AoPS Academy
be a positive integer. There are two numbers written on a blackboard, one red and one blue. Initially, both are 0. We define the following procedure: at each step, we choose a nonnegative integer 
(not necessarily distinct from the previously chosen ones), and, if the red and blue numbers are 
and 
respectively, we replace them with 
and 
, which we color blue and red (in this order). We keep doing this procedure until the blue number is at least 
.
are positiv real number, such that 
prove inequality 
is a set of 
points in the plane that are not all collinear. Prove that contains three points that form a triangle whose circumcentre is not a point in .
- such points, that all another points lies on line 
but not on segment or in the one side from line . There are some points that are not on line . Let point 
is not on and 
has minimal circumradius among triangles with side , and 
is circumcenter of . But then circumradius of 
is less then circumradius of 
- contradiction.
is 
. As 
lie on the perpendicular bisector of and is on same side as , then it means that 
less 
where is altitude of .
and 
so 
so the circumradius of is not necessary the great but the idea remain good : Fortunately ,except in the case of equilateral triangle, one of the angles has less than 
we consider its opposite side instead of .
and the circumcenter and the circumcenter of which ends the proof
and 
.
