Difference between revisions of "1994 IMO Problems/Problem 5"

Latest revision as of 01:27, 22 November 2023

Problem

Let $S$ be the set of real numbers strictly greater than $-1$ . Find all functions $f:S \to S$ satisfying the two conditions:

1. $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$ for all $x$ and $y$ in $S$ ;

2. $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1<x<0$ and $0<x$ .

Solution

The only solution is $f(x) = \frac{-x}{x+1}.$

Setting $x=y,$ we get

$f(x+f(x)+xf(x)) = x+f(x)+xf(x).$

Therefore, $f(s) = s$ for $s = x+f(x)+xf(x).$

Note: If we can show that $s$ is always $0,$ we will get that $x+f(x)+xf(x) = 0$ for all $x$ in $S$ and therefore, $f(x) = \frac{-x}{x+1}.$

Let $f(s)=s.$ Setting $x=y=s,$ we get

$f(s + f(s) + sf(s)) = f(2s + s^2) = 2s+s^2.$

If $t=2s+s^2,$ we have $f(t)=t$ as well.

Consider $s > 0.$ We get $t = 2s + s^2 > s.$ Since $\frac{f(x)}{x}$ is strictly increasing for $x>0$ and $t > s$ in this domain, we must have $\frac{f(t)}{t} > \frac{f(s)}{s}$ but since $f(s) = s$ and $f(t) = t,$ we also have that $\frac{f(s)}{s} = 1 = \frac{f(t)}{t}$ which is a contradiction. Therefore $s \leq 0$

Consider $-1 < s < 0.$ Using a similar argument, we will get that $t = 2s + s^2 < s$ but also $\frac{f(s)}{s} = 1 = \frac{f(t)}{t},$ which is a contradiction.

Hence, $s$ must be $0.$ Since $s=0,$ we can conclude that $x+f(x)+xf(x) = 0$ and therefore, $f(x) = \frac{-x}{x+1}$ for all $x$ .

@@ Line 30: / Line 30: @@
 Hence, <math>s</math> must be <math>0.</math> Since <math>s=0,</math> we can conclude that <math>x+f(x)+xf(x) = 0</math> and therefore, <math>f(x) = \frac{-x}{x+1}</math> for all <math>x</math>.
+==See Also==
+{{IMO box|year=1994|num-b=4|num-a=6}}

1994 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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Difference between revisions of "1994 IMO Problems/Problem 5"

Latest revision as of 01:27, 22 November 2023

Problem

Solution

See Also