[/quote]
,
board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
,
to be the largest value in row 
,
.
to be the largest value in column 
.
?
and 
.
,
lie on a line in this order and satisfy 
.
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
be a positive integer and let 
be real numbers.![\[
\left(\sum_{i,j=1}^{n}|x_i-x_j|\right)^2\le\frac{2(n^2-1)}{3}\sum_{i,j=1}^{n}(x_i-x_j)^2.
\]](http://latex.artofproblemsolving.com/8/1/e/81ef4493de1ab4a1ac30124fe291256917760ca1.png)
is an arithmetic sequence.
and 
.
and 
intersect the circumcircle of 
and 
,
and 
such that 
and 
are parallelograms (with 
,
)
and 
.
,
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
.
be a point on side 
such that 
.
,
satisfying 
.
that satisfy the following condition: for all 
and 
relatively prime to 
we have ![\[a \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.\]](http://latex.artofproblemsolving.com/e/8/5/e85c5b249ac8e75b45a525b8786414798a4921fe.png)
be distinct prime numbers. Let ![$\alpha_i=\sqrt{a}, \,\mathrm{A}=\mathbb{Q}[\alpha_1,\alpha_2,\hdots,\alpha_b]$](http://latex.artofproblemsolving.com/0/e/0/0e06919fe05d9f17760890bb2aebb9cfa85d54ce.png)
![$[\mathrm{A}:\mathbb{Q}]=2^b$](http://latex.artofproblemsolving.com/b/e/1/be12fe0e1becd8a9dcddd1eece210f2df8ecd026.png)
![$\mathrm{A}=\mathbb{Q}[\gamma];$](http://latex.artofproblemsolving.com/f/6/3/f639e791853133a444708bf47517457379d347e8.png)
and deduce that the minimum polynomial 
of 
over 
has degree 
![$\mathbb{Z}_a[X]$](http://latex.artofproblemsolving.com/e/9/9/e994e6e9083e52f44aef280b4346f6994a9d9f02.png)
into a product of polynomials of degree 
either of degree 
be a function from the set of integers to the set of positive integers. Suppose that, for any two integers 
and 
is divisible by 
.
,
is divisible by 
.
be a positive integer, and let 
denote the integers modulo 
satisfying 
for all 
.
such that 
and 
for all reals 
.
such that the following equation holds for every real ![\[ f(f(x) + y) = \lfloor x + f(f(y)) \rfloor. \]](http://latex.artofproblemsolving.com/3/f/b/3fbbb68e9f9c104fe0f701c8ad07e7fbbf583f1e.png)
Note: 
denotes the greatest integer not greater than 
.
such that
holds for all 
.
Y by
Let 
be an acute triangle, 
fixed, 
moves on the big arc 
of 
. Let 
be the circumcenter of 
are not collinear, 
, 
is the incircle of triangle . tangents to at 
. Let 
be the -excenter of triangle . 
cuts 
at 
. Let 
lies on such that 
.
a)
cuts 
at 
. Prove that 
.
b) Let
lies on the circle 
go through 
such that 
. 
cuts again at 
. Prove that the midpoint 
of 
lies on a fixed circle.
be an acute triangle, 
fixed, 
moves on the big arc 
of 
. Let 
be the circumcenter of 
are not collinear, 
, 
is the incircle of triangle . tangents to at 
. Let 
be the -excenter of triangle . 
cuts 
at 
. Let 
lies on such that 
.a)
cuts 
at 
. Prove that 
.b) Let
lies on the circle 
go through 
such that 
. 
cuts again at 
. Prove that the midpoint 
of 
lies on a fixed circle.This post has been edited 1 time. Last edited by 799786, Mar 5, 2022, 6:48 AM
be contact points of 
be Bevan point of 
.
,
.
be 
- excenter, 
so 
and 
.
and orthocenter 
be orthogonal projection of 
such as 
.
intersects 
again at 
be orthogonal projection of 
then it's easy to see that 
are homothetic. So 
which means 
.
we have 
is harmonic quadrilateral. Hence if we let 
with 
or 
are collinear
be midpoint of arc 
be intersection of 
.
which is fixed. We consider the following configuration
with 
are fixed, 
intersects 
intersects the line through 
tangents at 
be 
intersect 
.
then 
Then 
.
are fixed then 
are fixed, hence 
