[/quote]
,
such that the relation ![\[|z^{2n}+z^nw^n+w^{2n} | = 2^{2n}+2^n+1 \]](http://latex.artofproblemsolving.com/1/2/2/1229a609403904449c447225094b3d97e063efd4.png)
holds for all positive integers 
.
,
,
is a positive integer.
of positive integers 
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 
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 
which is parallel to line 
.
and 
,
at point 
meet line 
at point 
and the internal bisector of 
at point 
.
are concyclic.
and 
.
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 
and 
,
.
Y by Physicsknight, WizardMath, amar_04, Geometry013, Purple_Planet
Let 
and 
be two distinct points in the plane. Let 
be the midpoint of the segment 
, and let 
be a circle that goes through and . Let 
be a point on such that the line 
is tangent to . Let 
be a point (other than ) on the line such that 
, and let 
be the foot of the perpendicular from onto the line .
Prove that the lines
and 
are parallel.
Navneel Singhal, India
and 
be two distinct points in the plane. Let 
be the midpoint of the segment 
, and let 
be a circle that goes through and . Let 
be a point on such that the line 
is tangent to . Let 
be a point (other than ) on the line such that 
, and let 
be the foot of the perpendicular from onto the line .Prove that the lines
and 
are parallel.Navneel Singhal, India
,
is isosceles and 
hence 
is cyclic so 
the lines 
is cyclic , so 
=
. We get that 
=
(
. After squaring and using pop it reduces to (1).
be the midpoint of 
,
where the first equality holds because 
and the last equality holds because 
.
where the last equality holds since 
and 
,
to the segment 
,
and 
,
therefore, this homothety also sends 
be the midpoint of 
and 
,
.
.
,
where the last equality holds due to the tangency of 
,
,
the point 
therefore, 
where the last equality holds due to the tangency of 
is equivalent to 
.
,
which by looking at the power of 
being concyclic.
,
lie on a circle with midpoint 
and therefore 
.
,
.
.
and 
.
.
.
.
we have 
.
which implies 
.
Since 
and 
which implies 
![[asy]
size(7cm);
pair A=(-8.86,2.91);
pair B=(1.8499163754792471,2.9844830809920997);
pair M=(-3.505041812260376,2.9472415404960497);
pair T=(-2.0504048164735122,9.47612561960746);
pair X=(-5.860059676194846,2.9308633560035102);
pair Y=(-0.957891074132805,7.657760327427519);
filldraw(A--M--T--cycle,cyan,blue);
draw(circumcircle(A,M,T),green);
draw(B--T,purple);
draw(B--M,purple);
draw(circumcircle(T,M,X),orange+dashed);
draw(A--Y,purple);
draw(X--Y,purple);
draw(Y--M,purple);
draw(T--X,purple);
dot("$A$",A,S);
dot("$M$",M,S);
dot("$T$",T,N);
dot("$Y$",Y,NE);
dot("$X$",X,S);
dot("$B$",B,S);
[/asy]](http://latex.artofproblemsolving.com/8/9/4/8942629ae1b3da3a7c82798f2769df4e5d600929.png)
is cyclic
hence 
.
and we are done 
and we are done 
.
,
.
( since 
)
is tangent to 
,
.
.
which implies that 
and we are done.
cyclic
is cyclic.
which implies the conclusion.
- cyclic. So by Reim 
is cyclic.
then 
we find that 
which gives that 
so 
This gives 
as desired.![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(7cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -11.596925851065945, xmax = 9.250987235646235, ymin = -2.347242959423775, ymax = 7.058189585802859; /* image dimensions */
pen ffqqtt = rgb(1,0,0.2); pen dbwrru = rgb(0.8588235294117646,0.3803921568627451,0.0784313725490196); pen ffttww = rgb(1,0.2,0.4); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961);
/* draw figures */
draw((-4,-0.45)--(-0.5584965546539985,-0.416629514997293), linewidth(0.4) + ffqqtt);
draw(circle((-3.1738984537761103,3.09305821011176), 3.6380908791444244), linewidth(0.4) + dbwrru);
draw((0.25494455728530113,1.8770233214637557)--(-0.9535063381410938,-1.5304351697975988), linewidth(0.4) + ffqqtt);
draw((-4,-0.45)--(-0.9535063381410938,-1.5304351697975988), linewidth(0.4) + ffqqtt);
draw((-4,-0.45)--(0.25494455728530113,1.8770233214637557), linewidth(0.4) + ffqqtt);
draw((-0.9535063381410938,-1.5304351697975988)--(1.112709278027912,-0.4004246934529864), linewidth(0.4) + ffqqtt);
draw((1.112709278027912,-0.4004246934529864)--(0.25494455728530113,1.8770233214637557), linewidth(0.4) + ffttww);
draw((-2.279248277326999,-0.4333147574986465)--(-0.9535063381410938,-1.5304351697975988), linewidth(0.4) + ffqqtt);
draw((-2.279248277326999,-0.4333147574986465)--(0.25494455728530113,1.8770233214637557), linewidth(0.4) + ffqqtt);
draw((-0.5584965546539985,-0.416629514997293)--(1.112709278027912,-0.4004246934529864), linewidth(0.4) + ffqqtt);
draw(circle((-0.5898191880272617,0.25860068246196543), 1.8256280080640814), linewidth(0.4) + linetype("2 2") + dtsfsf);
/* dots and labels */
dot((-4,-0.45),dotstyle);
label("$A$", (-3.9420814276242977,-0.2942805256941039), NE * labelscalefactor);
dot((-0.5584965546539985,-0.416629514997293),dotstyle);
label("$B$", (-0.48864849646663094,-0.2624516507525586), NE * labelscalefactor);
dot((-2.279248277326999,-0.4333147574986465),linewidth(4pt) + dotstyle);
label("$M$", (-2.2233221807808507,-0.31019496316487654), NE * labelscalefactor);
dot((0.25494455728530113,1.8770233214637557),linewidth(4pt) + dotstyle);
label("$T$", (0.4503033143089558,1.8541685328602033), NE * labelscalefactor);
dot((-0.9535063381410938,-1.5304351697975988),linewidth(4pt) + dotstyle);
label("$Y$", (-1.2525414950637186,-1.8538953978298233), NE * labelscalefactor);
dot((1.112709278027912,-0.4004246934529864),dotstyle);
label("$X$", (1.182367437964498,-0.246537213281786), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/b/2/7/b272d7a7d80a8b2bb76bf46b7db2cb972054707e.png)
,
and 
.
is cyclic.
,
