AoPS Academy
has an incenter 
Suppose 
. Let 
be the midpoint of 
. Suppose that 
. 
meets 
again at point 
. Let points 
and 
be such that is the midpoint of 
and is the midpoint of 
. Point 
lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at 
and 
. Prove that 
.
with another positive integer 
, as long as 
is a perfect square. For example, if the number on the blackboard is 
, Carmela can replace it with 
, because 
, then replace it with 
, because 
. If the number on the blackboard is initially 
, determine all integers that Carmela can write on the blackboard after finitely many operations.
is a polygon which has a center of symmetry such that if belongs to , then so does 
, where is the midpoint of 
. Show that there is a parallelogram 
containing such that the midpoint of each side of lies on the border of .
is a polygon which has a center of symmetry such that if belongs to , then so does , where is the midpoint of . Show that there is a parallelogram containing such that the midpoint of each side of lies on the border of .
. of minimal area containing the polygon 
(and which has the center of symmetry 
of as its own center of symmetry). Let's prove that such a polygon satisfies the requirements.
of 
(respectively) do not belong to the border 
of . The intersection between and 
is a closed (maybe degenerate) segment 
which lies either in 
or ![$(MD]$](http://latex.artofproblemsolving.com/5/c/0/5c08767537684885e7b0a1f37b37a78df2cfff23.png)
. Suppose it's the former. Then, the intersection 
will be a closed sgement 
which lies in ![$(NC]$](http://latex.artofproblemsolving.com/0/9/b/09bad316970444b8a98fafee81496c64b9e13f1d.png)
. Now keep and the lines 
fixed, and move 
slightly, putting 
, until no longer meets , but such that still contains (a drawing should make all of this clear). Then 
will not meet either, and the new parallelogram will have the same area as the old one. We can slightly shrink it along the direction of 
so that it still contains , and this contradicts the minimality of the area of the initial parallelogram.
such that 
such that 
and 
