Difference between revisions of "2012 USAMO Problems/Problem 6"
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Revision as of 17:14, 3 July 2013
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
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
.
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.