[/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 that line 
is not parallel to line 
is perpendicular to 
such that 
.
and 
be the points on the circumcircle of triangle 
is a diameter, and 
and 
are parallel. Let 
.
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, Mango247
Let 
be an acute triangle.
denote its circumcenter.Points 
,
,
are the midpoints of the sides 
,
,and 
.Let 
be a point on the side . 
. 
Prove that 
lie on 
where is the symmetrical point of thought the midpoint of 
.
be an acute triangle.
denote its circumcenter.Points 
,
,
are the midpoints of the sides 
,
,and 
.Let 
be a point on the side . 
. 
Prove that 
lie on 
where is the symmetrical point of thought the midpoint of 
.This post has been edited 1 time. Last edited by microsoft_office_word, Feb 12, 2018, 5:12 AM
as the unit circle and denote point 
as the complex number 
we can write
And since 
are collinear 
which gives us 
Now 
is a parallelogram, so 
which is the same as 
after some simplification
since it's conjugate multiplies to 
which means that 
which means 
as desired 
