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 ?
(
) intersect at the point 
. The points 
and 
are marked on the line 
such that 
lies on the segment 
, 
lies on the segment 
, 
and 
. Prove that the lines 
and 
intersect on the bisector of the angle 
.
and 
intersect at 
.Easy to notice that is excenter of 
hence 
is angle bisector of 
.Now let angle bisectors of 
and 
intersect at (I mean outside angles) 
same for 
is cyclic(
) and 
are collinear.Done.
and intersect at .Easy to notice that is excenter of hence is angle bisector of .Now let angle bisectors of and intersect at (I mean outside angles) same for is cyclic() and are collinear.Done.
and 
.
