of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
,
be two sequences of positive real numbers. Suppose 
is a sequence of positive real numbers such that 
for all 
.
.![\[
\left( \frac{M+z_2+\dots+z_{2n}}{2n} \right)^2
\ge
\left( \frac{x_1+\dots+x_n}{n} \right)
\left( \frac{y_1+\dots+y_n}{n} \right). \]](http://latex.artofproblemsolving.com/8/c/4/8c4a72f57363eb1ef114825c8d36191bd5f42cb7.png)
Y by mathematicsy, Adventure10
Let a triangle 
inscribed in a circle 
with center 
. Let 
the incenter of triangle and 
the contact points of the incircle with sides 
of triangle respectively . Let also 
the foot of the perpendicular line from 
to the line 
.Prove that line 
passes from the antidiametric point 
of 
in the circle .( 
is a diametre of the circle ).
inscribed in a circle 
with center 
. Let 
the incenter of triangle and 
the contact points of the incircle with sides 
of triangle respectively . Let also 
the foot of the perpendicular line from 
to the line 
.Prove that line 
passes from the antidiametric point 
of 
in the circle .( 
is a diametre of the circle ).This post has been edited 1 time. Last edited by stergiu, Sep 23, 2019, 12:08 PM
Reason: explanation
Reason: explanation
be the intersection of 
and 
.
and 
are collinear which completes the problem.
,of the circle 
? Thank you !
has invers the Euler circle of triangle 
and 
.
.
be the midpoint of arc 
in 
be the antipode of 
be the second intersection point of 
with 
.
be the inversion around 
.
Therefore,
Hence
But then
This way 
and so Claim 1 is proved.
are collinear.
and 
are similar because
Analogously, right triangles 
and 
are similar.
From Claim 1, we get that 
to 
.
,
and 
are similar. Consequently,
Therefore, 
.
and 
,
