Difference between revisions of "1997 AIME Problems/Problem 7"

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\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
  
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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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 ==

Revision as of 02:46, 29 June 2011

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

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

See also

1997 AIME (ProblemsAnswer KeyResources)
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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