[/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 Adventure10, Mango247
The inscribed circle of 
touches 
and 
at 
and 
respectively. Denote the perpendicular foots from 
to 
by 
respectively. Let the second intersection of these perpendiculars with the incircle be 
respectively. Show that 
by Mahdi Etesami Fard
touches 
and 
at 
and 
respectively. Denote the perpendicular foots from 
to 
by 
respectively. Let the second intersection of these perpendiculars with the incircle be 
respectively. Show that 
by Mahdi Etesami Fard
be the incenter of 
denote its inradius. Then 
.
so 
.
be the midpoint of segment 
.
,
is a rectangle. Note that 
,
.
.
.
and we are done.
denote? Area or semiperimeter?
,![\[\frac{{{S}_{\triangle BMD}}}{{{S}_{\triangle CND}}}=\frac{BD \cdot MK}{NL\cdot CD} = \frac{KD\cdot BD\cdot \cot(\beta/2)}{\cot(\gamma/2)\cdot CD}=\frac{KD}{DL}\]](http://latex.artofproblemsolving.com/f/2/d/f2d0cda975352e46c1e055362a7472d60176c3c5.png)
since ![\[\frac{\cot(\beta/2)}{\cot(\gamma/2)}=\frac{s-c}{s-b}\]](http://latex.artofproblemsolving.com/8/c/2/8c25ecd9215b80596466a68b480ee46e5bc0e849.png)
by law of cotangents.
