[/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)
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 
such that 
.
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
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
Let 
be an acute-angled triangle inscribed in the circle 
, 
the foot of the altitude of at 
and 
a point inside lying on the bisector of 
. The circle of diameter 
cuts again at 
. Let 
be the projection of on 
. Prove that if 
bisects 
then is the incenter of the triangle .
Lê Phúc Lữ
be an acute-angled triangle inscribed in the circle 
, 
the foot of the altitude of at 
and 
a point inside lying on the bisector of 
. The circle of diameter 
cuts again at 
. Let 
be the projection of on 
. Prove that if 
bisects 
then is the incenter of the triangle .Lê Phúc Lữ
.
meet 
at 
resp. By Reim 
.
.
.
- rectangle, 
.
