[/quote]
,
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
be the incentre of acute-angled triangle 
,
,
respectively. Let line 
intersect the circumcircle of the triangle at 
lies between 
and 
.
so that for every 
in 
.
such that 
.
and 
at 
.
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 
,
divides 
for every 
.
,
:
,
Y by AdithyaBhaskar, Davi-8191, Math-Ninja, parola, rg_ryse, MathLuis, Bubu-Droid, SSaad, HamstPan38825, Adventure10, Mango247, Rounak_iitr, Exponent11, and 1 other user
Let 
be the circumcircle of the triangle 
. The circle 
is tangent to the sides 
and 
, and it is internally tangent to the circle at the point 
. A line parallel to 
intersecting the interior of triangle is tangent to at 
.
Prove that
.
be the circumcircle of the triangle 
. The circle 
is tangent to the sides 
and 
, and it is internally tangent to the circle at the point 
. A line parallel to 
intersecting the interior of triangle is tangent to at 
.Prove that
.
denote the angle bisector of 
and 
be the reflections of 
and 
.
and 
are isogonal. Note that by homothety, 
is the tangency point of the 
.
and 
.
is the tangency point of the 
to side 
and 
respectively and let 
be the midpoint of 
.It is well known that 
bisects 
.
is the 
.Hence 
.Similarly 
bisects 
Let 
and 
intersect 
and 
respectively Hence Line 
bisects the arc 
not containig 
it also bisects arc 
not containing 
we have 
.Now Let 
.
.Hence it is easy to get the desired result
and 
.
intersect 
.
are the arc midpoint of 
respectively in 
implies 
is the arc midpoint of 
intersect 
.
and 
,
,
,
,
cyclic.
implies 
bisects 
.
,
is symmetric wrt 
[proved].
on 
.
.
centered at 
.![[asy]
pair excentre(pair A, pair B, pair C) {
return((-abs(B - C)*A + abs(C - A)*B + abs(A - B)*C)/(-abs(B - C) + abs(C - A) + abs(A - B)));
};
pair mixtilinearcirclecenter(pair A, pair B, pair C) {
pair I, D, E;
I = incenter(A,B,C);
D = extension(A, B, I, rotate(90)*(A - I) + I);
E = extension(A, C, I, rotate(90)*(A - I) + I);
return(extension(D, rotate(90)*(A - D) + D, E, rotate(90)*(A - E) + E));
};
real mixtilinearcircleradius(pair A, pair B, pair C) {
pair I, D, E, O;
I = incenter(A,B,C);
D = extension(A, B, I, rotate(90)*(A - I) + I);
E = extension(A, C, I, rotate(90)*(A - I) + I);
return(abs(mixtilinearcirclecenter(A,B,C) - D));
};
path mixtilinearcircle(pair A, pair B, pair C) {
return(Circle(mixtilinearcirclecenter(A,B,C),mixtilinearcircleradius(A,B,C)));
};
pair mixtilinearcircletangentpoint(pair A, pair B, pair C) {
pair[] R;
R[0] = abs(circumcenter(A,B,C) - A);
R[1] = mixtilinearcircleradius(A,B,C);
return((R[0]*mixtilinearcirclecenter(A,B,C) - R[1]*circumcenter(A,B,C))/(R[0] - R[1]));
};
unitsize(0.4 cm);
pair A, B, C, Cp, D, E, F, G, O, P, Q, R;
pair[] I;
A = (0,0);
B = (14,0);
C = (2,12);
I[0] = incenter(A,B,C);
I[1] = excentre(A,B,C);
I[2] = excentre(B,C,A);
I[3] = excentre(C,A,B);
D = (I[0] + I[3])/2;
E = (I[1] + I[2])/2;
P = mixtilinearcircletangentpoint(C,A,B);
Q = intersectionpoint(I[0]--E,mixtilinearcircle(C,A,B));
O = circumcenter(A,B,C);
Cp = intersectionpoint(interp(P,C,0.1)--C,mixtilinearcircle(C,A,B));
draw(incircle(A,B,C),gray(0.7));
draw(mixtilinearcircle(C,A,B),red);
draw(E--C--D);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
draw(C--P--E);
draw(C--Q--Cp);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NW);
dot("$C'$", Cp, SE);
dot("$D$", D, S);
dot("$E$", E, N);
dot("$I$", I[0], dir(0));
dot("$P$", P, SW);
dot("$Q$", Q, SE);
dot(mixtilinearcirclecenter(C,A,B));
[/asy]](http://latex.artofproblemsolving.com/2/9/a/29ae440130b782220dc55586d71540c516c466ea.png)
.
is parallel to 
.
touch 
and the symmetry wrt the bisector of 
.
and circle 
.
.
.
are midpoints of 
in 
is concyclic.
is concyclic.
.
.
.
.
we get 
is an isosceles trapezoid.
.
which implies the tangent at 
.
be the circumcentre of 
is cyclic.
is cyclic. Since 
we conclude 
bisects 
are isogonals of 
touch 
at 
a
r
intersects 
i
,
.
so 
.
and 
are similar from which the result follows.
because the circle of inversion is orthogonal to it. In other words, consider two points 
by Power of a Point, and yet 
and vice-versa.
is tangent to 
are concyclic and 
,
.
.
,
,
the midpoints of the arcs 
and the center of 
are collinear, and so are 
.
is cyclic, and so is 
,
;
we are done, since 
is the angle bisector of 
the touch-point of 
with 
respectively. By the homothety wrt 
.
be midpoint of 
are collinear. Let 
.
center of 
.
are collinear so 
which 
.
and 
is the 
(since 
)
inversion about 
denote its image. Then, 
,
is the excircle touchpoint in triangle 
.
to meet 
,
.
.
and 
are isogonal with respect to 
,
.
,
.
inversion. Indeed, this inversion swaps 
and 
,
,
.
at 
and let the touch points of 
.
.
and 
is,indeed,the touch point of 
in 
and 
collinear. Pascal on 
then tells us that 
collinear, and noting 
implies 
.
and 
are a median and symmedian, respectively, of 
,
.
,
.
.
.
would be the foot from 
to 
passes through 
,
.
at 
respectively. Now consider a homothety centered at 
,
at point 
are collinear due to the homothety at 
and 
.
at one point. Hence 
,
must lie on a circle tangent to both 
and to ![\[\angle ACP = \angle Q'CB = \angle QCB, \]](http://latex.artofproblemsolving.com/6/6/3/663f6388dc36f66e5b84f0f27fea29f920e331c4.png)
as desired. 
.
be the 
we get that 
where 
.
is tangent to 
and 
,
.
,
denotes the image of 
Let 
are collinear (It's done with homothety)
Then prove that 
around 
.
