Difference between revisions of "2019 USAJMO Problems/Problem 5"
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==Solution 2== | ==Solution 2== | ||
− | There are <math>\frac{(2n)!}{2^n}</math> ways to choose <math>S_{0,0}, S_{1,1} .... S_{n,n}</math>, since, there are <math>\binom{2n}{2}</math> ways to choose <math>S_{1,1}</math>, and you can keep going down the line, and you get that there are <math>\frac{(2n)!}{2^n}</math> ways to pick <math>S_{0,0}, | + | There are <math>\frac{(2n)!}{2^n}</math> ways to choose <math>S_{0,0}, S_{1,1} .... S_{n,n}</math>, since, there are <math>\binom{2n}{2}</math> ways to choose <math>S_{1,1}</math>, and you can keep going down the line, and you get that there are <math>\frac{(2n)!}{2^n}</math> ways to pick <math>S_{0,0}, S_{1,1} ... S_{n,n}</math> Then we can fill out the rest of the gird. First, let’s prove a lemma. |
===Lemma=== | ===Lemma=== | ||
− | If we know what <math>S_{a,b}</math> is and what <math>S_{a+1,b+1}</math> is, then there are 2 choices for both <math>S_{a,b+1}</math> and <math>S_{a+1,b}</math>. It’s easy to see why. Let A be a set that contains all the elements in S_{a+1,b+1} that are not in S_{a,b} By the problem statement, <math>S_{a,b} \subseteq S_{a+1,b}, S_{a,b+1} \subseteq S_{a+1,b+1}</math>, and using the fact that <math> S_{a,b+1}</math> is 1 element larger than <math>S_{a,b}</math> , and 1 element smaller than <math>S_{a+1,b+1}</math> and same with <math>S_{a+1,b}</math>, we get that, <math>S_{a+1,b}, S_{a,b+1} = S_{a,b}+</math> some element, and, <math>S_{a+1,b}, S_{b,a+1} = S_{a+1,b+1}-</math> some other element. We can combine the two equations and get that <math>S_{a,b}+</math>some element <math>= S_{a+1,b+1}-</math>some other element. Rearranging the sides, and main the substitution for A, we get, <math>A =</math> some element + some other element. And, there are obviously 2 ways to pick what the “some element” is, and that leads to 2 choices for both <math>S_{a,b+1}</math> and <math>S_{a+1,b}</math>. | + | If we know what <math>S_{a,b}</math> is and what <math>S_{a+1,b+1}</math> is, then there are 2 choices for both <math>S_{a,b+1}</math> and <math>S_{a+1,b}</math>. It’s easy to see why. Let A be a set that contains all the elements in <math>S_{a+1,b+1}</math> that are not in <math>S_{a,b}</math> By the problem statement, <math>S_{a,b} \subseteq S_{a+1,b}, S_{a,b+1} \subseteq S_{a+1,b+1}</math>, and using the fact that <math> S_{a,b+1}</math> is 1 element larger than <math>S_{a,b}</math> , and 1 element smaller than <math>S_{a+1,b+1}</math> and same with <math>S_{a+1,b}</math>, we get that, <math>S_{a+1,b}, S_{a,b+1} = S_{a,b}+</math> some element, and, <math>S_{a+1,b}, S_{b,a+1} = S_{a+1,b+1}-</math> some other element. We can combine the two equations and get that <math>S_{a,b}+</math>some element <math>= S_{a+1,b+1}-</math>some other element. Rearranging the sides, and main the substitution for A, we get, <math>A =</math> some element + some other element. And, there are obviously 2 ways to pick what the “some element” is, and that leads to 2 choices for both <math>S_{a,b+1}</math> and <math>S_{a+1,b}</math>. |
===Filling in the rest of the grid=== | ===Filling in the rest of the grid=== |
Revision as of 13:25, 20 April 2019
Let be a nonnegative integer. Determine the number of ways that one can choose
sets
, for integers
with
, such that:
1. for all , the set
has
elements; and
2. whenever
and
.
Proposed by Ricky Liu
Contents
Solution 1
Note that there are ways to choose
, because there are
ways to choose which number
is,
ways to choose which number to append to make
,
ways to choose which number to append to make
... After that, note that
contains the
in
and 1 other element chosen from the 2 elements in
not in
so there are 2 ways for
. By the same logic there are 2 ways for
as well so
total ways for all
, so doing the same thing
more times yields a final answer of
.
-Stormersyle
Solution 2
There are ways to choose
, since, there are
ways to choose
, and you can keep going down the line, and you get that there are
ways to pick
Then we can fill out the rest of the gird. First, let’s prove a lemma.
Lemma
If we know what is and what
is, then there are 2 choices for both
and
. It’s easy to see why. Let A be a set that contains all the elements in
that are not in
By the problem statement,
, and using the fact that
is 1 element larger than
, and 1 element smaller than
and same with
, we get that,
some element, and,
some other element. We can combine the two equations and get that
some element
some other element. Rearranging the sides, and main the substitution for A, we get,
some element + some other element. And, there are obviously 2 ways to pick what the “some element” is, and that leads to 2 choices for both
and
.
Filling in the rest of the grid
We used our proved lemma, and we can fill in then we can fill in the next diagonal, until all
are filled, where
But, we haven’t finished everything! But filling out the rest of the diagonals in a similar fashion is pretty simple. And, it’s easy to see that we have made
decisions, each with 2 choices, when filling out the rest of the grid, so there are
ways to finish off.
Finishing off
To finish off, we have ways to fill in the gird, which gets us
-Alexlikemath
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2019 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |