[/quote]
,
and incircle touching 
at 
.
be the antipode of 
in the circumcircle of 
is chosen on the internal angle bisector of 
such that 
.
be the midpoint of arc 
,
be the midpoint of 
.
x
to 
row and 
column get symbolized by square 
. Subset 
belong to 
with 
or 
are not belong to 
denote its incentre and circumcentre respectively. Let 
be the circle through 
and 
which is tangent to the incircle of the triangle 
and 
are defined similarly. The circles 
distinct from 
and 
are defined similarly. Prove that the lines 
and 
are concurrent at a point on the line 
.
defined as follows:![\[
a_1 = \frac{\sqrt{6}}{3},
\quad
a_{n+1} = a_n + \frac{1}{3a_n},
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/8/9/2/892a2676214a3c837d6b18a10a4f6f83bc099f74.png)
![\[
0 \le a_n \sqrt{6} - 2\sqrt{n}
\le
\frac{1}{4\sqrt{n}}
\Bigl(
1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
\Bigr),
\quad
\forall n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/f/0/8/f0824edd96ddfeaa107d729466134812ad7ea777.png)
b,For each 
,![\[
b_n = \frac{3a_n^2 - 2n - 1}{\ln(n+1)}.
\]](http://latex.artofproblemsolving.com/a/9/3/a937b379140df5aa51d7ede1a685d263c3229e96.png)
Compute the limit 
(
has a right angle at 
such that 
and 
lies between 
and 
.
is chosen so that 
and 
is the bisector of 
.
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，
satisfying: 
is such that 
for any positive integers 
.
for all positive integers 
.
.
,
,
,
,
in some order.
.
be the point such that the quadrilateral 
is a parallelogram. Let 
such that 
.
at 
.
.
Y by Adventure10, Mango247
Points 
and 
lie respectively on sides 
and 
of a triangle 
and 
. Segments 
and 
cross at 
. Circumscribed circles of triangles 
and 
cross again at point 
different from . Prove that point lies on the bisector of angle 
.
and 
lie respectively on sides 
and 
of a triangle 
and 
. Segments 
and 
cross at 
. Circumscribed circles of triangles 
and 
cross again at point 
different from . Prove that point lies on the bisector of angle 
.
,
and 
are similar and thus congruent because 
.
,
is cyclic so 
and thus 
And we are done. 
to 
.
Now, since 
are cyclic quadrilaterals. Hence, 
and 
.
since 
.
,
and 
be the circumcircles of 
and 
,
.
.
,
.
.
,
is cyclic. But since 
,
is the bisector of angle 
and 
are also concyclic, that means that we have that there exists a spiral similarity mapping 
to 
.
That shows us that 
is cyclic, and analogously, 
is cyclic too. Finally, due to the fact that 
.
so 
meaning 
so we are done.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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draw((-2.590375969105767,4.069271180546538)--(-6.948194499531714,-0.8114855735304873), linewidth(0.8) + blue);
draw((-6.948194499531714,-0.8114855735304873)--(0.4725479122793272,-3.824605814567835), linewidth(0.8) + blue);
draw((0.4725479122793272,-3.824605814567835)--(-2.590375969105767,4.069271180546538), linewidth(0.8) + blue);
draw((-6.948194499531714,-0.8114855735304873)--(-0.509674242709973,-1.2931877403271366), linewidth(0.8) + blue);
draw((0.4725479122793272,-3.824605814567835)--(-5.139763875478851,1.2139567254087056), linewidth(0.8) + blue);
draw(circle((-1.6598741152946308,-3.1957461366265285), 2.223215733621701), linewidth(0.8) + linetype("2 2") + red);
draw(circle((-4.730730831640612,-0.9713075989400678), 2.2232157336216987), linewidth(0.8) + linetype("2 2") + red);
draw((-2.590375969105767,4.069271180546538)--(-3.876433238801054,-3.0238338036672094), linewidth(0.8) + blue);
draw((-3.876433238801054,-3.0238338036672094)--(-6.948194499531714,-0.8114855735304873), linewidth(0.8) + blue);
draw((-3.876433238801054,-3.0238338036672094)--(-0.509674242709973,-1.2931877403271366), linewidth(0.8) + blue);
draw((-3.876433238801054,-3.0238338036672094)--(-5.139763875478851,1.2139567254087056), linewidth(0.8) + blue);
draw((-3.876433238801054,-3.0238338036672094)--(0.4725479122793272,-3.824605814567835), linewidth(0.8) + blue);
/* dots and labels */
dot((-2.590375969105767,4.069271180546538),dotstyle);
label("$A$", (-2.485788324375541,4.350661748511184), NE * labelscalefactor);
dot((-6.948194499531714,-0.8114855735304873),dotstyle);
label("$B$", (-6.841116672784102,-0.5251146415310723), NE * labelscalefactor);
dot((0.4725479122793272,-3.824605814567835),dotstyle);
label("$C$", (0.5821159210443259,-3.5382348825684216), NE * labelscalefactor);
dot((-5.139763875478851,1.2139567254087056),dotstyle);
label("$P$", (-5.03324452816168,1.4745015184300776), NE * labelscalefactor);
dot((-0.509674242709973,-1.2931877403271366),linewidth(4pt) + dotstyle);
label("$Q$", (-0.40399615784063136,-1.072954685356045), NE * labelscalefactor);
dot((-2.5141717081341906,-1.1432199318993854),linewidth(4pt) + dotstyle);
label("$R$", (-2.4036123178017945,-0.9359946743998018), NE * labelscalefactor);
dot((-3.876433238801054,-3.0238338036672094),linewidth(4pt) + dotstyle);
label("$S$", (-3.773212427364235,-2.7986508234047087), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/6/5/6/656004c73b8e97b0d7852cec087ce28ce035e92c.png)
be the spiral similarity centered at 
and that 
.
is cyclic.
meanins that 
.
.
.
(not 
) so 
which along with the condition makes the two triangles congruent, so 
and 
proves the desired result 
Since we also know 
we see that 
Therefore, 
and 
.
.
be the midpoints of 
.
,
together with 
implies 
.
and 
.
and 
are cyclic. Hence:
Hence 
are colinear. In a similar way we have 
colinear. Therefore 
and we are done since it follows that 
.
![[asy]
size(6cm);
defaultpen(fontsize(12pt));
pair A,B,C,P,Q,R,S;
A=dir(120);B=dir(210);C=dir(330);P=(-0.72,0.05);Q=(0.47,-0.1);R=extension(B,Q,C,P); S=intersectionpoints(circumcircle(P,B,R),circumcircle(C,Q,R))[1];
draw(A--B--C--cycle,blue);
draw(circumcircle(P,B,R),heavygreen);
draw(circumcircle(C,Q,S),heavygreen);
draw(B--Q,heavycyan);
draw(C--P,heavycyan);
draw(A--S,heavycyan);
draw(S--P,heavycyan);
draw(S--Q,heavycyan);
dot("$A$",A,dir(A));dot("$B$",B,dir(B));dot("$C$",C,dir(C));dot("$P$",P,NW);dot("$Q$",Q,NE);dot("$R$",R,dir(80));dot("$S$",S,dir(260));
draw(S--B,heavycyan);
draw(S--C,heavycyan);
draw(R--S,heavycyan);
draw(circumcircle(A,B,S),dashed);
[/asy]](http://latex.artofproblemsolving.com/b/f/4/bf4c67152ee035efea6538551700c074107d03ef.png)
to 
is in fact a spiral congruence, meaning that 
and 
are similar isosceles triangles.
, as needed.
so 
