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.
satisfying: 
+1 w
, 
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 
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.
be a graph with 
that does not contain 
as a subgraph. Prove that 
is 
-colorable.
is blue when people 
and 
know each other, and red otherwise. The problem is equivalent to showing that the edge-induced subgraph of blue edges in 
is -colorable. We start with a lemma.
are colored red or blue in such a way that there is no blue and if 
is the edge-induced subgraph of blue edges, then 
. Then all the red edges of form a cycle of length 
.
. It is well known that 
(see here for a proof), so there is a monochromatic . Label the vertices of 
. Suppose, for contradiction, that there is a red , which, WLOG, we suppose is 
. If some edge , where 
, is red, then the 
-partition 
, where 
, shows that 
, which is forbidden. Hence 
is a blue . 
.
, otherwise we would have a blue . Suppose, WLOG, that 
are red. Then 
is a red , so that 
is a blue from . If now 
is blue, then 
is a blue . Otherwise, is red, and we have the -partition 
, so that , and we have our contradiction., and our claim is true. form a cycle of length . WLOG, let 
be red. If all edges , where 
are red, then we have a -partition 
, which is forbidden. Hence, WLOG, let 
be blue. One of the edges in the vertex-induced subgraph 
must be red: WLOG, let this be 
. Then 
is blue (otherwise 
is a red ) and 
is blue (otherwise we have a -partition 
).
is red, then 
must be blue and 
is red. Then 
are all blue, so our red edges are 
, which form a cycle of length . is blue, then similar arguments produce a contradiction. Thus our proof of the lemma is complete. 
, with vertices 
. It is well known that 
(see here for a proof), and since there is no blue , there must be a red . WLOG, we let 
be the red . If the edge-induced subgraph of blue edges in the vertex-induced subgraph 
is -colorable, then we are done.
are red. For 
, we have four cases depending on how many of the 
are blue. are blue. Then we have the -partition 
. is blue (WLOG 
). Then at least three of the edges 
, for ![$i \in [1,5]$](http://latex.artofproblemsolving.com/f/d/a/fdabbebe31f9e9134b40e7c34cda90f84914c4e6.png)
are red, otherwise we have a blue . WLOG, suppose is red when 
, so that we have a -partition 
. are blue (WLOG 
). Then as before, for at least three of the and three of the 
are red. We can easily find the desired -partition. are blue. By similar arguments to before, we can find a -partition. is -colorable, as required.
such that :
