[/quote]
,
and 
intersect at 
,
and 
and 
.
are concyclic.
the centre of the circle passing through 
,
are difined similarly. Prove that 
,
,
are concurrent.
> 
that satisfy 
,
€ 
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 Rounak_iitr, buddyram, ItsBesi
Acute-angled triangle 
with circumcircle 
is given. Let 
be the midpoint of 
, 
be the foot of altitude from 
to 
, and 
be the intersection point of 
and 
. Point 
lies on the arc of (the one that does not contain ) such that 
. Prove that 
.
with circumcircle 
is given. Let 
be the midpoint of 
, 
be the foot of altitude from 
to 
, and 
be the intersection point of 
and 
. Point 
lies on the arc of (the one that does not contain ) such that 
. Prove that 
.
.
is an isosceles trapezoid.
.
be the symmetry of 
is a rectangle. Hence, 
are collinear.
.
.
.
lies on 
.
.
.
,
and 
be the antipode of 
are collinear.![[asy]
size(200);
pair A = dir(120),B=dir(-155),C=dir(-25),O=(0,0);
draw(unitcircle);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
pair Ap = dir(60), E = foot(A,B,C), D = 1/2*(A+C), F = extension(A,B,D,E),H = IP(E--2*E-Ap,unitcircle);
dot("$A'$",Ap,dir(Ap));
dot("$H$",H,dir(-90));
dot("$E$",E,dir(130));
dot("$D$",D,dir(D));
dot("$F$",F,dir(F));
pair L = 2*E - A - B - C, B0 = -B;
dot("$L$",L,dir(20));
dot("$B_0$",B0,dir(B0));
draw(B--C^^A--Ap,red);
draw(C--A--F,royalblue);
draw(B--B0^^L--Ap,purple);
draw(B--Ap^^F--D,green);
draw(A--L^^H--Ap,brown);
draw(F--B0,dashed);
[/asy]](http://latex.artofproblemsolving.com/2/9/0/2905eb7808cea9753c97680d22949c23909f1f6e.png)
is the reflection of 
is a diameter of 
.
gives that our problem is equivalent to 
.
which gives 
.
,
.
Let's set 
.
This implies that 
,
is cyclic. Now just note that:
Thus 
as desired.
be such that 
.
are collinear.
.
.
.
is the antipode of 
.
.
.
and 
as needed
as desired. 
be the second intersection of the circumcircles of 
and 
.
and 
,
.
.
.
Thus from condition of tangency we have 
is tangent to the line 
at 
Thus 
as desired.
,
,
is cyclic.So, done.
meet 
.
Note that 
.
is tangent to 
so 
.
.
then 
.
.
is an isosceles trapezoid with 
.
.
.
.
.
.
an inversion centered at 
followed by a reflection wrt to the 
for each point 
.
be the circumcenter of 
be the midpoint of 
,
,
and 
such that 
.
,
,
are collinear. Hence, 
.
,
so we get 
are concyclic. Thus, 
as desired.
.
and 
intersect at 
intersect 
.
.
.
.
,
,
by 
cyclic. then 
that means 
.
,
.
that 
and then 
.
and we are done.
