Difference between revisions of "1997 AIME Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | {{ | + | 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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| + | <cmath>\begin{eqnarray*} | ||
| + | \sqrt{\left(\frac{2}{3}t - \frac 12t\right)^2 + \left(110-\frac{t}{2}\right)^2} &\le& 51\\ | ||
| + | \frac{t^2}{36} + \frac{t^2}{4} - 110t + 110^2 le& 51^2\\ | ||
| + | \frac{5}{18}t^2 - 110t + 110^2 - 51^2 &\le& 0\\ | ||
| + | \end{eqnarray*}</cmath> | ||
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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>. | ||
== See also == | == See also == | ||
{{AIME box|year=1997|num-b=6|num-a=8}} | {{AIME box|year=1997|num-b=6|num-a=8}} | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 21:56, 21 November 2007
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
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 
.
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 | ||
