[/quote]
,
be a positive integer. A Japanese triangle consists of 
circles arranged in an equilateral triangular shape such that for each 
,
,
,
row contains exactly 
circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of 
,
such that in each Japanese triangle there is a ninja path containing at least 
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
be a triangle 
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
and 
.
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)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
.
Equality holds when 
or 
and 
,
cookies from one of the piles, of which he eats 
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
Y by
Let 
be the intersection point of two circles. From this point along each circle, clockwise, with constant speeds, points 
and 
begin to move. After one rotation, both points again appear in . Prove that there is always such a fixed point 
that the equality 
holds throughout the motion.
be the intersection point of two circles. From this point along each circle, clockwise, with constant speeds, points 
and 
begin to move. After one rotation, both points again appear in . Prove that there is always such a fixed point 
that the equality 
holds throughout the motion.
and radius 
.
and radius 
.
,
depends on 
.
.
,
from the rule of sines in 
.
.
.
.
.
are centers of circles, 
are radiuses of circles and point 
and 
so 
and so 
