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 reals such that 
. Prove that
and 
satisfies the following recursion formulas: 
, 
and 
. Find 
and 
, an arbitrary point 
is chosen on the side 
. In triangles 
and 
, the angle bisectors 
and 
are drawn, respectively. The point 
is the circumcenter of 
. Prove that the second intersection point of the circumcircles of triangles 
and 
lies on the line 
.
+1 w
such that ![\[f(f(x)f(y))=f(xf(y))+f(y)\]](http://latex.artofproblemsolving.com/4/f/4/4f40c133524db64428861c756773ee25659d048a.png)
for all integers 
and 
.
satisfying 
Prove that 
When does equality hold?
be a prime number and 
an integer polynomial of degree 
such that 
and 
is congruent to 
or 
modulo for every integer 
. Prove that 
.
such that 
is an integer for all 
, where 
are all the positive divisors of 
.
be three positive reals such that 
. Prove that![\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]](http://latex.artofproblemsolving.com/a/1/4/a14adf0f1e35df57c734c7df701d5a67da22dc3c.png)
real numbers 
, we can select a real number 
and transform the sequence into 
. This transformation can be performed multiple times, with each chosen real number potentially being different
where 
are lengths of parallel sides and 
are other two sides.
be a cyclic quadrilateral. Let 
, 
, 
be the feet of the perpendiculars from 
to the lines 
, 
, 
, respectively. Show that 
if and only if the bisectors of 
and 
are concurrent with 
.
?
,
is the midpoint of 
and 
is the point of trisection of 
nearer to 
.
find 
.