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.
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
+7 w
, define![\[
E(x,y) = x^{-\!y} + y.
\]](http://latex.artofproblemsolving.com/6/1/2/6124224ada5e77a6c03e4cfe24e69435dc996ec4.png)
, prove that![\[
E(x,y) + E(y,x) \geq 4.
\]](http://latex.artofproblemsolving.com/0/6/5/065d58fba8e8799a307655dafdcfdadb2e351eea.png)
, prove that![\[
\frac{2}{1+xy} + x + y < E(x,y) + E(y,x) < 2 + \frac{x}{y} + \frac{y}{x}.
\]](http://latex.artofproblemsolving.com/8/c/9/8c912cad7eeabd17e145a08fc87ecb6b5c5b2abc.png)
+1 w
so that:
with all real numbers 
satisfies 
is called 
if there exists a permutation 
of the numbers 
, such that:
and 
have different parities for every 
;
the sum 
is a quadratic residue modulo 
for every 
.
+2 w
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
+1 w
for some positive integer .
. Find all functions 
such that for all 
,![\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]](http://latex.artofproblemsolving.com/f/9/3/f93b7526ff4338eea89a1c59bc009aa001ffc15d.png)
be a positive integer. Do there exist distinct -digit positive integers 
and 
, with each of their digits being 
or 
, such that 
divides 
?
, 
be positive integers. Julia and Florian play a game on a 
board. Julia cells. such that Florian has a strategy to always deduce the entire tiling.
be a positive integer. A sequence of integers 
is called -pop if the following holds: for every 
, 
is equal to the number of distinct elements in the set 
. Determine, as a function of , how many -pop sequences there are.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
and 
on side 
of a convex quadrilateral 
are given such that 
The circumcircles of triangles 
and 
meet again at 
and those of 
and 
meet again at 
. Show that the points 
lie on a circle.
and on side of a convex quadrilateral are given such that The circumcircles of triangles and meet again at and those of and meet again at . Show that the points lie on a circle.
be the point of the intersection of line 
with line . Because lies on a radical axis of 
and 
it's easy to conclude from the given relation that is a midpoint of (since 
. Analogically, lies on a radical axis of 
and 
, and it's obvious that it lies also on the radical axis of and and on the radical axis of and . So it's clear that is a radical center of these circles and in particular we have:
is a cyclic quadrilateral (from the converse of the Power of the Point theorem).
isn't necessary.
the circumcircles of the triangles 
respectively.
and let 
be, the intersection points of 
In my drawing the point 
is between 
and 
between 
and side line of 
be, the intersection point of 
from the segment line 
as the radical axis of 
we conclude that the point 
lies on the segment line 
as the radical axis of the circles 
be, the intersection point of from the segment line 
as the radical axis of 
we conclude that the point 
lies also, on the segment line 
and that the segment lines are concurrent at one point, here the point lies on the side line 
we conclude that the points 
are concyclic and the proof is completed.
isn't necessary.
the radius isn't necessary for this problem hence the problem becomes like this Image not found and we know that 
is concyclic because 
becomes a circle by inversion. Then we will show that 
we know that 
because 
are concyclic and 
has equal power to circles and 
is the radical axis of the circles so line hence 
is collinear so 
are concyclic. is , is , is 
, 
is 
and 
. Observe that and are radical axises of 
and 
, respectively. It directly implies that 
and 
, we get the 
coincide (why?). Then by PoP 
which means 
as desired.
be non-negative number. Prove that 
Let 
.
and 
Prove that
