[/quote]
,
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to 
of positive integers such that for every central sequence 
there are infinitely many positive integers 
.
be prime numbers such that their sum isn't divisible by 
.
positive integer quadruples satisfy:
![\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]](http://latex.artofproblemsolving.com/5/6/0/560a9568b2c09512cac1d317a011931a439e6a02.png)
and the sets 
If 
has exactly three elements, prove that 
cannot be an integer.
satisfy:
![$[(x-2)^2+(y-4)^2][ (x-1)^2 + (y-2)^2] = 36$](http://latex.artofproblemsolving.com/b/1/0/b108dc2cac5b7491fb83418eb36c7184f6d5d30c.png)
such that
holds for all 
.
such that:
with all 
be a triangle, and let 
,
,
be collinear points such that 
,
,
.
,
,
are the reflections of 
,
,
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of 
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
and has exactly 
then
Y by
Two circles with centers at points 
intersect at points 
. A circle is drawn through the points 
which intersects the given circles at the points 
for the second time. Prove that 
is the bisector of the angle 
.
(Timoshkevich T.)
intersect at points 
. A circle is drawn through the points 
which intersects the given circles at the points 
for the second time. Prove that 
is the bisector of the angle 
.(Timoshkevich T.)
This post has been edited 2 times. Last edited by parmenides51, Mar 16, 2021, 9:02 AM
.
and 
- that's just angle chase.
