[/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.
Y by Adventure10
Let 
be a triangle and 
the circle with diameter 
. The bisectors of 
and 
intersect (also) at 
and 
, respectively. The incircle of meets 
and 
at 
and 
, respectively. Prove that 
and are collinear.
be a triangle and 
the circle with diameter 
. The bisectors of 
and 
intersect (also) at 
and 
, respectively. The incircle of meets 
and 
at 
and 
, respectively. Prove that 
and are collinear.This post has been edited 2 times. Last edited by parmenides51, Sep 2, 2022, 6:39 PM
of 
at 
,
and 
.
.
and 
and the result follows.
and 
intersect each other at 
,
is cyclic. Now, angles on the same arc are equal and by thales theorem 
,
are collinear and 
are collinear, so all points 
are collinear.
