[/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 
and orthocenter 
,
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 
.
and 
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 Adventure10
Let 
be a cyclic quadrilateral such that 
is a diameter of it's circumcircle. Suppose that and 
intersect at 
, 
and 
at 
, 
and 
at 
, and let 
be a point on . Show that 
is perpendicular to 
if and only if is the midpoint of .
be a cyclic quadrilateral such that 
is a diameter of it's circumcircle. Suppose that and 
intersect at 
, 
and 
at 
, 
and 
at 
, and let 
be a point on . Show that 
is perpendicular to 
if and only if is the midpoint of .
are right. Thus AC, AD are 2 altitudes of the triangle 
intersecting at its orthocenter K and the 3rd altitude 
.
.
,
intersecting at its orthocenter K. The 3rd altitude 
.
is the midpoint of AB, 
.
and 
,
intersecting at its orthocenter L different from K. Hence, the 3rd altitude 
is different from IK. This means that IK is not perpendicular to JN, because IL is.
