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
which intersect at points 
. Let 
be an arbitrary point on 
. Suppose that the lines 
meet 
again at points 
respectively. Prove that the circumcircles of all triangles 
have the same radius.
which intersect at points . Let be an arbitrary point on . Suppose that the lines meet again at points respectively. Prove that the circumcircles of all triangles have the same radius.
is of a fixed length and thus by the Law of Sines, we are done.
subtends is actually the sum of the measures of the arc 
in both circles, from angle chasing, which is obviously constant.
is an isosceles trapezium and hence 
that is length of is constant.
is fixed, 
is also fixed.
.
is fixed the circumradius of 
is also fixed. 
is constant regardless of where is on .
and 
be four points. We are only required to prove 
, which is obvious because of cyclic quadrilaterals. is constant.
where is the circumradius of 
. We are required to prove that is constant. We already know that 
is constant.
so 
is constant. This implies is constant, so is constant, implying the result.
and 
on .
is a constant chord, we must have that 
.
and 
respectively for the points and respectively,
and 
.
and 
have the same circumcricle, (and the same circumradius). This implies that 
and 
have the same circumradius too.
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 
.
