AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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 
be the set of all positive integers. Determine all subsets 
of 
such that
and 
intersect at 
and 
. Let 
be the circumcircle of the triangle 
; the circle intersect the circles centered at and at points 
and 
, respectively. The point is inside triangle 
and 
intersects at 
. The point 
on is such that 
. Let 
and 
meet at 
, prove that 
.
. Prove that:
?
be a positive integer. A Japanese triangle consists of 
circles arranged in an equilateral triangular shape such that for each 
, 
, 
, , the 
row contains exactly 
circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with 
, along with a ninja path in that triangle containing two red circles., find the greatest such that in each Japanese triangle there is a ninja path containing at least red circles.
, such that 
and 
.
such that 
is prime, 
with 
and ?
and 
. Let 
meet 
at , then 
. So 
. Or even better, 
.
and 
.
is the center of the circle into which quadrilateral 
is inscribed. If angles 
and 
are both equal to 
degrees and angle 
is greater than angle 
,
.