be positive real numbers such that 
.![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
,
.
be reals such that 
P
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 
in 
.
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 
and 
are the intersections of 
we have 
.Now Let 
intersect 
.
.Hence it is easy to get the desired result
and 
.
is the midpoint of 
.
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 arc 
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 
.
