# 1995 IMO Problems/Problem 4

## Problem

The positive real numbers $x_0, x_1, x_2,.....x_{1994}, x_{1995}$ satisfy the relations

$x_0=x_{1995}$ and $x_{i-1}+\frac{2}{x_{i-1}}=2{x_i}+\frac{1}{x_i}$

for $i=1,2,3,....1995$

Find the maximum value that $x_0$ can have.

## Solution

First we start by solving for $x_{i}$ in the recursive relation

$x_{i-1}+\frac{2}{x_{i-1}}=2x_{i}+\frac{1}{x_{i}}$

$\frac{x_{i-1}^{2}+2}{x_{i-1}}=\frac{2x_{i}^{2}+1}{x_{i}}$

$\left( x_{i} \right)\left( x_{i-1}^{2}+2 \right)=\left( x_{i-1} \right)\left( 2x_{i}^{2}+1 \right)$

$x_{i}x_{i-1}^{2}+2x_{i}=2x_{i-1}x_{i}^{2}+x_{i-1}$

$x_{i}x_{i-1}^{2}-2x_{i-1}x_{i}^{2}+2x_{i}-x_{i-1}=0$

$x_{i}x_{i-1}\left( x_{i-1} \right)-2x_{i}x_{i-1}\left( x_{i} \right)+2x_{i}-x_{i-1}=0$

$x_{i}x_{i-1}\left( x_{i-1} -2x_{i}\right)+2x_{i}-x_{i-1}=0$

$\left( 2x_{i}-x_{i-1}\right)\left( 1-x_{i}x_{i-1} \right)=0$

$x_{i}=\frac{x_{i-1}}{2},\;$ or $x_{i}=\frac{1}{x_{i-1}}$

So we have two recursive properties to chose from.

If we want to maximize $x_{0}$ then we can use $x_{i}=\frac{x_{i-1}}{2}$ from $i=1$ to $1994$. This will make $x_{0}$ the largest and $x_{1994}$ the smallest.

Then we can simply use $x_{i}=\frac{1}{x_{i-1}}$ to get $x_{1995}$ since the reciprocal will make it very large.

Then we use $x_{0}=x_{1995}$ and solve for $x_{0}$

This means that we can write $x_{i}$ as:

$x_{i}=\begin{cases} \frac{x_{0}}{2^{i}} & 1 \le i \le 1994 \\ \frac{1}{x_{i-1}} & i=1995\end{cases}$

Then $x_{1994}=\frac{x_0}{2^{1994}}$,

thus $x_{1995}=\frac{1}{x_{1994}}=\frac{2^{1994}}{x_0}={x_0}$

Solving for ${x_0}$ we get:

${x_0}^{2}=2^{1994}$

${x_0}=\pm\sqrt{2^{1994}}=\pm2^{997}$. Since all ${x_i}$ are defined as positive, ${x_0}=2^{997}$.

Therefore, the maximum value that $x_0$ can have is ${x_0}=2^{997}$

~ Tomas Diaz. orders@tomasdiaz.com