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 real numbers. Find all functions 
where there exist a real constant 
such that 
holds for all reals 
, 
, 
that satisfy 
.
, denote the sum of digits of in its decimal representation by 
. Find all polynomials 
with integer coefficients such that for any positive integer 
, the integer 
is positive and 
be a triangle with incentre 
and circumcircle 
. Let 
be the midpoints of the arcs 
of not containing 
respectively. Let 
be the point of diametrically opposite to 
. Show that 
and the midpoint 
of 
lie on a line.
such that, for every positive integer 
the integer
is a multiple of 
(Note that 
denotes the greatest integer less than or equal to 
For example, 
and 
)
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.
is orthodiagonal iff 
".
be two fixed points. Then the geometrical locus of the point 
for which 
(a real constant) is a line 
, where 
.
be the intersection point of 
and 
. It's in the interior of since is convex. Thus: 
.
be the intersection point of 
and 
. Again, it is in 
since is convex.
the angle . Then, if we use the cosine rule:
.
. But, since distances are positive, if 
, this means a negative number equals a positive number, contradiction! Thus 
and hence 
. Q.E.D.
and 
Prove that
.
and 
Prove that
Prove that
.
.
,
,
and 
.
.
is in interior of it.
then 
we bring the altitude 
.
we bring the altitude 
.
then 
and 
are on the ray 
of course.
,
.
If 
and 
we get 
If 
