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.
be the incentre of acute-angled triangle 
. Let the incircle meet 
, and 
at 
, and 
respectively. Let line 
intersect the circumcircle of the triangle at 
and 
, such that 
lies between 
and . Prove that 
.
so that for every 
in 
.
such that 
. The lines 
and 
intersect at a point and the line passing through which is parallel to , intersects 
at 
. Prove that
be the circumcircle of acute triangle . Two tangents of from 
and 
intersect at , 
and intersect at 
. Point , are on 
and such that 
and 
.
are concyclic.
the centre of the circle passing through . 
, 
are difined similarly. Prove that 
, 
, 
are concurrent.
> 
that satisfy 
, for all 
€ .
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 
such that 
.
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
such that has exactly 2000 prime divisors and divides 
?
and 
is a subset of 
such that if 
then 
Find the smallest of 
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of with 
, then 
divides 
for every 
.
, in which 
: 
, but the proof is important.
, we have 
and 
. Let 
be the mid-points of and 
respectively. Suppose is a point on the segment 
such that 
and 
. Find 
.
intersect at . Note that is cyclic and 
. Since 
, we have 
, so 
so 
is the angle bisector of 
. But since 
, is also the angle bisector of 
. But since 
and 
(hopefully), we have 
is cyclic, so 
.
and 
.
