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.
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 
, and there are both positive and negative such sums. What is the maximum possible value of ?
pairwise distinct complex numbers are written on a board. It turns out that there are exactly 760 ways to choose two numbers and 
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 
, 
, 
, and 
are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle , lie on the same sphere.
and 
with integer coefficients is called \emph{important} if from the divisibility of both differences 
and 
by 
, it follows that both 
and 
are divisible by 100. Does there exist such an important pair of polynomials 
, 
, such that the pair 
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 
be positive real numbers with 
. Prove that
and 
, such that for any positive integer 
, there is a subset 
of 
with cardinality 
, and for any 
with 
, the difference 
is not a perfect square.
in 
. Then let ![\[ u = |WA| = |AU|, \qquad v = |UB| = |BV|, \qquad w = |VC| = |CW|. \]](http://latex.artofproblemsolving.com/d/5/d/d5d5674f274fd362b304a3972aac9e0754d7e05a.png)
To prove 
, proceed as Tintarn did. For the rest we'll just resort to some length bashing. Let 
be the intersection of the altitude onto 
with .
.
in two ways. (Heron and 1/2 times base times altitude) 
as the midpoint of . By Pythagoras and 
it is easy to compute 
respectively. We can conclude 
. So there is a homothety with center that takes 
to 
. That implies 
, so we're done. (Watch out for different cases while length bashing.) 
and 
giving us 
. Compute 
by Pythagoras. Now we can use the law of cosines to prove 
, so ![\[ x_S = \frac{x_A+x_B+x_C}{3} = \frac{0+(u+v)+(2u-v)}{3}=u \]](http://latex.artofproblemsolving.com/7/b/3/7b36bf81fad098dbe6b7eb709f19f4c6a16deb2d.png)
with which the result follows. 
are positive real numbers, 
then
and 
Prove that
and 
and centroid 
.
,
and line 
is perpendicular to one of the sides of the triangle.
is equilateral. But clearly, this is impossible.
be the point where the incircle touches 
.
.
and 
and 
and 
and we are done.