Difference between revisions of "1997 AIME Problems/Problem 7"
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A car travels due east at <math>\frac 23</math> mile per minute on a long, straight road. At the same time, a circular storm, whose radius is <math>51</math> miles, moves southeast at <math>\frac 12\sqrt{2}</math> mile per minute. At time <math>t=0</math>, the center of the storm is <math>110</math> miles due north of the car. At time <math>t=t_1</math> minutes, the car enters the storm circle, and at time <math>t=t_2</math> minutes, the car leaves the storm circle. Find <math>\frac 12(t_1+t_2)</math>. | A car travels due east at <math>\frac 23</math> mile per minute on a long, straight road. At the same time, a circular storm, whose radius is <math>51</math> miles, moves southeast at <math>\frac 12\sqrt{2}</math> mile per minute. At time <math>t=0</math>, the center of the storm is <math>110</math> miles due north of the car. At time <math>t=t_1</math> minutes, the car enters the storm circle, and at time <math>t=t_2</math> minutes, the car leaves the storm circle. Find <math>\frac 12(t_1+t_2)</math>. | ||
| − | == Solution == | + | == Solution 1== |
We set up a coordinate system, with the starting point of the car at the [[origin]]. At time <math>t</math>, the car is at <math>\left(\frac 23t,0\right)</math> and the center of the storm is at <math>\left(\frac{t}{2}, 110 - \frac{t}{2}\right)</math>. Using the distance formula, | We set up a coordinate system, with the starting point of the car at the [[origin]]. At time <math>t</math>, the car is at <math>\left(\frac 23t,0\right)</math> and the center of the storm is at <math>\left(\frac{t}{2}, 110 - \frac{t}{2}\right)</math>. Using the distance formula, | ||
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Noting that <math>\frac 12(t_1+t_2)</math> is at the maximum point of the parabola, we can use <math>-\frac{b}{2a} = \frac{110}{2 \cdot \frac{5}{18}} = \boxed{198}</math>. | Noting that <math>\frac 12(t_1+t_2)</math> is at the maximum point of the parabola, we can use <math>-\frac{b}{2a} = \frac{110}{2 \cdot \frac{5}{18}} = \boxed{198}</math>. | ||
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| + | == Solution 2== | ||
| + | |||
| + | First do the same process for assigning coordinates to the car. The car moves <math>\frac{2}{3}</math> miles per minute to the right, so the position starting from <math>(0,0)</math> is <math>(\frac{2}{3}t, 0)</math>. | ||
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| + | Take the storm as circle. Given southeast movement, split the vector into component, getting position <math>(\frac{1}{2}t, 110 - \frac{1}{2}t)</math> for the storm's center. Getting radius 51 yields <math>(x - \frac{1}{2}t)^2 - (y -110 + \frac{1}{2}t)^2 = 51^2</math> | ||
== See also == | == See also == | ||
Revision as of 23:22, 2 June 2018
Contents
Problem
A car travels due east at 
mile per minute on a long, straight road. At the same time, a circular storm, whose radius is 
miles, moves southeast at 
mile per minute. At time 
, the center of the storm is 
miles due north of the car. At time 
minutes, the car enters the storm circle, and at time 
minutes, the car leaves the storm circle. Find 
.
Solution 1
We set up a coordinate system, with the starting point of the car at the origin. At time 
, the car is at 
and the center of the storm is at 
. Using the distance formula,
Noting that is at the maximum point of the parabola, we can use 
.
Solution 2
First do the same process for assigning coordinates to the car. The car moves 
miles per minute to the right, so the position starting from 
is 
.
Take the storm as circle. Given southeast movement, split the vector into component, getting position 
for the storm's center. Getting radius 51 yields 
See also
| 1997 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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