[/quote]
,
and 
for which there exists a natural numbe
is integer for all natural 
.
has 
as the midpoint of 
and orthocenter 
be an arbitrary point on the nine point circle of 
intersects 
.
perpendicular to 
intersects 
at 
is the midpoint of 
.
that, for all 
,![\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]](http://latex.artofproblemsolving.com/4/8/b/48b9fc4af5a20de7fd9dfc45a74d5aa27e615e46.png)
that satisfy the condition![\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]](http://latex.artofproblemsolving.com/2/7/d/27d266570e3a105a4b416bb7c55904ffabf7a07c.png)
Find the maximum value of the expression![\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]](http://latex.artofproblemsolving.com/b/a/f/bafc95e59d2f28dbb4e02771db38ce736e3fd45c.png)
.
.
-
numbers 
each equal to either 
or 
.
so that, for any 
and indices 
so that 
for all 
,
and 
such that
be reals such that 
and 
Find the value of 
players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least 
next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
and 
Find the value of 
,
be the midpoints of arcs 
,
on the circumcircle. Line 
passes through the feet of the perpendiculars from 
and 
.
passes through the feet of the perpendiculars from 
.
denote the intersection of lines 
and 
similarly. Prove that triangle 
and 
are similar to each other.
Y by anantmudgal09, Mathuzb, Number_Ninjas_2003, Adventure10, Mango247
Given an acute-angled triangle 
with orthocenter 
. Reflection of nine-point circle about 
intersects circumcircle at points 
and 
. Prove that is the external bisector of 
.
Proposed by Mohammad Javad Shabani
with orthocenter 
. Reflection of nine-point circle about 
intersects circumcircle at points 
and 
. Prove that is the external bisector of 
.Proposed by Mohammad Javad Shabani
This post has been edited 1 time. Last edited by Dadgarnia, Apr 15, 2019, 7:42 PM
and 
meet 
.
where 
and let 
.
are concyclic, so 
.
and 
be the reflections of 
lie on the circumcircle, ![\[HA\cdot 2HD=HA\cdot HD'= HX\cdot HH' = HX\cdot 2HY'\implies HX\cdot 2HZ=HA\cdot HD = HX\cdot HY'\]](http://latex.artofproblemsolving.com/3/e/7/3e7587f811bf87686730d198944adbca2ca85e26.png)
giving 
.
and 
be the reflection of 
,
finishing the problem.
as the nine-point circle WRT 
,
as its reflection over 
as the reflections of 
over 
are concyclic points. Also reflection of 
lies on 
and 
.
be the midpoint of 
Also, 
bisects 
,
is the reflection of 
be the circumcircle and 
be the 
-
,
be their reflection in 
to 
in 
.
lies on 
with 
,
the circumcenter and 
so 
yields that 
bisects angle 
;
.
meet 
.
.
.
,
.
,
denote foot of altitude from 
to 
and similarly 
,
.
are coaxal with radical axis 
.
lie on the radical axis. First, note that 
so the first two circles are done. To see the third circle, recall that the projection 
of 
lie on both 
.
,
be the reflection of 
.
is the reflection of 
.
.
,
so 
are concyclic. Hence we are done.
at 
.
,
is the apollonial circle preserving 
,
we have that 
.
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the nine-point circle, and 
their respective reflections across 
.
denote negative inversion at 
.
and 
is an isosceles trapezoid and 
be a line through the exsimilicenter 
.
of 
across 
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across 
.
inversion, 
is an isoceles trapezoid so 
the reflection of the nine-point circle respect 
the center of this circle.
describe a circle. 
all satisfy the equation. So this circle is 
.
and is such that 
be the projection of 
is the midpoint of 
because it's the projection also of the center of nine-point circle which is midpoint 
.
and 
is perpendicular to 
