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Difference between revisions of "2018 USAMO Problems/Problem 2"
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==Problem 2==
 
==Problem 2==
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Find all functions <math>f:(0,\infty) \rightarrow (0,\infty)</math> such that
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Find all functions <math>f:(0,\infty) \to (0,\infty)</math> such that
  
 
<cmath>f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1</cmath> for all <math>x,y,z >0</math> with <math>xyz =1.</math>
 
<cmath>f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1</cmath> for all <math>x,y,z >0</math> with <math>xyz =1.</math>
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==Solution==
 
==Solution==
  
−
The only such function is <math>f(x)=\frac13</math>.
+
Obviously, the output of <math>f</math> lies in the interval <math>(0,1)</math>. Define <math>g:(0,1)\to(0,1)</math> as <math>g(x)=f\left(\frac1x-1\right)</math>. Then for any <math>a,b,c\in(0,1)</math> such that <math>a+b+c=1</math>, we have <math>g(a)=f\left(\frac1a-1\right)=f\left(\frac{1-a}a\right)=f\left(\frac{b+c}a\right)</math>. We can transform <math>g(b)</math> and <math>g(c)</math> similarly:
  
−
Letting <math>x=y=z=1</math> gives <math>3f(2)=1</math>, hence <math>f(2)=\frac13</math>. Now observe that even if we fix <math>x+\frac1y=y+\frac1z=2</math>, <math>z+\frac1x</math> is not fixed. Specifically,
+
<cmath>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)</cmath>
−
<cmath>y=\frac1{2-x}</cmath>
 
−
<cmath>z=\frac1{2-y}=\frac{2-x}{3-2x}</cmath>
 
−
<cmath>z+\frac1x=\frac12+\frac1x+\frac1{2(3-2x)}</cmath>
 
−
This is continuous on the interval <math>x\in\left(0,\frac32\right)</math> and has an asymptote at <math>x=\frac32</math>. Since it takes the  value 2 when <math>x=1</math>, it can take on all values greater than or equal to 2. So for any <math>a\ge2</math>, we can find <math>x</math> such that <math>2f(2)+f(x)=1</math>. Therefore, <math>f(x)=\frac13</math> for all <math>x\ge2</math>.
 
  
−
Now, for any <math>0<k<2</math>, if we let <math>x=\frac k2</math>, <math>y=\frac1x</math>, and <math>z=1</math>, then <math>f(k)+2f\left(1+\frac2k\right)=1</math>. Since <math>1+\frac2k>2</math>, <math>f\left(1+\frac2k\right)=\frac13</math>, hence <math>f(k)=\frac13</math>. Therefore, <math>f(x)=\frac13</math> for all <math>x\ge0</math>.
+
Let <math>x=\frac ca</math>, <math>y=\frac ab</math>, <math>z=\frac bc</math>. We can see that the above expression is equal to <math>1</math>. That is, for any <math>a,b,c\in(0,1)</math> such that <math>a+b+c=1</math>, <math>g(a)+g(b)+g(c)=1</math>.
  
−
-- wzs26843645602
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(To motivate this, one can start by writing <math>x=\frac ab</math>, <math>y=\frac bc</math>, <math>z=\frac ca</math>, and normalizing such that <math>a+b+c=1</math>.)
  
−
==Solution 2==
 
  
−
Make the substitution <math>x = \frac{a}{b},</math> <math>y = \frac{b}{c},</math> <math>z = \frac{c}{a},</math> so that the given equation becomes <cmath>f\bigg(\frac{a+b}{c} \bigg) + f \bigg(\frac{a+c}{b} \bigg) + f\bigg(\frac{b+c}{a} \bigg) = 1</cmath> for all <math>a, b, c \in \mathbb{R}_{>0}.</math> Now, let <cmath>\bigg(X, Y, Z \bigg) = \bigg(\frac{a+b}{c}, \ \frac{a+c}{b}, \ \frac{b+c}{a} \bigg).</cmath> We will show that if we fix <math>Y</math> and <math>Z,</math> <math>X</math> can still vary to be any positive real number. Notice that <math>(a,b,c) = \bigg((X+1)(Y+1), \ (X+1)(Z+1), \ (Y+1)(Z+1) \bigg)</math> will be a solution to the system of equations <cmath>\begin{cases} \frac{a+b}{c} = X \\ \frac{a+c}{b} = Y \\ \frac{b+c}{a}  = Z, \\ \end{cases},</cmath> and so even if we fix <math>Y</math> and <math>Z</math> this system of equations will have a solution in <math>(a,b,c)</math> for any real <math>X \in (0, \infty).</math> Now, for any real <math>X \in (0, \infty),</math> and for a fixed <math>Y</math> and <math>Z,</math> we have <math>f(X) + f(Y) + f(Z)</math> equals <math>1</math>, which implies that <math>f</math> is a constant function over its domain <math>(0, \infty).</math> However, letting <math>f(x) = k</math> for all real <math>x > 0,</math> we see that <cmath>f(X) + f(Y) + f(Z) = 1, \ \ \implies \ \ 3k = 1,</cmath> so the only possible value for <math>k</math> is <math>\frac{1}{3}.</math> Thus, the only possible function is <math>\boxed{f(x) = \frac{1}{3}},</math> which obviously satisfies all necessary conditions.
+
For convenience, we define <math>h:\left(-\frac13,\frac23\right)\to\left(-\frac13,\frac23\right)</math> as <math>h(x)=g\left(x+\frac13\right)-\frac13</math>, so that for any <math>a,b,c\in\left(-\frac13,\frac23\right)</math> such that <math>a+b+c=0</math>, we have
  
−
~ Professor-Mom
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<cmath>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.</cmath>
 +
 
 +
Obviously, <math>h(0)=0</math>. If <math>|a|<\frac13</math>, then <math>h(a)+h(-a)+h(0)=0</math> and thus <math>h(-a)=-h(a)</math>. Furthermore, if <math>a,b</math> are in the domain and <math>|a+b|<\frac13</math>, then <math>h(a)+h(b)+h(-(a+b))=0</math> and thus <math>h(a+b)=h(a)+h(b)</math>.
 +
 
 +
 
 +
At this point, we should realize that <math>h</math> should be of the form <math>h(x)=kx</math>. We first prove this for some rational numbers. If <math>n</math> is a positive integer and <math>x</math> is a real number such that <math>|nx|<\frac13</math>, then we can repeatedly apply <math>h(a+b)=h(a)+h(b)</math> to obtain <math>h(nx)=nh(x)</math>. Let <math>k=6h\left(\frac16\right)</math>, then for any rational number <math>r=\frac pq\in\left(0,\frac13\right)</math> where <math>p,q</math> are positive integers, we have <math>h(r)=6p*h\left(\frac1{6q}\right)=\frac{6p}q*h\left(\frac16\right)=kr</math>.
 +
 
 +
Next, we prove it for all real numbers in the interval <math>\left(0,\frac13\right)</math>. For the sake of contradiction, assume that there is some <math>x\in\left(0,\frac13\right)</math> such that <math>h(x)\ne kx</math>. Let <math>E=h(x)-kx</math>, then obviously <math>0<|E|<1</math>. The idea is to "amplify" this error until it becomes so big as to contradict the bounds on the output of <math>h</math>. Let <math>N=\left\lceil\frac1{|E|}\right\rceil</math>, so that <math>N\ge2</math> and <math>|NE|\ge1</math>. Pick any rational <math>r\in\left(\frac{N-1}Nx,x\right)</math>, so that <cmath>0<x-r<N(x-r)<x<\frac13.</cmath>
 +
All numbers and sums are safely inside the bounds of <math>\left(-\frac13,\frac13\right)</math>. Thus
 +
<cmath>h(N(x-r))=Nh(x-r)=N(h(x)+h(-r))=N(h(x)-h(r))=kN(x-r)+NE,</cmath>
 +
but picking any rational number <math>s\in\left(N(x-r),\frac13\right)</math> gives us <math>|kN(x-r)|<|ks|</math>, and since <math>ks=h(s)\in\left(-\frac13,\frac23\right)</math>, we have <math>kN(x-r)\in\left(-\frac13,\frac23\right)</math> as well, but since <math>NE\ge1</math>, this means that <math>h(N(x-r))=kN(x-r)+NE\notin\left(-\frac13,\frac23\right)</math>, giving us the desired contradiction.
 +
 
 +
 
 +
We now know that <math>h(x)=kx</math> for all <math>0<x<\frac13</math>. Since <math>h(-x)=-h(x)</math> for <math>|x|<\frac13</math>, we obtain <math>h(x)=kx</math> for all <math>|x|<\frac13</math>. For <math>x\in\left(\frac13,\frac23\right)</math>, we have <math>h(x)+h\left(-\frac x2\right)+h\left(-\frac x2\right)=0</math>, and thus <math>h(x)=kx</math> as well. So <math>h(x)=kx</math> for all <math>x</math> in the domain. Since <math>h(x)</math> is bounded by <math>-\frac13</math> and <math>\frac23</math>, we have <math>-\frac12\le k\le1</math>. It remains to work backwards to find <math>f(x)</math>.
 +
 
 +
<cmath>\begin{align*}
 +
h(x) &= kx \\
 +
g(x) &= kx+\frac{1-k}3 \\
 +
f(x) &= \frac k{1+x}+\frac{1-k}3\quad\left(-\frac12\le k\le1\right).
 +
\end{align*}</cmath>
 +
- wzs26843545602

Revision as of 23:32, 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. Since $h(x)$ is bounded by $-\frac13$ and $\frac23$, we have $-\frac12\le k\le1$. 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\quad\left(-\frac12\le k\le1\right). \end{align*} - wzs26843545602

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