2000 AIME I Problem 13

Attempt Problem 13 of the 2000 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AIME I solutions, or check the answer key.

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13.

In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at 5050 miles per hour along the highways and at 1414 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 mn\frac{m}{n} square miles, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 731
Concepts:coordinate geometrytangent lineareaoptimization
Difficulty rating: 3160
Solution:

In six minutes the truck can drive 55 miles on a highway or 1.41.4 miles across the prairie, and an optimal route is a highway stretch followed by a straight prairie segment. Work in the first quadrant with the highways as axes. Driving to (d,0)(d, 0) takes d50\frac{d}{50} hours, leaving a prairie range of 14(110d50)=1.4(1d5)14\left(\frac{1}{10} - \frac{d}{50}\right) = 1.4\left(1 - \frac{d}{5}\right) miles. As dd runs from 00 to 5,5, these disks shrink linearly to a point, so their union is the "cone": the convex hull of the disk of radius 1.41.4 about the origin and the point (5,0),(5, 0), bounded by the tangent line from (5,0).(5, 0). The tangent length is 521.42=4.8,\sqrt{5^2 - 1.4^2} = 4.8, so the ratios are 7724242525 and the tangent line is 7x+24y=35.7x + 24y = 35. The yy-axis gives the mirror-image region bounded by 24x+7y=35.24x + 7y = 35.

The two tangent lines meet at P=(3531,3531),P = \left(\frac{35}{31}, \frac{35}{31}\right), which lies at distance 35231>1.4\frac{35\sqrt{2}}{31} \gt 1.4 from the origin — outside the circle — so in the first quadrant the reachable set is exactly the (non-convex) quadrilateral with vertices (0,0),(0, 0), (5,0),(5, 0), P,P, (0,5).(0, 5). Splitting it along the diagonal from the origin to PP gives two triangles, each with area 1253531,\frac{1}{2} \cdot 5 \cdot \frac{35}{31}, for a quadrant area of 17531.\frac{175}{31}.

The full region is four copies, with area 70031\frac{700}{31} square miles. Since gcd(700,31)=1,\gcd(700, 31) = 1, the answer is 700+31=731.700 + 31 = 731.

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