[/quote]
,
be an arbitrary fixed point and 
an abitrary fixed point on 
.
be an abritrary point on 
.
intersect 
a
.
intersect 
.
intersect 
a second time at 
.
passes through a constant point 
as 
passes through a fixed point 
.
denote the set of positive integers. For any positive integer 
,
is called 
for all 
.
is defined below. 
Show that for all nonnegative integer 
,
is a positive integer.
be a root of 
.
satisfy ![\[a+b+c+d=0,\qquad \frac1a+\frac1b+\frac1c+\frac1d=0.\]](http://latex.artofproblemsolving.com/9/5/8/95849c32951bb80891e85d4b04da6034d328818e.png)
How many equalities ![\[ab=cd,\qquad ac=bd,\qquad ad=bc\]](http://latex.artofproblemsolving.com/2/9/7/29730d243048a530d2078d80daf0ca2522272554.png)
can simultaneously hold? Determine all possibilities.
be an acute triangle, 
its orthocenter, 
circumcenter and 
midpoint of 
Denote 
the intersection of line 
and the circumcircle 
intersection of line 
and 
Finally, let 
be the intersection of line 
and the circumcircle of 
Show that 
of a set of 21 distinct real numbers such that ![\[20|x-y|<(x+1)(y+1)?\]](http://latex.artofproblemsolving.com/3/0/e/30e7612bb44e1c889ec778ce928d258ea0c381d0.png)
cells of 
square grid can be colored in a way where no two dyed squares share a point and there are exactly two dyed squares in every column and every row.
(mod 4), determine the number of ordered integer triplets 
such that
be a sequence of positive numbers such that ![\[a_{n+1}=a_n+\frac{1}{a_n}\]](http://latex.artofproblemsolving.com/0/9/e/09ee35216612db492e63b57252ca6080c2c28f41.png)
for all positive integers 
Determine the greatest integer 
such that exactly 
)
be a positive integer and 
its greatest prime divisor. For each non-empty subset of divisors of 
write the sum of its elements on the board. Assume that more than 
are written and any of them occurs at most once. Show that all numbers on the board are distinct.
of real numbers, let 
denote the sum of the elements of 
of 
given by three of these points such that 
Y by GeoKing
A finite set 
of positive integers is called cardinal if contains the integer 
where denotes the number of distinct elements in . Let 
be a function from the set of positive integers to itself such that for any cardinal set , the set 
is also cardinal. Here denotes the set of all integers that can be expressed as 
where 
. Find all possible values of 
Proposed by Sutanay Bhattacharya
of positive integers is called cardinal if contains the integer 
where denotes the number of distinct elements in . Let 
be a function from the set of positive integers to itself such that for any cardinal set , the set 
is also cardinal. Here denotes the set of all integers that can be expressed as 
where 
. Find all possible values of 
Proposed by Sutanay Bhattacharya
This post has been edited 2 times. Last edited by math_comb01, Jan 21, 2024, 12:48 PM
is trivially a solution. Assume non-constant functions 
is the only such solution. We induct on 
.
which forces 
.
.
.
is also cardinal, which forces 
.
for all 
and 
for 
for some 
work?
and 
?
which can be proved by lots of casework.
gives a cardinal set every time. This is rather easy to note. Let's take 
.
.
.
where 
.
or 
but not both. As 
can be any real number, this is claiming that 
which will be false the moment we consider 
.
as a solution.![$f(x)=1, x \in [1,2023] \cup [2025, \infty]$](http://latex.artofproblemsolving.com/8/b/9/8b95950174cb6b3f96099c460f23c43dbee44070.png)
and 
also works. We cannot repeat this for more numbers as the set 
will get violated.
and 
?
?
.
is 
,
is 
and 
is 
.
.
,
.
.
.
cardinal, so 
(because if 
it would have 
,
.
is also cardinal. This implies that 
for each 
.
,
is cardinal. Since 
are fixed points, we have 
is cardinal. If 
,
,
is cardinal, which is false since it has 
elements. Therefore, we have 
and 
.
,
,
and get 
is cardinal, which is absurd. If 
,
gives that 
is cardinal, which is false. If 
,
,
gives that 
must be cardinal, which is false.
,
and 
for all 
,
for all 
,
,
,
,
,
must either be 
,
must either be 
.
.
.
.
,
.
,
,
,
.
.
,
.
,
.
,
a contradiction. If 
,
,
is required.
, but probably yes if you aren't in 12th. If you are in 12th, I don't know.
:
if and only if 
.
.
classes, we can choose a set 
.
,
.
integers satisfying 
is cardinal, and its image under 
.
.
is cardinal 
(other than 2), we can say that 
is also cardinal forcing 
,
.
or 
is cardinal 
is also cardinal forcing 
or 
,
.
.
,
.
.
is cardinal. But that means the set 
is cardinal which is clearly false (only 
distinct elements). Contradiction, so 
.
is cardinal, so 
is cardinal.
is cardinal. This set has 
.
.
and 
or 
,
.
,
where 
and 
.
is cardinal. So, 
,
is not cardinal, a contradiction. 
.
.
such that 
.
,
is not cardinal, a contradiction. 
by lemma 1, and 
either 
.
,
where 
.
,
and 
.
,
and 
.
and 
for all 
.
.
and the latter cannot be a cardinal set, a contradiction. 
,
to be consistent with lemma 2.
,
but the latter is not cardinal, which is impossible.
,
but the latter is not cardinal, which is impossible.
,
but the latter is not cardinal, which is impossible.
as another possibility.
,
or 2 ,
or 2.
as if 
,
is not cardinal.
using stong induction on 
being trivial. ASSUME holds for 
.
,
.
is cardinal. If 
So, set 
is not cardinal, though it was to be cardinal as 
is cardinal. Thus 
in this case.
.
,
,
is clearly not cardinal.
,
,
,
is clearly not cardinal.
,
is clearly not cardinal.
as
and 
.
or 
.
,
. Clearly these constructions are valid, now let's prove 
,
means that 
exactly once. For example 
.
that satisfy
. Since 
is cardinal, the set 
is also cardinal. Clearly 
where there are infinitely many 
that satisfy 
.
that satisfy
. Note that 
is cardinal. If 
,
is also cardinal which forces 
,
,
yields 
.
,
such that
. Note that the set 
is cardinal. If 
for some 
is cardinal. A contradiction, that means 
for all 
.
is cardinal, 
.
is cardinal, so 
.
is cardinal, so 
.
.
.
is cardinal, so 
.
is cardinal, so 
.
,
are cardinal, 
and 
or 
is as well, which implies that 
.
since then we can take 
which would mean that 
is cardinal which is contradiction. So 
.
as "ascending".
is ascending until a term that is less than or equal to the previous term.
,
,![$1, 2, 3, \dots, k, 1, [\text{some sequence of 1's and 2's}], x$](http://latex.artofproblemsolving.com/2/4/9/24942d4baf3dbf49d4d943b318cf8564eef2d256.png)
for some 
.
makes a contradiction. qed
to get 
.
.
.
gives contradiction for 
,
gives contradiction. For 
,
.
.
has 
elements, so 
has at most ![\[
1\;\le\;f(n)\;\le\;n.
\]](http://latex.artofproblemsolving.com/5/b/c/5bc37bcc9dab0611663c49b1ffbdc61909f00c7e.png)
is injective, then 
(apply to 
)![\[
f(n)=n
\quad\forall n,
\]](http://latex.artofproblemsolving.com/4/a/f/4af5e1e65ffd483160b683c1836c0651aa91ac0f.png)
and in particular 
.
with![\[
f(a)=f(b)=r>1.
\]](http://latex.artofproblemsolving.com/3/1/0/310a283641ed97eaf39534560bf1780a0ec14537.png)
Consider the set![\[
T=\{a,a+1,\dots,b\},
\]](http://latex.artofproblemsolving.com/9/a/e/9ae6b18c7824ef9f56ed65dc8971b034ebdc80ae.png)
which has 
.
is at 
and 
,![\[
\bigl|f(T)\bigr| \;=\; |T|-1 \;=\; b-a.
\]](http://latex.artofproblemsolving.com/6/3/2/6322b16e69a0fa1df248e842a9c231d682c2a707.png)
But 
must be cardinal, so 
.
,![\[
b-a \;=\; r.
\]](http://latex.artofproblemsolving.com/2/b/7/2b7f5cdbd35d0262e729d023316dc675b39957a5.png)
Hence any repeated value must equal the distance between its two preimages. In particular, if 
then 
but one checks this clashes with the bound 
.![\[
r=1\quad\text{or}\quad r=2,
\]](http://latex.artofproblemsolving.com/4/2/0/4206c4e53486e58a489c578f0d405943d18d9ba9.png)
and so![\[
f(n)\in\{1,2\}
\quad\text{for all }n.
\]](http://latex.artofproblemsolving.com/a/d/b/adbbf9f686439e875f92c80b1838e74d21a6f76d.png)
In particular 
.![\[
\boxed{f(2024)\in\{1,2,2024\}}.
\]](http://latex.artofproblemsolving.com/5/0/e/50e5a7cfb945942dad3b5be52cf01b7d45ef71af.png)
