[/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 
.
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)
Y by huashiliao2020, Adventure10, Mango247
Let 
be a convex pentagon such that 
and 
a circle with center on side 
that touches the sides 
, 
, 
and 
at points 
, 
, 
and 
(different from vertices of the pentagon) respectively. Prove that lines 
and are parallel.
be a convex pentagon such that 
and 
a circle with center on side 
that touches the sides 
, 
, 
and 
at points 
, 
, 
and 
(different from vertices of the pentagon) respectively. Prove that lines 
and are parallel.
and 
yields the conclusion.
.
-the intersection of the segment 
with the circle 
-the intersection of the segment 
with the circle 
we get that 
.
we get that 
.
.
.
is an issoscelles trapezoid, so 
.
,
and 
are tangents and since tangents from the same point are equal, we get:
and 
.
.
.
as we found above
since they are both radii
since the angle between a tangent and a radius is a right angle
.
is isosceles since both 
and 
are radii.
and 
are equal to 
and since they are alternate we can now conclude that 
so 
,
.
is an isosceles trapezoid, and 
.
