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Difference between revisions of "2018 USAMO Problems/Problem 2"

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==Solution 2==
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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.

Revision as of 01:08, 14 October 2022

Problem 2

Find all functions $f:(0,\infty) \rightarrow (0,\infty)$ 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

The only such function is $f(x)=\frac13$.

Letting $x=y=z=1$ gives $3f(2)=1$, hence $f(2)=\frac13$. Now observe that even if we fix $x+\frac1y=y+\frac1z=2$, $z+\frac1x$ is not fixed. Specifically, \[y=\frac1{2-x}\] \[z=\frac1{2-y}=\frac{2-x}{3-2x}\] \[z+\frac1x=\frac12+\frac1x+\frac1{2(3-2x)}\] This is continuous on the interval $x\in\left(0,\frac32\right)$ and has an asymptote at $x=\frac32$. Since it takes the value 2 when $x=1$, it can take on all values greater than or equal to 2. So for any $a\ge2$, we can find $x$ such that $2f(2)+f(x)=1$. Therefore, $f(x)=\frac13$ for all $x\ge2$.

Now, for any $0<k<2$, if we let $x=\frac k2$, $y=\frac1x$, and $z=1$, then $f(k)+2f\left(1+\frac2k\right)=1$. Since $1+\frac2k>2$, $f\left(1+\frac2k\right)=\frac13$, hence $f(k)=\frac13$. Therefore, $f(x)=\frac13$ for all $x\ge0$.

-- wzs26843645602

Solution 2

Make the substitution $x = \frac{a}{b},$ $y = \frac{b}{c},$ $z = \frac{c}{a},$ so that the given equation becomes \[f\bigg(\frac{a+b}{c} \bigg) + f \bigg(\frac{a+c}{b} \bigg) + f\bigg(\frac{b+c}{a} \bigg) = 1\] for all $a, b, c \in \mathbb{R}_{>0}.$ Now, let \[\bigg(X, Y, Z \bigg) = \bigg(\frac{a+b}{c}, \ \frac{a+c}{b}, \ \frac{b+c}{a} \bigg).\] We will show that if we fix $Y$ and $Z,$ $X$ can still vary to be any positive real number. Notice that $(a,b,c) = \bigg((X+1)(Y+1), \ (X+1)(Z+1), \ (Y+1)(Z+1) \bigg)$ will be a solution to the system of equations \[\begin{cases} \frac{a+b}{c} = X \\ \frac{a+c}{b} = Y \\ \frac{b+c}{a} = Z, \\ \end{cases},\] and so even if we fix $Y$ and $Z$ this system of equations will have a solution in $(a,b,c)$ for any real $X \in (0, \infty).$ Now, for any real $X \in (0, \infty),$ and for a fixed $Y$ and $Z,$ we have $f(X) + f(Y) + f(Z)$ equals $1$, which implies that $f$ is a constant function over its domain $(0, \infty).$ However, letting $f(x) = k$ for all real $x > 0,$ we see that \[f(X) + f(Y) + f(Z) = 1, \ \ \implies \ \ 3k = 1,\] so the only possible value for $k$ is $\frac{1}{3}.$ Thus, the only possible function is $\boxed{f(x) = \frac{1}{3}},$ which obviously satisfies all necessary conditions.

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