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Difference between revisions of "2000 AIME I Problems/Problem 13"
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== Problem ==
 
== Problem ==
−
In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at <math>50</math> miles per hour along the highways and at <math>14</math> 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 <math>m/n</math> square miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+
In the middle of a vast prairie, a firetruck is stationed at the intersection of two [[perpendicular]] straight highways. The truck travels at <math>50</math> miles per hour along the highways and at <math>14</math> 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 <math>m/n</math> square miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
 
== 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(<ABO)=\frac{OA}{AB}=\frac{7}{24}</math>. The slope of line <math>AB</math> is therefore <math>\frac{7}{24}</math>.
+
Let the intersection of the highways be at the origin <math>O</math>, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.  
  
−
The area of the region is <math>\frac{700}{31}</math> so the answer is <math>700+31=\boxed{731}</math>.
+
After going <math>x</math> 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 circles are [[homothety|homothetic]] with respect to a center at <math>(5,0)</math>.
 +
<center><asy>
 +
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));
 +
return Q;
 +
}
 +
pointpen = black; pathpen = black+linewidth(0.7); size(250);
 +
pair B=(5,0), C=(35/31,35/31);
 +
D(D(B)--D(C)--D(B*dir(90))--D(C*dir(90))--D(B*dir(180))--D(C*dir(180))--D(B*dir(270))--D(C*dir(270))--cycle);
 +
D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); truck((1,0)); truck((2,0)); truck((3,0)); truck((4,0));
  
−
{{incomplete|solution}}
+
</asy>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asy>
 +
 
 +
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(7/24));
 +
D(D(B)--D(A)--D(O)); D(O--D(C)--D(B*dir(90))--D(A*dir(90))--O--D(C*dir(90))--D(B*dir(180))--D(A*dir(180))--O--D(C*dir(180))--D(B*dir(270))--D(A*dir(270))--O--D(C*dir(270))--B,linewidth(0.5)); D(CR(O,1.4));
 +
D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4));
 +
MP("A",A,N); MP("B",B); MP("(5,0)",B,N); D(MP("\left(\frac{35}{31},\frac{35}{31}\right)",(35/31,35/31),NE));
 +
D(rightanglemark(O,A,B));
 +
</asy></center>
 +
Now consider the circle at <math>(0,0)</math>. 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 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}</math>. Then <math>\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}</math>, so the [[slope]] of line <math>AB</math> is <math>\frac{-7}{24}</math>. Since it passes through <math>(5,0)</math> its equation is <math>y=\frac{-7}{24}(x-5)</math>.
 +
 
 +
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 <math>y=5-\frac{24}{7}x</math> bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is <math>\left(\frac{35}{31},\frac{35}{31}\right)</math>. The bounded region in Quadrant I is made up of a square and two 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>.
 +
 
 +
 
 +
== Intuition ==
 +
 
 +
Here's a bit of intuition behind the solution. If I wanted to travel to some point on the prairie, normally I would go the straight line distance. Right? However, here I would choose to travel on the highway the greatest distance, then drive the rest on the prairie. This is why we can easily use parametric equations to solve for the equations of lines that would yield those distances.
 +
 
 +
 
 +
-WhatdoHumanitariansEat
  
 
== 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|t=721508}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:25, 28 November 2024

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 $(5,0)$.

[asy] 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)); return Q; } pointpen = black; pathpen = black+linewidth(0.7); size(250); pair B=(5,0), C=(35/31,35/31); D(D(B)--D(C)--D(B*dir(90))--D(C*dir(90))--D(B*dir(180))--D(C*dir(180))--D(B*dir(270))--D(C*dir(270))--cycle); D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); truck((1,0)); truck((2,0)); truck((3,0)); truck((4,0)); [/asy]     [asy] 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(7/24)); D(D(B)--D(A)--D(O)); D(O--D(C)--D(B*dir(90))--D(A*dir(90))--O--D(C*dir(90))--D(B*dir(180))--D(A*dir(180))--O--D(C*dir(180))--D(B*dir(270))--D(A*dir(270))--O--D(C*dir(270))--B,linewidth(0.5)); D(CR(O,1.4)); D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); MP("A",A,N); MP("B",B); MP("(5,0)",B,N); D(MP("\left(\frac{35}{31},\frac{35}{31}\right)",(35/31,35/31),NE)); D(rightanglemark(O,A,B)); [/asy]

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 $700+31=\boxed{731}$.


Intuition

Here's a bit of intuition behind the solution. If I wanted to travel to some point on the prairie, normally I would go the straight line distance. Right? However, here I would choose to travel on the highway the greatest distance, then drive the rest on the prairie. This is why we can easily use parametric equations to solve for the equations of lines that would yield those distances.


-WhatdoHumanitariansEat

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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