and 
.
and 
.
Equality holds when 
or 
be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
intersect 
at 
.
and 
are either parallel or coincide.
.
Where 
Y by Amir Hossein, alibez, Adventure10
The pentagon 
is inscirbed in a circle 
. Suppose that 
are reflections of with respect to sides 
respectively. Let 
be the second intersection point of 
and define 
similarly. Prove that
![\[2\le \frac{S_{A'B'C'D'E'}}{S_{ABCDE}}\le 3,\]](//latex.artofproblemsolving.com/0/8/e/08e140c3717be0e06780198d2860f40b62fb04be.png)
where
denotes the surface of figure 
.
Proposed by Morteza Saghafian, Ali khezeli
is inscirbed in a circle 
. Suppose that 
are reflections of with respect to sides 
respectively. Let 
be the second intersection point of 
and define 
similarly. Prove that![\[2\le \frac{S_{A'B'C'D'E'}}{S_{ABCDE}}\le 3,\]](http://latex.artofproblemsolving.com/0/8/e/08e140c3717be0e06780198d2860f40b62fb04be.png)
where
denotes the surface of figure 
.Proposed by Morteza Saghafian, Ali khezeli
This post has been edited 2 times. Last edited by Amir Hossein, Apr 25, 2012, 6:09 PM
Reason: Fixed LaTeX.
Reason: Fixed LaTeX.
are the midpoints of 
.
.
.
,
for now), such that 
and 
are on opposite sides of 
and also 
and 
,
pairs with sum 
.
with no common vertex and in that case the angle left will be 
,
.
be the midpoints of 
.
and similar for the other. So its enough to prove that, 
.
.
and similarly define 
.
and other such 3 triangles. WLOG, suppose, 
has maximum area among them. The note that, 
and 
and similar for 
.
.
.
.
.
.
.
define 
.
since the area of 
is exponentially decreasing.
be the origin. Easy to see that 
,
and analogously. So now we can compute their surfaces and it turns out to be 
.
as negative, if 
.
,
.
and it can be proven that some four out of these five triangles have greater area than 
,
are circumcenters of 
,
,
are all parallelograms. So, if we assign complex numbers 
to 
and 
to 
and 
.
and
This means 
is a parallelogram. In particular, 
and 
.
are midpoints of 
,
is similar to 
with similarity ratio of 
.
and we need to prove that
If
then this inequality can be written as
Without loss of generality assume either 
or rays 
and 
intersect.
intersect or are parallel, then 
since altitude from 
.
Draw lines 
and 
and note that 
and 
.
Draw a line 
through 
parallel to 
intersects line 
.
and we are done. Otherwise, let 
and note that 
.
intersects or parallel to the line 
and we are done again. Otherwise, ray 
intersects line 
and 
intersects segment 
.
and 
,
and 
.
to 
which is equivalent to
But this is true because rays 
and 
intersect at 
.
Multiplying both sides by 
If 
.
.
This means when we move 
So, if we move 
This follows easily from
and
