[/quote]
,
,
where 
be the midpoint of the arc 
or 
.
and 
meet at 
and the lines 
and 
.
and 
satisfying the following condition: 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
and 
, where 
.
axis divides the positive side of the plane 
)
is given, such that 
)
to be expressed as difference of two perfect squares?
be 
.
and 
be arbitrary points on sides 
Prove that 
.
square is given, and a quadratic trinomial with a positive leading coefficient is placed in each of its cells. There are 
coefficents in total, and these coefficents are chosen from the set ![$[-66;47]$](http://latex.artofproblemsolving.com/d/d/9/dd92777f4ba30299d3326b285f1f73656aba544b.png)
,
satisfying the following two conditions:
such that, for all indices 
,
.![\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]](http://latex.artofproblemsolving.com/5/4/a/54ac17063104439114f3c7b07de8fd27450ea1eb.png)
is a multiple of 
![\[\frac{x^3+1}{(y-1)(z+1)}+\frac{y^3+1}{(z-1)(x+1)}+\frac{z^3+1}{(x-1)(y+1)}\]](http://latex.artofproblemsolving.com/5/c/6/5c6894b617b990ec69c62ed1cf7d062d4660af53.png)
where 
are reals.
be a triangle and let 
denote the circumcircle of 
is 
, respectively. Let 
intersect 
,
.
to the following system:
,
.
be 
g
is the greatest number on at least one of the two arcs 
.
such that regardless of the values of 
Y by Dadgarnia, little-fermat, Adventure10, Mango247
is a circumscribed and inscribed quadrilateral. 
is the circumcenter of the quadrilateral. 
and 
are the intersections of 
, 
and 
respectively. 
and 
are points on 
and 
such that 
and 
. 
and 
are points on 
and 
such that 
and 
. 
is the midpoint of arc 
of 
which contains 
.Prove that the circumcircles of triangles
and 
are coaxal with the circle with diameter 
.This post has been edited 1 time. Last edited by TheOverlord, May 11, 2015, 2:10 PM
is the incenter of 
then 
and 
Condition 
means that 
which is orthogonal to any circle through 
thus particularly orthogonal to the circumcircle 
of 
and similarly 
into 
takes the circle with diameter 
into 
(as 
If 
cut 
then from the complete quadrilateral 
it follows that 
But since 
bisect 
we have 
and 
and the circle with diameter 
be the Incenter and Circumcenter of 
.
and similarly 
.
of 
.
.
WRT 
.
.
where 
is the circumradius of 
.
around 
swaps 
(by Brocard), 
and 
remains fixed. So, 
.
and 
.
and 
and let 
concurs.
and 
respectively. So, 
.
is the Midpoint of 
.
must be the midpoint of Minor Arc 
.
are coaxial. 
