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Difference between revisions of "1997 AIME Problems/Problem 7"
(→‎Solution 2)
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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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== Solution 2(more formal explanations for Solution 1)==
  
 
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>.  
 
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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NOTE: This is literally the exact same as Solution 1...
 
NOTE: This is literally the exact same as Solution 1...
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== See also ==
 
== See also ==

Revision as of 18:47, 4 August 2021

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:

$-\frac{b}{a} = -\frac{-110}{\frac{5}{18}} = 110\cdot{\frac{18}{5}} = 396\cdot{\frac{1}{2}} = \boxed{198}$

NOTE: This is literally the exact same as Solution 1...

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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