[/quote]
,
of parallelogram 
is chosen. The perpendicular from 
to 
and the perpendicular from 
to 
intersect at 
.
is the midpoint of 
.
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the 
cars from the 
be a convex hexagon, such that the triangles 
and 
are equilateral and the diagonals 
and 
are concurrent. Prove that 
or 
for which 
is prime.
is almost squarefree if there exists a prime number 
such that 
and 
is squarefree. Prove that for any almost squarefree positive integer 
is an integer.
be a family of at most 
with the following property: we can consider 
and draw segments connecting these points such that points 
and 
are connected if and only if 
belongs to 
Determine the maximal value that the sum of the cardinalities of the subsets in 
can take.
?
such that![\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]](http://latex.artofproblemsolving.com/d/b/1/db14b64aa71cb3c8dbf53707f719f5d9a43bd7e4.png)
for any positive real numbers 
.
and for any non-empty subset 
she computes the sum ![\[s_A=\sum_{k
\in A}a_k.\]](http://latex.artofproblemsolving.com/7/8/a/78afe6d1036a006b2252023d8f9619285fe0c6d5.png)
She orders these sums 
Prove that there exists a subset 
with 
elements such that, regardless of the integers 
with 
and the circle 
of radius 
centered at 
Let 
The line 
intersects 
Let 
Let 
be the intersection of 
and 
,
,
be non-negative real numbers, no two of which are zero. Prove that :
with 
.
points on sides 
,
are angle bisectors on 
is right-angled isosceles.
Y by Davi-8191, EulersTurban, adityaguharoy, buratinogigle, jasperE3, Adventure10, Mango247, and 1 other user
A real number 
is assigned to each point 
in the space.
It is known that for any tetrahedron
with 
the center of the inscribed sphere, we have :
![\[ f(O)=f(A)f(B)f(C)f(D). \]](//latex.artofproblemsolving.com/2/e/b/2ebe41051888d6238a03f3a8db495c075e56c488.png)
Prove that
for all points .
Proposed by Aleksandar Ivanov
is assigned to each point 
in the space.It is known that for any tetrahedron
with 
the center of the inscribed sphere, we have :![\[ f(O)=f(A)f(B)f(C)f(D). \]](http://latex.artofproblemsolving.com/2/e/b/2ebe41051888d6238a03f3a8db495c075e56c488.png)
Prove that
for all points .Proposed by Aleksandar Ivanov
and choose 
inside 
such that 
.
such that 
is a regular tetrahedron with incentre 
and 
is a regular tetrahedron with centre 
.
and 
from these 2 tetrahedrons.Deviding 2 equation we get 
.
inside 
and 
such that 
.
.
.
and from the 1st equation we have 
.
,
be the incenter of 
be the incenter of 
,
,
,
defined similarly;
be the incenter of 
,
,
,
defined similarly; and
be the incenter of 
.
,
,
,
.
.
collinear points 
in that order with 
,
.
on the plane through 
is the center of the insphere of the tetrahedron 
,
.
and 
,
be two arbitrary points, consider points 
on segment 
such that 
,
and 
,
,
is constant so if 
,
,
for all points 
