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

Latest revision as of 06:50, 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, $f(c) = f(-c). \hspace{1cm} ... (3)$

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.

@@ Line 8: / Line 8: @@
 ==Solution==
-{solution}
+Let <math>f(0) = c </math>.
+Substituting <math>x = y = 0 </math>, we get:
+<cmath>f(-c) = f(c) + c - 1. \hspace{1cm}    ... (1) </cmath>
+Now if c = 0, then:
+<cmath>f(0) = f(0) - 1 </cmath> which is not possible.
+<math>\implies c \neq 0 </math>.
+Now substituting <math>x = f(y) </math>, we get
+<cmath>c = f(x) + x^{2} + f(x) - 1 </cmath>.
+Solving for <math>f(x) </math>, we get <cmath>f(x) = \frac{c + 1}{2} - \frac{x^{2}}{2}. \hspace{1cm}  ... (2) </cmath>
+This means <math>f(x) = f(-x) </math> because <math>x^{2} = (-x)^{2} </math>.
+Specifically, <cmath>f(c) = f(-c). \hspace{1cm}    ... (3) </cmath>
+Using equations <math>(1) </math> and <math>(3) </math>, we get:
+<cmath>f(c) = f(c) + c - 1 </cmath>
+which gives
+<cmath>c = 1 </cmath>.
+So, using this in equation <math>(2) </math>, we get
+<cmath>\boxed{f(x) = 1 - \frac{x^{2}}{2}} </cmath> 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