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Difference between revisions of "2000 AIME I Problems/Problem 2"

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== Solution ==
 
== Solution ==
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[[Image:2000_I_AIME-2.png]]
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<center><asy>
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pointpen = black; pathpen = linewidth(0.7) + black; size(180);
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pair A=(11,10), B=(10,11), C=(-10, 11), D=(-10, -11), E=(10, -11);
 +
D(D(MP("A\ (u,v)",A,(1,0)))--D(MP("B",B,N))--D(MP("C",C,N))--D(MP("D",D))--D(MP("E",E))--cycle);
 +
D((-15,0)--(15,0),linewidth(0.6),Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5));
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</asy></center> <!-- Asymptote replacement for Image:2000_I_AIME-2.png -->
  
 
Since <math>A = (u,v)</math>, we can find the coordinates of the other points: <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and <math>ABE</math> is a triangle. The area of <math>BCDE</math> is <math>(2u)(2v) = 4uv</math> and the area of <math>ABE</math> is <math>\frac{1}{2}(2u)(u-v) = u^2 - uv</math>. Adding these together, we get <math>u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41</math>. Since <math>u,v</math> are positive, <math>u+3v>u</math>, and by matching factors we get either <math>(u,v) = (1,90)</math> or <math>(11,10)</math>. Since <math>v < u</math> the latter case is the answer, and <math>u+v = \boxed{021}</math>.
 
Since <math>A = (u,v)</math>, we can find the coordinates of the other points: <math>B = (v,u)</math>, <math>C = (-v,u)</math>, <math>D = (-v,-u)</math>, <math>E = (v,-u)</math>. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and <math>ABE</math> is a triangle. The area of <math>BCDE</math> is <math>(2u)(2v) = 4uv</math> and the area of <math>ABE</math> is <math>\frac{1}{2}(2u)(u-v) = u^2 - uv</math>. Adding these together, we get <math>u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41</math>. Since <math>u,v</math> are positive, <math>u+3v>u</math>, and by matching factors we get either <math>(u,v) = (1,90)</math> or <math>(11,10)</math>. Since <math>v < u</math> the latter case is the answer, and <math>u+v = \boxed{021}</math>.

Revision as of 15:22, 15 June 2008

Problem

Let $u$ and $v$ be integers satisfying $0 < v < u$. Let $A = (u,v)$, let $B$ be the reflection of $A$ across the line $y = x$, let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$. Find $u + v$.

Solution

[asy] pointpen = black; pathpen = linewidth(0.7) + black; size(180); pair A=(11,10), B=(10,11), C=(-10, 11), D=(-10, -11), E=(10, -11); D(D(MP("A\ (u,v)",A,(1,0)))--D(MP("B",B,N))--D(MP("C",C,N))--D(MP("D",D))--D(MP("E",E))--cycle); D((-15,0)--(15,0),linewidth(0.6),Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5)); [/asy]

Since $A = (u,v)$, we can find the coordinates of the other points: $B = (v,u)$, $C = (-v,u)$, $D = (-v,-u)$, $E = (v,-u)$. If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\frac{1}{2}(2u)(u-v) = u^2 - uv$. Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41$. Since $u,v$ are positive, $u+3v>u$, and by matching factors we get either $(u,v) = (1,90)$ or $(11,10)$. Since $v < u$ the latter case is the answer, and $u+v = \boxed{021}$.

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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