is given. Let 
be set of all 
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 
where 
.
cannot be a perfect cube of a positive integer.
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)
be rational numbers such that 
is rational.
.
so division by 
is possible and hence
and both 
and 
Y by Adventure10, Mango247
Let E and D be points on A'C' and A'B' (A'B'C' is orthic triangle of ABC) so that AEA'D be a parallelogram.
Then K=BD ⋂ CE.
Also, it is a simple proof that K is between midpoint of a side and midpoint of an altitude (See: Ross Honsberger - Episodes in Nineteenth and Twentieth Century pp 65). In fact, parallelogram is actually rhombus, so ED || BC. Further, midpoints of ED and BC are collinear with K.
Proof: The parallelogram AEA'D is a rhombus (AA' is angle bisector of angle C'A'B') diagonals are perpendicular. It's easy to see that AD=DG and AE=EF (F=AE⋂BC, G=AD⋂BC). Since AG and AF are antiparallels with A'C' and A'B', K=BD ⋂ CE. From intercept theorem, symmedian point and midpoints of BC , DE are collinear.
It's pretty obvious that points E and D lying on midline of triangle ABC.
Then K=BD ⋂ CE.
Also, it is a simple proof that K is between midpoint of a side and midpoint of an altitude (See: Ross Honsberger - Episodes in Nineteenth and Twentieth Century pp 65). In fact, parallelogram is actually rhombus, so ED || BC. Further, midpoints of ED and BC are collinear with K.
Proof: The parallelogram AEA'D is a rhombus (AA' is angle bisector of angle C'A'B') diagonals are perpendicular. It's easy to see that AD=DG and AE=EF (F=AE⋂BC, G=AD⋂BC). Since AG and AF are antiparallels with A'C' and A'B', K=BD ⋂ CE. From intercept theorem, symmedian point and midpoints of BC , DE are collinear.
It's pretty obvious that points E and D lying on midline of triangle ABC.
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This post has been edited 9 times. Last edited by nikolinv, Jul 12, 2016, 11:38 AM
