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.
. 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 
.
then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
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 hypotenuse 
, you know the positions of points 
and 
as well as some point 
lying on the bisector of angle 
. Construct the triangle using a compass and ruler.
.
