[/quote]
,
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
,
.
with integer coefficients such that for any positive integer 
,
is positive and 
every next number (from the fifth number) is the unit number of the sum of the four numbers preceding it. Is there any cases that we get the numbers 
and 
in this series?
is a chord of a circle with center 
,
is the midpoint of 
and 
.
and 
,
= 
.
students. Each group consisting of three of the students meet, and choose one of the other 
students, A, to make him a gift. In this case, A considers each member of the group that offered him a gift as being his friend. Prove that there is a student that has at least 
friends.
such that 
.
dwarfes with weight 
.
,
at 
at Q, meet at 
.
intersects 
at 
called "available" for city 
,
Y by Adventure10, PikaPika999
Suppose that 
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
,
-
and 
are collinear; therefore, the perpendicularity of 
and 
is equivalent with the fact that 
.
(the power of 
and so 
.
.
and 
(since the circumcircle of 
is the nine-point center of 
cuts the circumcircle 
of 
and 
is a circle with center 
cuts 
is cyclic and the angle 
is right. Since 
is a diameter of 
and 
goes through 
)
is perpendicular to 
,we are done.
.
and 
have 
as the radical axis and as 
are concurrent, then 
is cyclic. So 
is 
to 
as 
is the diameter of 
.
;
w.r.t. 
we get (already proven 
are collinear): 
,
is cyclic, i.e. 
.
,
,
are collinear, and we are done.
is the midpoint of the arc 
is the 
intersects 
is the incenter of 
and 
,
and ![\[\angle B_{1}BI\\=\angle IBC+\angle B_{1}BC
\\=\angle IBC+\angle B_{1}AC
\\=\angle IBC+(\angle R-\frac{1}{2}\angle AB_{1}C)
\\=\angle R+(\angle IBC-\frac{1}{2}\angle ABC)
\\=\angle R\cdot \cdot \cdot (6)\]](http://latex.artofproblemsolving.com/6/7/7/677420e109fa1a2b2bed326bd115276ddb6cd335.png)
and 
,
and 
are collinea
,
be a point such that 
and 
,
are concyclic
and 
are concyclic
, 
and 
are concyclic
and 
,
and 
are collinear
(
is a radical ceneter)
,
,
and 
,
.
.
be the line through 
perpendicular to 
.
.
are collinear on the external angle bisector of 
are collinear by the definition. Now consider the circles 
.
while 
since 
and 
since 
is a diameter, where 
is the 
.
is 
,
is 
,
is 
,
are collinear, showing 
.
,
,
has coordinates 
.
(as you don't believe check here 
be the 
excenter, By Brokard's Theorem on Quadrilateral 
and 
inversion, 
swap, so we're done because then we must have
which means that 
.
.
.
is cyclic.
,
.
is cyclic too. As 
Its consequence leads to the fact that 
.
)
with the circumcircle of 
.![\[B_2B \cdot B_2B_1 = B_2C \cdot B_2A = B_2P \cdot B_2I_B.\]](http://latex.artofproblemsolving.com/9/6/4/96421bc720fe61b02e3919130ca77c79e47d729b.png)
Thus 
are concyclic. Now by the incenter/excenter lemma, we have that 
.![\[\angle I_BPI = 90 = \angle I_BBB_1 = \angle I_BPB_1.\]](http://latex.artofproblemsolving.com/a/2/7/a27b81ccfacba67e9030325fdeee756a43c97e95.png)
Therefore 
are collinear, and 
,
.![\[BI\cdot BI_B = BB_1\cdot BB_2.\]](http://latex.artofproblemsolving.com/e/1/f/e1f805bf6fd4471a5ebaeb9de8300a2c7d69bce5.png)
Let 
with 
,
.
and 
,
.![\[BB_1\cdot BB_2 = BD\cdot BM = BI\cdot BI_B\]](http://latex.artofproblemsolving.com/3/5/7/3579b2a835b31fdcea59a6196ca4462e6417e23f.png)
as desired.
is simply 
.![\[\measuredangle IAI_b=\measuredangle ICI_b=90^\circ\]](http://latex.artofproblemsolving.com/d/4/b/d4b74190fff0dbce1ca7c2ecc0aefacb690144b9.png)
and thus, 
.
,
also lies on 
.
.![\[B_2I\cdot B_2B_3' = B_2B \cdot B_2B_1 = B_2A\cdot B_2C\]](http://latex.artofproblemsolving.com/d/4/0/d400d91248e047569cc5dd80d05fe0750e552886.png)
Thus, 
must also lie on 
and indeed, 
as required.
![[asy]
import olympiad; size(400);
draw(unitcircle);
pair B = dir(130), A = dir(200), C = dir(-20), M = midpoint(A--C), O = origin, B1 = intersectionpoint(O--(O+3*(O-M)),unitcircle);
pair I = incenter(A,B,C), L = intersectionpoint(B--(I+3*(I-B)),unitcircle), Ib = L + L-I;
pair exB = rotate(90,B) * I, B2 = extension(exB, B, A, C);
dot("$A$",A); dot("$B$",B); dot("$C$",C); dot("$I$",I); dot("$O$",O); dot("$M$",M); dot("$L$",L); dot("$I_B$",Ib); dot("$B_1$",B1); dot("$B_2$",B2);
draw(C--B2--B--C^^A--B--Ib^^Ib--B1--L^^B2--extension(B2,I,B1,Ib));
draw(B--B1);
// draw(incircle(A,B,C);
[/asy]](http://latex.artofproblemsolving.com/7/e/4/7e4b00a850d18aa54a1429fef14ca203eb05877b.png)
(since 
is diameter), all 
,
(from right triangle 
)
(from cyclic quad 
)
so from similar triangles 
.
,
so 
.
wrt 
and 
be 
.
are concyclic by incenter/excenter lemma we have that 
and because 
,
are concyclic.Then it succifies to show that 
are collinear which is trivial by the radical axis concurrence lemma on 
,
are concurrent at 
and 
.
lie on the same line, and 
