Difference between revisions of "2012 USAMO Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
− | Let <math>x_i=x_1,x_2,...,x_n</math> It is evident that <math>x_i = \frac{(-1)^i}{\sqrt{n}}</math> for evens because of the second equation and <math>x_i=\frac{(-1)^i}{\sqrt{n-1}}</math> for odds(one term will be 0 to maintain the first condition) | + | Let <math>x_i=x_1,x_2,...,x_n</math> It is evident that <math>x_i = \frac{(-1)^i}{\sqrt{n}}</math> for evens because of the second equation and <math>x_i=\frac{(-1)^i}{\sqrt{n-1}}</math> 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 <math>\lambda = \frac{1}{\sqrt{n}}</math>: | We may then try and get an expression for the maximum number of sets that satisfy this which occur when <math>\lambda = \frac{1}{\sqrt{n}}</math>: | ||
since it will be | since it will be |
Revision as of 21:07, 12 December 2022
Contents
[hide]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 which gives us:
and
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
See Also
2012 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
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.