1997 AIME Problems/Problem 7

Problem

A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the storm circle, and at time $t=t_2$ minutes, the car leaves the storm circle. Find $\frac 12(t_1+t_2)$ .

Solution 1

We set up a coordinate system, with the starting point of the car at the origin. At time $t$ , the car is at $\left(\frac 23t,0\right)$ and the center of the storm is at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right)$ . Using the distance formula,

$\begin{eqnarray*} \sqrt{\left(\frac{2}{3}t - \frac 12t\right)^2 + \left(110-\frac{t}{2}\right)^2} &\le& 51\\ \frac{t^2}{36} + \frac{t^2}{4} - 110t + 110^2 &\le& 51^2\\ \frac{5}{18}t^2 - 110t + 110^2 - 51^2 &\le& 0\\ \end{eqnarray*}$

Noting that $\frac 12(t_1+t_2)$ is at the maximum point of the parabola, we can use $-\frac{b}{2a} = \frac{110}{2 \cdot \frac{5}{18}} = \boxed{198}$ .

Solution 2

First do the same process for assigning coordinates to the car. The car moves $\frac{2}{3}$ miles per minute to the right, so the position starting from $(0,0)$ is $(\frac{2}{3}t, 0)$ .

Take the storm as circle. Given southeast movement, split the vector into component, getting position $(\frac{1}{2}t, 110 - \frac{1}{2}t)$ for the storm's center. Getting radius 51 yields $(x - \frac{1}{2}t)^2 - (y -110 + \frac{1}{2}t)^2 = 51^2$

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1997 AIME Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See also