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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 
.
checkerboard satisfy the following conditions:
checkers have been placed on the board.
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 
.
+1 w
, 
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.
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.
of the convex quadrilateral be such that 
Let 
be the center of the circumcircle of the triangle 
. Prove that 
.
, we obtain 
. Additionally, since PB=PC and PD=PA, triangles PBD and PCA are congruent, leading to 
.
.So from 
. And since 
, we find 
.
be an acute triangle and and are the points such that 
, 
and 
, 
. Let be the circumcenter of 
. Then prove that 
. and be the second intersection points of 
, 
with . Since 
we have will have that 
. Also from 
we get that 
which implies that 
.
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)
