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.
, 
and the function 
defined by![\[f(n)=\left\lfloor \frac{an+b}{cn+d}\right\rfloor\text{ for all } n\in\mathbb{Z}_{\ge 0}.\]](http://latex.artofproblemsolving.com/8/f/b/8fb372df4b44c3d7c0690b310b12c684134fee74.png)
Prove that the following are equivalent:
is surjective;
, 
and 
.
and 
be on sides 
and 
respectively in 
such that 
. Let 
be the circumcenter of 
and let 
. Prove that 
, where 
is the set of real numbers, such that
be the n-th prime number in increasing order for 
. Prove that there exists a sequence of distinct prime numbers 
satisfying 
for all 
be an acute triangle. Let 
be the midpoint of side 
, and let and 
be the feet of the altitudes from 
and 
, respectively. Suppose that the common external tangents to the circumcircles of triangles 
and 
intersect at a point 
, and that lies on the circumcircle of . Prove that line 
is perpendicular to line .
such that ![\[f(f(x)) + xf(xy) = x + f(y)\]](http://latex.artofproblemsolving.com/a/6/f/a6f84dd53eec985b90bad3df6580eeaf29d73195.png)
for all positive real numbers 
and 
.
, with 
, and the circle 
with center 
and radius . Denote by the midpoint of side . The line 
intersects the circle for the second time in . Let be a point on the circle such that 
and 
. Show that 
.
such that for any real value of 
we have:
be a triangle. 
and 
are points, such that 
, 
= 
and 
( - inside 
and - oitside). Proove that line 
passes through the fix point.
Consider an 
-point set 
in the plane. An eligible set is a non-empty set of the form 
where is a closed disk in the plane. In terms of 
determine the smallest possible number of eligible subsets may contain.
and 
are coprime, there are positive integers 
and 
such that 
By choosing 
and 
for 
the equation becomes 
and 
for 
so we found an infinite family of solutions:
and 
for 
for which ensures that the numbers 
with 
are pairwise distinct.
then clearly 
for 
then
for
such that :
positive integers and 
,
,
with 
w
,
for 
