AoPS Academy
be a triangle, and let 
be a point on side 
. A line through intersects side 
at 
and ray 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle again at point 
distinct from point 
. The lines 
and 
intersect again at 
and 
respectively.
such that 
for all real number 
and 
be a scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let intersect 
and 
at and respectively and intersect 
and 
at and respectively .
, point 
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds. is an isosceles triangle.
such that, for any triangle 
, we can choose points 
trisecting the perimeter of the triangle into three equal-length sections so that the area of 
is exactly times that of 
.
such that, for any triangle , we can choose points trisecting the perimeter of the triangle into three equal-length sections so that the area of is exactly times that of .
is designed in such a way that either or coincides with 
or 
.Now we assume the case when coincides with . See that 
.Here the equality holds when 
i.e. in the limiting case where 
.Other cases can be dealt in similar manner.So 
,contradiction.
and then choose the triangle. Actually I guess 
work.
and 
.
trisects the perimeter of ![$\frac{[XYZ]}{[ABC]} = 1 - \frac{(BC.AY.AZ + CA.BZ.BX + AB.CX.CY)}{AB.BC.CA}$](http://latex.artofproblemsolving.com/6/8/a/68a15155d65211c44edb84861176c61d8a02b724.png)
.![$\frac{[XYZ]}{[ABC]} \ge 1 - \frac{BC+CA+AB}{4} = \frac{1}{4}$](http://latex.artofproblemsolving.com/d/3/3/d3362cb2aa622053be6b81f7ccff99df6197be9d.png)
.
.
So
is atmost a quadratic in ![$\frac{[XYZ]}{[ABC]} = lBX^2 + mBX +n = f(BX)$](http://latex.artofproblemsolving.com/b/f/d/bfd00eddf16b38b2f48422b19edb6ff01fe8b006.png)
for some constants 
independent of 
to 
We see that 
is a quadratic with positive leading coefficient(this requires some calculation)so it attains it's maxima at endpoint of the interval.