[/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 
.
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
An acute triangle 
is given which 
.
is the interior and the incircle of meets 
at 
. 
and 
meet at 
. Let 
be a tangent from D to the circumcircle of 
, and define 
and 
on 
and 
, respectively.
Prove
meet at one point.
is given which 
.
is the interior and the incircle of meets 
at 
. 
and 
meet at 
. Let 
be a tangent from D to the circumcircle of 
, and define 
and 
on 
and 
, respectively.Prove
meet at one point.This post has been edited 4 times. Last edited by Hip1zzzil, Mar 29, 2025, 10:12 PM
Reason: Mistake
Reason: Mistake
.
be a point on ray 
such that 
.
be the intersecting point of lines 
.
be midpoints of segments 
,
be the foot of perpendicular from 
to 
wrt line 
.
,
.
and 
.
are concurrent.
are isogonal lines wrt angle 
.
.
.
are collinear, if we consider inversion on 
are cyclic. Obviously 
are cyclic, so 
are cyclic. Therefore, 
.
.
.
and 
are homothetic, and 
are corresponding lines in this homothety. So if we consider the point 
such that 
,
is the centroid of 
.
meet at 
.
.
and 
.
.
is the symmedian point, whose complex coordinate is given in 
(All the summations without index will denote a cyclic sum.)
be the intersection of 
is symmetric w.r.t. 
.
,
,
,
are collinear, hence
.
-
which is symmetric. Hence the claim.
