# 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 (more formal explanations for Solution 1)

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. This circle with radius 51 yields $(x - \frac{1}{2}t)^2 + (y -110 + \frac{1}{2}t)^2 = 51^2$.

Now substitute the car's coordinates into the circle's: $(\frac{2}{3}t - \frac{1}{2}t)^2 + (-110 + \frac{1}{2}t)^2 = 51^2$.

Simplifying and then squaring: $(\frac{1}{6}t)^2 + (-110 + \frac{1}{2}t)^2 = 51^2$. $\frac{1}{36}t^2 + \frac{1}{4}t^2 - 110t + 110^2$

Forming into a quadratic we get the following, then set equal to 0, since the first time the car hits the circumference of the storm is $t_{1}$ and the second is $t_{2}$. $\frac{5}{18}t^2 - 110t + 110^2 - 51^2 = 0$

The problem asks for sum of solutions divided by 2 so sum is equal to: $-\frac{b}{a} = -\frac{-110}{\frac{5}{18}} = 110\cdot{\frac{18}{5}} = 396\cdot{\frac{1}{2}} = \boxed{198}$

## Solution 3 (No Coordinates)

We only need to know how the storm and car move relative to each other, so we can find this by subtracting the storm's movement vector from the car's. This gives the car's movement vector as $\left(\frac{1}{6}, \frac{1}{2}\right)$. Labeling the car's starting position A, the storm center B, and the right triangle formed by AB with a right angle at B and the car's path, we get the following diagram, with AD as our desired length since D is the average of the points where the car enters and exits the storm. $[asy] size(200,200); draw((0,0)--(0,110)); label("A",(0,0),S); dot((0,0)); dot((0,110)); label("B",(0,110),NE); draw(circle((0,110),51)); draw((0,0)--(161/3,161.0),EndArrow); draw((0,110)--(110/3,110.0)); label("C",(110/3,110.0),SE); dot((110/3,110.0)); label("D",(33,99),SE); dot((33,99)); draw((0,110)--(33,99)); markscalefactor=1; draw(rightanglemark((0,110),(33,99),(0,0))); [/asy]$ $AB = 110$, so $CB = \frac{110}{3}$. The Pythagorean Theorem then gives $AC = \frac{110\sqrt{10}}{3}$, and since $\bigtriangleup ABC \sim \bigtriangleup ADB$, $AD = (AB)\frac{AB}{AC} = 33\sqrt{10}$. The Pythagorean Theorem now gives the car's speed as $\sqrt{\frac{5}{18}}$, and finally $\frac{33\sqrt{10}}{\sqrt{\frac{5}{18}}} = \boxed{198}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 