Difference between revisions of "2007 IMO Shortlist Problems/A4"
(arrgh, didn't have time to finish writing this solution) |
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== Solution == | == Solution == | ||
− | We will show that <math>f(x) = 2x</math> is the unique solution to this equation. To this end, let <math>g(x) = f(x) - x</math>. | + | We will show that <math>f(x) = 2x</math> is the unique solution to this equation. To this end, let <math>g(x) = f(x) - x</math>. The given condition then translates to |
+ | <cmath> g(x+y+ g(y)) + x+y+g(y) = g(x+y) + x+y + g(y) + y , </cmath> | ||
+ | or | ||
+ | <cmath> g(x+y+g(y)) = g(x+y) + y .</cmath> | ||
'''Lemma 1.''' The function <math>g</math> is injective. | '''Lemma 1.''' The function <math>g</math> is injective. | ||
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'''Lemma 3.''' For all <math>a,b</math>, <math>g(a+b) = g(a) + g(b)</math>. | '''Lemma 3.''' For all <math>a,b</math>, <math>g(a+b) = g(a) + g(b)</math>. | ||
+ | |||
+ | ''Proof.'' Pick an arbitrary positive real <math>c > \max(g(a)+g(b), g(a))</math>. Then by Lemma 2, | ||
+ | <cmath> g(c+g(a)+g(b)) = g(c+g(a))+b = g(c) + a+b = g(c+g(a+b)) . </cmath> | ||
+ | Since <math>g</math> is injective, it follows that <math>c+g(a)+g(b) = c + g(a+b)</math>. The lemma then follows. <math>\blacksquare</math> | ||
+ | |||
+ | Now, let <math>x</math> be any positive real; pick some <math>y>x</math>. Then by Lemmata 3 and 2, | ||
+ | <cmath> g(x) + g(y) = g(x+y) = g(y) + x . </cmath> | ||
+ | Hence <math>g(x)= x</math> and <math>f(x) = g(x) + x = 2x</math>. Therefore the function <math>x \mapsto 2x</math> is the only possible solution to the problem. Since this function evidently satisfies the problem's condition, it is the unique solution, as desired. <math>\blacksquare</math> | ||
+ | |||
+ | == Resources == | ||
+ | |||
+ | * [[2007 IMO Shortlist Problems]] | ||
+ | * <url>viewtopic.php?p=1165901#1165901 Discussion on AoPS/MathLinks</url> | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] |
Latest revision as of 16:44, 4 September 2008
Problem
(Thailand)
Find all functions such that
for all
. (Symbol
denotes the set of all positive real numbers [sic].)
Solution
We will show that is the unique solution to this equation. To this end, let
. The given condition then translates to
or
Lemma 1. The function is injective.
Proof. Suppose . Then
as desired.
Lemma 2. If , then
.
Proof. Set ,
.
Lemma 3. For all ,
.
Proof. Pick an arbitrary positive real . Then by Lemma 2,
Since
is injective, it follows that
. The lemma then follows.
Now, let be any positive real; pick some
. Then by Lemmata 3 and 2,
Hence
and
. Therefore the function
is the only possible solution to the problem. Since this function evidently satisfies the problem's condition, it is the unique solution, as desired.
Resources
- 2007 IMO Shortlist Problems
- <url>viewtopic.php?p=1165901#1165901 Discussion on AoPS/MathLinks</url>