Difference between revisions of "2018 USAMO Problems/Problem 2"
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Latest revision as of 10:40, 27 August 2023
Problem 2
Find all functions such that
for all
with
Solution
Obviously, the output of lies in the interval
. Define
as
. Then for any
such that
, we have
. We can transform
and
similarly:
Let ,
,
. We can see that the above expression is equal to
. That is, for any
such that
,
.
(To motivate this, one can start by writing ,
,
, and normalizing such that
.)
For convenience, we define as
, so that for any
such that
, we have
Obviously, . If
, then
and thus
. Furthermore, if
are in the domain and
, then
and thus
.
At this point, we should realize that should be of the form
. We first prove this for some rational numbers. If
is a positive integer and
is a real number such that
, then we can repeatedly apply
to obtain
. Let
, then for any rational number
where
are positive integers, we have
.
Next, we prove it for all real numbers in the interval . For the sake of contradiction, assume that there is some
such that
. Let
, then obviously
. The idea is to "amplify" this error until it becomes so big as to contradict the bounds on the output of
. Let
, so that
and
. Pick any rational
, so that
All numbers and sums are safely inside the bounds of
. Thus
but picking any rational number
gives us
, and since
, we have
as well, but since
, this means that
, giving us the desired contradiction.
We now know that for all
. Since
for
, we obtain
for all
. For
, we have
, and thus
as well. So
for all
in the domain. Since
is bounded by
and
, we have
. It remains to work backwards to find
.
- wzs26843545602
2018 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |