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Difference between revisions of "2006 AIME A Problems/Problem 15"
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== Problem ==
 
== Problem ==
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Given that <math> x, y, </math> and <math>z</math> are real numbers that satisfy:
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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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<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 [[integer]]s 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 a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$

Solution

This solution requires you to disregard rigor. Additional solutions, or justification for the nonrigorous steps, would be appreciated.

Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all altitudes are on the interior of the triangle). Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then we have by two applications of the Pythagorean Theorem that $x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}$. As a function of $h_x$, the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \frac1{16}$ and so $h_x = \frac{1}4$ and similarly $\displaystyle h_y = \frac15$ and $h_z = \frac16$.

Since the area of the triangle must be the same no matter how we measure, $x\cdot h_x = y\cdot h_y = z \cdot h_z$ and so $\frac x4 = \frac y5 = \frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$. The semiperimeter of the triangle is $s = \frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have $A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}$. Thus $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = 009$.


Justification that there is a triangle with sides of length $x, y$ and $z$:

Note that $x, y$ and $z$ are each the sum of two positive square roots of real numbers, so $x, y, z \geq 0$. (Recall that, by AIME convention, all numbers (including square roots) are taken to be real unless otherwise indicated.) Also, $\sqrt{y^2-\frac{1}{16}} < \sqrt{y^2} = y$, so we have $x < y + z$, $y < z + x$ and $z < x + y$. But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.


Justification that this triangle is an acute triangle: Still needed.

See also

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