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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 not <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>.
  
 
== 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 23:38, 12 March 2014

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

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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