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 
.
with 
, then I believe the answer is 
for 
and 
for 
. The case 
is obvious, and the fact that works when 
can be seen by considering the 
vertices of two unit squares in the plane which share a side. From now on we assume .
points, then that circle must contain points.
, and two other points, 
, not belonging to that circle. In the group 
we must have four concyclic points, and one of 
must not belong to this group, so we may assume WLOG the group is 
. Now look at the group 
. The group of concyclic points that we can extract from must contain , so it cannot contain 
. This means that it's 
. Finally, look at the group 
. The group of four concyclic points extracted from here must contain , but cannot contain either or 
, so we get a contradiction. for which there is a configuration satisfying the hypothesis for which every circle passing through at least four points passes through no other points except for those four. We will show that no works. Obviously, in order to do this, it suffices to prove it for 
.
be four concyclic points in our configuration. Just as above, if 
are the other three points, then we can find, say, circles through the quadruples 
and 
. Now perform an inversion of pole . For a general point , we denote its image by 
. 
are still concyclic, and 
. The six points 
must still satisfy the hypothesis, but it's clear that no four of the points 
are concyclic.
