Difference between revisions of "2006 AIME I Problems/Problem 15"

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== Problem ==
 
== Problem ==
Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}|. </math>
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Given that <math> x, y, </math> and <math>z</math> are real numbers that satisfy:
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<center><math> x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} </math> </center>
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<center><math> y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}} </math></center>
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<center><math> z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}</math></center>
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and that <math> x+y+z = \frac{m}{\sqrt{n}}, </math> where <math> m </math> and <math> n </math> are positive integers and <math> n </math> is not divisible by the square of any prime, find <math> m+n.</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 14:51, 25 September 2007

Problem

Given that $x, y,$ and $z$ are real numbers that satisfy:

$x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}$
$y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}$
$z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}$

and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$

Solution

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See also

2006 AIME I (ProblemsAnswer KeyResources)
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Problem 14
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