[/quote]
,
be a point in the plane, distinct from the vertices of 
.
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 
,
.
real solutions.
be a parallelogram. Let line 
externally bisect 
and let 
be the line passing through 
which is parallel to line 
.
at point 
and 
,
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 
(not necessarily distinct from the previously chosen ones), and, if the red and blue numbers are 
and 
,
.
be the set of all positive integers. Determine all subsets 
of 
such that
Y by Adventure10, Mango247, ItsBesi
Around the triangle 
the circle is circumscribed, and at the vertex 
tangent 
to this circle is drawn. The line 
, which is parallel to this tangent intersects the lines 
and 
at the points 
and 
, respectively. Prove that the points 
belong to the same circle.
(Montenegro)
the circle is circumscribed, and at the vertex 
tangent 
to this circle is drawn. The line 
, which is parallel to this tangent intersects the lines 
and 
at the points 
and 
, respectively. Prove that the points 
belong to the same circle.(Montenegro)
.
is parallel to 
,
.
,
is circumscribable.
be a point on 
are on opposite sides of 
.
so 
be inscribed in a circle with center 
be tangent at 
,
than 
and 
is a point in the tangent through 
,
,
.
,
.
,
![\[ \angle{BDE}=\angle{BCF}=\angle{ACF}+\angle{BCA}=\angle{ABC}+\angle{BCA}=180-\angle{BAC} \]](http://latex.artofproblemsolving.com/9/9/1/9917c2249c9a33e4dcdcda329705e3be5d8c949a.png)
Thus we know that 
Since DE is parallel to the tangent line, OC is perpendicular to DE, and bisects B'A'. This means triangles B'XC and A'XC are congruent so angles <XB'C and <XA'C are equal. Let the aforementioned angle have measure x. <XB'C and <CAA' both intercept arc A'C so <CAA'=x and by similar reasoning <B'BC=x.
