Difference between revisions of "2002 USAMO Problems/Problem 4"
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== See also == | == See also == |
Latest revision as of 22:09, 18 September 2018
Problem
Let be the set of real numbers. Determine all functions
such that
for all pairs of real numbers and
.
Solutions
Solution 1
We first prove that is odd.
Note that , and for nonzero
,
, or
, which implies
. Therefore
is odd. Henceforth, we shall assume that all variables are non-negative.
If we let , then we obtain
. Therefore the problem's condition becomes
.
But for any , we may set
,
to obtain
.
(It is well known that the only continuous solutions to this functional equation are of the form , but there do exist other solutions to this which are not solutions to the equation of this problem.)
We may let ,
to obtain
.
Letting and
in the original condition yields
But we know , so we have
, or
.
Hence all solutions to our equation are of the form . It is easy to see that real value of
will suffice.
Solution 2
As in the first solution, we obtain the result that satisfies the condition
.
We note that
.
Since , this is equal to
It follows that must be of the form
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2002 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.