AoPS Academy
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 
.
children sit in a circle facing inward with 
, and each child starts with an arbitrary even number of pencils. Each minute, each child simultaneously passes exactly half of all of their pencils to the child to their right. Then, all children that have an odd number of pencils receive one more pencil.
checkerboard satisfy the following conditions:
checkers have been placed on the board.
, 
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 .
+1 w
and 
of triangle intersect in 
. A perpendicular 
from to 
is drawn. Circumcircles and 
intersect at 
and .
.
intersect at . Denote 
the antipodal points of in 
and 
, respectively. From 
we draw a line parallel to side intersecting again at 
. The main claim is that 
are collinear (Which proves the problem statement). That being said, we're going to point out that 
.
, 
should be collinear. A final observation would be that as and are diametrically opposite in 
, we should have 
, implying 
. Now, we will do some angle chasing:![\[ \angle QXL = \angle ZXL = \angle ZBL = \angle ZBC - \angle LBC = ... = 90^{\circ} - \angle ABC \]](http://latex.artofproblemsolving.com/5/2/4/524c0d16d3d921309924d55fcf2716f8c87b4072.png)
![\[ \angle QXT = \angle QHT = \angle THF - \angle QHF = \angle BHD - \angle ACF = ... = 90^{\circ} - \angle ABC \]](http://latex.artofproblemsolving.com/b/9/e/b9ecbafcc9960de22572b6d83f17238af90ac937.png)
And we should be done! 
isn't something you see everyday 
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 
.
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)
