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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(\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>.
 
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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{{incomplete|solution}}
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2000|n=I|num-b=12|num-a=14}}

Revision as of 08:55, 2 April 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 $50$ 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 $700+31=\boxed{731}$.

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
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