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.
are there such that 
and 
?
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.
, where 
is a real number. Prove that for all positive , 
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 
.
is the midpoint of 
is a point on such that 
is perpendicular to , and 
perpendicular to 
. Suppose 
. Find .
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 
.
then prove
![[asy]size(200);
pair A=(0,8sqrt(3)),B=(24,8sqrt(3)),C=(24,0),D=(0,0),E=(A+B)/2,F=foot(B,A,C);
draw(C--A--B--C--D--A^^D--B--F--E);
label("$A$",A,NW);label("$B$",B,NE);label("$C$",C,SE);label("$D$",D,SW);label("$E$",E,N);label("$F$",F,unit(F-B));
draw(rightanglemark(E,extension(B,D,E,F),D,25));draw(rightanglemark(B,F,C,25));[/asy]](http://latex.artofproblemsolving.com/5/a/4/5a43982fbfd18d5b56aa84e1161a74c0f7169fd0.png)
.
is a right angle, and 
.
.
,
.
,
Hence,