[/quote]
,
satisfies 
.
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
.
be real numbers such that 
.
.
into 
or 
can sometimes be quite complex. That's why I’ve written a program to assist with this process.
such that for all positive real numbers 
and 
,![\[
\frac{x + f(y)}{x f(y)} = f\left( \frac{1}{y} + f\left( \frac{1}{x} \right) \right)
\]](http://latex.artofproblemsolving.com/7/d/f/7df25755bc33bef970fffb392fb5b6b789390881.png)
cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from 
to 
cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?
,
,
by the recurrence![\[
a_{n+1} = a_n^2 + 1
\]](http://latex.artofproblemsolving.com/3/1/1/3114595ded8c8961d611afaab9ad19e254973001.png)
Prove that there is no natural number 
such that![\[
\prod_{k=1}^{n} \left( a_k^2 + a_k + 1 \right)
\]](http://latex.artofproblemsolving.com/a/4/f/a4f88ee7e3d4e53e369319f74cd9b12be26c9681.png)
is a perfect square.
written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are 
;
;
;
,
;
,
be a cyclic quadrilateral with circumcenter 
,
is not a diameter of its circumcircle. The lines 
and 
intersect at point 
,
lies between 
and 
lies between 
and 
is acute and let 
be its orthocenter. The points 
and 
on the lines 
and 
.
,
.
distinct positive integers and then writes in blue the 
sums of each two red numbers. At most how many of the blue numbers can be prime?
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]$](http://latex.artofproblemsolving.com/c/3/1/c3177f15f923811226dabc1e2556939f7542c751.png)
and 
be points in 
and 
,
and 
.
be the center of the excircle of 
and 
respectively 
)
is 
and radius 
.
and 
into 
.
is mapped into a circle tangent to a circle tangent to 
(because it is the reflection of 
over the bisector of 
the inverse of 
.
.
.
.
of diameter 
.
.
is the power of 
.
,
(because both points are on the bisector of 
,
and 
is known.
maps the incircle into 
and the result follows using the first equality.
denote the circumcircle of 
denote the internally tangent circle through 
.
denote the point of tangency between 
.
.
.
.
and by symmetry, 
.
.
is cyclic. Therefore 
.
and 
and 
,
.
,
and 
,
we get 
,
,
,
and 
.
and, with the relations 2-5:
,
,
and let the angle bisector of 
hits the 
contains 
then 
.
is cyclic
bisects 
.
and 
.
and 
so it follows that 
b
are
with center 
.
.
are similar, we have 
.
![$\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)
.
-
with the 
cyclic, where 
are the excentral touchpoints, which is obvious.
which implies the result.
,
,
as 
and 
.
and 
collinear. Pascal on 
then tells us that 
collinear, and noting 
implies 
.
