[/quote]
,
the set of all positive integers. Find all functions 
such that for all positive integers 
and 
,
is nonzero and divides 
.
be an odd integer. Determine all functions 
from the set of integers to itself, such that for all integers 
and 
the difference 
divides 
,
on the ray 
such that 
.
on the ray 
.
.
be a quadrilateral that has both an incircle and a circumcircle, with radii 
and 
respectively. If the diagonals 
are perpendicular, find the area of 
that satisfy the following equation for all real numbers 
:![\[
f\big(f(x - y)\big) = f(x) - f(y).
\]](http://latex.artofproblemsolving.com/c/3/c/c3c423f0ceb2fcb91064c5de218ec88f69c95e6d.png)
be a function satisfying![\[
f(f(x)) = x^2 - 6x + 6
\quad \text{for all} \quad x \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/8/a/e/8aebed1e260316d675ca7d104d83fabcd8b5717e.png)
Given that![\[
f(i) < f(i+1) \quad \text{for} \quad i = 0, 1, 2, 3, 4, 5,
\]](http://latex.artofproblemsolving.com/9/f/9/9f92574f9fe14e2e30c3e51e194c661fbd4ece07.png)
find the value of![\[
f(0) + f(1) + f(2) + \cdots + f(6).
\]](http://latex.artofproblemsolving.com/4/0/7/4075d7efa78dda4c46a5cf764fb8852e8c0fb09b.png)
of convex polygons (in the plane). It turns out that every point of the plane lies in the same number of elements of 
as elements of 
.
.
of a composite number 
and 
are such that the main divisors of 
.
and 
.
and 
intersect the circumcircle of 
and 
,
such that 
and 
are parallelograms (with 
,
)
be the point of intersection of lines 
and 
.
,
be the set of real numbers. Find all functions 
such that, for all real numbers 
and 
,
satisfying 
for all real numbers 
such that: 
for all 
.
Y by Adventure10, Mango247
In a plane the circles 
and 
with centers 
and 
, respectively, intersect in two points 
and 
. Assume that 
is obtuse. The tangent to in intersects again in 
and the tangent to in intersects again in 
. Let 
be the circumcircle of the triangle 
. Let 
be the midpoint of that arc 
of that contains . The lines 
and 
intersect again in 
and 
, respectively. Prove that the line 
is perpendicular to 
.
and 
with centers 
and 
, respectively, intersect in two points 
and 
. Assume that 
is obtuse. The tangent to in intersects again in 
and the tangent to in intersects again in 
. Let 
be the circumcircle of the triangle 
. Let 
be the midpoint of that arc 
of that contains . The lines 
and 
intersect again in 
and 
, respectively. Prove that the line 
is perpendicular to 
.
Similarly, we have 
Thus, 
and 
imply that 
Since 
then it follows that 
as desired.
suppose 
goes to 
.
is a parallelogram. Suppose, 
goes to 
.
.
,
.
.
,
is anti-parallel to 
.
So 
and 
(or AK), 
Follows 
Hence, 
.
Thus 
is tangent to 
is tangent to 
and 
.
,
.
,
.
implying that 
all lie on the perpendicular bisector of the segment 
all lie on the perpendicular bisector of the segment 
meaning that 
as desired.
,
