[/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 
such that 
.
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 and 1 other user
Two circles with different radii intersect in two points 
and 
. Let the common tangents of the two circles be 
and 
such that 
lie on the first circle, and 
on the second. Prove that the orthocenters of the triangles 
, 
, 
and 
are the four vertices of a rectangle.
and 
. Let the common tangents of the two circles be 
and 
such that 
lie on the first circle, and 
on the second. Prove that the orthocenters of the triangles 
, 
, 
and 
are the four vertices of a rectangle.
on their center line 
Let the line XA meet the circles 
The triangles 
are centrally similar with the similarity center X, hence, 
and 
The radical axis AB of the circles 
Translate the triangle 
by the distance 
and reflect it in the radical axis AB to obtain an oppositely congruent triangle 
such that 
Since 
we have 
and the quadrilateral ASA'T is cyclic with a circumcircle (O) and one diagonal 
be the orthocenters of the triangles 
respectively. The common circumcircle (O') of the triangles 
is the reflection of the circumcircle (O) in the line ST. Since the circles 
are congruent and 
(all perprendicular to ST), we have 
Thus 
is a parallelogram, 
and 
Transforming the triangle 
back to the oppositely triangle 
the orthocenter 
of the triangle 
of the triangle 
in the radical axis AB. Since the triangles 
are reflections of the triangles 
in the center line 
respectively, the orthocenters 
of the triangles 
of the triangles 
is a rectangle with the midlines 
and 
and 
are in the other.
be 
.
and 
are symmetric about this line, so our figure is an iscoceles trapezoid. I claim now that the midpoint of 
is the midpoint of 
,
is parellel to, and equal in length, to 
.
using Alternate Segment Theorem a few times, so using definition of orthocentre the quadrilaterals 
and 
are concyclic, so consider the circumcircles of 
and 
.
are reflection of each other in 
where 
are resp orthocentres of 
.
is a parallelogram. Also 
would be a reflection of 
wrt perp bisector of 
be the center of the first circle and 
be the center of the second one.
over line 
.
= 
.
,
.
is a parallelogram and it suffices to show that 
.
,
is a parallelogram. Do the same for 
.
.
