AoPS Academy
has 
as the midpoint of 
and orthocenter 
. Let 
be an arbitrary point on the nine point circle of 
. The line through perpendicular to 
intersects at 
. The line through 
perpendicular to 
intersects 
at 
. If 
is the midpoint of , show that 
.
that, for all 
and positive real numbers 
, there is![\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]](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)
-sequence is a sequence of 
numbers 
each equal to either 
or 
. Determine the largest 
so that, for any -sequence, there exists an integer 
and indices 
so that 
for all 
, and 
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 
and in a non-isosceles triangle with incenter 
intersect its circumcircle at 
and 
, respectively. The line through , parallel to 
, intersects 
at . Prove that 
is tangent to .
is a triangle with incenter ; 
the midpoint of arc not containing .Let the bissectors of 
cut respectively 
at 
then 
and 
are collinear and 
through cut 
at 
; it ' s clear that 
are isoceles besides 
we get 
thus 
is bisector of 
hence 
idem 
the midpoint of arc not containing 
; 
we need to show that 
then 
.
we get 
are collinear which closes the proof .
and 
- so 
is the perpendicular bisector of 
, whence 
. Hence 
(the latter is from the line parallel to ) and 
, which solves the problem.
.
.
, 
.
is tangent to 
, hence![\[PI^2 = PC_0 \times PA_0.\]](http://latex.artofproblemsolving.com/0/4/9/0495c60a93d89cb998267c141bd6dc870579922a.png)
But we notice that is the symmetric point of wrt 
. is the perpendicular bisector of , so 
.![\[PB^2 = PC_0 \times PA_0,\]](http://latex.artofproblemsolving.com/5/2/9/529c43329e92ef2c5e41d9f088cc82627aa3132c.png)
which implies that 
.
.
.
.
