[/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, Rounak_iitr
Circles 
, 
are externally tangent at point M and tangent internally with circle 
at points 
and 
respectively. Let 
and 
be the points that their common tangent at point 
of circles and intersect with circle 
Prove that if 
then the segment 
is diameter of circle
Theoklitos Paragyiou (Cyprus)
, 
are externally tangent at point M and tangent internally with circle 
at points 
and 
respectively. Let 
and 
be the points that their common tangent at point 
of circles and intersect with circle 
Prove that if 
then the segment 
is diameter of circle Theoklitos Paragyiou (Cyprus)
This post has been edited 1 time. Last edited by parmenides51, Dec 11, 2022, 6:40 PM
intersect on 
.
points are collinear and 
.
.
is a deltoid and so 
.
.
.
.
,
which implies ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
is diameter.
be the intersection of tangents to circles 
and 
from points 
and 
passes through 
,
,
.
is the angle bisector of 
.
,
is the angle bisector of 
.
of 
lies on the angle bisector of 
