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Difference between revisions of "2001 IMO Shortlist Problems/N2"

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== Problem ==
 
== Problem ==
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Consider the system <math>x + y = z + u,</math> <math>2xy & = zu.</math> Find the greatest value of the real constant <math>m</math> such that <math>m \leq x/y</math> for any positive integer solution <math>(x,y,z,u)</math> of the system, with <math>x \geq y</math>.
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Consider the system <math>x + y = z + u,</math> <math>2xy = zu.</math> Find the greatest value of the real constant <math>m</math> such that <math>m \leq x/y</math> for any positive integer solution <math>(x,y,z,u)</math> of the system, with <math>x \geq y</math>.
  
 
==Solution==
 
==Solution==
 
First consider the real solutions to the system. We have by AM-GM that <math>\frac{z+u}{2}\ge\sqrt{zu}</math> and substituting we get <math>\frac{x+y}{2}\ge\sqrt{2xy}</math>. Squaring and simplifying and dividing by <math>y^2</math>, we get the inequality <math>r^2-6r+1\ge0</math>, where <math>r=\frac{x}{y}</math>. Then <math>r^2-6r+9\ge8</math>, so <math>r\ge3+2\sqrt2</math> or <math>r\le3-2\sqrt2</math>. Since <math>r\ge1</math>, we discard the second inequality and have that <math>3+2\sqrt2</math> is a lower bound for <math>r</math>
 
First consider the real solutions to the system. We have by AM-GM that <math>\frac{z+u}{2}\ge\sqrt{zu}</math> and substituting we get <math>\frac{x+y}{2}\ge\sqrt{2xy}</math>. Squaring and simplifying and dividing by <math>y^2</math>, we get the inequality <math>r^2-6r+1\ge0</math>, where <math>r=\frac{x}{y}</math>. Then <math>r^2-6r+9\ge8</math>, so <math>r\ge3+2\sqrt2</math> or <math>r\le3-2\sqrt2</math>. Since <math>r\ge1</math>, we discard the second inequality and have that <math>3+2\sqrt2</math> is a lower bound for <math>r</math>
  
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This bound is also attainable for real values when <math>z=u</math>. Since <math>\mathbb{Q+}</math> is dense, it is always possible to assign rational values to <math>x, y, z,</math> and <math>w</math> so that <math>r</math> approaches <math>\sqrt3</math>, though equality is never reached. From any rational solution, it is possible to create an integer solution by multiplying by the least common multiple of the denominators and keep the same value of <math>r</math>. Thus, <math>\boxed{m=3+2\sqrt2}</math>.
+
This bound is also attainable for real values when <math>z=u</math>. Since <math>\mathbb{Q+}</math> is dense, it is always possible to assign rational values to <math>x, y, z,</math> and <math>w</math> so that <math>r</math> approaches <math>3+2\sqrt2</math>, though equality is never reached. From any rational solution, it is possible to create an integer solution by multiplying by the least common multiple of the denominators and keep the same value of <math>r</math>. Thus, <math>\boxed{m=3+2\sqrt2}</math>.
  
 
== Resources ==
 
== Resources ==

Latest revision as of 13:38, 1 September 2017

Problem

Consider the system $x + y = z + u,$ $2xy = zu.$ Find the greatest value of the real constant $m$ such that $m \leq x/y$ for any positive integer solution $(x,y,z,u)$ of the system, with $x \geq y$.

Solution

First consider the real solutions to the system. We have by AM-GM that $\frac{z+u}{2}\ge\sqrt{zu}$ and substituting we get $\frac{x+y}{2}\ge\sqrt{2xy}$. Squaring and simplifying and dividing by $y^2$, we get the inequality $r^2-6r+1\ge0$, where $r=\frac{x}{y}$. Then $r^2-6r+9\ge8$, so $r\ge3+2\sqrt2$ or $r\le3-2\sqrt2$. Since $r\ge1$, we discard the second inequality and have that $3+2\sqrt2$ is a lower bound for $r$

This bound is also attainable for real values when $z=u$. Since $\mathbb{Q+}$ is dense, it is always possible to assign rational values to $x, y, z,$ and $w$ so that $r$ approaches $3+2\sqrt2$, though equality is never reached. From any rational solution, it is possible to create an integer solution by multiplying by the least common multiple of the denominators and keep the same value of $r$. Thus, $\boxed{m=3+2\sqrt2}$.

Resources

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