[/quote]
,
such that
has side lengths 
,
,
.
inside 
,
,
and denote them as 
,
,
.
?
be non-negative real numbers such that 
Prove that
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of 
is interesting if 2018 divides 
(the number of positive divisors of 
such that there exists an infinite arithmetic progression with common difference 
with its center 
.
and one moving point 
on 
is 
.
is a fixed point on 
intersect
,
intersects 
at 
.
when 
such that:![\[
a^2 + b^2 \mid a^3 + b^3
\]](http://latex.artofproblemsolving.com/a/1/d/a1ddbd20c854e291da5670de449229d1664fa56c.png)
be an isosceles triangle with 
.
,
and 
,
at 
.
is tangent to 
.
be the midpoints of segments 
,
intersects 
,
.
such that for any 
:
units horizontally (left or right); or
to 
using at most 
moves.
,
are parallel and the angles 
and 
are acute. Show that it is possible to divide the triangle 
and the triangle 
into 4 disjoint triangles 
such that the triangles 
and 
are congruent for all 
.
be the set of points which lie strictly inside it. Also let 
be the set of circles in the plane. For a point 
,
be the reflection of 
respectively. Define a function 
such that 
is the circumcircle of 
.
,
Y by
In a 
board, each square is colored either white or black, with all the squares on the border of the board being black. Additionally, no 
square within the board has all four squares of the same color. Prove that the board contains a square colored like a chessboard.
board, each square is colored either white or black, with all the squares on the border of the board being black. Additionally, no 
square within the board has all four squares of the same color. Prove that the board contains a square colored like a chessboard.
grid in the interior of the board) we define a graph with the mentioned 
points as its vertices. we draw an edge between two vertices if on the board they have distance 
and the two squares of the grid with corners as the vertices don't have the same color. it is easy to see that each vertex has even degree. if a vertex has degree 
then the 
square around it have the same color which is a contradiction and if a vertex has degree 
(the all black outline means that this is correct for all of our vertices).
with nodes of given grid as vertices. Then 
is odd. Draw an edge between two nodes iff they are contained in same black square. Easy to see that every vertex has an odd degree, hence a contradiction. 
