AoPS Academy
numbers 
, calculates the list 
where 
is the number of times the number 
appears in the list. So, for example, for the list 
, the program returns the list 
.
numbers. Then he asked him to apply the program to the resulting list, and so on, until a number greater than or equal to 
appears in the list. Find the largest value of for which, whatever the initial list of positive integers 
, it is possible for Alexander to do what the teacher asked him to do.
of convex polygons (in the plane). It turns out that every point of the plane lies in the same number of elements of 
as elements of 
. Prove that 
. or .
that satisfy the equation
of radius 
composed with reflection in bisector of 
maps this "problem" to the same problem in a similar triangle; the closer intersection is mapped to the farther one in the new figure. Using this similarity we get the isogonality.
by factor 
. Then reflect everything wrt angle bisector of . Finally invert with power 
. It can be seen that after this transformation 
are interchanged with one another. So 
has gone to 
and we are done.
with radius 
where 
is the foot from to 
. This inversion maps the circle with diameter 
to that with diameter 
. Now, their intersections are mapped to each other! (Since they are inverses) Since the intersection points are not fixed (check!) we get that 
are isogonal.
with radius 
and reflect all trough the angle bisector of angle 
. We will prove that and swap. Let 
and 
be the projections of 
and 
onto 
and 
. The circle 
pass trough 
, and while 
pass trough 
, and . By simple segment calculating, we get that: 
thus, our inversion and reflection makes:
as we wanted.
,
and are collinear?
, 
and 
(where and are circles with diameter and ,respectively) and concur on 
and since this intersection point is 
.But how it can help? 
be the respective midpoints of 
. Then it's clear that 
, so the internal angle bisector of 
also bisects , this yields![\[\frac{NL}{LM}=\frac{CN}{NM}=(\bigstar)\]](http://latex.artofproblemsolving.com/4/0/f/40f87eddfeed2c10b75982c935732ca85485c89e.png)
Now let the external angle bisector of meet 
at 
,
bisects 
bisects 
are concyclic on a circle with diameter 
.
lie on the -Appolonian cirle of 
are concyclic.
bisects 
.
, as desired. 
such that 
Prove that 
(i dont remember exactly if it was 8032038 or some other number, so correct me on this)
(i posted this at 12:29) lol
?
The circles with diameters 
