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== | ||
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+ | 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
[hide]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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