AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be positive integers such that 
and 
. Find the maximum possible value of 
.
, let the incircle 
touch sides 
at 
, respectively. Let 
lie on the line through 
perpendicular to 
. Let 
be the intersections of 
with 
, respectively. Let 
be the projections of 
onto line . Let 
be the second intersections of 
with the incircle . Let 
be the intersection of 
and 
. Prove that the line 
bisects segment 
.
be the incentre of 
such that the incircle touches the sides at the points respectively. The perpendicular to at 
and 
meets at 
and 
respectively. 
and 
meet the incircle at and 
respectively. 
and 
intersect at 
Prove that 
bisects 
![$P\in\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/f/3/d/f3d763f4197792a2956273485eb1e94700a43044.png)
has degree 
having real roots in 
. Suppose 
is the leading coefficient of . Show that, 
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both and 
. Integers and are coprime if 
.
that satisfy 
are positive real numbers, then
such that 
is a perfect square for all integers and .
is not isosceles. The incircle of 
touches the sides , 
, 
in the points 
, 
, . The parallel with 
through meets 
at , the parallel with through meets at 
.
passes through the midpoint of 
.
is not isosceles. The incircle of touches the sides , , in the points , , . The parallel with through meets at , the parallel with through meets at . passes through the midpoint of .
and 
. Notice that 
is harmonic, so 
so 
and by Menelaus, we get that bisects , as desired.
and 
Prove that
then prove 
Prove that
Prove that
and 
Prove that
