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2018 USAMO Problems/Problem 2: Difference between revisions

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f(x) &= \frac k{1+x}+\frac{1-k}3.
f(x) &= \frac k{1+x}+\frac{1-k}3.
\end{align*}</cmath>
\end{align*}</cmath>
- wzs26843545602

Revision as of 22:31, 4 March 2023

Problem 2

Find all functions $f:(0,\infty) \to (0,\infty)$ such that

\[f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1\] for all $x,y,z >0$ with $xyz =1.$


Solution

Obviously, the output of $f$ lies in the interval $(0,1)$. Define $g:(0,1)\to(0,1)$ as $g(x)=f\left(\frac1x-1\right)$. Then for any $a,b,c\in(0,1)$ such that $a+b+c=1$, we have $g(a)=f\left(\frac1a-1\right)=f\left(\frac{1-a}a\right)=f\left(\frac{b+c}a\right)$. We can transform $g(b)$ and $g(c)$ similarly:

\[g(a)+g(b)+g(c)=f\left(\frac ca+\frac ba\right)+f\left(\frac ab+\frac cb\right)+f\left(\frac bc+\frac ac\right)\]

Let $x=\frac ca$, $y=\frac ab$, $z=\frac bc$. We can see that the above expression is equal to $1$. That is, for any $a,b,c\in(0,1)$ such that $a+b+c=1$, $g(a)+g(b)+g(c)=1$.

(To motivate this, one can start by writing $x=\frac ab$, $y=\frac bc$, $z=\frac ca$, and normalizing such that $a+b+c=1$.)


For convenience, we define $h:\left(-\frac13,\frac23\right)\to\left(-\frac13,\frac23\right)$ as $h(x)=g\left(x+\frac13\right)-\frac13$, so that for any $a,b,c\in\left(-\frac13,\frac23\right)$ such that $a+b+c=0$, we have

\[h(a)+h(b)+h(c)=g\left(a+\frac13\right)-\frac13+g\left(b+\frac13\right)-\frac13+g\left(c+\frac13\right)-\frac13=1-1=0.\]

Obviously, $h(0)=0$. If $|a|<\frac13$, then $h(a)+h(-a)+h(0)=0$ and thus $h(-a)=-h(a)$. Furthermore, if $a,b$ are in the domain and $|a+b|<\frac13$, then $h(a)+h(b)+h(-(a+b))=0$ and thus $h(a+b)=h(a)+h(b)$.


At this point, we should realize that $h$ should be of the form $h(x)=kx$. We first prove this for some rational numbers. If $n$ is a positive integer and $x$ is a real number such that $|nx|<\frac13$, then we can repeatedly apply $h(a+b)=h(a)+h(b)$ to obtain $h(nx)=nh(x)$. Let $k=6h\left(\frac16\right)$, then for any rational number $r=\frac pq\in\left(0,\frac13\right)$ where $p,q$ are positive integers, we have $h(r)=6p*h\left(\frac1{6q}\right)=\frac{6p}q*h\left(\frac16\right)=kr$.

Next, we prove it for all real numbers in the interval $\left(0,\frac13\right)$. For the sake of contradiction, assume that there is some $x\in\left(0,\frac13\right)$ such that $h(x)\ne kx$. Let $E=h(x)-kx$, then obviously $0<|E|<1$. The idea is to "amplify" this error until it becomes so big as to contradict the bounds on the output of $h$. Let $N=\left\lceil\frac1{|E|}\right\rceil$, so that $N\ge2$ and $|NE|\ge1$. Pick any rational $r\in\left(\frac{N-1}Nx,x\right)$, so that \[0<x-r<N(x-r)<x<\frac13.\] All numbers and sums are safely inside the bounds of $\left(-\frac13,\frac13\right)$. Thus \[h(N(x-r))=Nh(x-r)=N(h(x)+h(-r))=N(h(x)-h(r))=kN(x-r)+NE,\] but picking any rational number $s\in\left(N(x-r),\frac13\right)$ gives us $|kN(x-r)|<|ks|$, and since $ks=h(s)\in\left(-\frac13,\frac23\right)$, we have $kN(x-r)\in\left(-\frac13,\frac23\right)$ as well, but since $NE\ge1$, this means that $h(N(x-r))=kN(x-r)+NE\notin\left(-\frac13,\frac23\right)$, giving us the desired contradiction.


We now know that $h(x)=kx$ for all $0<x<\frac13$. Since $h(-x)=-h(x)$ for $|x|<\frac13$, we obtain $h(x)=kx$ for all $|x|<\frac13$. For $x\in\left(\frac13,\frac23\right)$, we have $h(x)+h\left(-\frac x2\right)+h\left(-\frac x2\right)=0$, and thus $h(x)=kx$ as well. So $h(x)=kx$ for all $x$ in the domain. It remains to work backwards to find $f(x)$.

\begin{align*} h(x) &= kx \\ g(x) &= kx+\frac{1-k}3 \\ f(x) &= \frac k{1+x}+\frac{1-k}3. \end{align*} - wzs26843545602

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