Difference between revisions of "2010 IMO Problems/Problem 1"
m (→See also) |
m (→See Also) |
||
(4 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
== 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 == | ||
− | == Solution 1 == | + | === Solution 1 === |
Put <math>x=y=0</math>. Then <math>f(0)=0</math> or <math>\lfloor f(0) \rfloor=1</math>. | Put <math>x=y=0</math>. Then <math>f(0)=0</math> or <math>\lfloor f(0) \rfloor=1</math>. | ||
Line 29: | Line 29: | ||
So, <math>f(x)=0, \ \forall x</math> or <math>f(x)=a, \ \forall x, \ a \in [1,2)</math>. ( By socrates[http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=10045&]) | So, <math>f(x)=0, \ \forall x</math> or <math>f(x)=a, \ \forall x, \ a \in [1,2)</math>. ( By socrates[http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=10045&]) | ||
− | == Solution 2 == | + | === Solution 2 === |
Substituting <math>y=0</math> we have <math>f(0) = f(x) [f(0)]</math>. | Substituting <math>y=0</math> we have <math>f(0) = f(x) [f(0)]</math>. | ||
Line 43: | Line 43: | ||
Then the only solutions are <math>f(x)=0</math> or <math>f(x)=c</math> where <math>c \in [1,2)</math>.( By m.candales [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=35756&]) | Then the only solutions are <math>f(x)=0</math> or <math>f(x)=c</math> where <math>c \in [1,2)</math>.( By m.candales [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=35756&]) | ||
− | == Solution 3 == | + | === Solution 3 === |
Let <math>y=0</math>, then <math>f(0)=f(x)\left\lfloor f(0)\right\rfloor</math>. | Let <math>y=0</math>, then <math>f(0)=f(x)\left\lfloor f(0)\right\rfloor</math>. | ||
Line 74: | Line 74: | ||
) | ) | ||
− | == Solution 4 == | + | === Solution 4 === |
Clearly <math>f(\left\lfloor x\right\rfloor y) = f(\left\lfloor \lfloor x \rfloor \right\rfloor y) = f(\lfloor x \rfloor)\left\lfloor f(y)\right\rfloor</math>, so <math>(f(x) - f(\lfloor x \rfloor))\left\lfloor f(y)\right\rfloor = 0</math> for all <math>x,y\in\mathbb{R}</math>. | Clearly <math>f(\left\lfloor x\right\rfloor y) = f(\left\lfloor \lfloor x \rfloor \right\rfloor y) = f(\lfloor x \rfloor)\left\lfloor f(y)\right\rfloor</math>, so <math>(f(x) - f(\lfloor x \rfloor))\left\lfloor f(y)\right\rfloor = 0</math> for all <math>x,y\in\mathbb{R}</math>. | ||
Line 89: | Line 89: | ||
− | == See | + | == See Also == |
{{IMO box|year=2010|before=First question|num-a=2}} | {{IMO box|year=2010|before=First question|num-a=2}} | ||
− | [[Category:Olympiad | + | [[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 |