Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (typos)
 
(3 intermediate revisions by 3 users not shown)
Line 21: Line 21:
  
  
−
<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>
+
<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>
  
  
Line 33: Line 33:
  
  
−
<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>
  
  
Line 41: Line 41:
  
  
−
<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>
  
  
Line 53: Line 53:
  
  
−
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.
  
  
Line 85: Line 85:
  
  
−
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>.
+
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.
+
Finally, plug each of these 2 functions into the functional equation and verify that they indeed are solutions.
  
  
Line 103: Line 103:
 
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_IMO_Problems/Problem_4&oldid=204531"