[/quote]
,
such that 
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 
.
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 midpoint of 
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 ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
be 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
and 
intersect 
and 
at 
.
intersect 
.
.
Y by
Point 
is the center of the circumscribed circle of an acute-angled triangle 
. On the ray 
lies point 
. The circumscribed circles of triangles 
and 
intersect sides 
and 
at points 
and 
, respectively. Prove that the midpoint of the segment 
is equidistant from vertices 
and 
.
is the center of the circumscribed circle of an acute-angled triangle 
. On the ray 
lies point 
. The circumscribed circles of triangles 
and 
intersect sides 
and 
at points 
and 
, respectively. Prove that the midpoint of the segment 
is equidistant from vertices 
and 
.This post has been edited 1 time. Last edited by parmenides51, Oct 11, 2020, 11:18 PM
