,
and 
are taken such that the point 
and 
.
and 
,
and 
are taken so that 
and 
.
.
and 
,
.
and 
be intersection points of lines 
and 
with circumcircle 
of triangle 
,
and 
intersection points of lines 
and 
with side 
.
,
with 
such that 
divides 
for all positive integers 
and 
with 
.
such that 
Follows for all reals 
.
denote the set of positive integers. Find all functions 
such that for positive integers 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
,
so that 
and 
.
so that 
.
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
Y by Inequality_forever, Adventure10, Mango247
Let 
be a fixed positive integer, and let 
be an infinite family of sets, each of size 
, no two of which are disjoint. Prove that there exists a set of size 
that meets each set in .
be a fixed positive integer, and let 
be an infinite family of sets, each of size 
, no two of which are disjoint. Prove that there exists a set of size 
that meets each set in .Attachments:
This post has been edited 1 time. Last edited by orl, Oct 25, 2004, 12:20 AM
.
which is not a subset of 
(since 
has at most 
.
,
.
and 
,
.
so every other element of 
are finite so there are infinitely set of 
and 
....so we can show them like 
a
...also 
(because 
and do this 
times ...i mean that each of 
have intersection with a subset of 
(maybe empty!) so infinitely of them have intersection with a subset of 
that 
so we show them as : 
and so 
and 
now repeat this fo
and we konw that 
thjat because of the last step and difference between the elements of 
and if 
is nonempty and because
are disjiont
this is contradiction!!
of distinct 
in 
in 
and we are done.
.
.
.
sets. Now some element must show up infinitely many times in 
.
is the infinite subfamily of 
is the infinite family of 
elements meets everything in 
be "good" if, for any 
.
.
.
.
,
.
.
.
for any 
,
.
and a subset 
,
denote the set of sets 
with 
.
is a subset of 
.
.
,
.
.
.
such that 
and 
.
exists by the previous paragraph, and let 
be the subset of 
be the set of those sets. Now, all sets in 
,
.
,
,
be an infinite family of sets, each of size 
elements. Prove that there exists a set of size 
and 
.
which lies in infinitely elements of 
,
to 
contains every element of 
still.
.
,
of sets in 
for 
,
is in finitely many guys. In particular, 
.![[asy] size(6cm); draw( (0,0)--(3,0)--(3,1)--(0,1)--cycle ); label("$X$", (3,1), dir(45)); dot( (0.5,0.5) ); dot( (1.5,0.5) ); dot( (2.5,0.5) ); dot( (-0.5,1.2), blue ); dot( (-1.2,1.2), blue ); dot( (-0.5,0.6), red ); dot( (-1.2,0.6), red ); dot( (-0.5,-0.2), heavygreen ); dot( (-1.2,-0.2), heavygreen ); [/asy]](http://latex.artofproblemsolving.com/7/2/6/726e6f38287fa759d6630877504c29c7b962e679.png)
must meet each 
in at least 
.
with such that the set 
containing all sets in 
.
in 
is infinite, there exists an infinite set 
such that all the elements in 
contain 
such that all sets in 
where 
.
such that all sets in 
where 
(For the case when 
,
)
which doesn't meet 
.
and any 
let 
be the set of all sets of size 
.
is always infinite, since every element of 
be the smallest 
.
for all 
,
,
must be finite. Suppose otherwise, so 
is infinite. Then there exists some 
so that 
,
not in any of the 
.
,
,
,
.
;
,
as desired.
and it cannot be bigger than 
)
.
be the underlying set. 
we choose a set 
of 
.
and a set 
with 
.
,
such that 
and 
.
.
is infinite we run the same process again and obtain mutually disjoint sets (we use the same notations again) 
,
,
elements, such that 
and 
each with 
elements and sets 
,
elements, such that 
and 
is finite. (Such moment certainly will come, since otherwise we finally get 
and disjoint set of 
of 
,
coincide with some 
.
satisfy the requirement. Indeed, suppose there exists 
.
for infinitely many 
does not meet 
,
,
is some ordinal. In case 
in a sequence and apply the same argument. But, if 
(when you take the limit, and 
,
,
of sets with 
as 
or rather 
(to mark it may depend on 
me the maximal possible 
with 
and 
,
.
is also countable, and arranging those sets in a row it easily follows 
.
when 
)
.
elements) do the job. Indeed let 
.
.
.
and 
is the minimal one, we get 
or 
.
,
denotes the collection of all sets of size at most 
there exists some 
which is contained in an infinite number of sets in 
,
.
,
contained in an infinite number of sets 
in 
,
satisfying the condition in the Claim, being contained in sets 
,
are all distinct. Suppose on the contrary that some set 
.
,
.
.
such that 
(
,
would be a larger set satisfying problem conditions.
.
.
,
.
be the maximal set, selected from a greedy algorithm, such that 
.
for all 
.
,
such that 
and 
only intersect at 
and 
aren't disjoint for some 
all be mutually disjoint except at 
