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 
?
be a monic polynomial with complex coefficients such that there exist a polynomial 
with complex coefficients for which ![\[P(x^2-2)=P(x)Q(x).\]](http://latex.artofproblemsolving.com/a/6/2/a62b06890b6bbcc5c9634f6de8c7ae5fe9f37828.png)
Determine all complex numbers that could be the root of .
such that for all 
,![\[xf(x+f(y))=(y-x)f(f(x)).\]](http://latex.artofproblemsolving.com/e/4/a/e4a3bbb8b91d2aa62d699c24df342fa59be71915.png)
. 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)
of integers that has the following property:
(possibly 
) then 
for every integer 
.
of nonzero integers such that the only admissible set containing both 
and 
is the set of all integers.
be the circumcenter and orthocenter of triangle 
, respectively. Let 
and 
intersect at point 
. If 
is an arbitrary point on, and 
and 
be the intersection of 
with 
and 
, respectively.
are concyclic.
with incenter 
. Line 
and 
intersects 
and 
at 
and 
, respectively. Let 
and 
be the midpoints of and . Line 
meets the external bisector of angle 
at 
and line 
meets the external bisector of angle 
at 
. Prove that 
are concylic.
be the nonnegative integers in 
where 
. Define 
. Prove that 
.
such that 
where 
, and the statement is still true when 
? such that 
where 
, and the statement is still true when ?.
is tangent to sides and 
at 
and respectively, and 
is the circumcenter of 
. Prove that 
.
of the circumcircle of with point "? ?, bisects the length of the lines 
and " ?
, but because there are two points defined that way, we consider the one closer to .
.
is the projection of on .
be on the circumcircle of such that 
is the diameter of that circle and let be the projection of on line .
so and radius 
intersect once again at 
it is is to prove that 
(using that is the center of arc on the circumcircle of without 
) so 
but 
is isosceles so 
now 
is cyclic with as it's diameter and the projections of and on diagonal are and .
which is a known problem(or at least for me). is cyclic with diameter it's easy to prove 
so 
so 
and from the last 2 we have 
and done
and 
respectively on and (perpendicular meets it at its extension). 
are by Simson's theorem collinear.
. By the problem condition 
and by some simple angle chase we get 
. So 
which gives 
, or equivalently, 
. Done 
and 
Prove that
Prove that
satisfying
Prove
of the circumcircle of 
.
.
through 
.
be the second point of intersection of 
it is enough to prove that 
.
thus 
and 
.
.
,
,
![\[\cos \alpha=\frac{\sin{\beta}+\sin{(2\alpha+\beta)}}{\sin{(\alpha+\beta)}},\]](http://latex.artofproblemsolving.com/0/d/5/0d564a76fa956eab38aed49b3589a200fd5819f2.png)
which is true because 
by Product-to-Sum.