2010 IMO Problems/Problem 1
Find all function such that for all the following equality holds
where is greatest integer not greater than
Author: Pierre Bornsztein, France
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)
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 )
Let , then .
Then is a constant. Let , then . It is easy to check that this are solutions.
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  )
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 )
|2010 IMO (Problems) • Resources)|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All IMO Problems and Solutions|