[/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 
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 
,
,
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
Let 
be a square and 
a circle with center at point 
passing through points 
and 
, and let 
be a point on the circle within the given square. Tangent to a circle at a point intersects the segments 
and 
respectively at points 
and 
. Let 
and 
be the intersections of the segments 
and 
with segment 
respectively . Prove that the lines 
and 
pass through the same point.
be a square and 
a circle with center at point 
passing through points 
and 
, and let 
be a point on the circle within the given square. Tangent to a circle at a point intersects the segments 
and 
respectively at points 
and 
. Let 
and 
be the intersections of the segments 
and 
with segment 
respectively . Prove that the lines 
and 
pass through the same point.This post has been edited 3 times. Last edited by parmenides51, May 12, 2024, 7:47 PM
with segment 
lies on 
are concyclic and 
are concyclic which means 
and 
.
, 
and 
are altitudes. So they are concurrent.
stands for small arc 
as 
is tangent of circle 
,
is one altitude in triangle 
and 
as 
are tangents of 
i
because 
when 
are tangents
are concyclic so 
so EH is the 2nd altitude in triangle 
.
