AoPS Academy
square is filled with numbers 
.The numbers inside four 
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
satisfying 
, find the biggest value the following equation could acquire:
+1 w![\[\frac{x^3+1}{(y-1)(z+1)}+\frac{y^3+1}{(z-1)(x+1)}+\frac{z^3+1}{(x-1)(y+1)}\]](http://latex.artofproblemsolving.com/5/c/6/5c6894b617b990ec69c62ed1cf7d062d4660af53.png)
where 
are reals.
so that for any 
, we have ![\[ f(xf(x^2)+yf(x))=xf(y+xf(x)). \]](http://latex.artofproblemsolving.com/1/d/4/1d4fbf3f80306aaae5cfc11cc0e4be7ea41491d7.png)
![$f : [0,1] \longrightarrow \mathbb{R}$](http://latex.artofproblemsolving.com/9/a/4/9a4b7997c35990f839742d29866122791ade8b7c.png)
is differentiable with 
. If 
for all ![$x \in [0,1]$](http://latex.artofproblemsolving.com/d/3/4/d3423a3200d0ad7660fc4265d59d5598a53d551d.png)
, then show that 
for all 
.
.
be a continuous function defined on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
with 
and 
. We define the point 
to be the projection of point 
on the x-axis. Prove that there exist points 
such that 
is a square.
increasing with
? increasing with
?
and the distribution function is 
then
and find the expression of 
over all triples 
of distinct real numbers such that
are built on the sides of the triangle 
to the outside. On the segment 
to the outer side of the triangle 
, an equilateral triangle 
is constructed. Prove that 
is the midpoint of the segment 
.
are built on the sides of the triangle to the outside. On the segment to the outer side of the triangle , an equilateral triangle is constructed. Prove that is the midpoint of the segment .
, 
, 
are collinear using Torrichelli point. In the second part we will be proving that 
using rotations, and are collinear
. Then 
is the Torrichelli point of triangle and is the Torichelli point of triangle 
. So , , are collinear. Analogously, points , and are collinear. This means that points , , all lie on line 
, hence are collinear
. Consider rotation centered at point taking 
to 
, then 
is taken to 
, meaning that the segment 
is taken to . Analogously, rotating around you can get 
. Thus, 
. Note that the same can be done with vertices of triangle proving 
. From this follows, that 
