x)}{5v} = \frac{x}{2v}</math> (notice that <math>v</math> cancels out on both sides). Solving the equation gives <math>x = \boxed{800}</math>. | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/XixU0JZ5FLk?t=1355 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2018|n=II|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 03:25, 16 January 2023
Problem
Points ,
, and
lie in that order along a straight path where the distance from
to
is
meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at
and running toward
, Paul starting at
and running toward
, and Eve starting at
and running toward
. When Paul meets Eve, he turns around and runs toward
. Paul and Ina both arrive at
at the same time. Find the number of meters from
to
.
Solution 1
We know that in the same amount of time given, Ina will run twice the distance of Eve, and Paul would run quadruple the distance of Eve. Let's consider the time it takes for Paul to meet Eve: Paul would've run 4 times the distance of Eve, which we can denote as . Thus, the distance between
and
is
. In that given time, Ina would've run twice the distance
, or
units.
Now, when Paul starts running back towards , the same amount of time would pass since he will meet Ina at his starting point. Thus, we know that he travels another
units and Ina travels another
units.
Therefore, drawing out the diagram, we find that , and distance between
and
is the distance Ina traveled, or
Solution 2
Let be the distance from
to
. Then the distance from
to
is
. Since Eve is the slowest, we can call her speed
, so that Ina's speed is
and Paul's speed is
.
For Paul and Eve to meet, they must cover a total distance of which takes them a time of
. Paul must run the same distance back to
, so his total time is
.
For Ina to reach , she must run a distance of
at a speed of
, taking her a time of
.
Since Paul and Ina reach at the same time, we know that
(notice that
cancels out on both sides). Solving the equation gives
.
Video Solution by OmegaLearn
https://youtu.be/XixU0JZ5FLk?t=1355
~ pi_is_3.14
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.