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.
, 
is the Gergon point and the incircle 
touch 
, 
, 
at 
, 
, 
respectively.
, 
, 
be 
, 
, 
respectively. in 
and then we get the circle 
in 
and then the circle 
in 
and then the circle 
, , are coaxial.
+1 w
and primes 
such that 
(there is at least 1 zero) can't be a perfect square.
such that![\[f((x\oplus f(y))+y)=(f(x)\oplus y)+y\]](http://latex.artofproblemsolving.com/d/3/5/d355f091797570bdc6c023eedb30e8ee9deac168.png)
Note: 
denotes the bitwise XOR operation. For example, 
.
is divisible by 2024.
be a scalene triangle such that 
is its incircle. is tangent to at . A point (
) is located on .
, 
, and 
be the incircles of the triangles 
, 
, and 
, respectively. and is also tangent to .
and 
such that 
is a prime number. and intersect at points 
and 
. Line 
is tangent to at 
and to at 
so that is closer to than . Let 
and 
be points on major arcs 
(on ) and 
(on ), respectively, such that 
. Extend segments 
and 
through to 
and 
, respectively, such that 
. Given that the circumcenter of triangle 
lies on line 
, prove that 
.
, define the sequence 
for 
as
Determine all values of 
such that there exists a number such that 
for infinitely many values of 
.
, 
, 
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
, the number
is an integer. Prove that there is a positive integer 
such that 
for all 
.
that satisfy the equation
has an incenter 
Suppose 
. Let be the midpoint of 
. Suppose that 
. 
meets 
again at point 
. Let points and be such that is the midpoint of 
and is the midpoint of 
. Point lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
at and . Prove that 
.
with sum 
are such that the equation 
has a positive root 
smaller than 
.
such that the inequality 
holds.
and![\[ f(x+y)+f(x^2+f(y))=f(f(x))^2+f(x)+f(y)+y,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/c/8/9/c895ae7fdf8d7f284ac9fc94cc077d6edad6cbf0.png)
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
it is true that 
has two roots 
and 
and 
.
then exists two positive roots 
it is true that 
has two roots with 
and 
so contradiction