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.
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Finally! The article on the Method of Moving Points is written! I'd like to say thanks to Marko Khasin from the Ukrainian IMO team who found some misprints in my work and sent a good problem for one of the examples, to user Fermat_Theorem for sending me problems and giving advice for searching them, and also Pluto1708 and Physicsknight for sending problems with the usage of this method.
This post has been edited 1 time. Last edited by Vlados021, Jul 29, 2019, 3:57 PM
is projective since 
are collinear (we project 
from 
on 
,
is the line parallel to 
through 
.
.
is also projective. So we just have to check three distinct choices of 
.
.
keeping 
fixed. Here's how it panned out. First we drop the obtuse angle condition.
be two points in the plane and 
a point on their perpendicular bisector, 
be the midpoint of 
.
,
,
.
is projective.
followed by reflection through the point 
and hence we are done.
is projective.
,
and 
,
,
Hence, ![\[X \mapsto X' \mapsto X'O' \mapsto X'O' \cap \overline{AC} \mapsto (NXO) \cap \overline{AC} \equiv B.\]](http://latex.artofproblemsolving.com/5/8/8/588f41f3ad9c574cf698991bbc7cddf441ff5c1a.png)
However, I'm not sure if even 
is projective.
on fixed line 
.
is projective by inverting about 
and prove 
is projective by intersecting perpendicular bisectors of 
but that's about it. I'm not quite sure how to proceed from here.
Thanks!
and 
and 
.
are projective so it suffices to show the result for 3 points.Now choosing 
can finish the problem
be moving along a conic with degrees 
,
,
.
.
works but what about the remaining case
