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 the set of all nonnegative integers. Find all the functions 
satisfying the relation![\[ f(f(f(n))) = f(n+1 ) +1 \]](http://latex.artofproblemsolving.com/8/b/9/8b953428105df2c2344963cbaf3eb81d8201e0df.png)
.
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
.
be an acute-angled triangle with 
. Let 
be the circumcircle of . Let 
be the midpoint of the arc 
of containing 
. The perpendicular from to 
meets 
at 
and meets again at 
. The line through parallel to meets line 
at 
. Denote the circumcircle of triangle 
by 
. Let meet again at 
. Prove that the line tangent to at 
meets line on the internal angle bisector of 
.
, such that for any 
, 
very convex if![\[ \frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|
\]](http://latex.artofproblemsolving.com/3/b/a/3ba7733caf51cc70a090920ca21f8b47aad9b1ec.png)
and 
. Prove that no very convex function exists.
satisfying
for all integers 
.
, and for any positive integer 
, 
means applied times to , and 
means 
applied times to .
satisfying 
.Prove that:
such that 
is interesting if 2018 divides 
(the number of positive divisors of ). Determine all positive integers such that there exists an infinite arithmetic progression with common difference whose terms are all interesting.
be integers and 
be a prime for which![\[ x^2-3xy+p^2y^2=12p \]](http://latex.artofproblemsolving.com/2/4/c/24cb51eb55144a9a1e0e489d115993496a052e79.png)
.
satisfying 
for all positive integers .
be an isosceles trapezoid with 
. Let 
be the midpoint of 
. Denote by and the circumcircles of the triangles 
and 
, respectively. Let be the crossing point of the tangent to at with the tangent to at . Prove that 
is tangent to .
the orthocenter is 
and the foot of the altitude from is . Point satisfies 
, and the line 
is tangent to 
. Line 
intersects lines 
at points 
respectively.
or 
.
and let 
be the mid-point of 
. We have 
which means 
since 
.
Hence, we get 
as desired.
over as , and the intersection of 
is 
.
is the reflection of over perpendicular bisector of and 
meets 
at 
.
, so is on 
.
is perpendicular to 
, so is the orthocenter of 
.
so lies on 
.
lie on a circle which center is .
is the reflection of over 
.
, so 
is the miquel point of 
, which kills the problem.
lie on the rays 
.
wrt. be 
and define 
similiarly. Delete points in the problem and define 
, 
., the other feets of the altitudes. Notice that
Similiarly,
Multiplying these two toghether and using Menelaus finishes the proof.
passes trough , let 
and 
, this is equivalent to
And since 
, we need to show that is tangent to 
.
at and at being parallel, which is
be the reflections of 
over , redefine as 
and 
,
are collinear.
such that![\[
2^a \cdot 3^b + 1 = 5^c.
\]](http://latex.artofproblemsolving.com/0/a/c/0ac1f22162eacf698847bceb33f9a337bd53c44b.png)
and 
.
.
is the circumcenter. 
is cyclic 
is the midpoint of 
.
yields that 
are cyclic. From radical axis theorem on 
;
are concurrent. Note that 
and by symmetry 
.
.
.
over 
to meet 
then 
,
and then project 
onto 
.
is the homothety which takes 
and 
,
and 
.
lay on both the line 
and perpendicular bisector of 
are cyclic.
.
as the center of 
.
is on dual of 
which is the line 
is the intersection of 
.
and similarly, 
.
inversion 
inversion 
reflection over angle bisector of 
.
lies on 
.
are concurrent, then 
and 
are concyclic. Let 
.
are collinear. Pascal at 
gives 
are collinear. Let 
.
,
and 
are homothetic so 
are collinear. This yields the concurrency of 
.