[/quote]
,
positive integers not exceeding 
,
?
table, several cells are marked. At each move, Cyril can get to know the number of marked cells in any checkered square inside the initial table, with side less than 
.
be triangle with circumcenter 
and orthocenter 
intersect 
respectively. Lines through 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
.
be an odd integer. Prove that there exists a prime number 
such that![\[
p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n.
\]](http://latex.artofproblemsolving.com/3/5/1/3514b64ecd5f7b2586d5bfeae59b7c9504d5b455.png)
and 
with equal radii intersect at P and Q. Points B and C are located on the circle
The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
and 
,
be the midpoint of 
.
on the extension of 
past 
and point 
on segment 
such that 
.
be on the opposite side of 
from 
such that 
and 
.
intersect the circumcircle of 
again at 
,
intersect the circumcircle of 
again at 
.
,
are collinear.
which satisfy ![\[f(x)\cdot f(y)\mid (1+2x)\cdot f(y)+(1+2y)\cdot f(x)\]](http://latex.artofproblemsolving.com/6/4/f/64f2720634f57b4d88db823b9931379386ac1c27.png)
for all non-negative integers 
and 
.
be natural numbers such that: 
and 
.
are taken on the legs 
,
and 
.
and 
.
be the real polonomial such that all roots are real and distinct. Prove that there is a rational number 
that all roots of 
are real numbers
from the reals to the reals such that ![\[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz) \]](http://latex.artofproblemsolving.com/6/1/0/610e13a6f27341b5eed393923bb98e182537e254.png)
for all real 
, such that 
Prove that :
Y by Adventure10, Mango247
Suppose 
is a square piece of cardboard with side length 
. On a plane are two parallel lines 
and 
, which are also units apart. The square is placed on the plane so that sides 
and 
intersect at 
and 
respectively. Also, sides 
and 
intersect at 
and 
respectively. Let the perimeters of 
and 
be 
and 
respectively.
Prove that no matter how the square was placed,
remains constant.
is a square piece of cardboard with side length 
. On a plane are two parallel lines 
and 
, which are also units apart. The square is placed on the plane so that sides 
and 
intersect at 
and 
respectively. Also, sides 
and 
intersect at 
and 
respectively. Let the perimeters of 
and 
be 
and 
respectively.Prove that no matter how the square was placed,
remains constant.
intersecting these 2 parallel lines at 
,
.
centered at an arbitrary point O on the segment 
.
inscribed in this circle intersecting both the lines 
in the way described in the problem. Let the angle 
be fixed and let the square circumcenter O move on the perpendicular p. The triangles 
remain similar for all positions of the circumcenter O and the sum of their A- and C-altitudes is constant, 
,
of their perimeters (proportional to their altitudes) is then also constant. Consequently, we only have to consider the square circumcenter O being the midpoint of the segment 
are congruent, i.e., their perimeters 
are equal, and we have to show that the perimeter 
be the foot of the bisector AO of the right angle 
.
,
.
and
is constant, iff 
or using 
.
Then, suppose that 
and 
(sorry for using the variable 
by e.g. dropping a perpendicular from 
Thus, intersecting this with line 
,
This means that 
Furthermore, note that 
due to the parallel lines, so the perimeter of 
Since the perimeter of 
is 
,
