2012 USAMO Problems/Problem 6
Contents
Problem
For integer , let
,
,
,
be real numbers satisfying
For each subset
, define
(If
is the empty set, then
.)
Prove that for any positive number , the number of sets
satisfying
is at most
. For what choices of
,
,
,
,
does equality hold?
Solution 1
For convenience, let .
Note that , so the sum of the
taken two at a time is
. Now consider the following sum:
Since , it follows that at most
sets
have
.
Now note that . It follows that at most half of the
such that
are positive. This shows that at most
sets
satisfy
.
Note that if equality holds, every subset of
has
. It immediately follows that
is a permutation of
. Since we know that
, we have that
.
Solution 2
Let It is evident that
for evens because of the second equation and
for odds(one term will be 0 to maintain the first condition).
We may then try and get an expression for the maximum number of sets that satisfy this which occur when
:
since it will be
for any choice of A we pick, it will have to be greater than
which means we can either pick 0 negative
or
negatives for j positive terms. Since we also have that there are
positive and negative terms for evens. Which then gives us:
and
For odd values, let it be the same as the last even valued sequence where n is even(i.e. the same as the sequence before it but with an extra 0 in one of the spots). Then, the following is apparent:
Thus, we may say that this holds to be true for all
since
grows faster than the sum. Note that equality holds when
for all i which occurs when
and
since
is the only choice for
Note:
The proof to the last inequality is as follows:
First we may rewrite this as being:
Thus,
(the second equation is because the sum of the binomial coefficient is
)
Since
for all
and
for all
and
, it is apparent that:
must be true for all
(because if we rewrite this we get
.) For all
however, we may use some logic to first layout a plan. Since for
and
,
, we may say that whole sum will be less than
because
for all
Plugging this inequality back in gives us:
because of the fact that
Mathiscool109 (talk) 21:22, 13 December 2022 (EST)
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