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

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== Problem ==
 
== Problem ==
Let <math>u</math> and <math>v</math> be integers satisfying <math>0 < v < u</math>. Let <math>A = (u,v)</math>, let <math>B</math> be the reflection of <math>A</math> across the line <math>y = x</math>, let <math>C</math> be the reflection of <math>B</math> across the y-axis, let <math>D</math> be the reflection of <math>C</math> across the x-axis, and let <math>E</math> be the reflection of <math>D</math> across the y-axis. The area of pentagon <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>.
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Let <math>u</math> and <math>v</math> be [[integer]]s satisfying <math>0 < v < u</math>. Let <math>A = (u,v)</math>, let <math>B</math> be the [[reflection]] of <math>A</math> across the line <math>y = x</math>, let <math>C</math> be the reflection of <math>B</math> across the y-axis, let <math>D</math> be the reflection of <math>C</math> across the x-axis, and let <math>E</math> be the reflection of <math>D</math> across the y-axis. The area of [[pentagon]] <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>.
  
 
== Solution ==
 
== Solution ==
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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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2000|n=I|num-b=1|num-a=3}}
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 16:38, 31 December 2007

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


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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 (ProblemsAnswer KeyResources)
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