[/quote]
,
be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer 
such that 
.
such that 
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
sastifying 
be a set of Natural numbers from 
to 
.
)
.
and 
,
is not less than the area of 
,
inequalities, these given comparisons satisfy the following three constraints:
and 
,
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
.
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
,
the orthocenter of 
be the point of intersection of the circumcircle 
of the triangle 
,
be the point of intersection (other than 
and the circle 
must hold.
the length of the line segment 
such that 
(where 
)
that satisfies 
for all 
is the function obtained by composing 
times, that is, 
)
,
be the projection from 
past 
.
such that 
are on the same side of ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
be 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
at 
intersect 
.
such that the relation ![\[|z^{2n}+z^nw^n+w^{2n} | = 2^{2n}+2^n+1 \]](http://latex.artofproblemsolving.com/1/2/2/1229a609403904449c447225094b3d97e063efd4.png)
holds for all positive integers 
Y by samrocksnature
Let 
be a triangle with 
and with incenter 
. Let 
be the midpoint of 
, and let 
be the midpoint of the circular arc 
. Lines through parallel to 
meet 
at 
and 
, respectively, and meet 
and 
at 
and 
, respectively. Show that lies on the radical axis of the circumcircles of triangles 
and 
.
Proposed by Andrew Wu
be a triangle with 
and with incenter 
. Let 
be the midpoint of 
, and let 
be the midpoint of the circular arc 
. Lines through parallel to 
meet 
at 
and 
, respectively, and meet 
and 
at 
and 
, respectively. Show that lies on the radical axis of the circumcircles of triangles 
and 
.Proposed by Andrew Wu
and in your diagram it's other way around
.
(in that order, namely 
ect.) and 
.
is a point s.t. 
.
and 
.
and 
.
,
.
.
.
and 
.
Collinearity of 
is equivalent with
Now, without calculating 
,
equation 
follows.
LHS trivially matches, not lets show that RHS matches.
Notice that 
and by using this we get:
Which is correct and we are done.
