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

Revision as of 18:35, 4 July 2013

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

Revision as of 02:46, 29 June 2011 (view source) Neutrinonerd3333 (talk \| contribs) m (→‎Solution) ← Older edit		Revision as of 18:35, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
Line 17:		Line 17:

	[[Category:Intermediate Geometry Problems]]		[[Category:Intermediate Geometry Problems]]
		+	{{MAA Notice}}

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

Page

Toolbox

Search

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

Revision as of 18:35, 4 July 2013

Problem

Solution

See also