[/quote]
,
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
with circumcenter 
,
be an arbitrary point on the circumcircle of triangle 
and that line 
is not parallel to line 
be the point on the circumcircle of triangle 
is perpendicular to 
be the point on line 
such that 
.
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
with 
,
intersects the extension of 
at point 
.
is the midpoint of 
,
intersects the circumcircle of 
.
.
be positive real numbers such that 
.![\[ \frac{a_{1} a_{2} \cdots a_{n} \left[ 1 - (a_{1} + a_{2} + \cdots + a_{n}) \right] }{(a_{1} + a_{2} + \cdots + a_{n})( 1 - a_{1})(1 - a_{2}) \cdots (1 - a_{n})} \leq \frac{1}{ n^{n+1}}. \]](http://latex.artofproblemsolving.com/6/5/5/65558feba82e32266d3d3cbdfea85e079483403f.png)
with integer coefficients such that for all positive number 
and prime 
satisfying 
,
.
be reals such that 
P
,
,
is __
satisfying 
denote its inradius and ex-radii opposite to the vertices 
respectively. If 
and 
,
is obtuse
is acute
,
is semi perimeter
,![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
satisfy that: 
and 
.
and 
Prove that
Y by nervy
Given is a parallelogram 
, and the variable points 
lie on the sides 
respectively, so that 
is cyclic with circumcenter 
, 
, and 
. Let 
be the intersection point of the lines 
and 
, and let 
be the intersection of the lines 
and 
. Prove that the circle 
passes through a fixed point.
, and the variable points 
lie on the sides 
respectively, so that 
is cyclic with circumcenter 
, 
, and 
. Let 
be the intersection point of the lines 
and 
, and let 
be the intersection of the lines 
and 
. Prove that the circle 
passes through a fixed point.
,
,
the midpoint of 
.
is a paralelogram, 
is also the midpoint of 
is the midpoint of 
,
,
= 
= 
as the unit circle. Then, we wish to show 
.
,
,
But notice that the conjugate of the last expression, after multiplying each fraction by 
,
which is the same as the original, so 
and thus we are done 
is a parallelogram, 
and 
are isogonal in 
and point 
,
(as 
)
(
Therefore, 
