AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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 
be a triangle and , 
, 
be the midpoints of arcs , 
, 
on the circumcircle. Line 
passes through the feet of the perpendiculars from to 
and 
. Line 
passes through the feet of the perpendiculars from to and 
. Let 
denote the intersection of lines and . Define points 
and 
similarly. Prove that triangle 
and 
are similar to each other.
such that : 
and 
be a convex quadrilateral such that 
. Suppose that a point on the side 
satisfies the equality![\[AE\cdot ED + BE^2=CD\cdot AE.\]](http://latex.artofproblemsolving.com/d/7/5/d75cc3a449a56b43eac574cc44e9932a2e63b33b.png)
. on the side 
such that 
then we have 
and 
, as required.
intersect 
at point 
. From PoP we have that 
. Thus original equality becomes:
and that 
. Consequently we have:
Dividing relations 
and 
we get that:
This implies that and thus the claim is proved.
as desired.
is clearly unique. Let 
be the reflection of over 
and define 
similarly. Then 
Because 
is tangent to 
by the angle condition, 
as needed.
about 
to get 
. Then the condition becomes 
, and what is to be proven is the Alternate Segment Theorem, so we're done.
, then there is will be unique which requires the given condition. So let's assume that , and prove the given condition. Let 
be the reflection of 
over . Let 
. 
. be the midpoint of 
. So it's clear that 
. On the other hand we have: 
. Which means that 
. On the other hand by Thales theorem: 
, and conclusion follows.
.
.
on 
.
thus 
.
,
.
we are done.