[/quote]
,
and 
be real numbers such that 
.
and 
.
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
be an acute-angled, non-isosceles triangle with circumcenter 
and incenter 
,![\[
\prod_{\text{cyc}} \left( \frac{1}{a+b-c} + \frac{1}{a+c-b} - \frac{2}{b+c-a} \right) \neq 0,
\]](http://latex.artofproblemsolving.com/7/8/1/78125759bf5c73149e4a56538e9b0c44309704a7.png)
,
,
.
,
,
be the excenters opposite to vertices 
,
,
,
,
,
be the Lemoine points of triangles 
,
,
,
,
,
all pass through a common point 
.
.
intersect at point 
.
and 
with their common external tangents lie on a circle 
.
and 
with their common external tangents lie on a circle 
.
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 on segment 
,
is on segment 
such that the circumcircles of triangles 
and 
are tangent to line 
.
,
.
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?
Y by Adventure10, Mango247
P1: Consider a convex quadrilateral ABCD. Let O be the intersection of AC and BD;M,N be the centroid of 
and 
and P,Q be orthocenter of 
and 
,respectively. Prove that 
P2: Consider a triangle ABC. For every point M
BC,we define N CA and P AB such that APMN is a parallelogram. Let O be the intersection of BN and CP. Find M BC such that 
and 
and P,Q be orthocenter of 
and 
,respectively. Prove that 
P2: Consider a triangle ABC. For every point M
BC,we define N CA and P AB such that APMN is a parallelogram. Let O be the intersection of BN and CP. Find M BC such that 
be 
and the midpoint of side 
be 
.
.Hence the problem is equivalent for proving that 
.
on 
be 
on 
be 
be 
and 
on 
be 
.
and 
respectively.Since 
lies on the radical axis of circles 
lies on the radical axis of the same two circles 
as desired
