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 point in the plane, distinct from the vertices of 
. Consider 
the reflections of with respect to lines 
and 
, in this order.
are collinear if and only if lies on the circumcircle of . does not lie on the circumcircle of and the centroids of triangles and 
coincide, prove that is equilateral.
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 
.
be the set of all positive integers. Determine all subsets 
of 
such that
be a right-angled triangle with 
, a square is externally to the triangle. If 
, 
and is the circumcenter of the square , determine the value of 
be a right-angled triangle with , a square is externally to the triangle. If , and is the circumcenter of the square , determine the value of be the midpoint of . Since that both 
and 
are right-angled triangles,
In 
, 
. The law of cosine gives that
Since 
, we obtain
which gives 
. Hence 
thereby 
. 
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 
.
