Difference between revisions of "2010 IMO Problems/Problem 1"
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== Problem == | == Problem == | ||
Find all function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> such that for all <math>x,y\in\mathbb{R}</math> the following equality holds | Find all function <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> such that for all <math>x,y\in\mathbb{R}</math> the following equality holds | ||
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<math>f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor </math> | <math>f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor </math> | ||
+ | where <math>\left\lfloor a\right\rfloor </math> is greatest integer not greater than <math>a.</math> | ||
− | + | ''Author: Pierre Bornsztein, France '' | |
== Solutions == | == Solutions == | ||
Line 93: | Line 93: | ||
[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Latest revision as of 08:49, 19 July 2016
Contents
Problem
Find all function such that for all the following equality holds
where is greatest integer not greater than
Author: Pierre Bornsztein, France
Solutions
Solution 1
Put . Then or .
If , putting we get , that is f is constant. Substituing in the original equation we find or , where .
If , putting we get or .
For , we set to find , which is a solution.
For , setting yields .
Putting to the original we get . However, from we have , so which contradicts the fact .
So, or . ( By socrates[1])
Solution 2
Substituting we have . If then . Then is constant. Let . Then substituting that in (1) we have , or . Therefore where or
If then . Now substituting we have . If then and substituting this in (1) we have . Then . Substituting we get . Then , which is a contradiction Therefore . and then for all
Then the only solutions are or where .( By m.candales [2])
Solution 3
Let , then .
Case 1:
Then is a constant. Let , then . It is easy to check that this are solutions.
Case 2:
In this case we conclude that
Lemma:If is such that ,
Proof of the Lemma: If we have that , as desired.
Let , so that we have:
, using the lemma.
If is not constant and equal to , letting be such that implies that .
Now it's enough to notice that any real number is equal to , where and , so that . Since was arbitrary, we have that is constant and equal to .
We conclude that the solutions are , where .( By Jorge Miranda [3]
)
Solution 4
Clearly , so for all .
If for all , then by taking we get , so is identically null (which checks).
If, contrariwise, for some , it follows for all .
Now it immediately follows , hence .
For this implies . Assume ; then , absurd.
Therefore , and now in the given functional equation yields for all , therefore constant, with , i.e. (which obviously checks).( By mavropnevma [4])
See Also
2010 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |