[/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 
,
,
with cardinality 
,
with 
,
is not a perfect square.
Y by Adventure10, Mango247
The 
triangle is given, point 
is the center of an exscribed circle touching the side 
, the point 
is the midpoint of the side , the point 
is the intersection point of the bisector of the angle 
of the triangle with the circumscribed circle around him. Prove that the area of the triangle 
is calculated by the formula 
, where 
is the semiperimeter of the triangle .
(Mykola Moroz)
triangle is given, point 
is the center of an exscribed circle touching the side 
, the point 
is the midpoint of the side , the point 
is the intersection point of the bisector of the angle 
of the triangle with the circumscribed circle around him. Prove that the area of the triangle 
is calculated by the formula 
, where 
is the semiperimeter of the triangle .(Mykola Moroz)
This post has been edited 2 times. Last edited by parmenides51, Nov 7, 2022, 9:16 PM
![$I_a=\left[-\frac{a}{b+c-a},\frac{b}{b+c-a};\frac{c}{b+c-a}\right]$](http://latex.artofproblemsolving.com/5/4/a/54afe28ce87c4ede04f047008eaa49fe09cfc577.png)
we get that 
iff 
![$M=\left[0,\frac{1}{2},\frac{1}{2}\right]$](http://latex.artofproblemsolving.com/0/7/3/073b6c4357f188b49a3b470b3961789f366ab153.png)
and ![$W=\left[-\frac{a^2}{(b+c)^2-a^2}, \frac{b(b+c)}{(b+c)^2-a^2}, \frac{c(b+c)}{(b+c)^2-a^2}\right]$](http://latex.artofproblemsolving.com/7/8/d/78d069d2e9d08288fdf5e4ebaf0e2f5659bf7b83.png)
![$$x=-\frac{a^2}{(b+c)^2-a^2}, y=\frac{a^2+b^2-c^2}{2[(b+c)^2-a^2]}, z=\frac{a^2+c^2-b^2}{2[(b+c)^2-a^2]}$$](http://latex.artofproblemsolving.com/0/0/c/00c7f4f1bba197905e7bc2bfc0b284b87cd374ae.png)
![$$MW^2=-a^2yz-b^2xz-c^2xy=\frac{a^2}{4}\left(\frac{-a^4-b^4-c^4+2a^2b^2+2a^2c^2+2b^2c^2}{[(b+c)^2-a^2]^2}\right)=\frac{a^2}{4}\cdot \frac{a^2-(b-c)^2}{(b+c)^2-a^2}$$](http://latex.artofproblemsolving.com/5/9/1/59139417fa048fa5e35e4666f66eae2c434903a8.png)
and 
be the the foot of perpendicular from 
and 
.
QED.
