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 
.
isn't the solution of the given equation, we can rewrite as follows.
when the eqality holds when 
has at least one real root at 
should satisfy this is given by 
and ![$|a|\geq 4\ \cdots [A],\ |b+2|\geq2|a|\ \cdots [B].$](http://latex.artofproblemsolving.com/f/d/1/fd1a9c688c6ef9a2345b65b5da8d53c082536b70.png)
![$[A]$](http://latex.artofproblemsolving.com/5/9/7/5974998abb6495b7e7cb9c32143687050121fe86.png)
or ![$[B]$](http://latex.artofproblemsolving.com/7/e/c/7eceadf44d26e334cca20d8894b48d390e7076ac.png)
on 
plane gives the answer, 
has the minimum value of 
when 
which are the intersection points between two graphs of 
and 
the function 
is increasing. Particularly, 
with equality iff 
we have 
, i.e. 
because 
, 
and 
,
Thus, 
Using the above lemma 
, obtain 
and the minimum points are 
because 
, 
is the equation of a mobile line 
in the analytical plane 
Thus, square of the distance 
, where 
is 
Therefore, 
and (from the above example), i.e. the minimum points are the solutions of the systems 
, where 
In conclusion, 
and 
, where 
isn't the solution of the given equation, we can rewrite as follows. when the eqality holds when then 
and the given equation becomes: 
![\[\frac{t^{4}-4t^{2}+4}{t^{2}+1}-\frac{4}{5}=\frac{(5t^{2}-4)(t^{2}-4)}{5(t^{2}+1)}\ge 0\quad (2)\]](http://latex.artofproblemsolving.com/b/b/b/bbbc7c89f474e796867e6ad3db53669cad807474.png)
and 
we have: 
when 
traverses all the pairs of real numbers for which the equation ![\[ x^{4} + ax^{3} + bx^{2} + ax + 1 = 0 \]](http://latex.artofproblemsolving.com/7/f/a/7fabcaedc6df5dbb2f37173c9a508e8172908e5c.png)
has at least one real root.
then the given equation has at least one real root.
![\[t^{2}=-(at+b-2)\implies t^{4}=\left(at+b-2\right)^{2}\le\left[a^{2}+(b-2)^{2}\right]\left(t^{2}+1\right)\\ \implies a^{2}+(b-2)^{2}\ge\frac{t^{4}}{t^{2}+1}>t^{2}-1>3 \]](http://latex.artofproblemsolving.com/1/8/a/18a46c887850362622fdf675b4bc92d6b5d434d7.png)
And if the given equation is: 
then 
