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Difference between revisions of "1995 IMO Problems/Problem 4"
 
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==Solution==
 
==Solution==
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First we start by solving for <math>x_{i}</math> in the recurssive relation
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First we start by solving for <math>x_{i}</math> in the recursive relation
  
 
<math>x_{i-1}+\frac{2}{x_{i-1}}=2x_{i}+\frac{1}{x_{i}}</math>
 
<math>x_{i-1}+\frac{2}{x_{i-1}}=2x_{i}+\frac{1}{x_{i}}</math>
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<math>x_{i}=\begin{cases} \frac{x_{0}}{2^{i}} & 1 \le i \le 1994 \\ \frac{1}{x_{i-1}} & i=1995\end{cases}</math>
 
<math>x_{i}=\begin{cases} \frac{x_{0}}{2^{i}} & 1 \le i \le 1994 \\ \frac{1}{x_{i-1}} & i=1995\end{cases}</math>
  
−
Then <math>x_{1994}=\frac{x_0}{2^{1994}}</math>, <math>x_{1995}=\frac{1}{x_{1994}}=\frac{2^{1994}}{x_0}={x_0}</math>
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Then <math>x_{1994}=\frac{x_0}{2^{1994}}</math>,
 +
 
 +
thus <math>x_{1995}=\frac{1}{x_{1994}}=\frac{2^{1994}}{x_0}={x_0}</math>
  
 
Solving for <math>{x_0}</math> we get:
 
Solving for <math>{x_0}</math> we get:
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<math>{x_0}^{2}=2^{1994}</math>
 
<math>{x_0}^{2}=2^{1994}</math>
  
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<math>{x_0}=\pm\sqrt{2^{1994}}=\pm2^{997}</math>
+
<math>{x_0}=\pm\sqrt{2^{1994}}=\pm2^{997}</math>.  Since all <math>{x_i}</math> are defined as positive, <math>{x_0}=2^{997}</math>.
  
−
Since all <math>{x_i}</math> are defined as pisitive then, <math>max\left( x_{0} \right)=2^{997}</math>
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Therefore, the maximum value that <math>x_0</math> can have is <math>{x_0}=2^{997}</math>
  
 
~ Tomas Diaz. orders@tomasdiaz.com
 
~ Tomas Diaz. orders@tomasdiaz.com
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 +
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 03:43, 26 November 2023

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


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


See Also

1995 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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