AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
prove that
satisfy 
. Prove that:
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)
a grid of 9*9, we choose a little square in of this grid three times, we can choose three times the same.
with orthocenter 
and a point 
lying on the Euler line of 
Prove that the midpoint of 
lies on the Thomson cubic of the pedal triangle of WRT
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist.
, prove that 
be an acute triangle and 
it's incenter. Let 
, 
and 
be the projections of onto 
, 
and 
respectively. 
and 
is the projection of onto 
. Let 
be the middle of the arc not containing 
and 
the second intersection of 
and the circumcircle of 
. If 
is the midpoint of 
, 
, 
and 
, prove that 
is cyclic.
be an even integer and let 
be real numbers satisfying 
.
and 
such that 
.
has a solution
![$A=[1,0,0]$](http://latex.artofproblemsolving.com/5/e/d/5edd90e5880a3e676c65b04ec622149ac1c9e6fd.png)
, ![$B=[0,1,0]$](http://latex.artofproblemsolving.com/8/0/8/808aee5270277243ef8023672d85c99f1f4c7c8d.png)
, ![$C=[0,0,1]$](http://latex.artofproblemsolving.com/5/3/5/5350fac5fb4b9fa81700a91f7f11ea10503deffb.png)
, 
, 
, 
, we get:
, we have 
and 
, so 
and , that is, is tangent to 
![[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A = origin, B = (2,0);
path El = ellipse((1,0), sqrt(2),1);
pair C = relpoint(El, 0.36);
pair M = (A+B)/2, P = foot(A,C,B);
real r = abs(B-C), s = abs(A-C);
pair T = C * (r+s)/s, N = (A+T)/2, D = s/(r+s)*B;
draw(A--B--C--cycle,rgb(0.1,0.1,0.8));
draw(C--N,rgb(0.1,0.5,0.1));
draw(C--D^^N--M,rgb(0.9,0.1,0.1));
draw(circumcircle(B,M,P),orange+linetype("3 3"));
draw(A--P^^rightanglemark(A,P,C,2),rgb(0.6,0.1,0.6));
clip((-2,-1)--(-2,1.6)--(2.2,1.6)--(2.2,-1)--cycle);
dot("$A$",A,SW);
dot("$B$",B,SE);
dot("$C$",C,NW);
dot("$D$",D,S);
dot("$P$",P,dir(A--P));
dot("$M$",M,S);
dot("$N$",N,NW);
[/asy]](http://latex.artofproblemsolving.com/8/2/8/82837c16b10c60f37621b4f091dcc4d3898f00bb.png)
be the point on ray for which 
(not pictured above for space reasons), and let be the midpoint of 
. Then the given length condition implies![\[
AN = \frac{AC+CB}2 = \frac{AB}{\sqrt 2},
\]](http://latex.artofproblemsolving.com/4/3/c/43cdd9f20d046d60f127129dc4e2e911b9bcc8b8.png)
or 
. This means that the circumcircle of 
is tangent to at , implying 
.
is cyclic. To prove this, let 
be the foot of the angle bisector from 
. Observe that is the reflection of 
across the external angle bisector of 
, so 
. Thus, a (simple?) angle chase yields
Thus quadrilateral 
is cyclic, as desired.
is tangent to at , which is exactly what we wanted.
, let be the midpoint of and 
the foot of the altitude to . Prove that if 
, then the circumcircle of triangle 
is tangent to .
and prove F,M,B,P is clycic,as shown in the diagram.
be a point on such that 
, We claim that it is the tangency point
tangents 
tangents 
. and and tangents at so 
are concyclic and we are done.
be a point on rey behind such that
, similar argument to the first case finishes.
Trivial
,
.
and from B be 
.
and
and 
.