[/quote]
,
is a positive integer. Call all positive divisors of 
beautiful divisors and difference of any 
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
for every pair 
and 
,
and 
.
externally tangent to 
and internally tangent to 
.
and 
intersect at a point lying on 
be inscribed in the circle 
.
intersects 
and 
at points 
and 
,
be the reflection of 
be the reflection of 
of triangle 
lies on the circle 
and 
coincide.
on the board. For each procces, the stutent deletes the number x and replaces it with either 
or 
or 
.
on the board?
satisfying
Prove
:![\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]](http://latex.artofproblemsolving.com/3/1/a/31a0eac3c7ca298c8fa221083dc94e127e705784.png)
and 
.
?
lies inside triangle 
and 
.
.
.
.
the number of positive integer 
satisfying 
is less than or equal to 
where 
is a constant independent of 
be different real numbers. Prove that![\[\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}=\left\{\begin{array}{ll}
0, & \text { if } n \text { is even; } \\
1, & \text { if } n \text { is odd. }
\end{array}\right.\]](http://latex.artofproblemsolving.com/6/5/2/6525fc6b451ff5a3c83ff2a0541b063a7cf065c3.png)
satisfying
for all real numbers 
and 
.
Y by mathtastic, Adventure10, Mango247
Sebastian has a certain number of rectangles with areas that sum up to 3 and with side lengths all less than or equal to 
. Demonstrate that with each of these rectangles it is possible to cover a square with side in such a way that the sides of the rectangles are parallel to the sides of the square.
Note: The rectangles can overlap and they can protrude over the sides of the square.
. Demonstrate that with each of these rectangles it is possible to cover a square with side in such a way that the sides of the rectangles are parallel to the sides of the square.Note: The rectangles can overlap and they can protrude over the sides of the square.
