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2000 AIME I Problems/Problem 13

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Problem

In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at 50 miles per hour along the highways and at 14 miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is m/n square miles, where m and n are relatively prime positive integers. Find m + n.

Solution

Let the intersection of the highways be at the origin O, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.

After going x miles, t=\frac{d}{r}=\frac{x}{50} hours has passed. If the truck leaves the highway it can travel for at most t=\frac{1}{10}-\frac{x}{50} hours, or d=rt=14t=1.4-\frac{7x}{25} miles. It can end up anywhere off the highway in a circle with this radius centered at (x,0). All these circles are homothetic with respect to a center at (5,0).

pair truck(pair P){ pair Q = IP(P--P+(7/10,24/10),(35/31,35/31)--(5,0));  D(P--Q,EndArrow(5)); D(CP(P,Q),linewidth(0.5)); ret...     pointpen = black; pathpen = black+linewidth(0.7); size(250);pair O=(0,0), B=(5,0), A=1.4*expi(atan(24/7)), C=1.4*expi(atan(7/...

Now consider the circle at (0,0). Draw a line tangent to it at A and passing through B (5,0). By the Pythagorean Theorem AB^2+AO^2=OB^2 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}. Then \tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}, so the slope of line AB is \frac{-7}{24}. Since it passes through (5,0) its equation is y=\frac{-7}{24}(x-5).

This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line y=5-\frac{24}{7}x bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is \left(\frac{35}{31},\frac{35}{31}\right). The bounded region in Quadrant I is made up of a square and two triangles. A=x^2+x(5-x)=5x. By symmetry, the regions in the other quadrants are the same, so the area of the whole region is 20x=\frac{700}{31} so the answer is 700+31=\boxed{731}.

See also

2000 AIME I (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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