[/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 
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 
.
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 HWenslawski, Adventure10, Rounak_iitr
Let 
be an acute triangle with orthocenter 
. The circle through 
, and 
intersects lines 
and 
at 
and 
respectively, and segment 
intersects 
and 
at 
and 
respectively. Two points 
and 
, both different from 
, are located on lines 
and 
respectively such that 
are concyclic and 
are concyclic. Show that lines 
and 
are parallel.
be an acute triangle with orthocenter 
. The circle through 
, and 
intersects lines 
and 
at 
and 
respectively, and segment 
intersects 
and 
at 
and 
respectively. Two points 
and 
, both different from 
, are located on lines 
and 
respectively such that 
are concyclic and 
are concyclic. Show that lines 
and 
are parallel.
since 
are congruent and 
hence 
is the bisector of 
idem 
is the bisector of 
besides from the power of 
we deduce 
are cyclic so they all lie on the circle of center 
;from 
and 
i
-bisector we get 
idem from 
and 
is in the 
so 
is the bisector of 
has same circumradius.
so 
is axis of symmetry, and so 
and 
Same way for 
and 
and so 
are concyclic.
are concyclic, 
and hence the proof.
are concylic.
be foot of perpendicular from 
.
are perpendicular bisector of 
Similarly, 
and isosceles trapeziums show 
concyclic, Now Apply Reim's theorem and done!
.
.
are concylic. 
.
at 
)
are concyclic. ![[asy]
size(200);
pair A=dir(115),B=dir(-150),C=dir(-30),H=A+B+C,D=2*foot(H,A,B)-A,E=2*foot(H,A,C)-A,P=extension(D,E,H,B),Q=extension(D,E,H,C),X=2*foot(H,A,P)-A,Y=2*foot(H,A,Q)-A;
draw(circumcircle(A,D,E)^^arc(circumcenter(A,H,P),P,A)^^arc(circumcenter(A,H,Q),A,E),orange);
draw(arc(circumcenter(B,H,C),C,D)^^arc(circumcenter(A,B,H),X,A)^^arc(circumcenter(A,H,C),A,C),purple);
dot("$A$",A,dir(A));
dot("$B$",B,dir(150));
dot("$C$",C,dir(C));
dot("$H$",H,dir(H));
dot("$D$",D,dir(-90));
dot("$E$",E);
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(70));
dot("$X$",X,dir(X));
dot("$Y$",Y,dir(Y));
draw(B--C^^X--Y,brown);
draw(C--A--D--E--P,fuchsia);
draw(P--foot(H,A,C)^^C--foot(H,A,B),green);
draw(P--A--Y,cyan);
[/asy]](http://latex.artofproblemsolving.com/2/1/b/21bd27987ebe7a95a01bb3c6bc69609a4e2e1b16.png)
and 
we have![\[AQ\cdot QY= HQ\cdot QC= DQ\cdot QE\]](http://latex.artofproblemsolving.com/0/9/4/09478beff52a2f71f621474cbec991e99001d6b7.png)
so 
cyclic, similarly 
cyclic.
it suffices to prove 
as ![\[\angle HCD=90^{\circ}-\angle ADC=90^{\circ}-(180^{\circ}-\angle BHC)=(180^\circ -\angle HBC-\angle HCB)-90^\circ=\angle HCA\]](http://latex.artofproblemsolving.com/a/2/e/a2ec2a9e1a16d693017819808a526a46c6a64b8f.png)
as desired, similarly 
and 
.
we obtain 
must be the line on infinity (by Claim 2) hence by Claim 3, 
