[/quote]
,
and two fixed points 
on that circle, let 
be a moving point on 
is acute and scalene. Let 
be the midpoint of 
and let 
be the three heights of 
,
such that 
.
be the intersection of 
and 
,
be the second intersection of 
and 
.
always goes through a fixed point.
intersects 
.
of 
,
such that 
.
be the center of 
.
always goes through a fixed point.
where 
where 
.
cannot be a perfect cube of a positive integer.
is given. Let 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
,
and 
are integers, 
.
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to 
with real coefficients obeying![\[P(x) P(x+1) = P(x^2 + x + 1)\]](http://latex.artofproblemsolving.com/b/e/f/beff3420fa8394f4f98c031d81eea6a5a3b8d28c.png)
for all real numbers 
.
and![\[(x-y)[f(x)+f(y)]\leqslant f(x^2-y^2), \forall x,y \in \mathbb{R}\]](http://latex.artofproblemsolving.com/8/0/b/80bb1d7a992c9683a0c2a07b2febe6271f4e60f5.png)
such that 
,
and 
.
is prime, 
with 
and 
?
be the circumcircle of isosceles triangle 
)
lie on 
.
to the incircle of 
(different from
such that:
is the Gergon point and the incircle 
touch 
,
at 
,
respectively.
,
,
be 
, 
, 
respectively.
and then we get the circle 
and then the circle 
and then the circle 
be non-negative real numbers such that 
Prove that
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After graduated from IMO, I planned to LaTeX a complete FE handout.
However, because of my laziness, I have only finished a part of this handout.
Hope this note will help some of AoPS users ~
Anyway, Happy Lunar New Year
However, because of my laziness, I have only finished a part of this handout.
Hope this note will help some of AoPS users ~
Anyway, Happy Lunar New Year
This post has been edited 1 time. Last edited by FEcreater, Feb 16, 2018, 3:24 AM
Reason: typo
Reason: typo
.
.![$f(2x)=2f(x)+6x\sqrt[3]{f(x)^2}$](http://latex.artofproblemsolving.com/a/2/9/a29b1667c9a1a022d05b295971e6a7b26e3a296e.png)
rather than ![$f(2x)=2f(x)+\sqrt[3]{f(x)^2}$](http://latex.artofproblemsolving.com/e/0/8/e0801a15e5683e42e692ad96fc611833b9f15279.png)
.
and 
(not 
)
is a constant.
surjective?
for all real 
is surjective.
of both sides, but this does not seem to result in the claimed condition.
by 
,
such that 
We do this by injectevity property. If 
for some 
we have 
So, 
Hence, f is injective. By substituting 
in the equation we get 
due to injectivety and by substituting 
in the equation we get 
Upto to this all things are easy understood. But after that he says "plugging back to the equation, 
"
?
and that 
,
.
