AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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 
. (
) distinct elements from this set.
. If 
and 
, show that the area of 
is not less than the area of 
, and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to 
inequalities, these given comparisons satisfy the following three constraints:
and 
, if the sums of elements of and are greater than those of subsets 
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 
, by 
the midpoint of , and by 
the orthocenter of . Let 
be the point of intersection of the circumcircle 
of the triangle and the half line 
, and 
be the point of intersection (other than 
) of the line 
and the circle . Prove that 
must hold.
the length of the line segment .)
such that 
(where 
), and a functional equation 
that satisfies 
for all , has a finite number of solutions. Find such .
is the function obtained by composing 
times, that is, 
)
be an acute triangle with 
, and let 
be the projection from onto 
. Let be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
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 .
be a regular 
-sided polygon that has all its vertices in one circle with center 
and radius 
. The diagonals and 
intersect at and the diagonals 
and 
intersect at . Calculate the measure of the segment .
be a regular -sided polygon that has all its vertices in one circle with center and radius . The diagonals and intersect at and the diagonals and intersect at . Calculate the measure of the segment .
is a regular -sided polygon, the measure of the angles is 
.
and 
, 
is an isosceles trapezium. Then 
. With the same process in the quadrilateral 
, we get the same result, as both quadrilaterals are congruent.
and 
is a paralelogram, is a rhombus. With the same process we get that 
is a rhombus.
and 
, 
is an isosceles trapezium.
Then in 
.
be the circumcircle and 
;![\[-1 = (DI; HJ)_{\omega} \stackrel{A}{=} (XI; P\infty)\]](http://latex.artofproblemsolving.com/9/0/0/900947d38cf55336489f0c6a13629a1ec16f26cd.png)
so 
.
,
.