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.
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 
.
, 
, where 
is a positive integer.
of positive integers for which there exist injective functions 
such that 
.
denotes component-wise function multiplication)
be a positive integer. For a set of points in the plane 
, we call 
distinct points 
connected if the line 
contains exactly 
points from .
such that there exists a set of points in the plane with the property that any point 
is connected with exactly 
other points from .
be a fixed positive integer, and 
be a sequence of nonnegative real numbers such that, for all 
, we have that 
.
is bounded above.
is bounded above.
are positiv real number, such that 
prove inequality 
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
, solve the given equation in 
., the equation has at most real solutions.
be a parallelogram. Let line 
externally bisect 
and let 
be the line passing through 
which is parallel to line 
. Suppose that meets line at point 
and at point 
, and that meets the internal bisector of 
at point 
. Further let circle 
meet line 
at point 
and the internal bisector of meet circle 
at point 
., , 
, and 
are concyclic.
be an acute-angled triangle, with circumcenter 
, circumradius 
and orthocenter 
. Let 
be a point on 
such that 
. Define 
and 
similarly.
, prove that is equilateral.
be a positive integer. There are two numbers written on a blackboard, one red and one blue. Initially, both are 0. We define the following procedure: at each step, we choose a nonnegative integer (not necessarily distinct from the previously chosen ones), and, if the red and blue numbers are 
and 
respectively, we replace them with 
and 
, which we color blue and red (in this order). We keep doing this procedure until the blue number is at least 
.
, in which 
. Point is selected on the side . The point 
is such that is the midpoint of 
. It turned out that 
. Prove that 
.
and 
Prove 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 
.
