[/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 
at point 
.
meet line 
at point 
and the internal bisector of 
at point 
.
,
are concyclic.
and 
.
,
and orthocenter 
.
be a point on 
.
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
Triangle 
is an scalene triangle. Let 
the incenter, 
the circumcircle, 
the 
-excenter of triangle . Let 
the circle centered at and passes . and intersect at point 
, and the perpendicular line of 
which passes meets at point 
. 
is the perpendicular foot from to 
. Now if 
and 
intersects at 
, prove that 
.
is an scalene triangle. Let 
the incenter, 
the circumcircle, 
the 
-excenter of triangle . Let 
the circle centered at and passes . and intersect at point 
, and the perpendicular line of 
which passes meets at point 
. 
is the perpendicular foot from to 
. Now if 
and 
intersects at 
, prove that 
.This post has been edited 3 times. Last edited by Hypernova, Oct 11, 2022, 7:16 AM
Reason: Edit
Reason: Edit
.
and 
must be similar. These two triangles have common angle 
,
Also, let 
are concyclic, 
.
.
where 
and 
are collinear.
.
.
and points 
are collinear.
be the intersection of segments 
.
be the midpoint of arc 
we have 
Because 
above equality proves 
.
be the midpoint of segment 
.
thus 
.
and 
.
Hence (a,a,a) 
The rest is simple
QED
Notice that 
since 
is cyclic with center 
and 
Notice that 
and 
so 
Hence, 
and 
