[/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 
.
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)
Y by Adventure10, Mango247
In an equilateral trapezoid, the point 
is the midpoint of the base 
. A circle with a center at a point and a radius 
is tangent to a straight line 
. Let the segment 
intersect this circle at point 
, and let 
is a point such that 
is a parallelogram. The circumscribed circle of a triangle 
intersects the segment at a point 
. Prove that 
.
is the midpoint of the base 
. A circle with a center at a point and a radius 
is tangent to a straight line 
. Let the segment 
intersect this circle at point 
, and let 
is a point such that 
is a parallelogram. The circumscribed circle of a triangle 
intersects the segment at a point 
. Prove that 
.This post has been edited 1 time. Last edited by qweDota, May 4, 2018, 10:51 AM
is a parallelogram and 
is an equilateral trapezoid, we have the following equalities: 
, that is, the line 
touches the circumcircle of the quadrilateral 
. Whence, bearing in mind that 
touches the circumcircle of the triangle 
,
.
.
and 
are equal to the side 
and to the adjacent angles, that is, 
.
