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)
with 
is the midpoint of . Prove that if the quadrilaterals 
and 
are tangent, then the sides 
of the trapezoid satisfy the equalities 
and 
.
is tangential, we have 
or 
Similarly, because is tangential, we get 
or 
Now, 
is a median in triangle 
and 
is a median in triangle 
and therefore 
and 
Using the cosine law we get 
and 
From 
and 
we get 
and 
[Note: for the case 
we get 
and is a rectangle with 
in this case the equalities in question are obviously satisfied.] Substituting 
and 
into 
and 
and using the values of and from 
and 
we get the following system:
and 
we get the equivalent system
by 
and then adding and subtracting 
we get the equivalent system
and 
we get 
and after substituting this into we get 
From 
and 
we get 
which is equivalent to 
From and 
we get 
.
