[/quote]
,
are there such that 
and 
?
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.
, where 
is a real number. Prove that for all positive 
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 
.
then prove
be an acute-angled triangle, with circumcenter 
,
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 
,
.
Y by son7
i) Given three non-collinear points 
construct a fourth point 
such that there can be found a quadrilateral 
having 
as the midpoints of 
respectively.
ii) Given a convex pentagon
show how to construct five points 
having 
as midpoints of 
and 
respectively.
construct a fourth point 
such that there can be found a quadrilateral 
having 
as the midpoints of 
respectively.ii) Given a convex pentagon
show how to construct five points 
having 
as midpoints of 
and 
respectively.This post has been edited 1 time. Last edited by parmenides51, Jan 17, 2021, 12:20 PM
