[/quote]
,
be the set of all positive integers. Determine all subsets 
of 
such that
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)
Y by MazeaLarius, Adventure10, Mango247
Let 
be a triangle with 
; and let 
be the internal bisector of 
, 
. Denote by 
the midpoint of the arc 
which contains point 
. The circumscribed circle of the triangle 
intersects the segment 
at point 
. Let 
be the reflection of 
with respect to 
. If 
, prove that the points 
belong to the same circle.
be a triangle with 
; and let 
be the internal bisector of 
, 
. Denote by 
the midpoint of the arc 
which contains point 
. The circumscribed circle of the triangle 
intersects the segment 
at point 
. Let 
be the reflection of 
with respect to 
. If 
, prove that the points 
belong to the same circle.
.
and 
.
.
is cyclic. Thus
.
and therefore 
.
is cyclic. So,
.
is cyclic and we are done.
intersect 
at E
be midpoint of 
we have 
collinear
because 
and 
cyclic
and 
are internal and external bisectors of 
,
.
and since 
,
is the midpoint of 
(the latter sine 
is a midsegment in 
)
is cyclic. Therefore 
and 
