Problem 55: IMO Shortlist 2005, A2

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Joined: Mar 10, 2015

by henderson, Dec 2, 2016, 5:26 PM

${\color{red}\bf{Problem \ 55}}$ Let $\mathbb{R}^+$ denote the set of all positive real numbers. Determine all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ such that
$f(x)f(y)=2f(x+yf(x))$ for all positive real numbers $x$ and $y$ .

Proposed by Nikolai Nikolov, Bulgaria
${\color{red}\bf{Solution}}$ Replacing $x$ and $y$ in the given equation, we obtain
$f(x+yf(x))=f(y+xf(y)).$ So,
$\begin{eqnarray*}2f\left(x\right)f\left(y+2z\right)&=&f\left(x\right)f\left(y\right)f\left(\frac{2z}{f\left(y\right)}\right)\\ &=&2f\left(x+yf\left(x\right)\right)f\left(\frac{2z}{f\left(y\right)}\right)\\ &=&4f\left(x+yf\left(x\right)+\frac{2z}{f\left(y\right)}f\left(x+yf\left(x\right)\right)\right)\\ &=&4f\left(x+yf\left(x\right)+zf\left(x\right)\right)\\ &=&2f\left(x\right)f\left(y+z\right). \end{eqnarray*}$ Dividing both sides by $f\left(x\right)$ gives $f\left(y+z\right)=f\left(y+2z\right)$ for all positive reals $y,z.$ Now replacing $y$ with $y+z$ repeatedly gives $f\left(y+z\right)=f\left(y+nz\right)$ for all reals $y,z$ and positive integers $n.$ Now suppose $a,b$ are arbitrary positive real numbers with $a>b.$ Let $n$ be an integer greater than $a/b,$ and let
$y=\frac{nb-a}{n-1}, z=\frac{a-b}{n-1}$ Thus $y+nz=a$ and $y+z=b,$ so we have $f\left(a\right)=f\left(b\right).$ However, since $a,b$ were arbitrary, $f\left(x\right)\equiv c$ for some $c.$ Plugging this back into the given equation we get $f\left(x\right)\equiv 2.$ It's indeed the solution. $\square$