[/quote]
,
grid, taking turns. Petya goes first. On his turn, a player writes an uppercase Russian letter in an empty cell (each cell can contain only one letter). When all cells are filled, Petya is declared the winner if there are four consecutive cells horizontally spelling the word ``ПЕТЯ'' (PETYA) from left to right, or four consecutive cells vertically spelling ``ПЕТЯ'' from top to bottom. Can Petya guarantee a win regardless of Vasya's moves?
,
is marked. Point 
is on segment 
,
is on segment 
such that the circumcircles of triangles 
and 
are tangent to line 
.
and 
passing through the center of the circumcircle of triangle 
,
.
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 
.
marked points, he calculated the sum of the numbers on all 
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 
,
,
,
and 
with integer coefficients is called 
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 
![\[
\frac{(x+y+z)^5 - x^5 - y^5 - z^5}{(x+y)(y+z)(z+x)} = 0
\]](http://latex.artofproblemsolving.com/7/8/f/78f8e6c2570684631dc288144e13568d71e68ec8.png)
be reals such that 
. Prove that
Let 
. Prove that
Y by Adventure10
In a regular 
pentagon, we draw the diagonals 
and 
, which intersect at point 
We cut the triangle 
and get the hexagon 
. We have an infinite collection of pieces equal to the hexagon in size and shape. It shows that the plane can be tiled with these pieces.
NOTE: Tiling the plane means covering it with pieces without overlapping or leaving gaps.
pentagon, we draw the diagonals 
and 
, which intersect at point 
We cut the triangle 
and get the hexagon 
. We have an infinite collection of pieces equal to the hexagon in size and shape. It shows that the plane can be tiled with these pieces.NOTE: Tiling the plane means covering it with pieces without overlapping or leaving gaps.
This post has been edited 1 time. Last edited by parmenides51, Dec 18, 2022, 6:48 AM
Reason: latex
Reason: latex
