Difference between revisions of "1999 IMO Problems/Problem 6"
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− | Solution | + | ==Problem== |
− | + | ||
+ | Determine all functions <math>f:\Bbb{R}\to \Bbb{R}</math> such that | ||
+ | |||
+ | <cmath>f(x-f(y))=f(f(y))+xf(y)+f(x)-1</cmath> | ||
+ | |||
+ | for all real numbers <math>x,y</math>. | ||
+ | |||
+ | ==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== | ||
+ | |||
+ | {{IMO box|year=1999|num-b=5|after=Last Question}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Latest revision as of 06:50, 24 June 2024
Problem
Determine all functions such that
for all real numbers .
Solution
Let . Substituting , we get:
Now if c = 0, then:
which is not possible.
.
Now substituting , we get
.
Solving for , we get
This means because .
Specifically,
Using equations and , we get:
which gives
.
So, using this in equation , we get
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 |