[/quote]
,
be a positive integer. Find the number of permutations 
,
,
of the
,
,
, 
.
positive integer the equation: 
has positive integer solutions.
be a point inside of it such that the square root of the distance of 
of positive integers is called central if for every positive integer 
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)
be a triangle. On side 
take points 
and 
such that 
take points 
such that 
and on side 
take points 
such that 
Let 
and 
are concurrent.
that verify the following equation :
lie the following points: 
and 
on 
,
on 
,
on 
.![\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]](http://latex.artofproblemsolving.com/7/e/a/7ea90a0f1261ed7445bbac58fb0cdd76810259f3.png)
and let the line 
meet 
.
,
are collinear.
.
intersects 
and 
and 
,
of triangle 
lies on the circle 
and 
coincide.
be a triangle and 
and 
be two points on 
,
such that 
and let 
.
to the interior angle bisector of 
intersect 
in 
.
.
,
at 
,
to 
, respectively. Rays 
intersect 
at 
,
touches incircle at 
Y by Funcshun840, Kizaruno, Rounak_iitr
Let 
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.This post has been edited 1 time. Last edited by ErTeeEs06, Apr 26, 2025, 11:15 AM
Reason: latex
Reason: latex
has orthocenter 
then 
is a parallelogram, hence 
bisects 
,
,
,
.
,
,
,
,
,
.
with orthocenter 
