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Difference between revisions of "2008 IMO Problems/Problem 4"

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== Problem ==
 
== Problem ==
 +
 
Find all functions <math>f: (0, \infty) \mapsto (0, \infty)</math> (so <math>f</math> is a function from the positive real numbers) such that
 
Find all functions <math>f: (0, \infty) \mapsto (0, \infty)</math> (so <math>f</math> is a function from the positive real numbers) such that
 +
 
<center>
 
<center>
 +
 
<math>\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}</math>
 
<math>\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}</math>
 +
 
</center>
 
</center>
−
for all positive real numbes <math>w,x,y,z,</math> satisfying <math>wx = yz.</math>
+
 
 +
for all positive real numbers <math>w,x,y,z,</math> satisfying <math>wx = yz.</math>
 +
 
 +
 
  
 
== Solution ==
 
== Solution ==
 +
 
Considering <math>w=1</math> and <math>z=y=\sqrt{x}</math> which satisfy the constraint <math>wx=yz</math> we get the following equation:
 
Considering <math>w=1</math> and <math>z=y=\sqrt{x}</math> which satisfy the constraint <math>wx=yz</math> we get the following equation:
  
  
−
<cmath>\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \frac{1+x^2}{x+x} \Leftrightarrow x((f(1))^2 + (f(x))^2}) = (1+x^2)f(x)</cmath>
 
  
−
At once considering <math>x=1</math> we get <math>(f(1))^2 = f(1)</math> and kowing that <math>f : \mathbb{R}^+ \rightarrow \mathbb{R}^+</math> the only possible solution is <math>f(1)=1</math> since <math>f(1)=0</math> is impossible.
+
 
 +
 
 +
<cmath>\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \frac{1+x^2}{x+x} \Leftrightarrow x((f(1))^2 + (f(x))^2) = (1+x^2)f(x)</cmath>
 +
 
 +
 
 +
 
 +
At once considering <math>x=1</math> we get <math>(f(1))^2 = f(1)</math> and knowing that <math>f : \mathbb{R}^+ \rightarrow \mathbb{R}^+</math> the only possible solution is <math>f(1)=1</math> since <math>f(1)=0</math> is impossible.
 +
 
 +
 
  
 
So we get the quadratic equation:
 
So we get the quadratic equation:
  
−
<cmath>  x(f(x))^2} - (1+x^2)f(x) + x = 0 </cmath>
+
 
 +
 
 +
<cmath>  x(f(x))^2 - (1+x^2)f(x) + x = 0 </cmath>
 +
 
 +
 
  
 
Solving for <math>f(x)</math> as a function of <math>x</math> we get:
 
Solving for <math>f(x)</math> as a function of <math>x</math> we get:
  
−
<cmath> f(x) = \frac{1+x^2 \pm \sqrt{(1+x^2)^2-4x^2}}{2x} = \frac{1+x^2 \pm \sqrt{(1-x^2)^2}}{2x}</cmath>
+
 
 +
 
 +
<cmath> f(x) = \frac{ 1+x^2 \pm \sqrt{(1+x^2)^2-4x^2} }{2x} = \frac{ 1+x^2 \pm \sqrt{(1-x^2)^2} }{2x}</cmath>
 +
 
 +
 
  
 
At once we see that for one value of <math>x</math>, <math>f(x)</math> can only take one of 2 possible values:
 
At once we see that for one value of <math>x</math>, <math>f(x)</math> can only take one of 2 possible values:
 +
 +
  
 
<cmath> f(x) = x \vee f(x) = \frac{1}{x}</cmath>.
 
<cmath> f(x) = x \vee f(x) = \frac{1}{x}</cmath>.
  
−
Take into consideration that <math>f(2) = 2</math> but <math>f(3) = \frac{1}{3} </math> verifies the quadratic equation and thus so far we can't say that <math>f(x)=x \, \forall_{x \x \in \mathbb{R}^+}</math> or alternatively <math>f(x)=\frac{1}{x} \, \forall_{x \x \in \mathbb{R}^+\\}</math>. This is indeed the case but we haven't proved it yet.
+
 
 +
 
 +
Take into consideration that <math>f(2) = 2</math> but <math>f(3) = \frac{1}{3} </math> verifies the quadratic equation and thus so far we can't say that <math>f(x)=x \, \forall_{x \in \mathbb{R}^+}</math> or alternatively <math>f(x)=\frac{1}{x} \, \forall_{x \in \mathbb{R}^+}</math>. This is indeed the case but we haven't proved it yet.
 +
 
 +
 
  
 
To prove the previous assertion consider 2 values <math>a,b \in \mathbb{R}^+</math> such that <math>a\ne 1 \wedge b \ne 1 \wedge a \ne b</math> while having <math>f(a) = a \wedge f(b)=\frac{1}{b}</math>
 
To prove the previous assertion consider 2 values <math>a,b \in \mathbb{R}^+</math> such that <math>a\ne 1 \wedge b \ne 1 \wedge a \ne b</math> while having <math>f(a) = a \wedge f(b)=\frac{1}{b}</math>
 +
 +
  
 
Consider now the original functional equation with <math>w=a,\ x=b,\ y=z=\sqrt{ab}</math> which verifies the constraint. Substituting we have:
 
Consider now the original functional equation with <math>w=a,\ x=b,\ y=z=\sqrt{ab}</math> which verifies the constraint. Substituting we have:
 +
 +
  
 
<cmath> \frac{ (f(a))^2 + (f(b))^2}{ f(ab) + f(ab)} = \frac{a^2 + b^2}{ab + ab} \Leftrightarrow f(ab) = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2}</cmath>
 
<cmath> \frac{ (f(a))^2 + (f(b))^2}{ f(ab) + f(ab)} = \frac{a^2 + b^2}{ab + ab} \Leftrightarrow f(ab) = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2}</cmath>
 +
 +
  
 
Now either <math>f(ab)=ab</math> or <math>f(ab)=\frac{1}{ab}</math>. (notice that <math>ab \ne b \wedge ab\ne b</math> by hypothesis)
 
Now either <math>f(ab)=ab</math> or <math>f(ab)=\frac{1}{ab}</math>. (notice that <math>ab \ne b \wedge ab\ne b</math> by hypothesis)
 +
 +
 +
 +
  
  
  
 
If <math>f(ab)=ab</math> then we have <math> ab = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow b^4=1</math> and since <math>b>0</math> the only solution is <math>b=1</math>.
 
If <math>f(ab)=ab</math> then we have <math> ab = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow b^4=1</math> and since <math>b>0</math> the only solution is <math>b=1</math>.
 +
 +
  
 
If <math>f(ab)=\frac{1}{ab}</math> then we have <math> \frac{1}{ab} = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow  a^2+b^2 = a^4b^2 +a^2 \Leftrightarrow a^4=1</math> and since <math>a>0</math> the only solution is <math>a=1</math>.
 
If <math>f(ab)=\frac{1}{ab}</math> then we have <math> \frac{1}{ab} = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow  a^2+b^2 = a^4b^2 +a^2 \Leftrightarrow a^4=1</math> and since <math>a>0</math> the only solution is <math>a=1</math>.
  
−
So the only solutions are <math>a=1</math> or <math>b=1</math> in which case both alternatives imply <math>f(1)=1</math>. Thus we conclude that solutions to the functional equation are a subset of <math>\left\{f(x)=x \ \forall_{x \x \in \mathbb{R}^+},\ f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+} \right\}</math>.
 
  
−
Finally plug each of these 2 functions into the functional equation and verify that they indeed are solutions.
+
 
 +
So the only solutions are <math>a=1</math> or <math>b=1</math> in which case both alternatives imply <math>f(1)=1</math>. Thus we conclude that solutions to the functional equation are a subset of <math>\left\{f(x)=x \ \forall_{x \in \mathbb{R}^+},\ f(x)=\frac{1}{x}\ \forall_{x \in \mathbb{R}^+} \right\}</math>.
 +
 
 +
 
 +
 
 +
Finally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.
 +
 
 +
 
  
 
This is trivial since <math>f(x)=x</math> is an obvious solution and for <math>f(x)=\frac{1}{x}</math> we have:
 
This is trivial since <math>f(x)=x</math> is an obvious solution and for <math>f(x)=\frac{1}{x}</math> we have:
  
−
<cmath> \frac{ \frac{1}{w^2} + \frac{1}{x^2}  }{ \frac{1}{y^2} + \frac{1}{z^2}} = \frac{ \frac{w^2+x^2}{(wx)^2}  }{ \frac{y^2+z^2}{(yz)^2} } = \frac{w^2+x^2}{y^2+z^2} </cmath> provided that <math>(wx)^2 = (yz)^2</math> which is the original constraint.
+
 
 +
 
 +
<cmath> \frac{ \frac{1}{w^2} + \frac{1}{x^2}  }{ \frac{1}{y^2} + \frac{1}{z^2}} = \frac{ \frac{w^2+x^2}{(wx)^2}  }{ \frac{y^2+z^2}{(yz)^2} } = \frac{w^2+x^2}{y^2+z^2} </cmath> provided that <math>(wx)^2 = (yz)^2</math> which verifies the original constraint.
 +
 
 +
 
  
 
So the functional equation has 2 solutions:
 
So the functional equation has 2 solutions:
−
<cmath>f(x) = x\ \forall_{x \x \in \mathbb{R}^+}\ \vee\  f(x)=\frac{1}{x}\ \forall_{x \x \in \mathbb{R}^+}</cmath>
+
 
 +
<cmath>f(x) = x\ \forall_{x \in \mathbb{R}^+}\ \vee\  f(x)=\frac{1}{x}\ \forall_{x \in \mathbb{R}^+}</cmath>
 +
 
 +
== Video Solution ==
 +
 
 +
https://youtu.be/wb2gp8uoGfM [Video Solution by little fermat]
 +
 
 +
==See Also==
 +
 
 +
{{IMO box|year=2008|num-b=3|num-a=5}}
 +
 
 +
[[Category:Olympiad Algebra Problems]]
 +
[[Category:Functional Equation Problems]]

Latest revision as of 00:10, 19 November 2023

Problem

Find all functions $f: (0, \infty) \mapsto (0, \infty)$ (so $f$ is a function from the positive real numbers) such that

$\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}$

for all positive real numbers $w,x,y,z,$ satisfying $wx = yz.$


Solution

Considering $w=1$ and $z=y=\sqrt{x}$ which satisfy the constraint $wx=yz$ we get the following equation:



\[\frac{(f(1))^2 + (f(x))^2}{f(x) + f(x)} = \frac{1+x^2}{x+x} \Leftrightarrow x((f(1))^2 + (f(x))^2) = (1+x^2)f(x)\]


At once considering $x=1$ we get $(f(1))^2 = f(1)$ and knowing that $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ the only possible solution is $f(1)=1$ since $f(1)=0$ is impossible.


So we get the quadratic equation:


\[x(f(x))^2 - (1+x^2)f(x) + x = 0\]


Solving for $f(x)$ as a function of $x$ we get:


\[f(x) = \frac{ 1+x^2 \pm \sqrt{(1+x^2)^2-4x^2} }{2x} = \frac{ 1+x^2 \pm \sqrt{(1-x^2)^2} }{2x}\]


At once we see that for one value of $x$, $f(x)$ can only take one of 2 possible values:


\[f(x) = x \vee f(x) = \frac{1}{x}\].


Take into consideration that $f(2) = 2$ but $f(3) = \frac{1}{3}$ verifies the quadratic equation and thus so far we can't say that $f(x)=x \, \forall_{x \in \mathbb{R}^+}$ or alternatively $f(x)=\frac{1}{x} \, \forall_{x \in \mathbb{R}^+}$. This is indeed the case but we haven't proved it yet.


To prove the previous assertion consider 2 values $a,b \in \mathbb{R}^+$ such that $a\ne 1 \wedge b \ne 1 \wedge a \ne b$ while having $f(a) = a \wedge f(b)=\frac{1}{b}$


Consider now the original functional equation with $w=a,\ x=b,\ y=z=\sqrt{ab}$ which verifies the constraint. Substituting we have:


\[\frac{ (f(a))^2 + (f(b))^2}{ f(ab) + f(ab)} = \frac{a^2 + b^2}{ab + ab} \Leftrightarrow f(ab) = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2}\]


Now either $f(ab)=ab$ or $f(ab)=\frac{1}{ab}$. (notice that $ab \ne b \wedge ab\ne b$ by hypothesis)




If $f(ab)=ab$ then we have $ab = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow b^4=1$ and since $b>0$ the only solution is $b=1$.


If $f(ab)=\frac{1}{ab}$ then we have $\frac{1}{ab} = ab\frac{a^2 + \frac{1}{b^2}}{a^2 + b^2} \Leftrightarrow a^2+b^2 = a^4b^2 +a^2 \Leftrightarrow a^4=1$ and since $a>0$ the only solution is $a=1$.


So the only solutions are $a=1$ or $b=1$ in which case both alternatives imply $f(1)=1$. Thus we conclude that solutions to the functional equation are a subset of $\left\{f(x)=x \ \forall_{x \in \mathbb{R}^+},\ f(x)=\frac{1}{x}\ \forall_{x \in \mathbb{R}^+} \right\}$.


Finally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.


This is trivial since $f(x)=x$ is an obvious solution and for $f(x)=\frac{1}{x}$ we have:


\[\frac{ \frac{1}{w^2} + \frac{1}{x^2} }{ \frac{1}{y^2} + \frac{1}{z^2}} = \frac{ \frac{w^2+x^2}{(wx)^2} }{ \frac{y^2+z^2}{(yz)^2} } = \frac{w^2+x^2}{y^2+z^2}\] provided that $(wx)^2 = (yz)^2$ which verifies the original constraint.


So the functional equation has 2 solutions:

\[f(x) = x\ \forall_{x \in \mathbb{R}^+}\ \vee\ f(x)=\frac{1}{x}\ \forall_{x \in \mathbb{R}^+}\]

Video Solution

https://youtu.be/wb2gp8uoGfM [Video Solution by little fermat]

See Also

2008 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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