[/quote]
,
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
points are marked, no three of which are collinear. All possible segments between them are drawn. Grisha assigned to each drawn segment a real number with absolute value no greater than 1. For every group of 6 marked points, he calculated the sum of the numbers on all 15 connecting segments. It turned out that the absolute value of each such sum is at least 
,
pairwise distinct complex numbers are written on a board. It turns out that there are exactly 760 ways to choose two numbers 
from the board such that:![\[
a^2 + b^2 + 1 = 2ab
\]](http://latex.artofproblemsolving.com/8/f/c/8fc5df8dbc8e1647edfaa96dcfa8a0dfdac0c961.png)
Ways that differ by the order of selection are considered the same. Prove that there exist two numbers 
and 
from the board such that:![\[
c^2 + d^2 + 2025 = 2cd
\]](http://latex.artofproblemsolving.com/2/7/0/2706d72797c5e392b3f068a8174bd257e4d93fa1.png)
is given. It is known that triangles 
,
,
,
are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle 
and 
with integer coefficients is called \emph{important} if from the divisibility of both differences 
and 
by 
,
and 
are divisible by 100. Does there exist such an important pair of polynomials 
,
,
and 
is also important?
is given. A cube with side length 
is made up of 
unit cubes, each of which is either black or white. It turns out that among any 
cubes that share a common vertex and form a 
cube, there are at most 
black cubes. What is the maximum number of black cubes that could have been used?
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
are positive real numbers, 
then
be positive real numbers with 
.
and 
Prove that
and 
,
,
of 
with cardinality 
,
with 
,
is not a perfect square.
Y by ItsBesi
Let 
be an isosceles triangle with base 
. Points 
and 
are chosen on the sides and 
, respectively, such that 
. Let 
and 
be the midpoints of 
and 
, respectively. The circumcircle of triangle 
intersects 
at point 
, whereas the circumcircle of triangle 
intersects 
at 
. The line through parallel to intersects 
at 
. Let 
{
}. Prove that 
and are collinear points.
be an isosceles triangle with base 
. Points 
and 
are chosen on the sides and 
, respectively, such that 
. Let 
and 
be the midpoints of 
and 
, respectively. The circumcircle of triangle 
intersects 
at point 
, whereas the circumcircle of triangle 
intersects 
at 
. The line through parallel to intersects 
at 
. Let 
{
}. Prove that 
and are collinear points.
is cyclic. Keeping this and mind (as well as the cyclicity of 
and the fact that 
is a midline in triangle 
)
,
(midlines), and similarly 
,
is a parallelogram. Similarly we obtain that 
is a parallelogram whose diagonals intersect at 
,
,
be 
respectively. Then 
are concurrent.
.
is a parallelogram, we have 
.
is a parallelogram implying that 
pass through the midpoint of 
be the midpoint of 
.
and 
,
are collinear. Because 
,
is cyclic.
and 
be the midpoints of 
is cyclic implying 
.
is cyclic implying 
.
,
pass through a common point which is 
as desired. 
is the midpoint of the 
=
is the midpoint of the 
=
=
=
is the midsegment of the 
with 
we get that 
IJKH is a parallelogram 
and points 
are collinear 
