Difference between revisions of "1999 IMO Problems/Problem 6"

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==Problem==
http://www.4shared.com/document/krpZ5Oeg/IMO_1999_-_6_Solution.html
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Determine all functions <math>f:\Bbb{R}\to \Bbb{R}</math> such that
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<cmath>f(x-f(y))=f(f(y))+xf(y)+f(x)-1</cmath>
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for all real numbers <math>x,y</math>.
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==Solution==
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Let <math>f(0) = c </math>.
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Substituting <math>x = y = 0 </math>, we get:
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<cmath>f(-c) = f(c) + c - 1. \hspace{1cm}    ... (1) </cmath>
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Now if c = 0, then:
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<cmath>f(0) = f(0) - 1 </cmath> which is not possible.
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<math>\implies c \neq 0 </math>.
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Now substituting <math>x = f(y) </math>, we get
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<cmath>c = f(x) + x^{2} + f(x) - 1 </cmath>.
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Solving for <math>f(x) </math>, we get <cmath>f(x) = \frac{c + 1}{2} - \frac{x^{2}}{2}. \hspace{1cm}  ... (2) </cmath>
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This means <math>f(x) = f(-x) </math> because <math>x^{2} = (-x)^{2} </math>.
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Specifically, <cmath>f(c) = f(-c). \hspace{1cm}    ... (3) </cmath>
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Using equations <math>(1) </math> and <math>(3) </math>, we get:
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<cmath>f(c) = f(c) + c - 1 </cmath>
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which gives
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<cmath>c = 1 </cmath>.
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So, using this in equation <math>(2) </math>, we get
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<cmath>\boxed{f(x) = 1 - \frac{x^{2}}{2}} </cmath> as the only solution to this functional equation.
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==See Also==
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{{IMO box|year=1999|num-b=5|after=Last Question}}
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[[Category:Olympiad Algebra Problems]]
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[[Category:Functional Equation Problems]]

Latest revision as of 06:50, 24 June 2024

Problem

Determine all functions $f:\Bbb{R}\to \Bbb{R}$ such that

\[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]

for all real numbers $x,y$.

Solution

Let $f(0) = c$. Substituting $x = y = 0$, we get:

\[f(-c) = f(c) + c - 1. \hspace{1cm}    ... (1)\] Now if c = 0, then:

\[f(0) = f(0) - 1\] which is not possible.

$\implies c \neq 0$.

Now substituting $x = f(y)$, we get

\[c = f(x) + x^{2} + f(x) - 1\].

Solving for $f(x)$, we get \[f(x) = \frac{c + 1}{2} - \frac{x^{2}}{2}. \hspace{1cm}  ... (2)\]

This means $f(x) = f(-x)$ because $x^{2} = (-x)^{2}$.

Specifically, \[f(c) = f(-c). \hspace{1cm}    ... (3)\]

Using equations $(1)$ and $(3)$, we get:

\[f(c) = f(c) + c - 1\]

which gives

\[c = 1\].

So, using this in equation $(2)$, we get

\[\boxed{f(x) = 1 - \frac{x^{2}}{2}}\] as the only solution to this functional equation.

See Also

1999 IMO (Problems) • Resources
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Last Question
All IMO Problems and Solutions