IMO Shortlist 2009 A7

Discussion on International Mathematical Olympiad (IMO)
User avatar
asif e elahi
Posts:185
Joined:Mon Aug 05, 2013 12:36 pm
Location:Sylhet,Bangladesh
IMO Shortlist 2009 A7

Unread post by asif e elahi » Sat Mar 29, 2014 9:09 pm

Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$, $y$ the identity \[ f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2\]

Nirjhor
Posts:136
Joined:Thu Aug 29, 2013 11:21 pm
Location:Varies.

Re: IMO Shortlist 2009 A7

Unread post by Nirjhor » Mon Sep 08, 2014 2:02 am

Let \(P(x,y)\) denote the given equation.

1. \(f(0)=0\). If \(f(0)\neq 0\) then \(P\left(0,\frac{y}{f(0)}\right)\Longrightarrow f(y)=f(0)\) so \(f\) is constant. But no constant function works. So \(f(0)=0\). If \(f(n)=0\) for some \(n\) then \(P(n,0)\Longrightarrow 0=f(nf(n))=n^2\) so \(n=0\).

2. \(f\) is Bijective.

(I) Surjectivity: \(P(x,0)\Longrightarrow f\left(xf(x)\right)=x^2\) and \(P(x,-x)\Longrightarrow f\left(-xf(x)\right)=-x^2\). Therefore \(\forall ~ y\in\mathbb{R}~\exists~ x\in\mathbb{R}\mid f(x)=y\), hence \(f\) is surjective.

(II) Injectivity: Suppose \(f(x)=f(y)\) then \(f(xf(x))=f(xf(x+(y-x)))+x^2=x^2\). This leads to \(f((y-x)f(x))=0\). Hence \((y-x)f(x)=0\Longrightarrow x=y\). So \(f\) is injective.

This completes proving bijectivity of \(f\).

3. \(f\) is odd. Notice that \(f(-nf(n))=-n^2=-f(nf(n))\) hence \(f(-x)=-f(x)\) by bijectivity.

4. \(f(xf(y))=xy\). Firstly, \(P(y,x)\Longrightarrow f(xf(y))=f(yf(x+y))-y^2\). Now notice that
\[f(yf(x+y))-y^2=(x+y)^2-y^2-\left((x+y)^2+f(-x(x+y))\right)\] and by the given equation this is \(x^2+2xy-f\left((x+y)f(x)\right)\). But now \[x^2+2xy-f\left((x+y)f(x)\right)=\left((-x)^2+f((x+y)f(-x))\right)+2xy=2xy-f(xf(y)).\] So we finally have \(f(xf(y))=2xy-f(xf(y))\) from where the result follows.

5. \(f(x)\in\{x,-x\}\). Since bijective, take \(n\) such that \(f(n)=1\). Then last result implies \(f(x)=nx\). Setting in the equation \(f(xf(x))=x^2\) gives \(n^2x^2=x^2\Longrightarrow n\in\{1,-1\}\). Hence we have two solutions \(f(x)=x\) and \(f(x)=-x\). Verify that each solution works. \(\square\)
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

mutasimmim
Posts:107
Joined:Sun Dec 12, 2010 10:46 am

Re: IMO Shortlist 2009 A7

Unread post by mutasimmim » Mon Sep 08, 2014 8:26 am

Elaborate the surjectivity part, please.

Nirjhor
Posts:136
Joined:Thu Aug 29, 2013 11:21 pm
Location:Varies.

Re: IMO Shortlist 2009 A7

Unread post by Nirjhor » Mon Sep 08, 2014 4:16 pm

We have \(f(xf(x))=x^2\) and \(f(-xf(x))=-x^2\). Shift \(x\to \sqrt x\) to get \(f\left(\sqrt xf\left(\sqrt x\right)\right)=x\) and \(f\left(-\sqrt x f\left(\sqrt x\right)\right)=-x\), true for all \(x\in \mathbb{R}^+\). Also we have \(f(0)=0\). So for all \(x\in\mathbb{R}\), we have some \(n\in\mathbb{R}\) (namely \(n=\pm\sqrt x f\left(\sqrt x\right)\)) so that \(f(n)=x\), proving surjectivity.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

Post Reply