Difference between revisions of "2000 AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | Place the intersection of the highways at the origin <math>O</math> and let the highways be the x and y axis. We consider the case where the truck moves in +x. After going x miles, <math>t=\frac{d}{r}=\frac{x}{50}</math> hours has passed. If the truck leaves the highway it can travel for at most <math>t=\frac{1}{10}-\frac{x}{50}</math> hours, or <math>d=rt=14t=1.4-\frac{7x}{25}</math> miles. It can end up anywhere off the highway in a circle with this radius centered at <math>(x,0)</math>. All these circle are homothetic with center at <math>(5,0)</math>. Now consider the circle at (0,0). Draw a line tangent to it at <math>A</math> and passing through <math>B (5,0)</math>. By the Pythagorean Theorem <math>AB^2+AO^2=OB^2</math> so <math>AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}</math>. <math>tan( | + | Place the intersection of the highways at the origin <math>O</math> and let the highways be the x and y axis. We consider the case where the truck moves in +x. After going x miles, <math>t=\frac{d}{r}=\frac{x}{50}</math> hours has passed. If the truck leaves the highway it can travel for at most <math>t=\frac{1}{10}-\frac{x}{50}</math> hours, or <math>d=rt=14t=1.4-\frac{7x}{25}</math> miles. It can end up anywhere off the highway in a circle with this radius centered at <math>(x,0)</math>. All these circle are homothetic with center at <math>(5,0)</math>. Now consider the circle at (0,0). Draw a line tangent to it at <math>A</math> and passing through <math>B (5,0)</math>. By the Pythagorean Theorem <math>AB^2+AO^2=OB^2</math> so <math>AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}</math>. <math>tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}</math>. The slope of line <math>AB</math> is therefore <math>\frac{-7}{24}</math>. Since it passes through <math>(5,0)</math> its equation is <math>y=\frac{-7}{24}(x-5)</math>. The line and the x and y axis bound the region the truck can go if it moves in +x. Similarly, the line <math>y=5-\frac{24}{7}x</math> bounds the region the truck can go if it moves in +y. The intersection of these 2 lines is <math>(\frac{35}{31},\frac{35}{31})</math>. The bounded region in Quadrant I is made up of a square and 2 triangles. <math>A=x^2+x(5-x)=5x</math>. By symmetry, the regions in the other quadrants are the same, so the area of the whole region is <math>20x=\frac{700}{31}</math> so the answer is <math>700+31=\boxed{731}</math>. |
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| − | The area of the region is <math>\frac{700}{31}</math> so the answer is <math>700+31=\boxed{731}</math>. | ||
{{incomplete|solution}} | {{incomplete|solution}} | ||
Revision as of 20:56, 28 March 2008
Problem
In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at 
miles per hour along the highways and at 
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 
square miles, where 
and 
are relatively prime positive integers. Find 
.
Solution
Place the intersection of the highways at the origin 
and let the highways be the x and y axis. We consider the case where the truck moves in +x. After going x miles, 
hours has passed. If the truck leaves the highway it can travel for at most 
hours, or 
miles. It can end up anywhere off the highway in a circle with this radius centered at 
. All these circle are homothetic with center at 
. Now consider the circle at (0,0). Draw a line tangent to it at 
and passing through 
. By the Pythagorean Theorem 
so 
. 
. The slope of line 
is therefore 
. Since it passes through its equation is 
. The line and the x and y axis bound the region the truck can go if it moves in +x. Similarly, the line 
bounds the region the truck can go if it moves in +y. The intersection of these 2 lines is 
. The bounded region in Quadrant I is made up of a square and 2 triangles. 
. By symmetry, the regions in the other quadrants are the same, so the area of the whole region is 
so the answer is 
.
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
