[/quote]
,
be a monic polynomial with complex coefficients such that there exist a polynomial 
with complex coefficients for which ![\[P(x^2-2)=P(x)Q(x).\]](http://latex.artofproblemsolving.com/a/6/2/a62b06890b6bbcc5c9634f6de8c7ae5fe9f37828.png)
Determine all complex numbers that could be the root of 
with 
and 
.
and 
intersect at 
,
and 
intersect at 
.
be the midpoints of sides 
,
and 
be points on segment 
and 
,
is the angle bisector of 
and 
is the angle bisector of 
.
is parallel to 
if and only if 
divides 
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x + y^2) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/3/a/a/3aa20083835682dbd81be692eba65cab19e923e5.png)
-
and 
such that 
.
kawaii if it satisfies 
is the number of integers not more than 
such that the equation![\[f(xf(x)+y) = f(y) + x^2\]](http://latex.artofproblemsolving.com/2/0/a/20a245460634f0455fc703d7ce41c475c1ae4970.png)
holds for all rational numbers 
and 
.
denotes the set of rational numbers.
the centroid of a tetrahedron 
is divisible by 
for every positive integer 
.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
,
,
be nonzero real numbers such that 
and 
.
.
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be a function satisfying
for all integers 
and 
for all integers 
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
of positive integers with 
prime, 
,
,
is a multiple of 
.
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Let 
be a triangle, and 
an interior point such that 
, 
, 
and 
. Prove that the triangle is isosceles.
be a triangle, and 
an interior point such that 
, 
, 
and 
. Prove that the triangle is isosceles.This post has been edited 5 times. Last edited by Amir Hossein, Sep 5, 2011, 5:10 PM
...
,
,
.
(
,
.
,
,
.
,
,
,
,
.
.
.![\[ \mathrm {A\ happy\ New\ Year,\ Mathlinks\ !} \]](http://latex.artofproblemsolving.com/9/1/e/91e2f085a5312f497d2f6fa9a397d56833952323.png)
,
.
,
and this cant work.
,
.
,
as desired
is given, provided that 
then, 
and 
.
,
.
.
Easily we can prove that 
and 
.
on 
satisfying 
on 
such that 
is equilateral
on 
such that 
is equilateral
such that 
and 
.
,
is cyclic. Thus, 
,
and 
,
,
.
Let 
,
so at most one value of 
.
angle chasing gives 
and 
using law of sines on 
and 
yields 
now 
note that 
so this is 
using law of cosines in 
gives 
hence 
and triangle 
and 
.
,
.
,
.
and 
we obtain
Hence
Therefore it suffices to prove that 
,
.
as wanted.
.
.
.
.
,
