[/quote]
,
be the circumcircle of acute triangle 
and 
intersect at 
and 
are on 
such that 
and 
.
are concyclic.
the centre of the circle passing through 
,
are difined similarly. Prove that 
,
,
are concurrent.
> 
,
€ 
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 
.
and 
.
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
?
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 
,
divides 
for every 
.
,
:
,
and 
Prove that
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be non-negative real numbers such that 
Prove that
Y by Adventure10, Mango247
Given a circle 
and two fixed points 
on 
and an arbitrary point 
on such that the triangle 
is acute. 
lies on ray 
lies on ray 
such that 
and 
Let 
be the intersection of 
and 
intersects 
at 
a) Prove that
are collinear.
b)
is the midpoint of 
Let 
be the intersection of 
and 
is the line passing through and perpendicular to 
is the intersection of and 
intersects at 
Prove that 
passes through a fixed point.
and two fixed points 
on 
and an arbitrary point 
on such that the triangle 
is acute. 
lies on ray 
lies on ray 
such that 
and 
Let 
be the intersection of 
and 
intersects 
at 
a) Prove that
are collinear.b)
is the midpoint of 
Let 
be the intersection of 
and 
is the line passing through and perpendicular to 
is the intersection of and 
intersects at 
Prove that 
passes through a fixed point.This post has been edited 1 time. Last edited by CSS-MU, Jan 5, 2014, 3:24 PM
which means 
is cyclic,we have 
.
cyclic, so 
is the radical center of the three circles 
,
intersect 
,
be points of intersection of 
with 
with 
with 
and 
.
passes trough midpoint of 
is cyclic and 
hence 
which means 
is midpoint of 
is cyclic and 
lie on a line, so we can conclude that 
;
is the orthocentre of 
.
is the common chord of 
,
which implies 
are collinear.
is tangent to 
.
is tangent to 
.
.
,
,
are concyclic.Moreover
gives that 
are concyclic.Denote by 
the second point of intersection of 
,
due to 
.
are on the opposite sides of 
,
,
.
and 
.
,
.
,
is harmonic,which implies that 
,
are lines of tangency of the circumcircle of 
is tangent to 
,
;
is cyclic and 
.
;
;
we get 
and 
defined? 
. I hope you will get it. 
so 
is cyclic. Note that 
Then from cyclic quadrilaterals 
we have 
so 
as desired.
Hence 
and 
is tangent to 
Now it is well known that 
harmonic, and since B,C, and 
are fixed, 
is fixed. But by a well known property of a symmedian, 
so 
.
followed by Radical Axes on 
,
and 
thus, Proving the First part! Let 
Similar angle chasing yields, 
.
S
Which clearly implies 
are concyclic. Call this circle 
and 
are concurrent, but 
,
be the circumcentre of 
so the claim is true.
and 
already passes through 
and 
then show that 
,
and 
.
intersects 
.
