Question

A car travels due east at $\frac{2}{3}$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac{1}{2}\sqrt{2}$ 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 $\frac{1}{2}(t_1 + t_2).$

答案

Solution:

Put the car at the origin at $t = 0,$ with east as the positive $x$ -direction and north as the positive $y$ -direction. At time $t$ the car is at $\left(\frac{2t}{3}, 0\right),$ and the storm center, moving southeast at speed $\frac{\sqrt{2}}{2}$ (components $\frac{1}{2}$ east and $\frac{1}{2}$ south), is at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right).$

The car is on the storm boundary when the squared distance is $51^2:$ $\left(\frac{2t}{3} - \frac{t}{2}\right)^2 + \left(110 - \frac{t}{2}\right)^2 = 51^2,$ that is $\frac{t^2}{36} + \frac{t^2}{4} - 110t + 12100 - 2601 = 0,$ or $\frac{5}{18}t^2 - 110t + 9499 = 0.$

The roots are $t_1$ and $t_2,$ so by Vieta's formulas $t_1 + t_2 = \frac{110 \cdot 18}{5} = 396,$ and $\frac{1}{2}(t_1 + t_2) = 198.$

1997 AIME Problem 7

Solution:

Problem 7 in Other Years