[/quote]
,
,
,
is a positive integer.
of positive integers 
for which there exist injective functions 
such that 
.
denotes component-wise function multiplication)
and 
Prove that
,
distinct points 
connected if the line 
contains exactly 
points from 
such that there exists a set 
is connected with exactly 
other points from 
.
the reflections of 
and 
,
are collinear if and only if 
coincide, 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 
.
be a fixed positive integer, and 
be a sequence of nonnegative real numbers such that, for all 
,
.
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 
,
.
be a parallelogram. Let line 
externally bisect 
and let 
be the line passing through 
.
,
at point 
.
meet line 
at point 
and the internal bisector of 
at point 
.
,
are concyclic.
and 
.
,
and orthocenter 
.
be a point on 
such that 
.
and 
similarly.
,
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 
and 
,
.
Y by Adventure10, Mango247
Let 
be a triangle with 
. Let 
and 
be the feet of the perpendiculars from 
to the external angle bisectors of 
and 
, respectively. Let 
be the circumcenter of the triangle . Prove that the circumcircles of the triangles 
and 
are tangent to each other.
be a triangle with 
. Let 
and 
be the feet of the perpendiculars from 
to the external angle bisectors of 
and 
, respectively. Let 
be the circumcenter of the triangle . Prove that the circumcircles of the triangles 
and 
are tangent to each other.
be the excenters of triangle 
.
be the incenter. We will show that 
are tangent. Obviously, 
are cyclic and so are 
.
.
are collinear, the line joining the centers of the two circles 
Lies on the common point 
,
are cyclic, we conclude that 
are tangent. 
.
.
and 
so 
Because 
,
is cyclic. Hence, 
.
so 
is cyclic. Therefore, 
is the common tangency point of 
and 
,
and circumcenter of 
be 
.
is cyclic.
is cyclic.
is cyclic.
and 
are equialteral.
are collinear.
and 
touches at 
