[/quote]
,
with parallel sides 
and 
,
on line 
outside segment 
inside segment 
such that 
.
the point of intersection of 
,
the point of intersection of 
be the midpoint of segment 
if and only if 
.
such that 
is the square of an integer.
prove that
satisfy 
.
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 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
a grid of 9*9, we choose a little square in 
,
.
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 
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
,
be an acute triangle and 
it's incenter. Let 
,
and 
be the projections of 
,
and 
respectively. 
and 
is the projection of 
.
be the middle of the arc 
and 
the second intersection of 
and the circumcircle of 
.
is the midpoint of 
,
,
and 
,
is cyclic.
be an even integer and let 
be real numbers satisfying 
.
and 
such that 
.
Y by Adventure10, junioragd, and 4 other users
Let 
be a fixed positive integer. Given a set 
of 
points in the plane such that no three are collinear and no four concyclic, let 
be the number of circles 
that contain 
in their interior, and let ![\[m(S)=a_1+a_2+\cdots + a_n.\]](//latex.artofproblemsolving.com/4/9/b/49b928b1f20e2d1799a1234c7f25a2bc3d62d0dd.png)
Prove that there exists a positive integer 
depending only on 
such that the points of 
are the vertices of a convex polygon if and only if 
be a fixed positive integer. Given a set 
of 
points in the plane such that no three are collinear and no four concyclic, let 
be the number of circles 
that contain 
in their interior, and let ![\[m(S)=a_1+a_2+\cdots + a_n.\]](http://latex.artofproblemsolving.com/4/9/b/49b928b1f20e2d1799a1234c7f25a2bc3d62d0dd.png)
Prove that there exists a positive integer 
depending only on 
such that the points of 
are the vertices of a convex polygon if and only if 
This post has been edited 1 time. Last edited by djmathman, Oct 3, 2016, 3:25 AM
Reason: changed formatting to match imo compendium
Reason: changed formatting to match imo compendium
satisfies the requirements.
of the four points 
,
,
and 
be the number of points 
where 
such that ![\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d)\]](http://latex.artofproblemsolving.com/6/c/7/6c78c7cbbcd3042a8d8a7ec879080cc40baacc52.png)
.
and 
.
if and only if![\[\angle{ABC} > 180 - \angle{ADC} \quad \Leftrightarrow \quad \angle{ABC} + \angle{ADC} > 180\]](http://latex.artofproblemsolving.com/8/9/4/894e4aba933ad668d180b56845fbcfce5b221f19.png)
lies within the circumcircle of 
.
is 
,![\[\angle{ABC} + \angle{ADC} < 180 \quad \Rightarrow \quad \angle{BAD} + \angle{ACD} > 180\]](http://latex.artofproblemsolving.com/2/c/c/2cc326f2a246c2fb15611dd8d5b9aba849d12165.png)
and 
lies within the circumcircle of 
.
.
.
form a convex ![\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d) = 2 \binom{n}{4}\]](http://latex.artofproblemsolving.com/7/7/f/77f447a54854ca31b08dc50ddaf1fa7b95e84aa9.png)
![\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d) < 2 \binom{n}{4}\]](http://latex.artofproblemsolving.com/1/a/0/1a066690a3911afbcdaf78e1bce768a3adf6a4b0.png)
if and only if the points in 
with equality if and only if it is convex,
and 
is greater than 
and the other is less than 
.
contributes 
and 
(from the lemma) and thus contributes two to 
.
.
and thus 
.
,
,
,
,
.
.
.
and 
.
it follows that 
.
.
,
,
.
.
.
,
,
,
and that 
,
Equality holds here iff each of the quadrilaterals 
is convex, which occurs iff the points of 
consists of four points 
.
is 
must be greater than 
,
.
degrees, there will be exactly two points that are contained in the circumcircle of the other three.
be the convex hull. Then it is clear that the only point that is contained in the circumcircle of the other three is 
because both quantities count the amount of ordered pairs containing a single point of 
iff 
be the number 
It is easy to see that 
.
contains 
and 
Clearly, when 
then 
and when it is nonconvex 
.
if and only if 
where 
or 
but not both. However, since 
,
.
count the value 
.
.
.
,
than 
.
we have 
.
.
,
,
.
means 
and 
,
.
yields the contradiction 
.
has 
by claim 2, so 
.
,
for each sub-quadrilateral with points 
and we don't have sub-quadrilaterals with 
,
points 
if they form a convex polygon there are 
where 
and 
is the circumcircle of the other points. However if they form a non-convex polygon there is 
so there is a bijection.
quadruples are convex and 
are concave, 
But 
so 
and the desired result follows. QED
