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

Place the intersection of the highways at the origin $O$ and let the highways be the x and y axis. We consider the case where the truck moves in +x. 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 circle are homothetic with center at $(5,0)$. 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 $AB^2+AO^2=OB^2$ so $AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}$. $tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$. The slope of line $AB$ is therefore $\frac{-7}{24}$. Since it passes through $(5,0)$ its equation is $y=\frac{-7}{24}(x-5)$. The line and the x and y axis bound the region the truck can go if it moves in +x. Similarly, the line $y=5-\frac{24}{7}x$ bounds the region the truck can go if it moves in +y. The intersection of these 2 lines is $(\frac{35}{31},\frac{35}{31})$. The bounded region in Quadrant I is made up of a square and 2 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 $700+31=\boxed{731}$.

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See also

2000 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions