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Difference between revisions of "1986 IMO Problems/Problem 5"

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==Problem==
 
Find all (if any) functions <math>f</math> taking the non-negative reals onto the non-negative reals, such that
 
Find all (if any) functions <math>f</math> taking the non-negative reals onto the non-negative reals, such that
  
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(c) <math>f(x) \neq 0</math> for every <math>0 \leq x < 2</math>.
 
(c) <math>f(x) \neq 0</math> for every <math>0 \leq x < 2</math>.
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==Solution 1==
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For all <math>x+y\ge2</math>, there exists a nonnegative real <math>t</math> such that <math>t+2=x+y</math> and <math>f(x+y)=f(t+2)=f(tf(2))f(2)=0</math>. Hence, <math>f(x)=0</math> for all <math>x\ge 2</math>.
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For <math>x<2</math>, take <math>0=f(2)=f((2-x)+(x))=f((2-x)f(x))f(x)</math>.
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Since <math>f(x)\neq0</math> for <math>x<2</math>, then <math>f((2-x)f(x))=0</math>.
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Thus, <math>(2-x)f(x)\ge2\Leftrightarrow f(x)\ge\frac{2}{2-x}</math> for all <math>x<2</math>.
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Suppose <math>f(a)>\frac{2}{2-a}</math> so that <math>f(a)=\frac{2}{2-a-\epsilon}</math> for some <math>\epsilon>0</math> and <math>0\le a<2</math>.
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Then <math>f\left((2-a-\epsilon)f(a)\right)f(a)=f(2-\epsilon)</math>. The left hand side is equivalent to <math>f\left((2-a-\epsilon)\left(\frac{2}{2-a-\epsilon}\right)\right)f(a)=f(2)=0</math>, but the right hand side is nonzero, since <math>2-\epsilon<2</math> and <math>f(x)\neq0</math> for all <math>x<2</math>.
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Hence, <math>f(x)=\frac{2}{2-x}</math> for <math>0\le a<2</math>.
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For <math>x+y<2</math>,\begin{eqnarray*}f(xf(y))f(y) &=& f\left((x)\left(\frac{2}{2-y}\right)\right)\left(\frac{2}{2-y}\right)\\ &=&\left(\frac{2}{2-\frac{2x}{2-y}}\right)\left(\frac{2}{2-y}\right)\\ &=&\frac{4}{4-2y-2x}\\ &=&\frac{2}{2-y-x}\\ f(x+y)&=&\frac{2}{2-y-x}.\end{eqnarray*}
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Indeed, <math>f(xf(y))f(y)=f(x+y)</math>, so the desired function is
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<cmath>f(x)=\begin{cases}\frac{2}{2-x},& 0\le x<2\\0,& x\ge2.\end{cases}  </cmath>
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This solution was posted and copyrighted by Shen Kislay Kai. The original thread for this problem can be found here: [https://aops.com/community/p397033]
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==Solution 2==
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Taking <math>y=2</math> gives <math>f(x+2)=0 \forall x\ge 0</math>, or equivalently <math>f(x)=0  \forall x\ge 2</math>. Taking <math>y=2-x</math> for <math>0\le x<2</math> yields<cmath>f(xf(2-x))f(2-x)=f(2)=0</cmath>Since <math>0<2-x\le 2</math>, we must have <math>f(xf(2-x))=0</math>, which forces<cmath>xf(2-x)\ge 2\implies f(x)\ge \frac{2}{2-x}\forall 0\le x<2.</cmath>Now taking <math>x=2/f(y)</math>, for <math>0\le y<2</math> gives,<cmath>f(y+2/f(y))=f(2)f(y)=0\implies y+2/f(y)\ge 2.</cmath>Thus,<cmath>f(y)\le \frac{2}{2-y}\forall 0\le y<2</cmath>Hence, this forces <math>f(x)=2/(2-x)</math> for <math>0\le x<2</math>, giving a final answer of<cmath> f(x)=\begin{cases} 0 & \text{ if }x \ge 2 \\ \frac{2}{2-x} & \text{ if } 0 \le x < 2 . \end{cases} </cmath>
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This solution was posted and copyrighted by winnertakeover. The original thread for this problem can be found here: [https://aops.com/community/p11534549]
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==Solution 3==
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Let <math>P(x,y)</math> be the assertion. Note that <math>P(x,2)\implies f(x+2)=0.</math>
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Since <math>x</math> is defined for non-negative reals, we have <math>x+2\geq 2.</math> So, <math>\boxed{f(x)=0\qquad \forall x\geq2}.</math>
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Now, <math>P(2-y,y)\implies f((2-y)f(y))f(y)=0.</math>
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Here, we are working with <math>0\leq y<2.</math> Since we may not have <math>f(y)=0</math> due to condition (iii), we must have <math>f((2-y)f(y))=0.</math>
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From our first solution, it is imperative that <math>(2-y)f(y)\geq 2.</math> So, <math>f(y)\geq \dfrac{2}{2-y} \qquad (1).</math>
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Now consider <math>P(\dfrac{2}{f(y),y}</math>. We get,<cmath>0=f\left(\dfrac{2+yf(y)}{f(y)}\right).</cmath>Similar to the previous step, we have <math>\dfrac{2+yf(y)}{f(y)}\geq 2.</math> Re arranging this, we get <math>f(y)\leq \dfrac{2}{2-y}\qquad (2).</math>
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Since <math>f(y)</math> satisfies both <math>(1)</math> and <math>(2),</math> we must have <math>\boxed{f(x)=\dfrac{2}{2-x}\qquad \forall x\in [0,2)}.</math>
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Of course, we plug both solution back to see if they satisfy the three conditions. Sure enough, they do!
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This solution was posted and copyrighted by proshi. The original thread for this problem can be found here: [https://aops.com/community/p11129743]
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== See Also == {{IMO box|year=1986|num-b=4|num-a=6}}
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[[Category:Olympiad Algebra Problems]]
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[[Category:Functional Equation Problems]]

Latest revision as of 12:11, 3 September 2024

Problem

Find all (if any) functions $f$ taking the non-negative reals onto the non-negative reals, such that

(a) $f(xf(y))f(y) = f(x+y)$ for all non-negative $x$, $y$;

(b) $f(2) = 0$;

(c) $f(x) \neq 0$ for every $0 \leq x < 2$.

Solution 1

For all $x+y\ge2$, there exists a nonnegative real $t$ such that $t+2=x+y$ and $f(x+y)=f(t+2)=f(tf(2))f(2)=0$. Hence, $f(x)=0$ for all $x\ge 2$.

For $x<2$, take $0=f(2)=f((2-x)+(x))=f((2-x)f(x))f(x)$. Since $f(x)\neq0$ for $x<2$, then $f((2-x)f(x))=0$. Thus, $(2-x)f(x)\ge2\Leftrightarrow f(x)\ge\frac{2}{2-x}$ for all $x<2$.

Suppose $f(a)>\frac{2}{2-a}$ so that $f(a)=\frac{2}{2-a-\epsilon}$ for some $\epsilon>0$ and $0\le a<2$. Then $f\left((2-a-\epsilon)f(a)\right)f(a)=f(2-\epsilon)$. The left hand side is equivalent to $f\left((2-a-\epsilon)\left(\frac{2}{2-a-\epsilon}\right)\right)f(a)=f(2)=0$, but the right hand side is nonzero, since $2-\epsilon<2$ and $f(x)\neq0$ for all $x<2$.

Hence, $f(x)=\frac{2}{2-x}$ for $0\le a<2$. For $x+y<2$,\begin{eqnarray*}f(xf(y))f(y) &=& f\left((x)\left(\frac{2}{2-y}\right)\right)\left(\frac{2}{2-y}\right)\\ &=&\left(\frac{2}{2-\frac{2x}{2-y}}\right)\left(\frac{2}{2-y}\right)\\ &=&\frac{4}{4-2y-2x}\\ &=&\frac{2}{2-y-x}\\ f(x+y)&=&\frac{2}{2-y-x}.\end{eqnarray*} Indeed, $f(xf(y))f(y)=f(x+y)$, so the desired function is \[f(x)=\begin{cases}\frac{2}{2-x},& 0\le x<2\\0,& x\ge2.\end{cases}\]

This solution was posted and copyrighted by Shen Kislay Kai. The original thread for this problem can be found here: [1]

Solution 2

Taking $y=2$ gives $f(x+2)=0 \forall x\ge 0$, or equivalently $f(x)=0 \forall x\ge 2$. Taking $y=2-x$ for $0\le x<2$ yields\[f(xf(2-x))f(2-x)=f(2)=0\]Since $0<2-x\le 2$, we must have $f(xf(2-x))=0$, which forces\[xf(2-x)\ge 2\implies f(x)\ge \frac{2}{2-x}\forall 0\le x<2.\]Now taking $x=2/f(y)$, for $0\le y<2$ gives,\[f(y+2/f(y))=f(2)f(y)=0\implies y+2/f(y)\ge 2.\]Thus,\[f(y)\le \frac{2}{2-y}\forall 0\le y<2\]Hence, this forces $f(x)=2/(2-x)$ for $0\le x<2$, giving a final answer of\[f(x)=\begin{cases} 0 & \text{ if }x \ge 2 \\ \frac{2}{2-x} & \text{ if } 0 \le x < 2 . \end{cases}\]

This solution was posted and copyrighted by winnertakeover. The original thread for this problem can be found here: [2]

Solution 3

Let $P(x,y)$ be the assertion. Note that $P(x,2)\implies f(x+2)=0.$

Since $x$ is defined for non-negative reals, we have $x+2\geq 2.$ So, $\boxed{f(x)=0\qquad \forall x\geq2}.$

Now, $P(2-y,y)\implies f((2-y)f(y))f(y)=0.$

Here, we are working with $0\leq y<2.$ Since we may not have $f(y)=0$ due to condition (iii), we must have $f((2-y)f(y))=0.$

From our first solution, it is imperative that $(2-y)f(y)\geq 2.$ So, $f(y)\geq \dfrac{2}{2-y} \qquad (1).$

Now consider $P(\dfrac{2}{f(y),y}$. We get,\[0=f\left(\dfrac{2+yf(y)}{f(y)}\right).\]Similar to the previous step, we have $\dfrac{2+yf(y)}{f(y)}\geq 2.$ Re arranging this, we get $f(y)\leq \dfrac{2}{2-y}\qquad (2).$

Since $f(y)$ satisfies both $(1)$ and $(2),$ we must have $\boxed{f(x)=\dfrac{2}{2-x}\qquad \forall x\in [0,2)}.$

Of course, we plug both solution back to see if they satisfy the three conditions. Sure enough, they do!

This solution was posted and copyrighted by proshi. The original thread for this problem can be found here: [3]

See Also

1986 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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