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

Revision as of 06:48, 24 June 2024

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 $x,y$ .

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,$ (Error compiling LaTeX. Unknown error_msg)f(c) = f(-c). \hspace{1cm} ... (3) $$ (Error compiling LaTeX. Unknown error_msg)

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

$$ (Error compiling LaTeX. Unknown error_msg)\boxed{f(x) = 1 - \frac{x^{2}}{2}} $ as the only solution to this functional equation.

@@ Line 11: / Line 11: @@
 Substituting <math>x = y = 0 </math>, we get:
-<cmath>f(-c) = f(c) + c - 1.    ... (1) </cmath>
+<cmath>f(-c) = f(c) + c - 1. \hspace{1cm}    ... (1) </cmath>
 Now if c = 0, then:
@@ Line 22: / Line 22: @@
 <cmath>c = f(x) + x^{2} + f(x) - 1 </cmath>.
-Solving for f(x), we get <math>f(x) = \frac{c + 1}{2} - \frac{x^{2}}{2}.   ... (2) </math><math>
+Solving for f(x), we get <math>f(x) = \frac{c + 1}{2} - \frac{x^{2}}{2}. \hspace{1cm}  ... (2) </math><math>
 This means </math>f(x) = f(-x) <math> because </math>x^{2} = (-x)^{2} <math>.
-Specifically, </math>f(c) = f(-c).     ... (3) <math></math>
+Specifically, </math>f(c) = f(-c). \hspace{1cm}    ... (3) <math></math>
 Using equations <math>(1) </math> and <math>(3) </math>, we get:

1999 IMO (Problems) • Resources
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