[/quote]
,
and 
intersect at 
and 
.
be the circumcircle of the triangle 
;
and 
,
and 
intersects 
.
on 
.
and 
meet at 
.
.
?
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 
.
)
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)
Y by buratinogigle, anantmudgal09, Adventure10, Mango247
Let 
be angle bisector of acute triangle 
.Point 
choosen on such that 
.point 
lies on the extention of from 
such that 
.Point 
is diametrically opposite the point on the circumcircle of 
.Prove that 
passes through midpoint of arc .(S. Berlov)
be angle bisector of acute triangle 
.Point 
choosen on such that 
.point 
lies on the extention of from 
such that 
.Point 
is diametrically opposite the point on the circumcircle of 
.Prove that 
passes through midpoint of arc .(S. Berlov)
my 100th post This post has been edited 1 time. Last edited by MRF2017, Mar 14, 2016, 4:02 PM
to intersect 
gives us the following equivalence:
.
such that 
as 
such that 
and 
are collinear.
,
in 
such that it corresponds to 
are isogonal wrt 
because 
are isogonal. Thus 
.
,
,
.
.
,
similarly. Since 
are angle bisectors in similar triangles 
respectively, therefore![\[\frac {FX}{XH}=\frac {YE}{YH}\implies XY||EF\]](http://latex.artofproblemsolving.com/3/6/b/36b9b833b99a2cbfcaa24a01256541a2ef719bee.png)
It follows that 
are homothetic about 
are collinear and we are done.
(external) meet 
and let 
meet 
at point 
.
.
maps 
we have that 
.
and we get the claim. Now, clearly 
implies that 
passes through the anti-pose of 
.
are collinear.
is the intersection of 
and 
.
