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 | ||
− | |||
<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 == |
Revision as of 11:01, 9 April 2012
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 |