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 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 ?
be a function such that for every regular 
-gon 
we have 
. Prove that 
for all reals 
.
be a triangle (
). 
are the altitudes of the triangle. The bisector of 
intersects 
at 
. Let be a point such that 
and 
. Prove that 
passes through the midpoint of 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
is tangent to side 
at point 
, and to side at point 
. On the side 
there is a point 
such that 
. Prove that 
is tangent to side at point , and to side at point . On the side there is a point such that . Prove that 
be the touch point of the incircle with . Because lies outside of the incircle we have
and 
be the incenter of 
. Then
and 
meet again at 
and 
, then
hence 
is a parallelogram as desired. 
so
as desired.
.
