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)
satisfying 
is inscribed in circle 
.
denote its incenter and 
its 
-
denote the midpoint of arc 
.
meets 
.
at 
and 
meets 
a second time at 
,
meets 
a second time at 
.
passes through the antipodal to 
that satisfy 
with 
,
and area 
.
with diameter 
at 
and extend to meet 
and 
is the diameter of 
of 
,
of 
and 
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
.
lie on 
respectively, such that 
.
to meet the circumcircle of 
respectively. Let the circumcircles of 
meet at points 
.
to 
,
,![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
with 
,
to meet the circumcircle of 
at 
,
.
Y by Adventure10, Mango247, Rounak_iitr, and 1 other user
Let 
be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at 
. The midpoints of 
and 
are 
and 
respectively, and 
is the line through parallel to . The feet of the perpendiculars from E onto the lines and are 
and 
, respectively. Prove that the lines 
and 
are perpendicular.
be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at 
. The midpoints of 
and 
are 
and 
respectively, and 
is the line through parallel to . The feet of the perpendiculars from E onto the lines and are 
and 
, respectively. Prove that the lines 
and 
are perpendicular.
WRT 
it follows that 
is the E-circumdiameter of 
Let 
be the orthogonal projection of 
cuts 
From the cyclic quadrilateral 
(due to its right angles <EKG and <EHG) we deduce that 
Since 
are isogonals WRT 
then 
are also isogonals WRT 
From 
we have then 
are collinear, i.e. 
.
.
.
and 
are similar and the angles 
are these that are formed by the median and the height to these similar triangles so they are equal and we are done.
,
.
,
,
.
The condition can be written as 
,
are cocyclic, so 
.
,
so 
Denote 
.
Denote 
,
,
,
,
(
.
.
.
.
,
.
THen 
is cyclic.
and 
,
as desired.
perpendicular to 
by Similarity
is parallel to 
so 
!
wrt 
and 
and 
and it is so easy that we can proof by 
so 
and 
problem proof! 
bring the symmedian of triangle 
.
and 
![\[\frac{sin \ \alpha}{sin \ \beta}=\frac{sin \ \lambda}{sin \ (\alpha+\beta+\lambda)} \ (\star)\]](http://latex.artofproblemsolving.com/3/1/b/31b500f686afdc58684e270964507a6d173a2ea7.png)
and similarly in triangle ![\[\frac{sin \ \alpha}{sin \ \beta}=\frac{sin \ \phi}{sin \ (\alpha +\beta + \phi)} \ (\star\star)\]](http://latex.artofproblemsolving.com/7/d/d/7dd993247d44cfcb55a1270e4a9003c53961a214.png)
Combinig 
and 
we get ![\[\frac{sin \ \phi}{sin \ (\alpha+\beta+\phi)} = \frac{sin \ \lambda}{sin \ (\alpha +\beta + \lambda)}\]](http://latex.artofproblemsolving.com/4/e/f/4ef1a24f91b539ee6f54faa8ca113628880ae217.png)
Solving the equation we get 
(Which is impossible) then 
By angle Chasing we get 
Where 
and since 
which follows 
and let 
and 
we have 
Since 
we get 
which means that 
be the intersection of 
and let 
be the intersection of 
with 
since the quadrilateral 
and, 
(since, the triangles 
and 
are simillar). hence, the triangles 
and 
are simillar and, 
now, we see that the quadrilateral 
is Cyclic since, 
therefore, 
and thus, the quadrilateral 
is Cyclic. hence, 
since, 
.
