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 for a vector 
in 
and permutation 
.We have 
and 
is injective and if 
and 
differ in more than 
places then 
and 
differ in more than places. and vector 
that 
be the number of differences between and 
then the number of 
such that 
and 
only depends on .
's in (1) is 
.(
) Then
into another !
or 
, ![$[{n\over 2}]=2k+1$](http://latex.artofproblemsolving.com/4/0/b/40b2f323c069e738daa652c97aca3f49f159a3c0.png)
so ![$[{n\over 2}]+1=2k$](http://latex.artofproblemsolving.com/4/d/b/4db15835b25e01ab2b45e8842730181ac6bd488a.png)
.
. The number of 's that are good (like the part 2) to and are . Suppose 
. We have that 
is also good to and and also the inverse. By part 2 we have 
.![$D(x,y)=[{n\over 2}]+1$](http://latex.artofproblemsolving.com/1/2/c/12cb2bb3f796664fc289c8dce792302e32a2635d.png)
. Because it's an even number then ![$D(f(x),f(y))=[{n\over 2}]$](http://latex.artofproblemsolving.com/a/4/f/a4fbf276e8ee5dec24ece8bd41eae36276963d5b.png)
or ![$[{n\over 2}]+1$](http://latex.artofproblemsolving.com/0/c/2/0c2fe0894db9a503299ec583ce4a3ca8f3d0f641.png)
. But its more than 
. So it is exactly . So 
![$D(x,y)=[{n\over 2}]$](http://latex.artofproblemsolving.com/1/4/5/1453815eacf9f59c1b2cad072a0374f59029bb4a.png)
then 
.
and 
.There exist a such that ![$D(x,y)=[{n\over2}]+1$](http://latex.artofproblemsolving.com/5/9/5/595c50fdd1699c22672602cd586a4c3f551a7418.png)
and and differs with in the 
'th member.
such that it only differs with in 'th member. and 
.
is the set of indexes that and differ in.
like the for and 
.
. If it is not then it is easily seen that and differ in more than 2 members. But 
so 
.
. we can make the to become all the members of . changing one member causes one change to .. And suppose that changing the 'th member causes the 
'th member of to change. We have 
is a permutation because 
is injective. is uniqe for all .
and such that a change to in causes the 
'th member of to change, and a change to causes the 'th member. is 
and the changed is .
and to obtain 
and .
xor 
and the other one is 
..
.(Because in other ways we have 
) we get .(Like the last part) is uniqe for all . to every by changing members, we have that the problem is proved!
