[/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 Adventure10, Mango247
Acute scalene triangle 
has 
as its centroid and 
as its circumcenter. Let 
be the projections of 
on respective opposite sides and 
be the midpoints of 
in that order. 
intersect 
at 
respectively.
a. Prove that the circle
pass through the midpoint of 
b. Let
be the midpoints of 
respectively. Prove that 
are concurrent.
has 
as its centroid and 
as its circumcenter. Let 
be the projections of 
on respective opposite sides and 
be the midpoints of 
in that order. 
intersect 
at 
respectively.a. Prove that the circle
pass through the midpoint of 
b. Let
be the midpoints of 
respectively. Prove that 
are concurrent.This post has been edited 2 times. Last edited by gausskarl, Jan 27, 2018, 11:49 AM
and 
are circumcevian triangle of 
are bound a triangle which is perpsective to 
meets 
then 
so 
and 
are isogonal in 
Hence 
then 
cu
at 
then
be the reflections of 
through midpoints of 
From midline we see 
and 
Thus the homothety center 
swap line 
to line 
Similarly with lines 
and 
so we will prove that 
are concurrent. Consider triangle 
and two points 
and 
Then 
is circumcevian triangle of 
is bound by 
which is perspective to triangle 
We are done.
.
