[/quote]
,
and 
.
Is there any generalizations?
and 
Prove that
![$$ a+b+kc^2\geq\frac{3\sqrt[3]{k}}{4}$$](http://latex.artofproblemsolving.com/5/5/b/55be0378562556d1aa4400bef11d1510ff722854.png)
Where 
be a function from the set of integers to the set of positive integers. Suppose that, for any two integers 
and 
,
is divisible by 
.
,
is divisible by 
.
and 
be positive integers with 
such that 
divides 
.
and 
;
and 
.
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
.
be the second intersection of the circumcircles of triangles 
and 
.
,
,
are concurrent.
in the expansion of 
have the same incenter 
.
is concyclic iff 
is concurrent
a perfect square?
with the following property: there exists positive integers 
such that 
and 
are divisible by 
where 
.
be positive integers such that 
and 
.
be the set of values attained by 
as 
be an integer. A sequence 
is defined by 
,
,
,
Y by
An acute triangle 
is given on the plane. Let the points 
, 
and 
be on the sides 
, 
and 
, respectively. (, and are different from the vertices 
, 
and 
) Also the points 
, 
and 
are taken such that 
is an equilateral hexagon which the opposite sides are parallel. Suppose that the circumcenters of and 
are coincident. Then determine the least possible value of:
Note: 
area of 
is given on the plane. Let the points 
, 
and 
be on the sides 
, 
and 
, respectively. (, and are different from the vertices 
, 
and 
) Also the points 
, 
and 
are taken such that 
is an equilateral hexagon which the opposite sides are parallel. Suppose that the circumcenters of and 
are coincident. Then determine the least possible value of:
Note: 
area of 
This post has been edited 1 time. Last edited by Ege_Saribass, Apr 27, 2025, 11:10 AM
Reason: missing information
Reason: missing information
be the circumcenter of 
.
.
,
.
.
and 
.
.
hence 
thus 
.
.
.
goes to 
via inversion.
,
,
.
and 
and 
.
.
are linear.![$[E'F']$](http://latex.artofproblemsolving.com/b/8/1/b8142d200cf7f09b6a3c009b17fe198a406538ec.png)
intersects 
,
and 
.
,
and 
.
which is a contradiction.
.
and 
.
.
and 
.
and circumradius of the 
.
and 
,
be the circumcenter of 
hence 
.
