[/quote]
,
with orthocenter 
,
lies on the McCay cubic. Let 
be the pedal triangle of 
wrt 
.
be the second intersection of 
with 
and let 
be the Poncelet point of 
.
are concurrent.
and let 
be the second intersection of 
be the pedal triangle of 
wrt 
.
be the circumcevian triangle of 
and 
are homothetic.
.
.
and 
,
is the insimilicenter of 
and 
.
of 
,
are collinear. \qquad \blacksquare $$[/quote]
be real numbers . Prove that
![$$-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}\leq \frac{2+\sqrt[3] 2+\sqrt[3] {4}}{12} $$](http://latex.artofproblemsolving.com/9/9/9/9997397eb38cd4bceec2ff8c85951c78e2797d0f.png)
, 
.
such that 
and there exists an integer 
such that 
is a multiple of 
.
is chosen on the side 
of a rectangle 
.
,
is dropped to the segment 
.
be a non-constant function from the set of positive integers into the set of positive integer, such that 
divides 
for all distinct positive integers 
,
.
such that 
for some positive integer 
.
and 
.![\[ab+bc+ac\leqslant 27\]](http://latex.artofproblemsolving.com/c/b/5/cb5e2a1efd462abd2354f4136f5fa64408f3138c.png)
of degree 
has exactly 
real roots, counted with multiplicity. We know that the coefficient of 
is 
,
.
take?
lies inside triangle 
and 
.
is the midpoint of segment 
.
lies on segment 
with 
.
.
satisfy,
Where 
}
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
such that:
For all real numbers 
.
-
,
inequalities, these given comparisons satisfy the following three constraints:
and 
,
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
Y by Adventure10, Mango247, Rounak_iitr
In a triangle 
we have 
A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides 
in the points 
respectively 
Prove that the midpoint of 
is the center of the inscribed circle of the triangle 
we have 
A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides 
in the points 
respectively 
Prove that the midpoint of 
is the center of the inscribed circle of the triangle 
the triangle semiperimeter and area. Let S, T be the tangency points of the incircle (I) with the sides AB, AC. Let K, L be the midpoints of of BC, PQ and M the midpoint of the arc BC of the circumcircle (O) opposite to the vertex A. The isosceles triangles 
are all centrally similar with the homothety center A. The homothety coefficient for the triangles 
is 
.
is 
.
are also centrally similar with the homothety center A and the same homothety coefficient as the triangles 
,
.
is then 
Denote h' = AL the A-altitude of the triangle 
.
,
and 
,
for an isosceles triangle)
is identical with the incenter of the triangle 
,
and the circle 
interior tangent to the circumcircle in the point 
and tangent to the sides 
in the points 
respectively. From the a wellknown property, the points 
,
belong to the circumcircle. Thus, the quadrilaterals 
,
are cyclic and
,
.
is true in any triangle ![$[AB],[AC]$](http://latex.artofproblemsolving.com/b/5/c/b5c8682c48eeeea06d0a348f2f442ec67b91fbd3.png)
in the points 
of the triangle ![$[PQ]$](http://latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
and 
,
belong to the circumcircle of the triangle 
results that the points 
,
,
are collineary, i.e. 
be the tangency of circles and 
be the center of inner circle. 
and 
.
.
.
at 
and 
at 
. 
is cyclic. 
.
is cyclic. 
.
.
is angle bisector.
is internally tangent (in the point 
)
and is also tangent to the sides ![$ [AB],[AC]$](http://latex.artofproblemsolving.com/1/a/d/1ada2a442db83b27aec868dd31b04d8ef6651553.png)
in the points 
.
of the triangle ![$ [PQ]$](http://latex.artofproblemsolving.com/c/3/1/c3177f15f923811226dabc1e2556939f7542c751.png)
and 
be points in 
and 
,
and 
.
be the center of the excircle of 
and 
respectively 
)
is 
and radius 
.
and 
into 
.
is mapped into a circle tangent to a circle tangent to 
(because it is the reflection of 
over the bisector of 
the inverse of 
.
.
.
.
of diameter 
.
.
is the power of 
.
,
(because both points are on the bisector of 
,
and 
is known.
maps the incircle into 
and the result follows using the first equality.
denote the circumcircle of 
denote the internally tangent circle through 
denote the point of tangency between 
denote the center of 
.
,
.
.
.
and by symmetry, 
.
.
is cyclic. Therefore 
.
and 
and 
,
.
,
and 
,
we get 
,
,
,
and 
.
and, with the relations 2-5:
,
,
and let the angle bisector of 
hits the 
contains 
then 
.
is cyclic
bisects 
.
and 
.
and 
so it follows that 
b
are
with center 
.
.
are similar, we have 
.
![$\sqrt[2]{bc} $](http://latex.artofproblemsolving.com/8/8/d/88d5de0ab1469b1304aaa3e7fec3a9c067460806.png)
inversion. The mixitilinear Incircle becomes the excircle and the incenter the excenter.
.
.
and 
.
.
,
is a kite, done.
;
;
but 
too and 
so a contradiction thus 
(hopefully not a wrong solution)
.
-
cyclic, where 
are the excentral touchpoints, which is obvious.
which implies the result.
,
as 
,
.
and 
collinear. Pascal on 
then tells us that 
collinear, and noting 
implies 
.
