[/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 
.
.
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 Adventure10, Mango247
Circle 
, described about the triangle 
, intersects the sides 
and 
of the parallelogram 
, for the second time in points 
and 
respectively. Let 
be the intersection point of lines 
and 
. Let 
- diameter , and the point 
is symmetric to the center of with respect to . Prove that the lines 
and 
are perpendicular.
, described about the triangle 
, intersects the sides 
and 
of the parallelogram 
, for the second time in points 
and 
respectively. Let 
be the intersection point of lines 
and 
. Let 
- diameter , and the point 
is symmetric to the center of with respect to . Prove that the lines 
and 
are perpendicular.This post has been edited 1 time. Last edited by izat, Mar 22, 2015, 10:14 AM
and 
are trapezoids inscribed in circle thus they are both isosceles.Thus 
.
is the perpendicular bisector of 
.
thus 
lies on the perpendicular bisector of 
.
since 
we have 
which gives that 
is the perpendicular bisector of 
which means that 
is the circumcentre of 
.
and 
.
bisect each other,this is a parallelogram which gives 
.
is rhombus since its diagonals are perpendicular to each other and bisect each other.Thus 
.
.
and 
thus it's perpendicular to their common segment.
is 
obviously passes through 
.
which gives that 
q.e.d.
and 
since the quadrilaterals 
are trapezoids and con-cyclic. Thus, 
is the circumcenter of 
be the point such that 
.
are con-cyclic.
is the circumcenter of 
must be on the radical axis of the circumcircles of 
,
