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.
of parallelogram 
a point 
is chosen. The perpendicular from 
to 
and the perpendicular from 
to 
intersect at 
. Point 
is the midpoint of 
. to passes through the center of parallelogram .
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 grid.
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid. that guarantees that Kritesh's job is possible?
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 the ratio 
is an integer.
be a positive integer and 
be a family of at most distinct subsets of the set 
with the following property: we can consider distinct points in the plane, labelled 
and draw segments connecting these points such that points 
and 
are connected if and only if 
belongs to subsets in for any 
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 
.
be a positiv integer. Ana chooses the positive integers 
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 chosen by Ana, these can be determined by only knowing the sums 
with 
with 
and the circle 
of radius 
centered at 
Let be the midpoint of 
The line 
intersects a second time at 
Let be a point on such that 
Let 
be the intersection of 
and Prove that 
, 
, 
be non-negative real numbers, no two of which are zero. Prove that :
with 
. Let 
points on sides 
, respectively, such that 
are angle bisectors on . is isosceles if and only if 
is right-angled isosceles.
with 
, points and are taken on the rays 
and 
respectively such that 
. The circumcircles of triangles 
and 
intersect at 
and 
. Prove that 
be the circumcenters of the triangles 
and 
a circle with diameter centered at the midpoint 
of this diagonal of the quadrilateral . Let the lines 
intersect the circle at points 
different from the point . The angles 
, 
, 
, 
are all right, because they span the diameter of the circle . The angles 
are all equal to 
:
are all isosceles and consequently, the perpendicular bisectors of the segments 
pass through the points . Let the perpendicular bisectors of the segments meet the circle at points 
. The angles 
are equal, hence, they span the same arc 
of the circle and the points 
are identical. Similarly, the angles 
are equal, hence, they span the same arc 
of the circle and the points 
are identical. Since these angles are all equal to , the points 
are diametrally opposite points of the circle , the points 
are collinear and 
. The perpendicular bisectors 
of the segments 
are parallel, being both perpendicular to the same line 
and they pass through the circumcenters , respectively. Similarly, the perpendicular bisectors 
of the segments 
are parallel, being both perpendicular to the same line 
and they also pass through the circumcenters , respectively. The quadrilateral 
is a parallelogram and the center of the circle is the intersection of its diagonals, being the midpoint of the diagonal 
. Consequently, the centers of the circles 
are collinear. Since these 3 circles intersect at the point , they belong to the same elliptic pencil and they all intersect at one other point, the intersection of the circles 
. Since the angles 
spans the diameter of the circle , it is right and 
.
?
be the midpoints of 
,
we get 
,
concyclic and we are done.