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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.
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let 
be the point on the circumcircle of triangle such that 
is perpendicular to , and let 
be the point on line 
such that 
. Let 
and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
and 
. This means that we are done by Thales.
and 
with 
on 
(possibly extended) to conclude that
.So 
.
only!
be a trapezium with 
. Let be a point on such that 
is between 
and ; and let 
be the midpoints of 
respectively. Let 
intersect in 
and 
intersect 
in . Prove that 
.
be the point at infinity of 
.Let 
cut at 
respectively![\[-1=(A,B,X,\infty)\overset{P}{=}(B,C,N,H)\]](http://latex.artofproblemsolving.com/8/5/9/859261ccbd431a9b9cf19074ec749a26115592da.png)
![\[-1=(D,C,Y,\infty)\overset{P}{=}(D,A,M,G)\]](http://latex.artofproblemsolving.com/0/4/2/0426dadc578a76f74e26304821686300de70ca22.png)
.Clearly we have 
, since ratio is bijective whence 
passes through 
such that 
and 
and 
and 
;
be a quadrilateral. Let 
and 
such that the lines 
concur at a point 
.
on line 
,
and 
.
concur at a point 
.
.
and 
.![\[(B,C;H,N) \overset{P}{=} (B,A;E,X) \overset{F}{=} (C,D;E,Y) \overset{P}{=} (A,D;G,M)\]](http://latex.artofproblemsolving.com/5/e/e/5eed979ed264321d3ac3422cc72328626d1f5ae9.png)
But also,![\[(B,C;H,N) \overset{E}{=} (A,D;G,EN\cap AD)\]](http://latex.artofproblemsolving.com/8/d/e/8de7cd25c22dc973af4f86184aac00d151976c60.png)
Therefore, 
and we are done.