Difference between revisions of "2000 AIME I Problems/Problem 13"
(→Solution) |
(cleanup and asymptote) |
||
Line 1: | Line 1: | ||
== 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 == | ||
− | + | 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. | |
+ | |||
+ | 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 [[homethecy|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)); | ||
+ | |||
+ | </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></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]] |
Revision as of 14:42, 16 June 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
Let the intersection of the highways be at the origin , 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 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 circles are homothetic with respect to a center at
.
![[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]](http://latex.artofproblemsolving.com/b/1/e/b1e4a60f3081861da720715755129e0abc341cb9.png)
![[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]](http://latex.artofproblemsolving.com/b/1/d/b1dbbfe9154757443da1f35c5fce02513e7afdd6.png)
Now consider the circle at . Draw a line tangent to it at
and passing through
. By the Pythagorean Theorem
. Then
, so the slope of line
is
. Since it passes through
its equation is
.
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 bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is
. The bounded region in Quadrant I is made up of a square and two 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 |