AoPS Academy
, the points 
and 
are taken such that the point lies between the points 
and , and 
. On segments 
and 
, points 
and 
are taken so that 
and 
. Prove that 
.
and points and on sides 
and 
, respectively, such that 
. Let 
and 
be intersection points of lines 
and 
with circumcircle 
of triangle 
, and 
and 
intersection points of lines 
and 
with side 
. If 
, prove that circle contains intersection point of angle bisector of 
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 and 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](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 
, how many orderly colorings are there? is selected inside acute 
so that 
and 
. Point is chosen on ray 
so that 
. Let be the midpoint of . is tangent to the circumcircle of triangle 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
) . and it is this :
be natural numbers such that 
) .find the maximum of the numbers of the subset 
form the set 
which sum of the each elements of dont be equal to 
.
Well we can easily show that the answer is 
where 
is the biggest natural number such that is 
and by my supprise the answer is exactly 
and the proof is by the induction 
on the .
will do.
is a set of 
elements and has a minimum sum (with 
distinct elements) of 
, so 
.also for a set of size atleast we can form pairs of distinct elements that sum to 
and pigeonhole 
there exists a pair of distinct elements with sum .once we remove such a pair from the 
-set and we are still left with an 
set from which another pair,this time with sum 
(again pigeonhole) can be extracted and so we are through.
of size contains a pair of distinct elements which sum up to .this is independent of what i do before getting to that -set. the more important thing is to get the sum to first because choosing elements from will not guarantee 
distinct elements to add up to , for eg:
.
.
(with 
.