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 | ||
<cmath>x_1 + x_2 + ... + x_n </cmath> | <cmath>x_1 + x_2 + ... + x_n </cmath> | ||
− | for any choice of A we pick, it will have to be greater than <math>\frac{1}{\sqrt{n}}</math> which means we can either pick 0 negative <math>x_m</math> or <math>\frac{ | + | for any choice of A we pick, it will have to be greater than <math>\frac{1}{\sqrt{n}}</math> which means we can either pick 0 negative <math>x_m</math> or <math>j-1</math> negatives for j positive terms. Since we also have that there are <math>\frac{n}{2}</math> positive and negative terms for evens. Which then gives us: |
− | <cmath>\sum_{k= | + | <cmath>\sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k}\sum_{i=0}^{k-1} \binom{\frac{n}{2}}{i}</cmath> |
and | and | ||
− | <cmath>\sum_{k= | + | <cmath>\sum_{k=1}^{\frac{n}{2}}\binom{\frac{n}{2}}{k}\left(2^{\frac{n}{2}} - \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i}\right)</cmath> |
− | + | 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: <cmath>\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\binom{\lfloor \frac{n}{2}\rfloor}{k}\left(2^{\lfloor\frac{n}{2}\rfloor} - \sum_{i=k}^{\lfloor\frac{n}{2}\rfloor}\binom{\lfloor \frac{n}{2}\rfloor}{i}\right) \le n2^{n-3}</cmath> | |
− | + | Thus, we may say that this holds to be true for all <math>n \ge 2</math> since <math>n2^{n-3}</math> grows faster than the sum. Note that equality holds when <math>S_A \in \{\lambda,0,-\lambda\}</math> for all i which occurs when <math>x_i= \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,....</math> and <math>\lambda = \frac{1}{\sqrt{2}}</math> since <math>S_A=\frac{1}{\sqrt{2}}</math> is the only choice for <math>S_A \ge \lambda</math> | |
− | 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: | ||
− | <cmath>2^{\lfloor\frac{n}{2}\rfloor}- | ||
− | Thus, we may say that this holds to be true for all <math>n \ge 2</math>. Note that equality holds when <math>S_A \in \{\lambda,0,-\lambda\}</math> for all i which occurs when <math>x_i= \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,....</math> | ||
<math>\blacksquare</math> | <math>\blacksquare</math> | ||
+ | |||
+ | ==Note:== | ||
+ | The proof to the last inequality is as follows: | ||
+ | First we may rewrite this as being: | ||
+ | <cmath>\sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \left( 2^{\frac{n}{2}} - \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i}\right) \le n2^{n-3}</cmath> | ||
+ | Thus, | ||
+ | <cmath>2^n - 2^{\frac{n}{2}} - \sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i} \le n2^{n-3}</cmath> | ||
+ | (the second equation is because the sum of the binomial coefficient is <math>2^{\frac{n}{2}} - 1</math>) | ||
+ | <cmath>2^n - 2^{\frac{n}{2}} \le n2^{n-3} + \sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i}</cmath> | ||
+ | Since <math>2^n - 2^{\frac{n}{2}} \le n2^{n-3}</math> for all <math>n \ge 8</math> and <math>\sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i} \ge 0</math> for all <math>k</math> and <math>i</math>, it is apparent that: | ||
+ | <cmath>\sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \sum_{i=0}^{k-1} \binom{\frac{n}{2}}{i} \le n2^{n-3}</cmath> | ||
+ | must be true for all <math>n \ge 8</math>(because if we rewrite this we get <math>2^{\frac{n}{2}}\left(8-n\right) \le 8</math>.) For all <math>2 \le n < 8</math> however, we may use some logic to first layout a plan. Since for <math>n=6,n=4,</math> and <math>n=2</math>, <math>2^{n} - 2^{\frac{n}{2}} = n2^{n-3} + 1</math>, we may say that whole sum will be less than <math>n2^{n-3}</math> because <math>\sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i} \ge 1</math> for all <math>n \ge 2</math> Plugging this inequality back in gives us: | ||
+ | <cmath>2^{n} - 2^{\frac{n}{2}} - \sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i} \le 2^{n} - 2^{\frac{n}{2}} - 1 = n2^{n-3}</cmath> | ||
+ | because of the fact that <math>- \sum_{k=1}^{\frac{n}{2}} \binom{\frac{n}{2}}{k} \sum_{i=k}^{\frac{n}{2}} \binom{\frac{n}{2}}{i} \le -1</math> | ||
+ | |||
==See Also== | ==See Also== | ||
*[[USAMO Problems and Solutions]] | *[[USAMO Problems and Solutions]] |
Latest revision as of 12:37, 21 June 2023
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. 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
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.